%% Scott Van Thuong.
%%
%% Note from the author:  My dissertation document was created with this file. It passed the general formatting guidelines for PhD dissertations
%% of the Graduate College at the University of Oklahoma. I accept no responsibility in connection with the use of this file. Use it under your own risk.
%%

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% To be removed later
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%       Math definitions
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%4d geometries
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%% THESIS %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{document}

\title{Classification, Cobordism, and Curvature of Four-Dimensional Infra-Solvmanifolds}
\author{Scott Van Thuong}
\date{July, 2013}

\pagenumbering{roman}
{%% PAGE NUMBERING ROMAN

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% TITLE PAGE  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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{\singlespacing

\newpage
\thispagestyle{empty}
\begin{center}
{ %\large
UNIVERSITY OF OKLAHOMA
\par
\vspace{0.16in}
GRADUATE COLLEGE
\par
\vspace{1.2in}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5
CLASSIFICATION, COBORDISM, AND CURVATURE OF
\par
\vspace{0.17in}
FOUR-DIMENSIONAL INFRA-SOLVMANIFOLDS
\par
\vspace{1.2in}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
A DISSERTATION
\par
\vspace{0.17in}
SUBMITTED TO THE GRADUATE FACULTY
\par
\vspace{0.17in}
in partial fulfillment of the requirements for the
\par
\vspace{0.17in}
Degree of
\par
\vspace{0.17in}
DOCTOR OF PHILOSOPHY
\par
\vfill
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
By
\par
\vspace{0.17in}
SCOTT VAN THUONG
\par
%\vspace{0.05in}
Norman, Oklahoma
\par
%\vspace{0.05in}
2014
}
\end{center}


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%% SIGNATURE PAGE  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\newpage
\thispagestyle{empty}
\begin{center}
{%\large
%\par
\vspace{0.5in}
CLASSIFICATION, COBORDISM, AND CURVATURE OF
\par
%\vspace{0.17in}
FOUR-DIMENSIONAL INFRA-SOLVMANIFOLDS
\par
\vspace{0.9in}
A DISSERTATION APPROVED FOR THE

%\vspace{0.17in}
DEPARTMENT OF MATHEMATICS
\par
\vspace{0.9in}
BY
\par
\vfill
\begin{flushright}
\begin{tabular}{cr}
\hline
\  \  \  \  \  \  \  & Dr. Kyung-Bai Lee, Chair\\
 \\ &  \\
  \\ &  \\
\hline
  & Dr. Michael Jablonski \\
\\ & \\
 \\ &  \\
\hline
 & Dr. Krishnan Shankar \\
\\   & \\
 \\ &  \\
\hline
 & Dr. Gerard Walschap \\
 \\ &  \\
  \\ &  \\
\hline
   & Dr. Sowon Hahn \\
\end{tabular}
\end{flushright}
}
\end{center}


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%% COPYRIGHT PAGE  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newpage
\thispagestyle{empty}
\   \
\par
\vfill
\begin{center}
\copyright \    Copyright by SCOTT VAN THUONG \   2014

All Rights Reserved.
\end{center}
}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% ACKNOWLEDGEMENTS PAGE  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\chapter*{Acknowledgements}
{\doublespacing

First and foremost, I express my gratitude
to Professor Kyung-Bai Lee, who has been not just a dissertation adviser to me,
but also a mentor in both mathematics and life. Our first paper written together formed the foundation for this dissertation. My gratitude also extends to Jonathan Hillman 
and Jim Davis for their enlightening exchanges with me. 
I also wish to thank the other members of my committee for their invaluable guidance: Professors Michael Jablonski, 
Krishnan Shankar, Gerard Walschap, and Sowon Hahn.

I also thank my colleagues at OU, as well as Jean Verrette; our conversations helped keep my morale up. The generous financial
support of the OU Mathematics Department and the OU Graduate College was also very helpful.
Last but not least, I thank my parents for their unconditional
love and support throughout my life. I could not ask for better parents.

}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% TABLE OF CONTENTS PAGE  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\renewcommand{\contentsname}{Table of Contents}
{\singlespacing
\tableofcontents
\listoftables
}
\chapter*{Abstract}
{\doublespacing

Crystallographic groups of solvable Lie groups generalize the crystallographic groups of Euclidean space.
The quotient of a solvable Lie group $G$ by the action of a torsion-free crystallographic group of $G$ is an infra-solvmanifold of $G$. Infra-solvmanifolds are aspherical manifolds that generalize both closed flat manifolds and closed almost flat manifolds (that is, infra-nilmanifolds).

Here we complete the classification of 4-dimensional infra-solvmanifolds by classifying torsion-free crystallographic groups of certain 4-dimensional solvable Lie groups. The classification also includes those crystallographic groups with torsion.

We prove that every $4$-dimensional infra-solvmanifold is the boundary of a compact $5$-dimensional manifold by constructing an involution on certain $4$-dimensional infra-solvmanifolds which is either free, or has $2$-dimensional fixed set.

The Ricci signatures (that is, signatures of the Ricci transformation) of $4$-dimensional Lie groups have been classified. A Ricci signature can be realized on an infra-solvmanifold $M$ of $G$ if $M$ is a compact isometric quotient of $G$, where $G$ has left invariant metric with prescribed Ricci signature. We classify which Ricci signatures can be realized on certain $4$-dimensional infra-solvmanifolds. 

}

\newpage

} %% END PAGE NUMBERING ROMAN

\pagenumbering{arabic}

\chapter{Introduction}

This dissertation is a study of the crystallographic groups of 4-dimensional
solvable Lie groups and their corresponding 4-dimensional infra-solvmanifolds.
Crystallographic groups of solvable Lie groups generalize the well known crystallographic
groups of $\bbr^n$. Infra-solvmanifolds are aspherical manifolds that generalize closed flat manifolds. This document is organized as follows: Chapter 1
provides background material and summarizes the main results, Chapter 2 establishes that every 4-dimensional infra-solvmanifold is the boundary of a compact 5-dimensional manifold, Chapter 3 completes the classification of the crystallographic groups of the solvable 4-dimensional geometries, and Chapter 4 explores Ricci signature properties of the 4-dimensional geometries.
Finally, we provide some topics for further exploration in Chapter 5.

\section{Definitions and Notation}

\step[{\bf Crystallographic Groups and Infra-Solvmanifolds}]

For our purposes, we will assume all Lie groups to be connected and simply connected.
Let $G$ be a solvable Lie group. The group of \emph{affine diffeomorphisms} of $G$ is:
\[\aff(G)  = G \rx \aut(G),\]
which has group operation $(a,A)(b,B) = (aA(b),AB)$.
There is an action of $\aff(G)$ on $G$.
For $(a,A) \in \aff(G)$ and $g \in G$,
\[(a,A).g = aA(g).\]
Under this action, the subgroup $G \cong \{(g, \id) \mid g \in G\} \subset \aff(G)$ acts on $G$ as left translations.

\begin{definition}
\label{def1}
Let $K$ be a maximal compact subgroup of $\aut(G)$. A discrete subgroup $\varPi$ of
$G \rx K \subset \aff(G)$ is called a \emph{crystallographic group} of $G$ when

{\rm (1)} The quotient $\varPi \bs G$ is compact.

{\rm (2)} The translation subgroup $\Gamma:=\varPi \cap G$ is of finite index in $\varPi$.

\end{definition}
\medskip

The translation subgroup $\Gamma$ is normal in $\varPi$, and
we refer to the finite group $\Phi:=\varPi/\Gamma$ as the \emph{holonomy group}
of $\varPi$. Thus, a crystallographic group $\varPi$ fits the diagram of short exact sequences
\[
\CD
1 @>>> \Gamma @>>> \varPi @>>> \Phi @>>> 1\\
@. @VVV @VVV @VVV @.\\
1 @>>> G @>>> G \rx K @>>> K @>>> 1.\\
\endCD
\]
Condition (2) of Definition \ref{def1} implies that $\Gamma$ is a discrete subgroup
of $G$ and that $\Gamma \bs G$ is compact. That is, $\Gamma$ is a \emph{cocompact lattice} of $G$.
For simply connected solvable Lie groups,
a result of Mostow \cite[Theorem 6.2]{mostow2} implies that $\Gamma \bs G$ has finite volume
if and only if $\Gamma \bs G$ is compact.  So, the terms ``lattice" and
``cocompact lattice" are equivalent for our purposes.
Since $\Gamma$ acts as left translations on $G$, it acts freely on $G$, and
we say that the quotient $\Gamma \bs G$ is a \emph{solvmanifold} when $G$ is solvable, and a
\emph{nilmanifold} when $G$ is nilpotent.
The quotient $\varPi \bs G$ is a closed
manifold
precisely when $\varPi$ acts freely on $G$. This is equivalent to $\varPi$ being torsion-free.

\begin{definition}
\label{def2}
Let $\varPi$ be a torsion-free crystallographic group of $G$. The quotient
$\varPi \bs G$ is an \emph{infra-solvmanifold} when $G$ is solvable, and an
\emph{infra-nilmanifold} when $G$ is nilpotent.
\end{definition}
\medskip

In \cite{KurokiYu},
the various definitions of infra-solvmanifold appearing in the literature are shown to all be equivalent. Here
we have adopted Definition 1 in \cite{KurokiYu}.

Note that with
the Euclidean metric, $\isom(\bbr^n) = \bbr^n \rx \on{O}(n,\bbr)$, so an infra-solvmanifold for $G=\bbr^n$ and
 $K=\on{O}(n,\bbr)$ is simply a closed flat manifold.

Topologically, Condition (2) of Definition \ref{def1} states that
an infra-solvmanifold $\varPi \bs G$ is finitely covered by the
solvmanifold $\Gamma \bs G$ with covering transformation group $\Phi$. This explains the prefix ``infra" and
is motivated by the classical result that any closed flat $n$-manifold is finitely covered by a flat torus $\bbz^n \bs \bbr^n$.
Thus, in our context, infra-solvmanifolds generalize closed flat manifolds, and solvmanifolds generalize
flat tori.

When $G$ is nilpotent, Auslander proved that Condition (1) of Definition \ref{def1} actually implies
Condition (2). When $G$ is abelian, this was due to Bieberbach.

\begin{theorem}[Bieberbach's First Theorem in the nilpotent case {\cite[Theorem 8.3.2]{seif-book}}]
\label{Bieberbach1}
Let $G$ be nilpotent and \[\varPi \subset G \rx K\] be discrete and such that $\varPi \bs G$ is compact.
Then the translation subgroup $\Gamma=\varPi \cap G$ is of finite index in $G$, and
the quotient $\Gamma \bs G$ is compact.
\end{theorem}

When $G$ is solvable, Condition (2) in Definition \ref{def1} is needed, because there
are examples of solvable $G$ and discrete subgroups $\varPi \subset G \rx K$ with $\varPi \bs G$ compact,
for which $\varPi \cap G$ is \emph{not} of finite index in $\varPi$.
In other words, Bieberbach's First Theorem
does \emph{not} extend to all solvable Lie groups. Dekimpe, Lee, and Raymond
provide sufficient criteria in \cite{DLR} for a solvable Lie group to satisfy
Bieberbach's First Theorem.

Crystallographic groups of solvable Lie groups of type (R) satisfy
a rigidity property:

\begin{theorem}[Rigidity of crystallographic groups {\cite[Theorem 8.4.3]{seif-book}}]
\label{Rigidity}
Let $G$ be a solvable Lie group of type (R) and $\varPi$, $\varPi'$ be finite extensions
of lattices in G. Then every isomorphism between $\varPi$ and  $\varPi'$ is conjugation by
an element of $\aff(G)$.
\end{theorem}

\noindent
Consequently, infra-solvmanifolds of solvable Lie groups of type (R) are rigid:

\begin{theorem}[Rigidity of infra-solvmanifolds {\cite[ Theorem 8.4.3]{seif-book}}]
\label{Rigidity2}
Homotopy equivalent infra-solvmanifolds of type (R) are affinely diffeomorphic.
\end{theorem}

\step[{\bf The 4-Dimensional Geometries}]

The classification of the 4-dimensional geometries appears in \cite{CTCWall}, analogous to Thurston's eight 3-dimensional geometries \cite{Thurston, scott}.

Recall that the
3-dimensional geometry $\Nil^3$ is the
group of upper triangular matrices
\[\left[ \begin{matrix} 1 & x & z \\
0 & 1 & y\\
0 & 0 & 1
\end{matrix}  \right],\]
and the 3-dimensional geometry $\Sol$ is the semidirect product
$\bbr^2 \rx_{\phi(u)} \bbr$ where $\phi(u) =
\left[ \begin{matrix} e^{-u} & 0 \\
0 & e^{u}
\end{matrix} \right]$.

\begin{notation}
Following Hillman's convention \cite{hillman1, hillman2}, the term  \emph{4-dimensional solvable geometry} will refer to the type (R), solvable, and unimodular groups
$\bbr^4$, $\Nilr$, $\Nilf$, $\Solr$, $\Solmn$, $\Solo$, and $\Solof$.
\end{notation}

In dimension 4, except for $\Solof$, which splits as $\Nil^3 \rx \bbr$,
all solvable geometries are of the form $\bbr^3 \rx_{\phi(u)} \bbr$
for $\phi: \bbr \ra \gl(3,\bbr)$:

\begin{align}
\label{listOfGeometries}
\begin{array}{lllll}
&\bbr^4: &\phi(u)= \left[ \begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{matrix}\right]
\hspace {1 cm}  &\Nilr: &\phi(u)=
\left[ \begin{matrix} 1 & u & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{matrix}\right]\\
&\Nilf: &\phi(u)= \left[ \begin{matrix} 1 & u & {\frac 1 2} u^2 \\ 0 & 1 & u\\ 0 & 0 & 1 \end{matrix}\right]
\hspace {1 cm}  &\Solr: &\phi(u)=
\left[ \begin{matrix} e^{-u} & 0 & 0\\ 0 & e^{u} & 0\\ 0 & 0 & 1 \end{matrix}\right]\\
&\Solo: &\phi(u)= \left[ \begin{matrix} e^{u} & 0 & 0\\ 0 & e^{u} & 0\\ 0 & 0 & e^{-2u} \end{matrix}\right]
\hspace {1 cm}  &\Solmn: &\phi(u)=
\left[ \begin{matrix} e^{\theta u} & 0 & 0\\ 0 & e^{u} & 0\\ 0 & 0 & e^{-(1+\theta)u} \end{matrix}\right]\\
&{\Solo}': &\phi(u)= \left[ \begin{matrix} e^{u} & u e^u & 0\\ 0 & e^{u} & 0\\ 0
    & 0 & e^{-2u} \end{matrix}\right].
\end{array}
\end{align}

For $\Solmn$, $\theta>1$ is such that $\phi(u)$ is conjugate to an element
of $\on{SL}(3,\bbz)$. This guarantees that $\Solmn$ has a lattice \cite{unimodular-groups}.
The characteristic polynomial of $\phi(u)$ is $x^3-mx^2+nx-1$ for $m,n \in \bbz$.
It is known that ${\Solo}'$ has no compact forms \cite{unimodular-groups}
and therefore does not appear in the list of 4-dimensional geometries in \cite{CTCWall}.

$\Solof$ can be described as the
multiplicative group of matrices
\[\left[ \begin{matrix} 1 & {e^{u} x} & z \\
0 & {e^u} & y\\
0 & 0 & 1
\end{matrix}  \right],\] which splits as the semidirect
product $\Nil^3 \rx \bbr$. It has 1-dimensional center ($u=x=y=0$ in the
above matrix). The quotient of $\Solof$ by its center
is $\Sol$; we have the short exact sequence
\[1 \ra \calz(\Solof) \ra \Solof \ra \Sol \ra 1.\]

All 4-dimensional solvable geometries are type (R) and
unimodular. Therefore, with a left
invariant metric on $G$,
\[
\isom(G) \subseteq G \rx K \subset \aff(G),
\]
where $K$ is a maximal compact subgroup of $\aut(G)$ \cite{GW}.
In fact, all of the 4-dimensional solvable geometries admit a left invariant metric
so that \[\isom(G) = G \rx K.\]
The discrete cocompact subgroups $\varPi$ of $\isom(G)$ acting freely on $G$ yield
the compact isometric quotients of $G$, denoted $\varPi \bs G$.

All 4-dimensional solvable geometries except
$\Solo$ satisfy the generalized First Bieberbach Theorem
\cite{DLR}: if $\varPi \subset \isom(G) \subseteq G \rx K$ ($G \neq \Solo$) is discrete and cocompact,
then the translation subgroup $\Gamma = \varPi \cap G$
is a lattice of $G$ and the quotient $\Phi = \varPi / \Gamma$ is a
finite subgroup of $K \subset \aut(G)$. Therefore, the compact isometric quotients of
a 4-dimensional solvable geometry $G$, excluding $\Solo$, are in fact infra-solvmanifolds of $G$.
The compact isometric quotients of $\Solo$ are known to be mapping tori of linear self diffeomorphisms
of the $3$-torus $T^3$. However,
they are not infra-solvmanifolds of $\Solo$.
Rather, they can be realized as infra-solvmanifolds of a different solvable 4-dimensional
Lie group. However, this Lie group is not of type (R), and it also is not a 4-dimensional solvable geometry. In fact, $\Solo$ does not even admit a lattice and therefore has no crystallographic groups (\cite[Corollary 8.5.1]{hillman2}
and \cite[Theorem 3.5, Theorem 4.2]{unimodular-groups}).

Conversely, Hillman has shown that every 4-dimensional infra-solvmanifold $M$ is \emph{geometric}. That is, $M$ is diffeomorphic
to a compact isometric quotient of one of the solvable 4-dimensional geometries equipped with left invariant metric \cite[Theorem 8]{hillman1}. Thus, we have

\begin{theorem}[{\cite[Theorem 8]{hillman1}}]
\label{allAreGeometric}
{\rm (1)} Any 4-dimensional infra-solvmanifold is diffeomorphic to a compact
isometric quotient of a 4-dimensional solvable geometry.

{\rm (2)} The compact isometric quotients of $G$, where $G$ is one of the six 4-dimensional geometries
$\bbr^4$, $\Nilr$, $\Nilf$, $\Solr$, $\Solmn$, $\Solof$, are precisely the infra-solvmanifolds
of $G$.
\end{theorem}

\section{Main Results}

It is a remarkable theorem of Hamrick and Royster that
for every closed flat $n$-manifold, $M$, there is an $(n+1)$-dimensional
compact manifold $W$ with $\partial W = M$ \cite{HR}. That is,
$M$ \emph{bounds}. By the works of Gromov and Ruh, a closed manifold has an almost flat structure
if and only if it is diffeomorphic to an infra-nilmanifold $\varPi \bs G$ \cite{Gromov, Ruh}.
Thus, it seems natural to ask if
 infra-nilmanifolds bound.

\begin{conjecture}[{\cite[Conjecture 1]{FZ}}]
\label{almostFlatBound}
{\it
If $M$ is an $n$-dimensional infra-nilmanifold,
then there exists a compact $(n+1)$-dimensional manifold $W$ with $\partial W = M$.}
\end{conjecture}

Some partial results are known \cite{Upadhyay}.
Hillman has asked if
all 4-dimensional infra-solvmanifolds bound \cite{hillman3}.
Dimension 4 is the first dimension of interest,
as the only 2-dimensional infra-solvmanifolds are the torus and Klein bottle, and both
are boundaries. Also, it is well known that all closed 3-dimensional manifolds bound.
We shall answer Hillman's question affirmatively:

%\begin{theorem*}[Theorem \ref{mybound}]
%\label{mybound}

%\end{theorem*}

\noindent
{\bf Theorem~\ref{mybound}.}
{\it
If $M$ is a 4-dimensional infra-solvmanifold, then $M$ bounds.
That is, there is a compact 5-dimensional manifold $W$ with $\partial W = M$.
}

%\noindent
%{\bf Theorem~\ref{LABEL}.}
%{\it
%... state theorem (Copy the contents of the theorem)
%}

Classifying the crystallographic groups of 4-dimensional solvable geometries is important.
The crystallographic groups of $\bbr^4$ have been classified; see for example \cite{4DBieberbachGroup},
while Dekimpe has classified those of $\Nilr$ and $\Nilf$ \cite{Dekimpe}.
%The classification of crystallographic groups of $\Solof$ is joint with K.B. Lee
%\cite{solof-geo}. 
The infra-solvmanifolds of $\Solmn$ have been classified by Hillman \cite{hillman2}. This is equivalent to a classification of the torsion-free crystallographic groups of $\Solmn$. Partial classifications of the infra-solvmanifolds of $\Solr$ (torsion-free crystallographic groups of $\Solr$) are presented in \cite{Cobb, hillman3}.
Here we will complete the classification of the crystallographic groups of $\Solof$, $\Solr$, and $\Solmn$. Easily checked criteria for such groups to be torsion-free are also provided.

\noindent
{\bf Theorem~\ref{ClassificationSolof-geometry}.}
{\it
There are 14 families of $\Solof$ crystallographic groups. Let $\varPi$ be a crystallographic group of $\Solof$. Any subgroup of $D_4$ can be the holonomy group of $\varPi$,
where $D_4$ is the dihedral group of 8 elements. The maximal holonomy of a torsion-free $\Solof$ crystallographic group, so that $\varPi \bs G$ is an infra-solvmanifold, is $D_4$.
}

\noindent
{\bf Theorems~\ref{0family}-\ref{7ifamily}.}
{\it
There are 60 families of $\Solr$ crystallographic groups. Let $\varPi$ be a crystallographic group of $\Solr$.
Any subgroup of $D_4 \x \bbz_2$ can be the holonomy group of $\varPi$.
If $\varPi$ is torsion-free, so that $\varPi \bs G$ is an infra-solvmanifold, $\varPi$ must have holonomy
$\{e\}$, $\bbz_2$, $\bbz^2_2$, $\bbz^3_2$, $\bbz_4$, or $D_4$. In particular, there is no infra-solvmanifold of $\Solr$ with maximal holonomy $D_4 \x \bbz_2$.
}

\noindent
{\bf Theorem~\ref{ClassificationSolmn-geometry}.}
{\it
There are 7 families of $\Solmn$ crystallographic groups. Let $\varPi $ be a crystallographic group of $\Solmn$. The possible holonomy groups of $\varPi$ are the subgroups of $\bbz_2 \x \bbz_2$. In particular, the maximal compact subgroup of $\aut(\Solmn)$, $\bbz_2^3$, cannot be the holonomy of a $\Solmn$ crystallographic group.
If $\varPi$ is torsion-free, so that $\varPi \bs G$ is an infra-solvmanifold, $\varPi$ must have holonomy
$\{ e \}$ or $\bbz_2$.
}



%\begin{corollary}[\bf{T.}]
%Let $\varPi $ be a crystallographic group of $\Solmn$. The possible holonomy groups of $\varPi$ are the subgroups of $\bbz_2 \x \bbz_2$. The maximal compact subgroup of $\aut(\Solmn)$, $\bbz_2^3$, cannot be the holonomy of a $\Solmn$ crystallographic groups.
%If $\varPi$ is torsion-free, so that $\varPi \bs G$ is an infra-solvmanifold, $\varPi$ must have holonomy
%$\{ e \}$ or $\bbz_2$.
%\end{corollary}

%\begin{corollary}[\bf{T.}]
%
%\end{corollary}

We now consider curvature of the 4-dimensional solvable geometries.
Let $G$ be a 4-dimensional solvable geometry.
With arbitrary left invariant metric on $G$,
\[\isom(G) \subseteq G \rx K \subset \aff(G),\]
for a maximal compact subgroup $K \subset \aut(G)$ \cite{GW}.
Depending
on choice of left invariant metric, a 4-dimensional solvable geometry can have different
isometry groups and
Ricci signatures.
Here
are the possible Ricci signatures on some of the 4-dimensional
solvable geometries
\cite{KremlevNikonorov}:

\begin{table}[H]
\caption{Ricci signatures}
\begin{center}
  \begin{tabular}{| l | l |}
    \hline
    $G$ & Possible Ricci signatures \\ \hline
    $\bbr^4$ & $(0,0,0,0)$ \\ \hline
    $\Nilr$ & $(0,+,-,-)$ \\ \hline
    $\Nilf$ & $(0,+,-,-), (+,+,-,-), (+,-,-,-) $ \\ \hline
    $\Solr$ & $(0,0,0,-), (0,+,-,-), (+,+,-,-), (+,-,-,-) $ \\ \hline
    $\Solmn$ & $(0,0,0,-), (0,+,-,-), (+,+,-,-), (+,-,-,-) $ \\ \hline
    $\Solo$ & $(0,0,0,-), (0,+,-,-)$ \\ \hline
  \end{tabular}
\end{center}
\end{table}

Suppose we only consider left invariant metrics on $G$ that induce a
particular Ricci signature. The compact isometric quotients of $G$ may not
account for all infra-solvmanifolds of $G$, as
$\isom(G)$ may be a \emph{proper} subgroup of $G \rx K$,
for a maximal compact subgroup $K$ of $\aut(G)$.
Given an infra-solvmanifold $M$ of $G$, a Ricci signature
can be \emph{realized} on $M$ if there is a left invariant metric on $G$ with prescribed Ricci signature
such that $M=\varPi \bs G$, for some $\varPi \subset \isom(G)$.

\noindent
{\bf Theorems~\ref{NilrSig}, \ref{SolmnSig}, \ref{SolrSig}.}
{\it
{\rm (1)} If $M$ is an infra-nilmanifold of $\Nilf$, then any of the three Ricci signatures
$(0,+,-,-)$, $(+,+,-,-)$, $(+,-,-,-)$ can be realized on $M$.

{\rm (2)} Every infra-solvmanifold of $\Solmn$ is the mapping torus
of a linear self diffeomorphism $\cals$ of $T^3$; $\cals$ has three distinct real eigenvalues. If all eigenvalues have the same sign, then any of $(0,0,0,-)$,
 $(0,+,-,-)$, $(+,+,-,-)$, $(+,-,-,-)$
can be realized on $M$. Else, only $(0,0,0,-)$ and $(0,+,-,-)$ can be realized on $M$.

{\rm (3)} If $M$ is an infra-solvmanifold of $\Solr$ which has an order 4 element in its
holonomy, then only $(0,0,0,-)$ can be realized on $M$. 

%{\rm (4)}  If $M$ is a compact isometric quotient of $\Solo$, then $M$ is the mapping torus
%of a linear self diffeomorphism $\cals$ of $T^3$; $\cals$ has two complex eigenvalues $z, \bar z$,
%and one real eigenvalue $\alpha$. The signature $(0,+,-,-)$ cannot be realized on $M$ when
%$\mbox{Re}(z)>0$ and $\alpha<0$, or when $\mbox{Re}(z)<0$  and $\alpha > 0$.
}


\chapter{Cobordism Classes of 4-Dimensional Infra-Solvmanifolds}

In this section, we shall prove that every compact 4-dimensional infra-solvmanifold
bounds.
That is, if $M$ is a 4-dimensional infra-solvmanifold, then there is a compact 5-dimensional manifold $W$ with $\partial W = M$. In other words, the corbordism class of a
$4$-dimensional infra-solvmanifold
in the unoriented Thom cobordism group $\Omega_4^{O}$ is trivial. In light of Hillman's result
 that 4-dimensional infra-solvmanifolds are geometric (Proposition \ref{allAreGeometric}),
 it suffices to show that
 all compact isometric quotients of the solvable 4-dimensional geometries bound.
Theorem \ref{mybound} is established by exhibiting an involution (if not free)
with 2-dimensional fixed set on infra-solvmanifolds with  $\Nilr$, $\Nilf$, $\Solr$, or $\Solof$ geometry.
We will show that the Stiefel-Whitney number $\omega^4_1(M)$ vanishes by a result
of R.E. Stong \cite{Stong}, and from this, it will follow that all Stiefel-Whitney numbers vanish.

\section{Translational Involutions}

We construct an involution on an infra-solvmanifold
$M = \varPi \bs G$
when the center of $G$, $\calz(G)$, is non-trivial. The
involution is induced by left translation.
This technique
was used to show closed flat $n$-manifolds bound \cite{HR, MGor}.

\begin{lemma}
\label{involution}
Let $M = \varPi \bs G$ be an infra-solvmanifold with $\calz(G)$
non-trivial. Note $\Gamma \cap \calz(G)$ is a lattice of $\calz(G)$.
Let $t$ be a free generator of $\Gamma \cap \calz(G)$ and set $s = t^{\frac 1 2}$.
Translation by
$s$ induces an involution on $M$ if and only if $A(s) = s$ modulo $\Gamma \cap \calz(G)$, for all $A \in \Phi$. That is, translation by $s$ commutes with the action
of $\Phi$ on $\Gamma \bs G$.
\end{lemma}
\begin{proof}
Since $s$ commutes with $\Gamma$, translation by
$s$
defines a free involution on the solvmanifold $\Gamma \bs G$.
To induce an involution on $M$,
translation by $s$ must normalize the action of $\Phi$ on $\Gamma \bs G$.
For any $(a,A) \in \varPi$, we have
\begin{align*}
(s,\id) (a,A) (-s, \id)
&=(s\cdot a\cdot A s\inv, A)\\
&=((I-A)s \cdot a, A)\text{ (since $s\in\calz(G)$)}.
\end{align*}
Therefore, $s$ induces an involution on $M$
when $(I-A)s\in \Gamma \cap \calz(G)$; that is,
$A(s) = s$ modulo $\Gamma \cap \calz(G)$, for all $A \in \Phi$. Note that $\calz(G) \cong \bbr^k$ for some $k$, and $s \in \calz(G)$. Note that $A$ restricts to a linear map
on $\calz(G)$, so that we may write $s\cdot a\cdot A s\inv = (I-A)s$.
\qedhere
\end{proof}


\section{Structure of Fixed Set}

Let $\hat M$ denote the solvmanifold $\Gamma \bs G$.
We have the coverings
$$
\CD
G @>{q}>> \mhat @>{p}>> M.\\
\endCD
$$
We refer
to the involutions induced by translation by $s$ as \emph{translational involutions}.
Let ${ \hat i}_s : \hat M \ra \hat M$ denote the induced involution on $\hat M$,
$i_s : M \ra M$ denote the induced involution on $M$, and
$F$ denote the fixed set of $i_s$ on $M$.

\begin{lemma}
\label{struc1}
The preimage of $F$ in $\hat M$ is a finite disjoint union of closed, connected, submanifolds.
We can write
\begin{align*}
p^{-1}(F) = \bigcup_{\eta} E_{\eta},
\end{align*}
where the union is over all possible injective homomorphisms
$\eta: \angles{{ \hat i}_s} \cong \bbz_2 \ra \Phi$ and
\[
E_{\eta} = \{\xhat \in \mhat\ |\ s(\xhat) = \eta({ \hat i}_s)(\xhat) \}.
\]
Each $E_{\eta}$ is a finite disjoint union of components of $p \inv (F)$.
\end{lemma}
\begin{proof}
The fixed set $F$ of the translational involution must be a finite disjoint union of closed connected submanifolds
\cite[p.72]{CF}. Since $p$ is a finite sheeted covering, $p^{-1}(F)$ also admits the structure of a finite
disjoint union of closed connected submanifolds.

If $\xhat \in p^{-1}(F)$, then $s(\xhat) = (a,A)(\xhat)$
for some unique $A \in \Phi$ where $(a,A) \in \varPi$.
Thus,
\[
\xhat = s^2(\xhat) = (a,A)^2(\xhat).
\]
Since the deck transformation group acts freely,
$(a,A)^2 \in \Gamma$, and thus $A^2=I$.
So $\eta({ \hat i}_s)=A$ defines an injective homomorphism
$\eta:\bbz_2 \ra \Phi$. We warn the reader the action of
$\eta({ \hat i}_s)=A \in \Phi$ on $\mhat$ is induced not just by the automorphism $A$, but rather by
the affine transformation
$(a,A)$.
The preimage of $F$ in $\mhat$ is indexed by all
possible injective homomorphisms $\eta: \bbz_2 \ra \Phi$. That is,
\begin{align*}
p^{-1}(F) = \bigcup_{\eta} E_{\eta}.
\end{align*}
Note that $E_{\eta_1} = E_{\eta_2}$ when $\eta_1 = \eta_2$ and
  $E_{\eta_1} \cap E_{\eta_2} = \emptyset$ otherwise.

The actions of $s$ and $\eta({ \hat i}_s)$ commute on $\hat M$ by Lemma \ref{involution}.
By definition, $E_\eta$ is the fixed set of the involution $\eta({ \hat i}_s) \inv \circ { \hat i}_s =
\eta({ \hat i}_s) \circ { \hat i}_s$ on $\hat M$.  So it must be a finite disjoint union of
closed connected submanifolds \cite[p.72]{CF}, and therefore must be a finite disjoint union of
components of  $p^{-1}(F)$.
\end{proof}

When $\exp: \mathfrak g \ra G$ is a diffeomorphism,
for $A \in \aut(G)$, we have $\fix(A) = \exp(\fix(A_*))$,
where $A_*$ is the automorphism of $\mathfrak g$ induced from $A$.
When $G$ is a 4-dimensional solvable geometry, $G$ is type (E)
and $\exp$ is a diffeomorphism.
So $\fix(A)$ is always diffeomorphic to $\bbr^k$. We have the diagram
of coverings, where the vertical arrows are inclusions.
$$
\CD
G @>{q}>> \mhat @>{p}>> M\\
@AAA @AAA @AAA\\
 \bigcup_{\eta} q^{-1}(E_{\eta}) @>{q}>> \bigcup_{\eta} E_{\eta} @>{p}>> F.\\
\endCD
$$
Now
we analyze $q^{-1}(E_{\eta})$.




\begin{lemma}
\label{struc2}
Assume that $\exp: \mathfrak g \ra G$ is a diffeomorphism. The preimage of $E_\eta$ in $G$ under $q: G \ra \hat M$
is a disjoint union of submanifolds of $G$. In fact,
if $\eta({ \hat i}_s)=A \in \Phi$ with $(a,A) \in \varPi$, then any component of $q^{-1}(E_\eta)$
is $\fix(\gamma s \inv  a,A) $ for some $\gamma \in \Gamma$.
Consequently,  the preimage of $E_{\eta}$ in $G$ is
\begin{align*}
q^{-1}(E_\eta) = \bigcup_{\gamma \in \Gamma} \fix(\gamma s \inv  a,A).
\end{align*}
Further, each $\fix(\gamma s \inv  a,A)$
is a left translate of the connected subgroup $\fix(A)$ of $G$ and is
diffeomorphic to $\bbr^n$, where $n = \dim(\fix(A))= \dim(\fix(A_*))$.
\end{lemma}
\begin{proof}
Because $E_\eta$ is a disjoint union of closed submanifolds and $q$ is a
covering, $q^{-1}(E_\eta)$ is a (possibly not connected) submanifold of $G$ without boundary.

Let $A
= \eta({ \hat i}_s)$ and let $(a,A) \in \varPi$. An element $\tilde x \in G$
projects to $\xhat \in E_{\eta}$ if and only if there exists
$\gamma \in \Gamma$ such that
$s({\tilde x}) = \gamma (a,A)(\tilde x)$, or equivalently,
\[
{\tilde x} = (\gamma s\inv a,A)(\tilde x).
\]
That is, $\tilde x$ must be in the fixed set of the affine transformation
$(\gamma s\inv a,A)$.
Consequently,  the preimage of $E_{\eta}$ in $G$ is
\begin{align*}
q^{-1}(E_\eta)=\bigcup_{\gamma \in \Gamma} \fix(\gamma s \inv  a,A).
\end{align*}

Some sets in the above union may be empty.
The fixed set of an affine transformation, if non-empty,
is just a translation of the fixed subgroup of its automorphism part; that is,
if $x_{0} \in \fix(b,B)$, then \[\fix(b,B) = x_{0} \fix(B).\]
To see this, note that if $x_{0} y \in x_{0} \fix(B)$ (so $y \in \fix(B)$),
then
\begin{align*}
(b,B)(x_{0}y) &= bB(x_{0})B(y)\\
&= (b,B)(x_{0}) y \\
&= x_{0} y.
\end{align*}
For the other inclusion, note that if $z \in \fix(b,B)$, then $z=x_{0} x_{0} \inv z$. We claim that $x_{0} \inv z$ is in $\fix(B)$:
\begin{align*}
B(x_{0} \inv z) &= B(x_{0}) \inv B(z)\\
&= B(x_{0}) \inv b \inv b B(z) \\
&=  \left[ (b,B)(x_{0}) \right] \inv (b,B)(z)\\
&= x_{0} \inv z.
\end{align*}

Any two left translates of $\fix(A)$ are either disjoint or equal.
Since $\exp$ is a diffeomorphism, any left translate of $\fix(A)$ is a submanifold of $G$
diffeomorphic to $\bbr^n$, where $n=\dim(\fix(A))$.
Since $\Gamma$ is countable, $q^{-1}(E_\eta)$ is expressed as a countable
union of submanifolds of $G$,  each of which has dimension $\dim(\fix(A))$. This forces
each component of the submanifold $q^{-1}(E_\eta)$ to have dimension equal to
that of $\fix(A)$.

In fact, we claim a component $\tilde E_\eta$ of $q^{-1}(E_\eta)$ is equal
to $\fix(\gamma s \inv  a,A)$ for some $\gamma \in \Gamma$.
The argument above shows that $\tilde x \in \tilde E_\eta$ belongs to
$\fix(\gamma s \inv  a,A)$ for some $\gamma \in \Gamma$.
Since  $\fix(\gamma s \inv  a,A)$
is connected, \[\fix(\gamma s \inv  a,A) \subset \tilde E_\eta.\] Also, $\fix(\gamma s \inv  a,A) $
is closed in $\tilde E_\eta$, since it is closed in $G$. Note that the inclusion
$\fix(\gamma s \inv  a,A) \hookrightarrow \tilde E_\eta$
is open by invariance of domain, as both manifolds have the same dimension.
Consequently, $\fix(\gamma s \inv  a,A)=\tilde E_\eta$.
\end{proof}


An important consequence of Lemma \ref{struc2} is that
all components of $F$ lifting
to $E_\eta$ must have the same dimension equal to that of $\fix(\eta({ \hat i}_s))$,
where $\eta({ \hat i}_s) \in \Phi$ is the unique automorphism of $G$ coming from $\eta$.

%2.3
\begin{lemma}
\label{empty}
Let $\varPi \backslash G$ be an infra-solvmanifold
with translational involution induced by $s$.
Suppose that $A \in \Phi$ has order 2  and
$A(s) = s \inv$. Let $\eta: \bbz_2 \ra \Phi$ be
the homomorphism $\eta({ \hat i}_s)=A$. Then $E_{\eta} = \emptyset$.
\end{lemma}
\begin{proof}
Let $\alpha = (a,A) \in \varPi$ and
define $\varPi' = \angles{\Gamma, \alpha}$. Note that $\varPi' \bs G$ is an infra-solvmanifold
with $\bbz_2$ holonomy and translational involution $i'_s$ induced by $s$.

We claim that the group generated by $\varPi'$ and $s$, $\angles{\varPi',s}$,
is torsion-free.
A general element of $\angles{\varPi',s}$ with holonomy $A$ is of the form
$(s\gamma a, A)$, where $\gamma\in \Gamma$. We have to check that $(s\gamma a, A)^2$ is not the identity, for any $\gamma \in \Gamma$. Now
\begin{align*}
(s\gamma a, A)^2 &= (s \gamma a A(s) A(\gamma) A(a),\id)\\
&= (\gamma a A(\gamma a), \id),\ \text{ (since $s\in\calz(G)$)}\\
&=(\gamma a, A)^2 \neq (e, \id),
\end{align*}
and this cannot be the identity because $(\gamma a, A) \in \varPi$ and $\varPi$ is torsion-free.

Consequently, $\angles{\varPi',s}$ is torsion-free and hence $\angles{\varPi',s}$ acts freely
on $G$.
Therefore $s$ acts as a free involution on
the infra-solvmanifold $\varPi' \bs G$.
Note that the preimage of $\fix(i'_s) \subset \varPi' \bs G$
in $\Gamma \bs G$ under the double covering $\Gamma \bs G \ra \varPi' \bs G$
is precisely $E_{\eta}$. Hence $E_{\eta}$ must be empty.
\qedhere
\end{proof}


\section{Dimension of Fixed Set}
\label{dimfix}
Given a 4-dimensional infra-solvmanifold $\varPi \bs G$ with
translational involution $i_s$ induced by translation by $s$ as defined
in Lemma \ref{involution}, the fixed set $F$
will be a disjoint union of submanifolds. We will need
to compute the dimension of $\fix(i_s)$.
By Lemma \ref{struc2}, a component of $F$
lifts to $G$ as a left translate of $\fix(\eta({ \hat i}_s))$, where
$\eta({ \hat i}_s)$ is an involution in $\aut(G)$.
Every involution in $\aut(G)$ belongs to a maximal
compact subgroup $K$ of $\aut(G)$.
When $G$ is one of the 4-dimensional solvable geometries,
$\aut(G)$ has finitely many components, as $\aut(G)$ is algebraic.
A result of Mostow \cite[Theorem 3.1]{mostow} implies that all
maximal compact subgroups of $\aut(G)$ are conjugate.
Therefore, we can
fix a maximal compact subgroup $K$ and compute $\dim(\fix(A))$
for each involution $A$ in $K$.
% 
\begin{lemma}[{\cite[p.135]{hillman2}}]
\label{dimNilR}
{\rm (1) }
A maximal compact subgroup
of $\aut(\Nilr)$ is \[\om(2,\bbr) \x \bbz_2. \]

{\rm (2) }
If $A \in \om(2,\bbr) \x \bbz_2$ restricts to the identity on $\bbr$ and has order 2,
then \[\dim(\fix(A))=2.\]
\end{lemma}
\begin{proof}
A maximal compact subgroup of $\aut(\Nil^3)$ is $\om(2,\bbr)$ and
acts as
{\small
$$
A=
\kern-4pt
\left[\begin{matrix}
a&b\\ c&d
\end{matrix}\right]
:\
\left[
\begin{matrix} 1 &x &z\\
               0 &1 &y\\
               0 &0 &1 \end{matrix}
\right]
\longmapsto
\left[
\begin{matrix} 1
&\kern-7pt (a x+b y)
&\kern-7pt
\frac12 (a b x^2+2 b c x y+c d y^2+2 \det(A)z)\\
               0 &1 &(c x+d y)\\
               0 &0 &1 \end{matrix}
\right].
$$}
The induced action of $A \in \om(2,\bbr)$ on $\calz(\Nil^3)$
is multiplication by $\det(A)$. Therefore, $\fix(A)$ is
1-dimensional on $\Nil^3$ for all $A \in \on{O}(2,\bbr)$.
A maximal compact subgroup of $\aut(\Nilr)$ is
$\om(2, \bbr) \x \bbz_2$, where $\om(2, \bbr) \subset \aut(\Nil^3)$
and $\bbz_2$ acts as a reflection on $\bbr$. Thus, if $A \in \om(2, \bbr) \x \bbz_2$
restricts to the identity on $\bbr$, then $\dim(\fix(A))=2.$\qedhere
\end{proof}


\begin{lemma}
\label{dimNilF}
{\rm (1)}
A maximal compact subgroup of $\aut(\Nilf)$ is $\bbz_2 \x \bbz_2$.

{\rm (2)}
Let $A \in \aut(\Nilf)$ have order 2. If
$A$ restricts to the identity on $\calz(\Nilf)$, then $\dim(\fix(A))=2.$

\end{lemma}
\begin{proof}
Recall the splitting of $\Nilf$ as the semidirect product
$\bbr^3 \rx \bbr$. Letting $\mathfrak g$ denote the Lie algebra
of $\Nilf$, we have $\mathfrak g \cong \bbr^3 \rx \bbr$,
where $\bbr$ acts by the matrix
\[
\left[\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{matrix}\right].
\]
With standard bases $\bfe_1,\bfe_2,\bfe_3$ of $\bbr^3$
and $\bfe_4$ of $\bbr$, $\mathfrak g$
has relations
\[
\left[ \bfe_4, \bfe_2 \right] = \bfe_1 ,
\left[ \bfe_4, \bfe_3 \right] = \bfe_2.
\]
For $A \in \aut(\mathfrak g)$, $A$ induces an action
on the quotient
$\mathfrak g / \left[\mathfrak g,\mathfrak g \right] \cong \angles{\bfe_3,\bfe_4}$,
denote this action by
\[B=\left[\begin{matrix} a & b \\ c & d \end{matrix}\right].\]
We will see that $B$ determines the action of $A$
on $\bfe_1$ and $\bfe_2$.
The relation $\left[\bfe_4, \bfe_3\right] = \bfe_2$, implies
$A(\bfe_2) = \det(B) \bfe_2$. Compactness forces
 $\det(B)=\pm 1$. We also compute
\begin{align*}
A(\bfe_1)&=A(\left[\bfe_4, \bfe_2\right]) =\left[A(\bfe_4), A(\bfe_2)\right]\\
&=\left[ b \bfe_3 + d \bfe_4, \det(B) \bfe_2 \right] = \det(B) d \bfe_1.
\end{align*}
Again, compactness implies $d = \pm 1$. Since $\left[\bfe_3, \bfe_2\right]$ vanishes,
 we have
 \begin{align*}
 0&=A(\left[\bfe_3, \bfe_2\right]) =\left[A(\bfe_3), A(\bfe_2)\right]\\
 &=\left[ a \bfe_3 + c \bfe_4, \det(B) \bfe_2 \right] = \det(B) c \bfe_1.
 \end{align*}
 Thus $c$ vanishes and $B$ must be upper triangular of the form
 \[B=\left[\begin{matrix} \pm 1 & b \\ 0 & \pm 1 \end{matrix}\right].\]
By conjugation we can set $b=0$. Thus, a maximal compact subgroup of $\aut(\Nilf)$
cannot be larger than $\bbz_2 \x \bbz_2$.
Conversely, we see that
\begin{align}
\label{maxnilf}
\bbz_2 \x \bbz_2=\{(\pm I_3,1),(\pm J,-1)\}\subset\aut(\bbr^3 \rx \bbr),
\end{align}
where
\[
J=
\left[\begin{matrix}
1&0&0\\ 0&-1&0\\ 0&0&1
\end{matrix}\right],
\]
defines a subgroup of $\aut(\Nilf)$. It now follows that a maximal compact subgroup
of $\aut(\Nilf)$ is $\{(\pm I_3,1),(\pm J,-1)\} \cong \bbz_2 \x \bbz_2$.


Note that $(-I_3,1)$ and $(-J,-1)$ do not act as the identity on $\calz(\Nilf)$.
The remaining involution $(J,-1)$ restricts to the identity on $\calz(\Nilf)$
and has 2-dimensional fixed subgroup.
\end{proof}


\begin{lemma}
\label{dimSolof}
{\rm (1)}
A maximal compact subgroup of $\aut(\Solof)$ is $D_4$.

{\rm (2)}
Let $A \in \aut(\Solof)$ have order 2. If
$A$ restricts to the identity on $\calz(\Solof)$, then \[\dim(\fix(A))=2.\]
\end{lemma}
\begin{proof}
Both $\aut(\Sol)$ and $\aut(\Solof)$ have $D_4$ as their maximal compact subgroup (Propositions \ref{autosol} and \ref{autosolof}),
\[ D_4= \left\langle \left[\begin{matrix} 0 & -1 \\ 1 & 0
\end{matrix}\right],
\left[\begin{matrix} 1 & 0 \\ 0 & -1
\end{matrix}\right] \right\rangle = \bbz_4 \rx \bbz_2. \]
For $A \in D_4$, let $\abar$ be $+1$ if $A$ is diagonal, and $-1$
if $A$ is off-diagonal. Then $A$ acts on $\Sol=\bbr^2 \rx \bbr$ as

\[
A: \left(\xy{x}{y},u\right) \longmapsto \left(A\xy{x}{y}, \abar u\right),
\]

\noindent and on $\Solof$ as

{\small
$$
A=
\kern-4pt
\left[\begin{matrix}
a&b\\ c&d
\end{matrix}\right]
:\
\left[
\begin{matrix} 1 &e^u x &z\\
               0 &e^u &y\\
               0 &0 &1 \end{matrix}
\right]
\longmapsto
\left[
\begin{matrix} 1
&\kern-7pt e^{\abar u}(a x+b y)
&\kern-7pt
\frac12 (a b x^2+2 b c x y+c d y^2+2 \det(A)z)\\
               0 &e^{\abar u} &(c x+d y)\\
               0 &0 &1 \end{matrix}
\right].
$$}

For both $\Sol$ and $\Solof$, $\abar$ is the induced action of $A$ on
$\Sol /\bbr^2 \cong \Solof /\Nil \cong \bbr$. Note that
multiplication by $\det(A)$
is the induced action of $A$ on $\calz(\Solof)$. Thus, the only involution
in $D_4$ restricting to the identity on $\calz(\Solof)$ is
$\left[ \begin{matrix} -1 & 0 \\ 0 & -1\end{matrix} \right]$ and evidently it
has 2-dimensional fixed subgroup.
\qedhere
\end{proof}

\begin{lemma}
\label{dimSolR}
{\rm (1)} A maximal compact subgroup of $\aut(\Solr)$ is \[D_4 \x \bbz_2,\]
where $D_4 \subset \aut(\Sol)$ and $\bbz_2$ acts as a reflection on $\bbr$.

{\rm (2)} Let $A \in D_4 \subset D_4 \x \bbz_2 \subset \aut(\Solr)$ have order 2.

If
$A = \left[ \begin{matrix} -1 & 0 \\ 0 & -1\end{matrix} \right]$ or
$\pm \left[ \begin{matrix} 0 & 1 \\ 1 & 0\end{matrix} \right]$, then
\[ \dim(\fix(A)) =2.\]

If
$A =
\pm \left[ \begin{matrix} 1 & 0 \\ 0 & -1\end{matrix} \right]$, then
\[ \dim(\fix(A)) =3.\]


\end{lemma}
\begin{proof}
The first statement is clear since a maximal
compact subgroup of $\aut(\Sol)$ is $D_4$. For
the second statement, recall that the induced
action of $A$ on the quotient $\Sol / \bbr^2 \cong \bbr$
is $+1$ if $A$ is diagonal and $-1$ otherwise.
\qedhere
\end{proof}

%Whenever $G$ is one of $\Nilr, \Nilf, \Solof$ or $\Solr$,
%we can equip $G$ with left invariant metric so that
%\[\isom(G)=G \rx K,\]
%where $K$ is one of the maximal compact subgroups of
%$\aut(G)$ described in Lemmas \ref{dimNilR}, \ref{dimNilF},
%\ref{dimSolof}, \ref{dimSolR}.




\section{Proof of Bounding}

The following relations among Stiefel-Whitney classes for 4-manifolds are known.
%See for example, \cite[p. 96]{} or \cite[Theorem 4.10]{milnorCC}.
\begin{lemma}[{\cite{StongBook}}]
For any 4-manifold $M$,

{\rm (1)} $\omega_{1}^2 \omega_{2} = \omega_{1} \omega_{3} = 0$

{\rm (2)} $\omega_{2}^2 = \omega_{1}^4 + \omega_{4}$

Therefore, $M$ is a boundary if and only if the Stiefel-Whitney numbers
$\omega_{1}^{4}(M)$ and $\omega_{4}(M)$ are $0$. 
\end{lemma}

A solvmanifold $\Gamma \bs G$ is parallelizable since
one can project a framing of left invariant vector fields
from $G$ to $\Gamma \bs G$. Hence the Euler characteristic
$\chi(\Gamma \bs G)$ vanishes. Since any infra-solvmanifold $\varPi \bs G$
is finitely covered by a solvmanifold, $\chi(\varPi \bs G)= 0$.
Therefore, the mod 2 Euler characteristic, which is $\omega_{4}(M)$,
vanishes. Hence the only Stiefel Whitney consider is $\omega_{1}^{4}(M)$.

\begin{proposition}[{\cite[Proposition 9.2]{Stong}}]
\label{Stong}
A manifold $M^n$ is unoriented cobordant to a manifold $M'$ with
differentiable involution having a fixed set of dimension $n-2$
if and only if $\omega_{1}^n (M)=0$.
\end{proposition}

We will use this result to show that $\omega_{1}^4 (M)=0$ for a 4-dimensional infra-solvmanifold, by constructing an involution with 2-dimensional fixed set. It is well known that any closed manifold with free involution bounds. So, henceforth we assume our involution to have non-empty fixed set.

We will also need the following result on
the crystallographic groups of $\Sol = \bbr^2 \rx \bbr$.
Note that the nil-radical of $\Sol$ is $\bbr^2$.
Let $\varPi \subset \isom(\Sol)=\Sol \rx D_4$ be a crystallographic group with
lattice $\Gamma$ and holonomy $\Phi$. Recall the action of
$D_4$ as automorphisms of $\Sol$ from Lemma \ref{dimSolof}.
Let \[\pr_{1}:\Sol \ra \Sol / \bbr^2 \cong \bbr.\] denote the quotient map.
If $\Gamma$ is a lattice of $\Sol$, then $\Gamma$ meets the nil-radical
in a lattice $\Gamma \cap \bbr^2
\cong \bbz^2$ and $\pr_{1}(\Gamma) \cong \bbz$ is a lattice
of $\bbr$ \cite[Corollary 8.28]{rag}.


\begin{proposition}[Lemma \ref{PPM-A30}]
\label{force}
Let $\varPi \subset \isom(\Sol)=\Sol \rx D_4$ be crystallographic
and let $v$ denote a generator of $\pr_{1}(\Gamma) \cong \bbz$.

\noindent
If $(b,B) \in \varPi$ where $B= \pm \left[
\begin{matrix}
1 & 0 \\
0 & -1
\end{matrix}
\right] \subset D_4$, then $\pr_{1}(b) = v^{\frac 1 2}.$
\end{proposition}
\noindent
%Also see \cite[Lemma 3.4]{solof-geo}.


For $\Nilr$ geometry manifolds,
we need to study the holonomy representation
\[\rho: \Phi \ra \aut(\calz(\Nilr))\]
closely.


\begin{lemma}
\label{holonomyRep}
Let $M=\varPi \bs G$ with  $G=\Nilr$. Then there is a set of generators
$t_1, t_2$ for $\Gamma \cap \calz(G) \cong \bbz^2$,
$t_1 \in \calz(\nil^3)$, so that with respect to the basis $t_1, t_2,$

\noindent
\rm{(1) }$\rho(\Phi)  \subset
\left\langle
\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right],
\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \end{matrix} \right]
\right\rangle
$, or

\noindent
\rm{(2)} $\rho(\Phi)  \subset
\left\langle
\left[ \begin{matrix} 1 & 1 \\ 0 & -1 \end{matrix} \right],
\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \end{matrix} \right]
\right\rangle.
$
\end{lemma}
\begin{proof}
We have $\calz(G) = \calz(\Nil^3) \x \bbr$.
Since \[ \left[ G , G \right]  = \calz(\Nil^3)=\bbr,\]
$\calz(\Nil^3)$ is invariant
under any automorphism of $G$. Further, $\Gamma \cap \calz(\Nil^3)$ is
a lattice of $\calz(\Nil^3)$ \cite[Lemma 1.2.5]{Dekimpe}.

Note that $\rho(\Phi)$ has $\calz(\Nil^3)$ as an invariant subspace.
Because $\rho(\Phi)$ can be
conjugated (over $\gltr$) into $\on{O}(\calz(G))$, we can assume that
it leaves the orthogonal complement of $\calz(\Nil^3)$
invariant as well.
 The maximal compact subgroup of $\on{O}(\calz(G))$ leaving $\calz(\Nil^3)$ invariant
is
\[\bbz_2 \x \bbz_2 =\left\langle
 \left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right],
 \left[ \begin{matrix} -1 & 0 \\ 0 & -1 \end{matrix} \right]
\right\rangle.\]
 Thus, over $\gltr$, $\rho(\Phi)$ can be conjugated into this $\bbz_2 \x \bbz_2$.
 But over $\gltz$, there is one more case.

 Let $t_1, t_2$ be two generators of $\varPi \cap \calz(G) \cong \bbz^2$,
 where $t_1$ generates $\varPi \cap \calz(\Nil^3)$.
 It is known that an involution $A \in \gltz$ with
 vanishing trace is $\gltz$ conjugate to either
$\left[ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right]$ or
 $\left[ \begin{matrix} 1 & 1 \\ 0 & -1 \end{matrix} \right]$ \cite{CampTrouy}.
 Using this, it is not hard to see that we can
 keep $t_1$ the same, but change $t_2$ to $t'_2 = a t_1 \pm t_2$ for $a \in \bbz$,
 to put $\rho(\Phi)$ in the desired form.
\qedhere
\end{proof}


\noindent
We are now ready to prove the main theorem.

\begin{theorem}
\label{mybound}
All 4-dimensional infra-solvmanifolds are boundaries.
\end{theorem}
\begin{proof}

The flat 4-dimensional manifolds $M= \varPi \bs \bbr^4$
are all boundaries by Hamrick-Royster \cite{HR}.

When $G=\Solmn$ or $\Solo$, any compact form $M = \varPi \bs G$ is
a mapping torus of $T^3$, and is therefore a $T^3$
bundle over $S^1$ (\cite[Corollary 8.5.1]{hillman2}
and \cite[Theorem 3.5, Theorem 4.2]{unimodular-groups}). Because $T^3$ is orientable, $\omega_1(M)$ is induced
from the base of the fibration.
That is, let $p$ denote the projection
$p:M \ra S^1$. Now $\omega_1(M) = p^*(c)$ for some class
$c \in H^1(S^1; \bbz_2) \cong \bbz_2$. To see that $\omega_1$ is induced from a class in
the base of the fibration, note that
\[
H^1(M;\bbz_2) \cong \hom(H_1(M;\bbz),\bbz_2) \cong \hom(\pi_1(M),\bbz_2).
\]
It is known that $\omega_1(M)$, under the above isomorphism,
is the cohomology class which assigns $1$ to an element of $\gamma \in \pi_1(M)$ if
the tangent bundle is non-orientable when restricted to $\gamma$ and $0$ if the tangent
bundle of $M$ is orientable. Since the fiber of $M$ is orientable, the
$\omega_1(M)$ must come from the base. Then:
\[
\omega_1(M)^4 = (p^* (c) )^4 = p^* (c^4) = p^*(0) = 0.
\]
Of course, $\omega_4 (M)$ vanishes as well.

\medskip
\noindent
Case $G=\Nilf,\Solof:$
\medskip

For $M=\varPi \bs G$ when $G$ is $\Nilf$ or $\Solof$,
let $s=t^{\frac12}$ where $t$ is a generator of $\Gamma\cap \calz(G)$.
We consider the translational involution defined in Lemma \ref{involution}.
Our explicit computation of maximal compact subgroups of
$\aut(\Nilf)$ and $\aut(\Solof)$ shows that $A(s)=\pm s$ for any
holonomy $A \in \Phi$ (recall that $s$ is central). This also follows since $\calz(G)$ is
invariant under any automorphism and 1-dimensional. Let $\eta:\bbz_2=\angles{s}\ra\Phi$ be an injective homomorphism.
If $\eta(s)(s)=-s$, then $E_\eta=\emptyset$ by Lemma \ref{empty}.
If $\eta(s)(s)=s$, then
$\eta(s)$ acts as the identity on
$\calz(G)$ and $\eta(s)$ has 2-dimensional fixed subgroup on $G$
(Lemmas \ref{dimNilF} and \ref{dimSolof}). By Lemma \ref{struc2},
$E_\eta$ is 2-dimensional.
Therefore, $\fix(s)$ is 2-dimensional.

\medskip
\noindent
Case $G=\Solr:$
\medskip

Now consider a $\Solr$ geometry manifold $\varPi \bs G$.
Let $s=t^{\frac12}$ where $t$ is a generator of $\Gamma\cap \calz(G)$.
Since $A(s)=\pm s$ for all $A \in \Phi$, $s$ defines an involution
on $\varPi \bs G$.
Let $\eta:\bbz_2=\angles{s}\ra\Phi$ be an injective homomorphism.
If $\eta(s)(s)=-s$, then $E_\eta=\emptyset$ by Lemma \ref{empty}.

In the $\Solr$ geometry case, not all involutions in $\Phi$
inducing the identity on $\calz(\Solr)$ have 2-dimensional fixed subgroup
(Lemma \ref{dimSolR}).
When $\eta(s)=A$, where $A$ is one of
$\left[
\begin{matrix}
-1 & 0 \\
0 & -1
\end{matrix}
\right]$ or $\pm \left[
\begin{matrix}
0 & 1 \\
1 & 0
\end{matrix}
\right]$,
$E_\eta$ is 2-dimensional, since $\fix(A)$
is 2-dimensional. But for \[B=\pm \left[
\begin{matrix}
1 & 0 \\
0 & -1
\end{matrix}
\right], \]
$\fix(B)$ is 3-dimensional.
However, we shall see that when
$\eta(s)=B$, $E_\eta$ is empty.
Note that the nil-radical of $\Solr$ is $\bbr^3$ with quotient $\bbr$.
Let \[\pr:\Solr \ra \ \bbr\] denote the quotient homomorphism by the nil-radical of $\Solr$.
If we let $\pr_{2}:\Solr \ra \Sol$ denote the quotient of $\Solr$
by its center
and let $\pr_{1}: \Sol \ra \bbr$ denote the quotient
of $\Sol$ by its nil-radical, then $\pr$ factors
as $\pr_{1} \circ \pr_{2}$,

\[
\CD
\pr: \Solr @>{\pr_{2}: /\calz(\Solr)}>> \Sol @>{\pr_{1}: / \bbr^2}>> \bbr. \\
\endCD
\]

Now $\pr(\Gamma)$ is a lattice of $\bbr$. Let $v$ denote a generator of $\pr(\Gamma)$. By
Lemma \ref{struc2}, the preimage of $E_\eta$ in $\Solr$ is given by,
\begin{align*}
\bigcup_{\gamma \in \Gamma} \fix(\gamma s \inv  b,B) \text{ for } (b,B) \in \varPi.
\end{align*}
However, all sets $\fix(\gamma s \inv  b,B)$ are empty for any $\gamma \in \Gamma$.
To see this, suppose $x \in \Solr$ satisfies
\[
\gamma s \inv  b B(x) = x.
\]
We will apply $\pr=\pr_{1} \circ \pr_{2}$ to both sides.
Note that $\pr(b) = v^{\frac 1 2}$ by Proposition \ref{force},
$\pr(\gamma) = v^n$ for some $n \in \bbz$, $\pr(s)=0$, and \[\pr(B(x)) = {\bar B} (\pr(x)) =
\pr(x)\]
(since $B$ is diagonal, ${\bar B}=+1$).
Thus, application of $\pr$ yields
\[
v^{n + \frac 1 2} +\pr(x)= \pr(x),
\]
which is a contradiction. This shows that $E_\eta$ is empty
when $\eta(s) = B$.

Therefore, $\fix(s)$ has no 3-dimensional
components and is 2-dimensional
in the $\Solr$ geometry case.

\medskip
\noindent
Case $G=\Nilr:$
\medskip

Finally, consider a $\Nilr$ geometry manifold
$\varPi \bs G$. Now $\calz(\Nilr) = \calz(\Nil^3) \x \bbr$.
We will use either $\calz(\Nil^3)$ or $\bbr$ to induce an involution
on $\varPi \bs G$  depending on which case of Lemma \ref{holonomyRep}
occurs.

Suppose
 we can take $t_1, t_2$ with $t_1 \in \calz(\Nil^3)$,
 as a generating set of $\Gamma \cap \calz(G)$ so
 that $\rho(\Phi)$ is diagonal for this generating set (case \rm{(1)}
 of Lemma \ref{holonomyRep}).
 Take $s= t^{\frac 1 2}_2$ for our involution on $M$.
 Lemma \ref{empty} implies that $E_\eta$ is non-empty only
 when $\eta(s)(s)=s$. This conditions means that $\eta(s)$ fixes
 the $\bbr$ factor in $\calz(\Nilr) = \calz(\Nil^3) \x \bbr$.
But all such $\eta(s)$ fixing $\bbr$ must
have a 2-dimensional fixed set on $\Nilr$ by Lemma \ref{dimNilR}.

Now suppose case \rm{(2)} of
Lemma \ref{holonomyRep} occurs. This time, we must take
$s = t^{\frac 1 2}_1$  for our involution.
If $\eta(s)$ acts as a reflection
on  $\calz(\nil^3)$, Lemma \ref{empty} implies $E_\eta$
is empty.

We claim further that $E_\eta$ is empty when  $\eta(s) \in \Phi$ with
$\rho(\eta(s))=\left[ \begin{matrix} 1 & 1 \\ 0 & -1 \end{matrix} \right].$
To see this,
note that translation by $s = t^{\frac 1 2}_1$ is also induced
from translation by $s' = t^{\frac 1 2}_1 t^{-1}_2$. But $\eta(s)(s')=s'^{-1}$.
Hence Lemma \ref{empty} implies that $E_\eta$ is empty
in this case.

Thus we conclude that the only non-empty components of
$\fix(s)$ can arise from $\eta(s) \in \Phi$ with $\rho(\eta(s)) = \id$.
But Lemma \ref{dimNilR} implies that all such $\eta(s)$
have 2-dimensional fixed set. Therefore, when $G=\Nilr$, in either case
of Lemma \ref{holonomyRep}, we have an involution
with fixed set of constant dimension 2.

For any manifold with $\Nilr, \Nilf, \Solof$, or
$\Solr$ geometry, we have constructed an involution
with 2-dimensional fixed set. By Stong's result (Proposition
\ref{Stong}), $\omega_1^4(M)=0$.
Thus, all Stiefel-Whitney numbers are zero and we
have established that all 4-dimensional infra-solvmanifolds bound.
\qedhere
\end{proof}


\chapter{Crystallographic Groups of the 4-Dimensional Solvable Geometries}

In the sections below, we compute $\aut(G)$, and the maximal compact subgroup of $\aut(G)$, where $G$ is one of $\Sol$, $\Solr$, $\Solmn$, $\Solo$, and $\Solof$. 

\section{Computation of $\aut(G)$}

To understand $\aut(G)$ when $G$ is one of
$\Sol = \bbr^2 \rx \bbr$, $\Solmn = \bbr^3 \rx \bbr$,
or $\Solo=\bbr^3 \rx \bbr$, first note that
 $\Sol$ has nilradical $\bbr^2$, and both $\Solmn$ and $\Solo$ have nilradical
 $\bbr^3$.
The nilradical is invariant under automorphisms,
and we will apply the following lemmas.

\begin{lemma}
\label{autolemma}
Let $G = \bbr^n \rx_{\phi} \bbr$ and suppose that $\bbr^n$ is invariant under
all automorphisms of $G$. Then $\alpha \in \aut(G)$ restricts to an automorphism
$A$ of $\bbr^n$, and hence, induces an automorphism $\abar$ of the quotient
$\bbr$. There is a homomorphism
$$
\CD
\aut(\bbr^n \rx \bbr) @>>> \aut(\bbr^n)\x\aut(\bbr)\\
\alpha     @>>> (A,\abar)
\endCD
$$

\noindent
Furthermore, $A$ and
$\abar$ must satisfy
\begin{align}
\label{E-action}
\phi(\abar u)&=A \phi(u) A\inv
\end{align}
\end{lemma}
\begin{proof}
Let $f:\bbr\ra\bbr^n$ be the map (not a homomorphism in general) defined by
$\alpha({\mathbf 0},u)=(f(u),\abar(u))\in\bbr^n \rx_\phi \bbr$.
 Then
$
\alpha(\bfx,u)=(A\bfx+f(u),\abar(u)).
$
The equality
$\alpha\left((\bfx,u)(\bfy,v)\right)=\alpha(\bfx,u) \alpha(\bfy,v)$ yields
\begin{equation*}
\label{homo-cond}
f(u)+\phi(\abar u)(A \bfy +  f(v))= A (\phi(u)(\bfy)) + f(u+v).
\end{equation*}
For this to hold for every $\bfx,\bfy,u$, and $v$, we must have
$f(0)=\bzero$, which yields the desired result.
\qedhere
\end{proof}

\noindent
Next, we calculate the kernel of the map
$\aut(\bbr^n \rx_\phi \bbr) \lra \aut(\bbr^n)\x\aut(\bbr).$

\begin{lemma}
\label{kernelaut}
The kernel of the map
$\aut(\bbr^n \rx_\phi \bbr) \lra \aut(\bbr^n)\x\aut(\bbr)$
can be identified with the group of crossed homomorphisms $Z^1(\bbr,\bbr^n)$.
The crossed homomorphism $\eta \in Z^1(\bbr,\bbr^n)$ action is given by
\[
f(\bfx,u)=(\bfx+\eta(u),u),
\]
and satisfies the cocycle condition
\begin{align}
\label{1-cocycle}
\eta(u+v)=\eta(u)+\phi(u) (\eta(v)).
\end{align}
\end{lemma}
\begin{proof}
Let $f:\bbr^n \rx_\phi \bbr \ra \bbr^n \rx_\phi \bbr$ be an
automorphism which induces the identities on $\bbr^n$ and $\bbr$.
Then $f(\bfx,u)=(\bfx+\eta(u),u)$ for a map $\eta: \bbr \ra \bbr^n$.
We compute
\begin{align*}
f(\bfx,u) f(\bfy, v) &= (\bfx + \eta(u) + \phi(u)(\bfy) + \phi(u)(\eta(v)),u+v), \\
f(\bfx + \phi(u)(\bfy),u+v) &= (\bfx + \phi(u)(\bfy) + \eta(u+v),u+v),
\end{align*}
which yields the desired conclusion.
\qedhere
\end{proof}

\begin{remark}
\label{principal}
When $H^1(\bbr, \bbr^n)=0$, every crossed homomorphism is principal. This occurs when $\det(I-\phi(u))\neq 0$.
That is, for $\eta \in Z^1(\bbr, \bbr^n)$, there exists $\bfx_0 \in \bbr^n$ such that
$\eta(u) = \delta \bfx_0 = (I-\phi(u)) \bfx_0$. So when $H^1(\bbr, \bbr^n)=0$,
the correspondence  $\eta \ra \bfx_0$ gives an isomorphism between
the group of crossed homomorphisms and $\bbr^n$.
\end{remark}


\begin{proposition}
\label{autosol}
We have $\aut(\Sol) \cong \Sol \rx (\bbr^+ \x D_4)$, where $D_4$ is the dihedral group with 8 elements.
Under this isomorphism, $\Sol$ acts as inner automorphisms, and  $(\bbr^+ \x D_4)$ is identified
with the group of matrices
$\bbr^+ \x  D_4= \left\langle k \left[\begin{matrix} 0 & -1 \\ 1 & 0
\end{matrix}\right],
k \left[\begin{matrix} 1& 0 \\ 0 & -1
\end{matrix}\right] \right\rangle$, $k > 0$, ($k=1$ yields $D_4$), $A \in \bbr^+ \x  D_4$  acts on $\Sol$ as
\[
A: \left(\xy{x}{y},u\right) \longmapsto \left(A\xy{x}{y}, \abar u\right).
\]
($\abar = +1$ if $A$ is diagonal,
$\abar = -1$ otherwise.)
\end{proposition}
\begin{proof}
Any automorphism $\alpha$ restricts to an automorphism of the nilradical $\bbr^2$
of $\Sol$, so Lemma \ref{autolemma} applies.
Since eigenvalues are invariant under conjugation, $\phi(\abar u)$ and $A \phi(u) A\inv$ must have
the same eigenvalues. Hence $\abar = \pm 1$ is forced.
That is, we have
$$
\abar(u)=u\quad\text{or}\quad \abar(u)=-u.
$$
If $\abar(u)=u$ \Big($\abar(u)=-u$, respectively\Big), then $A$ must be of
the form
$$
A=
\left[\begin{matrix} a & 0\\
0 & d\end{matrix}\right] \in \gl(2,\bbr)
\left(A=
\left[\begin{matrix} 0 & b\\
c & 0\end{matrix}\right] \in \gl(2,\bbr), \text{ respectively}\right).
$$
Conversely, for any matrix $A$,
$$
A=
\left[\begin{matrix}
a&b\\ c&d
\end{matrix}\right]:\
\left(\xy{x}{y},u\right)
\longmapsto
\left(A\xy{x}{y},\abar u\right),
$$
where either $a=d=0$ or $b=c=0$, defines an automorphism of $\Sol$.

Thus the image of $\aut(\Sol)\ra\gl(2,\bbr)$ is
$$
\left\{
\left[\begin{matrix} a & 0\\ 0 & d\end{matrix}\right],
\left[\begin{matrix} 0 & b\\ c & 0\end{matrix}\right]:
a,b,c,d \in \bbr^*
\right\},
$$
and this group is isomorphic to
$\bbr^+\x(\bbr\rx D_4)$, where $\bbr^+$ is the scalar matrices with
positive entries, and
\[
\bbr=\left\{ E^c=
\left[\begin{matrix} e^{-c} & 0\\
0 & e^c\end{matrix}\right]
:\ c\in\bbr\right\}\quad\text{and}\quad
D_4=\left\{
\left[\begin{matrix} \pm 1 & 0\\
0 & \pm 1\end{matrix}\right],\
\left[\begin{matrix} 0 & \pm 1\\
\pm 1 & 0\end{matrix}\right]\right\}.
\]

By Lemma \ref{kernelaut}, we obtain a splitting
\[\aut(\Sol) \cong  Z_{\phi}^1(\bbr,\bbr^2) \rx  (\bbr^+\x(\bbr\rx D_4)).\] But on $\Sol$,
every crossed homomorphism is principal, so that $Z_\phi^1(\bbr, \bbr^2) \cong \bbr^2$
as described in Remark \ref{principal}. Thus we have,
\begin{align*}
\aut(\Sol) &\cong  \bbr^2 \rx  (\bbr^+ \x (\bbr\rx D_4))
\cong (\bbr^2 \rx \bbr) \rx  (\bbr^+ \x D_4) \\ &\cong \Sol \rx (\bbr^+\x D_4).
\end{align*}
For the last isomorphism, we observe that the action of $\bbr^2 \rx \bbr$
on $\Sol$ is by inner automorphisms. Since $\Sol$ has trivial center,
$\inn(\Sol) \cong \Sol$.
\qedhere
\end{proof}

\noindent
The computation of $\aut(\Solmn)$ and $\aut(\Solo)$ proceeds in an identical fashion.

\begin{proposition}
\label{autosolmn}
We have $\aut(\Solmn) \cong \bbr^3 \rx (\bbr^*)^3$. Here $(\bbr^*)^3$ is identified
with the group of diagonal matrices in $\gl(3, \bbr)$ and acts on the $\bbr^3$ factor linearly
in the semi-direct product. For $(\bfx_0, A) \in \bbr^3 \rx (\bbr^*)^3$, $(\bfx_0, A)$ acts on
$\Solmn$ as follows: for $(\bfx, u) \in \Solmn = \bbr^3 \rx_{\phi} \bbr$,
\[
(\bfx, u) \ra (A(\bfx) + (I-\phi(u))\bfx_0, u).
\]
\end{proposition}
\begin{proof}
Any automorphism $\alpha$ restricts to an automorphism of the nilradical $\bbr^3$
of $\Solmn$, so Lemma \ref{autolemma} applies.
Since eigenvalues are invariant under conjugation, $\phi(\abar u)$ and $A \phi(u) A\inv$ must have
the same eigenvalues. Hence $\abar = +1$ is forced. Thus
$\phi(u)$
and $A$ commute. Computation shows that $A$ must be diagonal.
Thus the image of $\aut(\Solmn)$ in $\aut(\bbr^3)$ consists only of diagonal matrices.
Conversely, when
$A$ is diagonal, $(\bfx, u) \ra (A\bfx, u)$ defines an automorphism of $\Solmn$.
So the group of invertible diagonal matrices $(\bbr^*)^3$ lifts back to $\aut(\Solmn)$.
By Lemma \ref{kernelaut}, we obtain a splitting
$\aut(\Solmn) \cong  Z_{\phi}^1(\bbr,\bbr^3) \rx (\bbr^*)^3 $. But on $\Solmn$,
every crossed homomorphism is principal. That is,  for
$\eta \in Z_{\phi}^1(\bbr,\bbr^3)$, $\eta(u) = \delta \bfx_0 = (I-\phi(u))\bfx_0$, for some $\bfx_0 \in \bbr^3$.
\qedhere
\end{proof}

Note that we could proceed as in Proposition \ref{autosol} and express $\aut(\Solmn)$
in terms of $\inn(\Solmn) \cong \Solmn$, but this will not be necessary.


\begin{proposition}
\label{autosolo}
We have $\aut(\Solo) \cong \bbr^3 \rx (\gl(2,\bbr) \x \bbr^*)$. Here $\gl(2,\bbr) \x \bbr^*$ is identified
with the group of block diagonal matrices in $\gl(3,\bbr)$ given by
\[\left[
\begin{matrix}
A & \bfzero \\
\bfzero & b
\end{matrix}
\right],\]
and acts on the $\bbr^3$ factor linearly
in the semi-direct product. For $(\bfx_0, A) \in \bbr^3 \rx (\gl(2,\bbr) \x \bbr^*)$, $(\bfx_0, A)$ acts on
$\Solo$ as follows: for $(\bfx, u) \in \Solo$,
\[
(\bfx, u) \ra (A(\bfx) + (I-\phi(u))\bfx_0, u).
\]
\end{proposition}
\begin{proof}
Any automorphism $\alpha$ restricts to an automorphism of the nilradical $\bbr^3$
of $\Solo$, so Lemma \ref{autolemma} applies.
Since eigenvalues are invariant under conjugation, $\phi(\abar u)$ and $A \phi(u) A\inv$ must have
the same eigenvalues. Hence $\abar = +1$ is forced. Thus
$\phi(u)$
and $A$ commute. Computation shows that $A$ must be block diagonal as in the statement of
the proposition.
Conversely,
for $B \in \gl(2,\bbr) \x \bbr^*$,
$(\bfx, u) \ra (B\bfx, u)$ defines an automorphism of $\Solo$.
Thus the image of $\aut(\Solo)$ in $\aut(\bbr^3)$ is $\gl(2,\bbr) \x \bbr^*$ and
$\gl(2,\bbr) \x \bbr^* $ lifts back to $\aut(\Solo)$.
By Lemma \ref{kernelaut}, we obtain a splitting
$\aut(\Solo) \cong  Z_{\phi}^1(\bbr,\bbr^3) \rx (\gl(2,\bbr) \x \bbr^*)$. But on $\Solo$,
every crossed homomorphism is principal, so that $Z_{\phi}^1(\bbr,\bbr^3) \cong \bbr^3$.
\qedhere
\end{proof}

Note that both the geometries $\Solr$ and $\Solof$ have center $\bbr$ with quotient
isomorphic to $\Sol$. So when $G=\Solr$ or $\Solof$, we have the short exact sequence
\[1 \ra \calz(G) \ra G \ra \Sol \ra 1.\]
When $G=\Solr$ or $\Solof$, an automorphism
$\hat{\alpha}$ of $G$, induces an automorphism $\alpha$ of $\Sol$, an automorphism $\ahat$ of the center, and an automorphism $\abar$ of the quotient $\bbr=\Sol / \bbr^2$.
\begin{align}
\label{quotientSol}
\CD
\aut(G) @>>> \aut(\Sol) \x \aut(\bbr)@>>> \aut(\bbr)\x \aut(\bbr^2)\x\aut(\bbr) \\
\hat{\alpha}     @>>> (\alpha,\ahat) @>>> (\abar,A, \ahat)
\endCD
\end{align}
Recall that $A$ determines $\abar$, according to whether $A$ is diagonal
or off-diagonal.


\begin{proposition}
\label{autosolr}
We have $\aut(\Solr) \cong \bbr \rx (\aut(\Sol) \x \bbr^*)$. Here $\aut(\Sol) \x \bbr^*$
acts on $\Solr$ in the obvious way as automorphisms on each factor. The $\bbr$ factor is the kernel of the homomorphism
$\aut(\Solr) \ra \aut(\Sol) \x \aut(\bbr)$, it acts as automorphisms as follows:
for $\left(   \left( \xy{x}{y}, u \right),  t    \right) \in \Solr$ and $k \in \bbr$,
 \[
\left(   \left( \xy{x}{y}, u \right),  t    \right)  \ra  \left(   \left( \xy{x}{y}, u \right),  t + ku   \right).
\]
The action of $\aut(\Sol) \x \bbr^*$ on $\bbr$ in $\bbr \rx (\aut(\Sol) \x \bbr^*)$ is given by multiplication
by $\ahat \abar$.
\end{proposition}
\begin{proof}
Clearly, $\aut(\Sol) \x \aut(\bbr) \cong  \aut(\Sol) \x \bbr^*$ lifts back to $\aut(\Solr)$.
The kernel of $\aut(\Solr)\ra\aut(\Sol)\x\aut(\bbr)$ is the group of crossed
homomorphisms $Z^1(\Sol,\bbr)$. Since $\Sol$ acts trivially on the center
$\bbr$, the crossed homomorphisms become genuine homomorphisms, and
$Z^1(\Sol,\bbr)= \hom(\Sol,\bbr)=\hom(\Sol/ \left[ \Sol, \Sol\right],\bbr)=\hom(\bbr,\bbr)=\bbr$, acting as translations of the central $\bbr$-factor of $\Sol \x \bbr$.
We therefore obtain a splitting
 $\aut(\Solr) \cong \bbr \rx (\aut(\Sol) \x \bbr^*)$.
\qedhere
\end{proof}

Given $\hat{\alpha} \in \aut(\Solof)$, $A$ determines $\ahat$.
That is, $\ahat = \det(A)$. Of course, $\abar$ is also determined.

\begin{proposition}
\label{autosolof}
We have $\aut(\Solof) \cong \bbr \rx \aut(\Sol)$.
The group $\bbr$ is the kernel of the homomorphism
\[\aut(\Solof) \ra \aut(\Sol).\] For $k \in \bbr$,
the action on $\Solof$ is given by
\[
\left[
\begin{matrix} 1 &e^u x &z\\
               0 &e^u &y\\
               0 &0 &1 \end{matrix}
\right]
\longmapsto
\left[
\begin{matrix} 1 &e^u x &z+ku\\
               0 &e^u &y\\
               0 &0 &1 \end{matrix}
\right].
\]
The action of $\aut(\Sol)$ on $\bbr$ in $\bbr \rx \aut(\Sol)$ is multiplication
by $\ahat \abar$.
\end{proposition}
\begin{proof}
We have seen that the image of $\aut(\Solof)$ under
\[\aut(\Solof) \ra \aut(\Sol) \x \aut(\bbr) \ra \aut(\bbr)\x \aut(\bbr^2)\x\aut(\bbr)\]
is determined by its image in $\aut(\bbr^2)$.
On the other hand, $\aut(\Sol)$ lifts back to $\aut(\Solof)$.
Recall the isomorphism $\aut(\Sol)\cong\Sol \rx (\bbr^+ \x D_4)$ given in Proposition \ref{autosol}.
First, $\Sol\subset\Sol\rx  (\bbr^+ \x D_4)$, corresponding to
the inner automorphisms of $\Sol$ lifts to the inner automorphisms of $\aut(\Solof)$.
Note that $\inn(\Solof)=\inn(\Sol)\cong\Sol$.

For the part $\bbr^+\x D_4$ of $\aut(\Sol)$, we have that
a diagonal or off-diagonal matrix $A \in \gl(2,\bbr)$ can be lifted to an automorphism of $\Solof$:
{\small
$$
A=
\kern-4pt
\left[\begin{matrix}
a&b\\ c&d
\end{matrix}\right]
:\
\left[
\begin{matrix} 1 &e^u x &z\\
               0 &e^u &y\\
               0 &0 &1 \end{matrix}
\right]
\longmapsto
\left[
\begin{matrix} 1
&\kern-7pt e^{\abar u}(a x+b y)
&\kern-7pt
\frac12 (a b x^2+2 b c x y+c d y^2+2(a d-b c)z)\\
               0 &e^{\abar u} &(c x+d y)\\
               0 &0 &1 \end{matrix}
\right].
$$}

\noindent
This formula is valid only for the cases when
either $a=d=0$ ($\abar=-1$) or $b=c=0$ ($\abar=+1$).

The kernel of $\aut(\Solof)\ra\aut(\Sol)$ is the group of crossed
homomorphisms $Z^1(\Sol,\bbr)$. Since $\Sol$ acts trivially on the center
$\bbr$, the crossed homomorphisms become genuine homomorphisms, and
$Z^1(\Sol,\bbr)=\hom(\Sol,\bbr)=\hom(\bbr,\bbr)=\bbr$.
Thus we have a splitting $\aut(\Solof) \cong  \bbr \rx \aut(\Sol).$\qedhere
\end{proof}
In summary,

\begin{table}[H]
\caption{\aut(G) of 4-dimensional solvable geometries}
\begin{center}
  \begin{tabular}{| l | l |}
    \hline
    $G$ & \aut(G) \\ \hline
    $\Sol$ & $\Sol \rx (\bbr^+ \x D_4)$ \\ \hline
    $\Solr$ & $\bbr \rx (\aut(\Sol) \x \bbr^*)$ \\ \hline
    $\Solmn$ & $\bbr^3 \rx (\bbr^*)^3$ \\ \hline
    $\Solo$ & $\bbr^3 \rx (\gl(2,\bbr) \x \bbr^*)$ \\ \hline
    $\Solof$ & $\bbr \rx \aut(\Sol)$\\ \hline
  \end{tabular}
\end{center}
\end{table}

With any left
invariant metric on $G$,
\[
\isom(G) \subset G \rx K \subset \aff(G),
\]
where $K$ is a maximal compact subgroup of $\aut(G)$ \cite{GW}.
In fact, all of the 4-dimensional solvable geometries admit a left invariant metric
so that \[\isom(G) = G \rx K,\] where $K$ is any of the maximal compact subgroups listed in the table below.
In the table below, we list $K$ for each of the 4-dimensional solvable geometries.

\begin{table}[H]
\caption{Maximal compact subgroup of \aut(G)}
\begin{center}
  \begin{tabular}{| l | l |}
    \hline
    $G$ & $K \subset \isom(G)$ \\ \hline
  %  $\bbr^4$ & $\on{O}(4,\bbr)$ \\ \hline
    $\Nilr$ & $\on{O}(2,\bbr) \x \bbz_2$ \\ \hline
    $\Nilf$ & $\bbz_2 \x \bbz_2$ \\ \hline
    $\Sol$ & $D_4$ \\ \hline
    $\Solr$ & $D_4 \x \bbz_2$ \\ \hline
    $\Solmn$ & $\bbz_2^3$ \\ \hline
    $\Solo$ & $\on{O}(2,\bbr)\x \bbz_2$ \\ \hline
    $\Solof$ & $D_4$\\ \hline
  \end{tabular}
\end{center}
\end{table}




%\begin{center}
%  \begin{tabular}{| l | r |}
%    \hline
%    $G$ & \aut(G) \\ \hline
%    $\bbr^4$ & $\gl(4,\bbr)$ \\ \hline
%    $\Nilr$ & $TODO$ \\ \hline
%    $\Nilf$ & $TODO$ \\ \hline
%    $\Sol$ & $\Sol \rx (\bbr^+ \x D_4)$ \\ \hline
%    $\Solr$ & $\bbr \rx (\aut(\Sol) \x \bbr^*)$ \\ \hline
%    $\Solmn$ & $\bbr^3 \rx (\bbr^*)^3$ \\ \hline
%    $\Solo$ & $\bbr^3 \rx (\gl(2,\bbr) \x \bbr^*)$ \\ \hline
%    $\Solof$ & $\bbr \rx \aut(\Sol)$\\ \hline
%  \end{tabular}
%\end{center}

%\begin{center}
%  \begin{tabular}{| l | r |}
%    \hline
%    $G$ & $K \subset \aut(G)$ \\ \hline
%    $\bbr^4$ & $\on{O}(4,\bbr)$ \\ \hline
%    $\Nilr$ & $\on{O}(2,\bbr) \x \bbz_2$ \\ \hline
%    $\Nilf$ & $\bbz_2 \x \bbz_2$ \\ \hline
%    $\Solr$ & $D_4 \x \bbz_2$ \\ \hline
%    $\Solmn$ & $(\bbz_2)^3$ \\ \hline
%    $\Solo$ & $\on{O}(2,\bbr)\x \bbz_2$ \\ \hline
%    $\Solof$ & $D_4$\\ \hline
%  \end{tabular}
%\end{center}




\section{Crystallographic Groups of $\Sol$}

In this section, we shall classify the crystallographic groups of $\Sol$. Given a $\Sol$ crystallographic group $Q \subset \Sol \rx D_4$, $Q \cap \Sol$ is a lattice $\Gamma$
with finite quotient group $\Phi \subseteq D_4$, and we have the short exact sequence
\[1 \ra \Gamma \ra Q \ra \Phi \ra 1.\] The first step will be to classify the lattices of
$\Sol$ up to isomorphism.  For $u\in \bbr$, let $E^{u}$ denote the action of $\bbr$ on $\bbr^2$ in $\Sol$. 
Our classification, as well as the notation conventions, will be needed in the 4-dimensional case. See \cite{HaLee} for another classification.

\step[{\bf Lattices of $\Sol$}]
\label{latsol}

In this section, we show the well known fact that there is a one-one
correspondence between the isomorphism classes of lattices (discrete,
cocompact subgroups) of $\Sol$ and the weak-conjugacy classes (see below)
of $\cals \in \sltz$  with trace, $\tr(\cals)>2$.

We say a matrix $H \in \gltz$ is \emph{hyperbolic}
if it has two distinct real eigenvalues, say $\lambda$ and $\frac 1 \lambda$,
with $\lvert \lambda \rvert + \lvert {\frac 1 \lambda} \rvert > 2$.
Note that $H$ induces a linear diffeomorphism of $T^2$ with directions of
both expansion and contraction.
Let
$
\cals=
\left[
\begin{matrix} \sigma_{11} &\sigma_{12}\\
               \sigma_{21} &\sigma_{22} \end{matrix}  \right]
\in\text{SL}(2,\bbz)
$
with $\tr(\cals) >2$. Such a matrix is a hyperbolic matrix with both
eigenvalues $\lambda, \frac 1 \lambda$ positive.

Let $\bbz^2\rx_\cals\bbz$
be the extension of $\bbz^2$ by $\bbz$ with the group operation
$$
(\bfx,m)\cdot(\bfy,n)=(\bfx+ \cals^m \bfy, m+n).
$$
Let
$$
\bfe_1=\left(\left[\begin{matrix} 1\\ 0 \end{matrix} \right],0\right),\quad
\bfe_2=\left(\left[\begin{matrix} 0\\ 1 \end{matrix} \right],0\right),\quad
\bfe_3=\left(\left[\begin{matrix} 0\\ 0 \end{matrix} \right],1\right).
$$
Then $\bbz^2\rx_\cals\bbz$ has relations
\begin{align}
\label{3Lattice-presentation}
\begin{split}
[\bfe_1,\bfe_2]=1,\ \ \
\bfe_3\cdot \bfe_1\cdot \bfe_3\inv
= \bfe_1^{\sigma_{11}}\cdot \bfe_2^{\sigma_{21}},\ \ \
\bfe_3\cdot \bfe_2\cdot \bfe_3\inv= \bfe_1^{\sigma_{12}}\cdot \bfe_2^{\sigma_{22}}.
\end{split}
\end{align}

\begin{notation}
For uniformity of statements, we always take
$$\Delta
=\left[
\begin{matrix} \tfrac{1}{\lambda}&0\\
               0&\lambda \end{matrix}
\right]
$$
with $\tfrac{1}{\lambda}<1<\lambda$.
\end{notation}


\begin{lemma}
\label{general-p}
Let $\cals\in\sltz $ with $\tr(\cals) >2$.
Let $P$ and $\Delta$ be matrices such that $P \cals P\inv=\Delta$ and
$\det(P)=1$. Then
$Q \cals Q\inv=\Delta$ (with $\det(Q)=1$) if and only if $Q$ is of the form
$\pm E^d\cdot P$ for $d\in\bbr$.
\end{lemma}
\begin{proof}
Recall $E^d$ is a diagonal matrix.
If $P$ and $Q$ both diagonalize $\cals$, then
$(Q P\inv)\Delta (Q P\inv )\inv$ $=\Delta$.
Since $\text{trace}(\cals) >2$, $\cals$ has two distinct positive real
eigenvalues: $\lambda, 1/\lambda$.
For $Q P\inv$ to leave the two eigenspaces
invariant, it should be a diagonal matrix.
Since $\det (Q P\inv) = 1$, $Q P\inv = \pm E^d$ for some $d \in \bbr$.
Conversely, Suppose $P \cals P\inv=\Delta$. Then
\[
(\pm E^d P) \cals (\pm E^d P)\inv=
E^d (P \cals P\inv)E^{-d}=
E^d \Delta E^{-d}=\Delta.
\qedhere
\]
\end{proof}

With such $P$ and $\Delta$ for $\cals$, we define a map
\begin{align}
\label{def-phi}
\phi: \bbz^2\rx_\cals\bbz &\lra \kern24pt\Sol\\
\notag \left(\xy{x}{y},u\right)
&\longmapsto
\notag \left(P\xy{x}{y},u\ln(\lambda)\right).
\end{align}
The relation $P \cals P\inv =\Delta$ guarantees that $\phi$ is
a homomorphism. Clearly, the image is a lattice of $\Sol$.
It maps the generators as follows.
\begin{align}
\label{sol-iso}
\notag
\bfe_1 &\mapsto \bft_1=P \bfe_1\\
\bfe_2 &\mapsto \bft_2=P \bfe_2\\
\notag
\bfe_3 &\mapsto \bft_3=(\bfzero,\ln(\lambda))
\end{align}
We denote image of $\bbz^2\rx_\cals\bbz$ as $\gamsig$:
$$
\gamsig=\angles{\bft_1,\bft_2,\bft_3} \subset\Sol.
$$

Conversely, we show any lattice of $\Sol$ is isomorphic to such a $\gamsig$,
as the following proposition shows. We say $\cals, \cals' \in \sltz$ are
\emph{weakly conjugate} if and only if  $\cals'$ is conjugate, via an
element of $\gltz$, to
$\cals$ or ${\cals} \inv$.

\begin{notation}
We shall refer to a lattice of $\Sol$ generated by $\bft_1, \bft_2, \bft_3$ of the form
in assignment (\ref{sol-iso}) in Subsection \ref{latsol} as a \emph{standard lattice} of $\Sol$.
\end{notation}

\begin{proposition}
\label{lattice-a=0}
There is a one-one correspondence between the isomorphism classes of
$\Sol$-lattices and the weak-conjugacy classes of
$\cals \in \sltz$ with $\tr(\cals)>2$.
Therefore, any lattice  of $\Sol$ is conjugate to $\gamsig$,
for some $\cals$, by an inner automorphism of $\Sol$.
\end{proposition}

\begin{proof}
Let $\cals\in\sltz$ with $\tr(\cals)>2$.
Then the assignment (\ref{sol-iso}) in Subsection \ref{latsol} yields a lattice $\gamsig\subset\Sol$.

We show that any lattice $\Gamma$ is conjugate to $\gamsig$ for some $\cals$.
The intersection of
$\Gamma$ with the nilradical $\bbr^2$, $\Gamma\cap\bbr^2$, is a lattice
of $\bbr^2$, \cite[Corollary 8.28]{rag}.
Let $\bft_1,\bft_2\in\bbr^2$ be a generating set for $\Gamma\cap\bbr^2$,
and suppose $\{\bft_1,\bft_2,\bft\}$ generates $\Gamma$.
Then $\bft$ is of the form $(\bfa,\ln(\lambda))$. We can assume that $\lambda>1$,
by taking the inverse of $\bft$, if needed.
But conjugation by $((E^{\ln(\lambda)}-I)\inv \bfa,0)$ maps this lattice to
$$
\Gamma=\angles{(\bft_1,0),\
(\bft_2,0),\
(\bfzero,\ln(\lambda))}.
$$
Define $P$ by $P \bfe_1 = \bft_1$ and $P \bfe_2 = \bft_2$, and set
$\cals=P\inv E^{\ln(\lambda)} P$.  We claim that $\cals\in\sltz$. Since
$\bft_3 \bft_1 \bft_3 \inv = E^{\ln(\lambda)} \bft_1 $ must be in $\Gamma$,
$E^{\ln(\lambda)} \bft_1 = \bft^{\sigma_{11}}_1 \bft^{\sigma_{21}}_2$,
for some integers $\sigma_{11},\sigma_{21}$,
so that $P \inv E^{\ln(\lambda)} P \bfe_1 = \sigma_{11} \bfe_1 + \sigma_{21}
\bfe_2$. Similarly, for integers $\sigma_{12},\sigma_{22}$,
$P \inv E^{\ln(\lambda)} P \bfe_2 =
\sigma_{12} \bfe_1 + \sigma_{22} \bfe_2$, which shows that $\cals$ is integral.

Notice that, in case $\det(P)<0$, we can make it positive by taking
$-\bft_1$ in place of $\bft_1$.
Furthermore, we can assume that $\det(P) = 1$. For conjugation by $(0,
{\frac {1} {\sqrt{\det(P)}}} I) \in \aff(\Sol)$ maps this lattice to
another which has $\det(P)=1$. Therefore our $\Gamma$ is, in fact,
conjugate to the lattice $\gamsig$.

It remains to show the isomorphism class statement.
If $\cals' = B {\cals}^{\epsilon} B \inv$ with $B\in\gltz$,
where $\epsilon=\pm 1$, then an isomorphism between $\bbz^2 \rx_{\cals}
\bbz$ and $\bbz^2 \rx_{\cals'} \bbz$ is given by $\bfe_i \ra B \bfe_i$ for
$i = 1,2$ and $\bfe_3 \ra \epsilon \bfe_3$.

Conversely,
suppose two lattices of $\Sol$ are isomorphic. We may assume these lattices
are of the form $\Gamma_{\cals}$ and $\Gamma_{\cals'}$.
Our goal is to show that $\cals$ and $\cals'$ are weakly conjugate.
We have an isomorphism
\[
\phi: \bbz^2 \rx_{\cals} \bbz \lra \bbz^2 \rx_{\cals'}\bbz.
\]
Since $\bbz^2$ is the discrete nil-radical (maximal normal nilpotent
subgroup) in both groups, $\phi$ restricts to an isomorphism on the
$\bbz^2$ factors, say $B\in\gltz$. Also, $\phi$ induces an isomorphism
$\epsilon$ on the quotients. Then $\epsilon=\pm 1$.
%By composing our $\phi$ with the obvious isomorphism
%\[
%\CD
%\bbz^2 \rx_{\cals'}\bbz @>>> \bbz^2 \rx_{{\cals'}\inv}\bbz\\
%(\bfx,m) @>>> (\bfx,-m)
%\endCD
%\]
%\boxt{if necessary, we may assume that $\epsilon=1$.: Is this necessary?
%See $\epsilon$s below.}
Therefore, $\phi$ is of the form
\[
\phi(\bfx,m)=(B\bfx +\eta(\bfx,m),\epsilon m).
\]
However, one can quickly see that $\eta$ is independent of $\bfx$.
So $\eta(\bfx,m)=\eta(m)$.
Then applying $\phi$ to $(0,1)(\bfx,0)(0,-1)\in \bbz^2 \rx_{\cals} \bbz$
in two different ways, we get
\begin{align*}
\phi((0,1)(\bfx,0)(0,-1))
&=\phi(\cals\bfx,0)
=(B(\cals\bfx),0)\\
\phi(0,1) \phi(\bfx,0) \phi(0,-1)
&=(\eta(1),\epsilon)(B\bfx,0)(\eta(1),\epsilon)^{-1}
%&=(\cals' B\bfx,0)\quad\text{(since $\epsilon=+1$).}
=(\cals'^{\epsilon} B\bfx,0)\quad
\end{align*}
Thus, $B\cals\bfx=\cals'^{\epsilon} B\bfx$ for every $\bfx$, and we have
$\cals'^{\epsilon}=B\cals B\inv$.
\end{proof}


\step[{\bf Classification of $\Sol$ Crystallographic Groups}]
\label{classSol}
% % % % % % % % % % % % % % % % % % % % % % % % % % % %5


For our classification, we first show that $Q$ can be conjugated in $\aff(\Sol)$
in such a way that the lattice $Q \cap \Sol$ is of the form $\gamsig$ for some $\cals$, see assignment (\ref{sol-iso}) in Subsection \ref{latsol}.

\begin{proposition}
\label{puttingToLattice}
Any crystallographic group $Q'$ of $\Sol$ can be conjugated in $\aff(\Sol)$
to $Q \subset \Sol \rx D_4$ so that:

\noindent
{\rm{(1)}}
We have $Q \cap \Sol = \gamsig$. That is, the translation subgroup of $Q$ is a standard lattice of $\Sol$, generated by $\bft_1$, $\bft_2$, and $\bft_3$ as in {\rm (\ref{sol-iso})} in {\rm Subsection \ref{latsol}}.

\noindent
{\rm{(2)}} The holonomy group $\Phi$ is generated by at most two elements of $D_4$, and therefore $Q$ is generated by $\angles{\bft_1, \bft_2, \bft_3}$, and at most two isometries of the form $(\bft_1^{a_1} \bft_2^{a_2} \bft_3^{a_3}, A)$, for $A \in D_4$ and real numbers $a_i$.

\end{proposition}
\begin{proof}
Any maximal compact subgroup of $\aut(\Sol)$ is conjugate to $D_4 \subset \aut(\Sol)$. Then
we observe that the conjugations performed in the proof of Proposition \ref{lattice-a=0} leave $D_4$ invariant.
\end{proof}

For ease of computation, we will assume our $\Sol$ crystallographic group is embedded in $\Sol \rx D_4$ as in Proposition \ref{puttingToLattice}.
Note that $Q \cap \bbr^2 = \angles{\bft_1, \bft_2}$ is a lattice of $\bbr^2$, isomorphic to $\bbz^2$.
Denote the quotient $Q/\Gamma$ by $\bbzphi$ so that we have the diagram:
\begin{equation}
\label{coreClass}
\CD
@. 1 @. 1 @.  @.\\
@. @VVV @VVV @. @.\\
@. \bbz^2 @=  \bbz^2 @. @.\\
@. @VVV @VVV @. @.\\
1 @>>> \gamsig @>>> Q @>>> \Phi @>>> 1\\
@. @VV{/ \bbz^2}V @VV{/ \bbz^2}V @| @.\\
1 @>>> \bbz @>>> \bbzphi @>>> \Phi @>>> 1\\
@. @VVV @VVV @. @.\\
@. 1 @. 1 @.  @.\\
\endCD
\end{equation}

First we will classify all possible extensions $\bbzphi$ which can arise as quotients of $\Sol$ crystallographic groups.


\begin{lemma}
\label{abarMinus}
If $\abar=-1$, $a_3$ can be taken to be $0$.
Otherwise, $a_3=0$ or $\tfrac12$ {\rm  ($\mod \bbz$)}.
\end{lemma}
\begin{proof}
For $(\bft_3^{a_3}, A) \in \bbzphi$, when $\abar=-1$, we can assume that $a_3=0$.
That is, given $(\bft_1^{a_1} \bft_2^{a_2} \bft_3^{a_3}, A) \in Q$, conjugation by $(\bft_3^{-\frac{a_3}{2}}, I)$ sets $a_3=0$:
\[
(\bft_3^{-\frac{a_3}{2}}, I) (\bft_1^{a_1} \bft_2^{a_2} \bft_3^{a_3}, A)
(\bft_3^{\frac{a_3}{2}}, I) =   (\bft_1^{a'_1} \bft_2^{a'_2} \bft_3^{0}, A).
\]
Note that this conjugation keeps the lattice in standard form.

All elements in $D_4$ with $\abar=1$ are involutions, so when $\abar=1$, we compute in $\bbzphi$:
\[(\bft_3^{a_3}, A)^2 = (\bft_3^{a_3 + \abar(a_3)}, I) = (\bft_3^{2 a_3}, I).\]
For this to be in $\bbz = \angles{\bft_3}$, $a_3=0$ or $\tfrac12$ (\mod $\bbz$) is forced.
\qedhere
\end{proof}


Note that we only have to consider subgroups of $D_4 \subset \aut(\Sol)$ up to conjugacy. Using Lemma \ref{abarMinus}, we give a complete list of $\bbzphi$.
%It is important that conjugation by an element $(e, \alpha) \in \aff(\Sol)$ keeps the lattice in standard form.


\pagebreak
\begin{equation}
\label{bbzphi-list}
\text{--- List of $\bbzphi$ ---}
\end{equation}
%\hrule
\[
\begin{array}{lllll}
&&\Phi   &\bbzphi  &a_3,(\ b_3)\\
&&&&\\
(0)& %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
&\{1\}
&\bbz
&\\
(1)& %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
&\bbz_2 = \left< \matPpm \right>
&\bbz\x\bbz_2^{(\dagger)}
&a_3=0\\
&&&\bbz
&a_3=\frac12\\
&&&\\
(2)& %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
&\bbz_2 = \left< \matPmm \right>
&\bbz\x\bbz_2
&a_3=0\\
&&&\bbz
&a_3=\frac12\\
&&&\\
(3)& %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
&\bbz_2 = \left< \matNpp \right>
&\bbz\rx\bbz_2
&a_3=0\\
&&&\\
(4)& %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
&\bbz_4 = \left< \matNpm \right>
&\bbz\rx\bbz_4
&a_3=0\\
&&&\\
(5)& %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
&\bbz_2\x\bbz_2 = \left< \matPmm,\matPpm \right>
&(\bbz\x\bbz_2)\x\bbz_2^{(\dagger)}
&a_3=0,\ b_3=0\\
&&&\bbz\x\bbz_2,\
&a_3=0,\ b_3=\frac12\\
&&&\bbz\x\bbz_2^{(\dagger)},
&a_3=\frac12,\ b_3=0\\
&&&\\
(6)& %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
&\bbz_2\x\bbz_2 = \left< \matNpp,\matPmm \right>
&(\bbz\x\bbz_2)\rx\bbz_2
&a_3=0,\ b_3=0\\
&&&\bbz\rx\bbz_2
&a_3=0,\ b_3=\frac12\\
&&&\\
(7)& %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
&\bbz_4\rx\bbz_2 = \left< \matNpm,\matPpm \right>
&(\bbz\rx\bbz_4)\rx\bbz_2^{(\dagger)}
&a_3=0,\ b_3=0\\
&&&(\bbz\x\bbz_2)\rx\bbz_2
&a_3=0,\ b_3=\frac12
\end{array}
\]

The groups with $(^\dagger)$ mark will be shown to not lift to
crystallographic groups of $\Sol$. That is, there does not exist a
$\Sol$-crystallographic group whose quotient is such a $\bbzphi$. See Corollary \ref{noLift}.

Recall from diagram (\ref{coreClass})
that $Q$ is an extension
\[
1 \ra \bbz^2 = \angles{\bft_1, \bft_2} \ra Q \ra \bbzphi \ra 1.
\]
Having classified all possible $\bbzphi$, we now classify all abstract kernels
\[\varphi: \bbzphi \ra \gltz.\] Of course
\[
\varphi(\bft_3)=\cals
\]
is fixed already. For $\alphabar=(\bft_3^{a_3},A) \in \bbzphi$, if $\abar=+1$, $\varphi(\alphabar)$ must commute with $\cals$.
Otherwise, $\varphi(\alphabar)$ conjugates $\cals$ to its inverse.

Recall the embedding
\[
\phi: \bbz^2\rx_\cals\bbz \lra \Sol,
\]
of assignment (\ref{sol-iso}) in Subsection \ref{latsol}, where $\phi(\bfe_i)=P \bfe_i=\bft_i$ for $i=1,2$ and
$\phi(\bfe_3)=(\bfzero,\ln(\lambda))=\bft_3$. We compute, for $\xy{n_1}{n_2} \in \bbz^2$,
\begin{align*}
(\bft_1^{a_1} \bft_2^{a_2} \bft_3^{a_3}, A)(\bft_1^{n_1} \bft_2^{n_2}, I)
(\bft_1^{a_1} \bft_2^{a_2} \bft_3^{a_3}, A) \inv &=\\
(\bft_3^{a_3} A(\bft_1^{n_1} \bft_2^{n_2}) \bft_3^{-a_3}, I ).
\end{align*}
Such conjugation is independent of $a_1, a_2$ and defines
an element of $\gltz$. That is,
\[\bft_3^{a_3} A(\bft_1^{n_1} \bft_2^{n_2}) \bft_3^{-a_3} = \bft_1^{m_1} \bft_2^{m_2},\]
for $m_1, m_2 \in \bbz$. With the definition of $P$ and $\Delta$ in assignment (\ref{sol-iso}) in Subsection \ref{latsol},
\[
\xy{m_1}{m_2} = {P \inv} \Delta^{a_3} A P \xy{n_1}{n_2}.
\]
We define
\begin{align*}
& \wa = {P \inv }A P\\
& \cals^{a_3} = {P \inv} \Delta^{a_3} P,
\end{align*}
so that $S^{a_3} \wa = {P \inv} \Delta^{a_3} A P$.

\begin{theorem}
\label{abstract-kernel}
For each $\bbzphi$ in the list {\rm (\ref{bbzphi-list})},
the following is a complete list of homomorphisms
$\varphi:\bbzphi\ra\gltz$ with $\varphi(\bft_3)=\cals$ and
\begin{align*}
\varphi(\bft_3^{a_3},A)&=\cals^{a_3}\wa\\
\varphi(\bft_3^{b_3},B)&=\cals^{b_3}\wb,
\end{align*}
up to conjugation in $\gltz$.
\begin{enumerate}

\item[\gr{0}]
$\Phi$ is trivial,\parn
$\bbzphi=\bbz=\angles{\bft_3}$.\parn

\item[\gr{1}]
$\Phi=\bbz_2$: $A=
\left[\begin{matrix}
1 &0\\ 0& -1
\end{matrix}\right]$,\parn
$\bbzphi=\bbz=\angles{\bft_3,\bar\alpha=(\bft_3^{\frac12},A)}$.\parn
$\bullet$
$\vara=-K$ with $\det(K)=-1$,  $\tr(K)=n>0$, and $\cals=n K+I$.
%
\item[\gr{2a}]
$\Phi=\bbz_2$: $A=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,\parn
$\bbzphi=\bbz\x\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A)}$.\parn
$\bullet$
$\vara=A$, and $\cals\in\sltz$ with $\tr(\cals)>2$.
%
\item[\gr{2b}]
$\Phi=\bbz_2$: $A=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,\parn
$\bbzphi=\bbz=\angles{\bft_3,\bar\alpha=(\bft_3^{\frac12},A)}$.\parn
$\bullet$ $\vara=-K$ with $\det(K)=+1$,  $\tr(K)=n>2$, and $\cals=n K-I$.
%
\item[\gr{3}]
$\Phi=\bbz_2$: $A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,\parn
$\bbzphi=\bbz\rx\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A)}$.\parn
$\bullet$ $\vara=A$,
and $\cals\in\sltz$ with $\sigma_{12}=-\sigma_{21}$.
%
\item[\gr{3\ii}]
$\Phi=\bbz_2$: $A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,\parn
$\bbzphi=\bbz\rx\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A)}$.\parn
$\bullet$
$\vara=\left[\begin{matrix}
1&0\\ 0&-1
\end{matrix}\right]$, and
$\cals\in\sltz$
with $\sigma_{11}=\sigma_{22}$.
%
\par\noindent
\item[\gr{4}]
$\Phi=\bbz_4$: $A=
\left[\begin{matrix}
0 &1\\ -1& 0
\end{matrix}\right]$,\parn
$\bbzphi=\bbz\rx\bbz_4=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A)}$.\parn
$\bullet$ $\vara=A$, and
$\cals\in\sltz$, symmetric and $\tr(\cals)>2$.
%
\item[\gr{5}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
1 &0\\ 0& -1
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,\parn
$\bbzphi=\bbz\x\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{\frac12},A),
\bar\beta=(\bft_3^0,B)}$.\parn
$\bullet$ $\vara=-K$, $\varb=B$
\hfill \gr{1}$+$\gr{2a}\parn
$\bullet$
$\cals=n K+I$, where $K\in\gltz$, $\det(K)=-1$, and  $\tr(K)=n>0$.
%
\item[\gr{6a}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,\parn
$\bbzphi=(\bbz\x\bbz_2)\rx\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A),
\bar\beta=(\bft_3^0,B)}$.\parn
$\bullet$ $\vara=A$, $\varb=B$ \hfill \gr{3}$+$\gr{2a}\parn
$\bullet$
$\cals\in\sltz$ with $\sigma_{12}=-\sigma_{21}$ and $\tr(\cals)>2$.\parn
%
\item[\gr{6a\ii}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,
$B=\def\tr{\text{\rm{tr}}}
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,\parn
$\bbzphi=(\bbz\x\bbz_2)\rx\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A),
\bar\beta=(\bft_3^0,B)}$.\parn
$\bullet$ $\vara=
\left[\begin{matrix}
1 &0\\ 0& -1
\end{matrix}\right]$,\
$\varb=B$
\hfill \gr{3\ii}$+$\gr{2a}\parn
$\bullet$
$\cals\in\sltz$
with $\sigma_{11}=\sigma_{22}$ and $\tr(\cals)>2$.\parn
%
\item[\gr{6b}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,\parn
$\bbzphi=\bbz\rx\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A),
\bar\beta=(\bft_3^{\frac12},B)}$.\parn
$\bullet$ $\vara=A$, $\varb=-K$ \hfill \gr{3}$+$\gr{2b}\parn
$\bullet$
$\cals=n K-I$, where $K\in\sltz$ with $\tr(K)= n>2$; $k_{12}=-k_{21}$.
%
\item[\gr{6b\ii}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,\parn
$\bbzphi=\bbz\rx\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A),
\bar\beta=(\bft_3^{\frac12},B)}$.\parn
$\bullet$ $\vara=\left[\begin{matrix}
1 &0\\ 0& -1
\end{matrix}\right]$, $\varb=-K$ \hfill \gr{3\ii}$+$\gr{2b}\parn
$\bullet$
$\cals=n K-I$, where $K\in\sltz$ with $\tr(K)= n>2$;\
$k_{11}=k_{22}$.\parn
%
\item[\gr{7}]
$\Phi=\bbz_4\rx\bbz_2$:
$A=\matNpp, B=\matPpm,$\parn
$\bbzphi=(\bbz\x\bbz_2)\rx\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A),
\bar\beta=(\bft_3^{\frac12},B)}$.\parn
$\bullet$
$\vara=A$, $\varb=-K$ \hfill (includes \gr{6a})\qquad  \gr{3}$+$\gr{1}\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$, $\det(K)=-1$, $\tr(K)>0$;
$k_{12}=-k_{21}$.\parn
%
\item[\gr{7\ii}]
$\Phi=\bbz_4\rx\bbz_2$:
$A=\matNpp, B=\matPpm$,\parn
$\bbzphi=(\bbz\x\bbz_2)\rx\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A),
\bar\beta=(\bft_3^{\frac12},B)}$.\parn
$\bullet$
$\vara=\matPpm$, $\varb =-K$
\hfill (includes \gr{6a\ii})\qquad \gr{3\ii}$+$\gr{1}\parn
$\bullet$
$\cals=n K+I$,  $K\in\gltz$,  $\det(K)=-1$, $\tr(K)=n>0$;
$k_{11}=k_{22}$.\parn
\end{enumerate}
\end{theorem}

\noindent
The proof will be deferred to the end of the section.

\begin{lemma}
\label{PPM-A30}
For $\bbzphi=\bbz\rx\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A)}$,
where
$\Phi=\bbz_2$, $A=
\left[\begin{matrix}
1 &0\\ 0& -1
\end{matrix}\right]$, there is no homomorphism
$\varphi:\bbzphi\ra\gltz$ with \[\varphi(\bft_3)=\cals\in\sltz.\]
\end{lemma}
\begin{proof}
By applying the Rational Zero Theorem to the characteristic polynomial
of $\cals$, we see that $\lambda$ must be irrational. The matrix
\[
\cals \varphi(\alphabar) = P \inv \Delta P {P \inv} A P = P \inv \Delta A P
\]
must be integral. The sum of its eigenvalues, $\frac{1}{\lambda} - \lambda$, must
be an integer. On the other hand, since $\frac{1}{\lambda} + \lambda$ is an integer,
we infer that $2{\lambda}$ is an integer, so that $\lambda$ is rational, which is a contradiction.
\qedhere.
\end{proof}

%\begin{align*}
%\varphi(\alphabar)=
%\cals^{0}\wa=\wa=
%P\inv A  P
%&= P\inv\fmat{1}{0}{0}{-1} P\\
%&=
%-\tfrac{1}{\sqrt{T^2-4}}
%\left[
%\begin{array}{cc}
% {\sigma_{11}}-\sigma_{22} & {2\sigma_{12}} \\
% {2\sigma_{21}} & -\sigma_{11}+{\sigma_{22}}
%\end{array}
%\right],
%\end{align*}
%where $T=\sigma_{11}+\sigma_{22}$, the trace of $\cals$.
%This $\wa$ can never be an integral matrix because $T^2 - 4$ is never a
%perfect square.

\begin{corollary}
\label{noLift}
In the list of $\bbzphi$ {\rm(\ref{bbzphi-list})}, the 4 cases with
$(^\dagger)$ cannot be the quotient of a $\Sol$-crystallographic group.
\end{corollary}


\begin{proposition}[Procedure of finding $\cals$ and its diagonalizing
matrix $P$]
\label{simplifyAtilde}
With a group $\bbzphi$ in the list {\rm(\ref{bbzphi-list})},
qualifying $P$ and  $\cals$ can be found as follows:
\begin{enumerate}
\item[\rm(1)]
If $a_3=0$, $\ka=\cals^0\wa=\wa=P\inv A P$ should satisfy the following.
%there are 3 cases.
\item[\textcircled{a}]
If $A=\matPmm$, then $\wa=A$ (no condition on $\cals$).
\item[\textcircled{b}]
If $A=\matNpm$, then $\wa=A$, and $\sigma_{12}=\sigma_{21}$.
\item[\textcircled{c}]
If $A=\matNpp$, then $A$ can be $\sltz$-conjugate to either $A$ itself
or $\matPpm$. For $\wa=A$, $\sigma_{12}=-\sigma_{21}$;
for $\wa=\matPpm$, $\sigma_{11}=\sigma_{22}$.
\item[\rm(2)]
If $a_3=\frac12$, then $\abar=+1$. There are two cases.
\item[\textcircled{a}]
If $A=\matPmm$, the condition $\cals^{a_3}\circ P\inv A P\in\gltz$
forces $\cals=n K-I$ with $\det(K)=+1$, $\tr(K)=n>2$. Then
$\varphi(\alphabar)=\cals^{a_3}\circ P\inv A P=-K$.
\item[\textcircled{b}]
If $A=\matPpm$, the condition $\cals^{a_3}\circ P\inv A P\in\gltz$
forces $\cals=n K+I$ with $\det(K)=-1$, $\tr(K)=n>0$. Then
$\varphi(\alphabar)=\cals^{a_3}\circ P\inv A P=-K$.
\item[\rm(3)]
If $\Phi$ has two generators, negotiate so that both conditions are met.
\end{enumerate}
That is, $\varphi:\bbzphi\ra\gltz$ given by
\begin{align*}
\varphi(\bft_3)&=\cals\\
\varphi(\alpha)&=\cals^{a_3} \circ P\inv A P\\
\varphi(\beta)&=\cals^{b_3} \circ P\inv B P
\end{align*}
must be a homomorphism.
\end{proposition}

Once we found $\cals$ and $\ka=\cals^{a_3}\circ P\inv A P$, we form
$\alpha=(\bft_3^{a_3},A)$.
Then the unique extension $\angles{\bft_1,\bft_2,\bft_3, \alpha}$
satisfies the following:
\begin{align*}
&\bft_3(\bft_1^{n_1}\bft_2^{n_2})\bft_3\inv=\bft_1^{m_1}\bft_2^{m_2},
\text{ where } {\xy{m_1}{m_2}=\cals\xy{n_1}{n_2}},\\
&\alpha(\bft_1^{n_1}\bft_2^{n_2})\alpha\inv=\bft_1^{w_1}\bft_2^{w_2},
\text{ where } {\xy{w_1}{w_2}=\ka\xy{n_1}{n_2}}.\\
\end{align*}
\noindent
The proof requires a series of lemmas.

When $a_3=0$, $\ka = \cals^{0} \wa = \wa$. There are many matrices $\wa$,
depending on $P$. We show how to cut down the number of possibilities for
$\wa$. Essentially, we will be selecting a new set of generators for the
lattice $\Gamma_\cals$ to obtain a simpler formula for $\ka = \wa$.
The following is known (see e.g.,
\cite[p.182]{CampTrouy}).

\begin{lemma}%\cite[p.182]{CampTrouy}
\label{CamTr}
 Let $A \in\gltz$, $\tr(A) = 0$. Then $A$ is $\sltz$-conjugate
to one and only one of the following matrices:
\[
\pm\matNpm,\ \matNpp,\ \matPpm.
\]
The last two matrices are $\sltz$-conjugate to their own negatives.
\end{lemma}

However,
the first two (order 4 elements) are only conjugate by
a matrix of negative determinant. Thus, they are not $\sltr$-conjugate.
A very important fact for us is:

\begin{corollary}
$\matNpp$ and $\matPpm$ are $\gltr$-conjugate but not $\gltz$-conjugate.
\end{corollary}

For the lemma below, recall that our convention is $P\cals P\inv =\left[
\begin{matrix} \tfrac{1}{\lambda}&0\\
               0&\lambda \end{matrix}
\right]$ with $\lambda >1$.

\begin{lemma}
\label{czero}
Let $P\in\sltr$, $\cals=(\sigma_{ij}) \in\sltz$ with $\tr(\cals)>2$. Then
\begin{enumerate}
\item[(a)]
$P\inv \matNpm P = \matNpm$ if and only if
%$P\in\so(2)$.
$P=\pm \left[\begin{array}{rr}
\cos t&\sin t\\ \kern-6pt-\sin t& \cos t
\end{array}\right]$.\\
$\cals$ is diagonalized by such a $P$ if and only if $\sigma_{12}=\sigma_{21}$.
\item[(b1)]
$P\inv \matNpp P = \matNpp$ if and only if
$P=\pm\left[\begin{matrix}
\cosh t&\sinh t\\ \sinh t& \cosh t
\end{matrix}\right]$.\\
$\cals$ is diagonalized by such a $P$ if and only if $\sigma_{12}=-\sigma_{21}$.
\item[(b2)]
$P\inv \matNpp P = \matPpm$ if and only if
$P=\pm \frac{1}{\sqrt{2}}\left[\begin{array}{rr}
t& -\frac{1}{t}\\ t &\frac{1}{t}
\end{array}\right]$, $t\not=0$.\\
$\cals$ is diagonalized by such a $P$ if and only if $\sigma_{11}=\sigma_{22}$.
\end{enumerate}
\end{lemma}
\bigskip

\begin{proof}[Proof \rm{(of Proposition \ref{simplifyAtilde})}]

For cases (1)\textcircled{b} and (1)\textcircled{c}, let $\cals\in\sltz$, $P$ be such that
\begin{align*}
P\cals P\inv&=\Delta\\
P\inv A P&= B\in\gltz.
\end{align*}

For a $U\in\sltz$, set $\cals'= U \cals U\inv$ and $P'=P U\inv$.
We can embed $\bbz^2 \rx_{\cals'} \bbz$ into $\Sol$ using $P'$ so that the
following diagram commutes:
\[
\CD
\Sol @= \Sol
%\qquad
@.\phantom{AAAAA}
@. (Px,v\ln(\lambda))  @= (P U \inv (U x), v \ln(\lambda) )\\
@AAA            @AAA
@. @AAA        @AAA           \\
\bbz^2\rx_{\cals}\bbz              @>>>  \bbz^2\rx_{U \cals U \inv}\bbz
@. %\qquad
@. (x, v)  @>>>
(U x,v)
\endCD
\]

Therefore the image of both $\bbz^2 \rx_\cals \bbz$ and $\bbz^2
\rx_{\cals'} \bbz$ in $\Sol$ is $\Gamma_\cals$; we have not changed the
lattice. The effect is just a change of basis.
For this new embedding of our lattice, note that
\begin{align*}
{\wa}'={P'} \inv  A P' = U P \inv A P U \inv = U \wa U \inv.
\end{align*}

In case (1)\textcircled{b}, we use this fact and Lemma $\ref{CamTr}$ to find a $U$ such that
$ U \wa U \inv = \left[ \begin{matrix} 0 & 1 \\  -1 & 0 \end{matrix} \right]$; whereas
in case (1)\textcircled{c} we can find a $U$ such that
$ U \wa U \inv = \left[ \begin{matrix} 0 & 1 \\  1 & 0 \end{matrix} \right]$ or
$ U \wa U \inv = \left[ \begin{matrix} 1 & 0 \\  0 & -1 \end{matrix} \right]$.

%(1)\textcircled{b}

%y Lemma \ref{CamTr}, $B$ is $\sltz$-conjugate to $A$ itself, say
%$U B U\inv =A$ with $U\in\gltz$. Then
%$\cals'= U \cals U\inv$ and $P'=P U\inv$ satisfies all the conditions
%\begin{align*}
%P' A {P'}\inv &=\Delta\\
%{P'}\inv A P'&= A\in\gltz.
%\end{align*}
%Consequently,
%$Q=\angles{\Gamma_\cals, (\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3} , A)}$
%is conjugate to
%$Q'=\angles{\Gamma_{\cals'}, (\bft_1^{a'_1}\bft_2^{a'_2}\bft_3^{a_3} , A)}$
%(again $a_3=0$).
%By the case (a) of Lemma \ref{czero}, $\cals$ satisfies
%$\sigma_{12}=\sigma_{21}$.


%1)\textcircled{c}
%Suppose  $A=\matNpp$, and let $\cals\in\sltz$, $P$ be such that
%\begin{align*}
%P\cals P\inv&=\Delta\\
%P\inv A P&= B\in\gltz.
%\end{align*}
%From (b1) and (b2) of Lemma \ref{czero}, there are two cases: either
%$B$ is $\sltz$-conjugate to $A$ itself, or $B$ is conjugate to
%$A'=\matPpm$.
%With $B=A$, the above two equalities forces $\cals$ to satisfy
%$\sigma_{12}=-\sigma_{21}$.
%With $B=A'$, the above two equalities forces $\cals$ to satisfy
%$\sigma_{11}=\sigma_{22}$.
%The rest of the argument is the same as previous case.
\bigskip

(2) Now we look at the case $a_3=\frac12$.

(2)\textcircled{a}
Let $A=\matPmm$.
\begin{align*}
\cals^{\frac12}\wa
&=(P\inv \Delta^{\frac12} P) (P\inv A P)
= P\inv\fmat{-\tfrac{1}{\sqrt{\lambda}}}{0}{0}{-\sqrt{\lambda}} P\\
&=
-\tfrac{1}{\sqrt{T+2}}
\left[
\begin{array}{cc}
 {\sigma_{11}}+1 & {\sigma_{12}} \\
 {\sigma_{21}} & {\sigma_{22}}+1
\end{array}
\right].
\end{align*}
Note that this is one of the square roots of $\cals$. That is,
$$
\left(-\tfrac{1}{\sqrt{T+2}}
\left[
\begin{array}{cc}
 {\sigma_{11}}+1 & {\sigma_{12}} \\
 {\sigma_{21}} & {\sigma_{22}}+1
\end{array}
\right]\right)^2
=
\left[
\begin{array}{cc}
 {\sigma_{11}} & {\sigma_{12}} \\
 {\sigma_{21}} & {\sigma_{22}}
\end{array}
\right].
$$
The condition $\cals^{\frac12}\wa\in\gltz$ requires that
the trace of $\cals$ be of the form $n^2-2$ for some integer $n>0$.
Since $\sqrt{T+2}=n$, $\cals$ must be of the form
\[
\cals=n K-I, \quad\text{where }
K=\fmat{k_{11}}{k_{12}}{k_{21}}{k_{22}}.
\]
Then $\tr(\cals)=n^2-2$ and $\det(\cals)=1$ imposes
\begin{align*}
k_{11}+k_{22} &= n\not=0,\\
k_{11} k_{22} -k_{12} k_{21} &= 1.
\end{align*}
%
With such an $\cals$,
\[
\ka=\cals^{\frac12}\wa= -\tfrac{1}{\sqrt{T+2}}(\cals+I)
=-\frac{1}{n}(n K)=-K
\text{ (assuming $\tr(K)=n>2$)}.
\]
%Note that $K$ and $-K$ (that is, when  $\tr(K)=n<0$) yield the same
%$\cals$.

(2)\textcircled{b}
Let $A=\matPpm$.
\begin{align*}
\cals^{\frac12}\wa
&=(P\inv \Delta^{\frac12} P) (P\inv A P)
= P\inv\fmat{\tfrac{1}{\sqrt{\lambda}}}{0}{0}{-\sqrt{\lambda}} P\\
&=
-\tfrac{1}{\sqrt{T-2}}
\left[
\begin{array}{cc}
 {\sigma_{11}}-1 & {\sigma_{12}} \\
 {\sigma_{21}} & {\sigma_{22}}-1
\end{array}
\right].
\end{align*}
Note that this is one of the square roots of $\cals$. That is,
$$
\left(-\tfrac{1}{\sqrt{T-2}}
\left[
\begin{array}{cc}
 {\sigma_{11}}-1 & {\sigma_{12}} \\
 {\sigma_{21}} & {\sigma_{22}}-1
\end{array}
\right]\right)^2
=
\left[
\begin{array}{cc}
 {\sigma_{11}} & {\sigma_{12}} \\
 {\sigma_{21}} & {\sigma_{22}}
\end{array}
\right].
$$
The condition $\cals^{\frac12}\wa\in\gltz$ requires that
the trace of $\cals$ be of the form $n^2+2$ for some integer $n>0$.
Since $\sqrt{T-2}=n$, $\cals$ must be of the form
\[
\cals=n K+I, \quad\text{where }
K=\fmat{k_{11}}{k_{12}}{k_{21}}{k_{22}}.
\]
Then $\tr(\cals)=n^2+2$ and $\det(\cals)=1$ imposes
\begin{align*}
k_{11}+k_{22} &= n\not=0,\\
k_{11} k_{22} -k_{12} k_{21} &= -1.
\end{align*}
%
With such an $\cals$,
\[
\ka=\cals^{\frac12}\wa= -\tfrac{1}{\sqrt{T-2}}(\cals-I)
=-\frac{1}{n}(n K)=-K
\text{ (assuming $\tr(K)=n>0$)}.
\]
%Note that $K$ and $-K$ (that is, when  $\tr(K)=n<0$) yield the same
%$\cals$.
\bigskip

This finishes the proof of Proposition \ref{simplifyAtilde}, and hence of
Theorem \ref{abstract-kernel}. For Theorem \ref{abstract-kernel}, when
there is more than one generator, apply criteria from both
cases \rm(1) and \rm(2) from Proposition \ref{simplifyAtilde}.
\end{proof}

Theorem \ref{H1-cokernel-theorem} below will be needed in our classification
of both the crystallographic groups of $\Sol$ and  $\Solmn$, and we state it in suitable generality.

 \begin{theorem}
\label{H1-cokernel-theorem}
For each homomorphism $\varphi: \bbzphi\ra \on{GL}(k,\bbz)$, where $\bbzphi$ is 
an extension of $\bbz = \angles{\bft_{k+1}}$ by a finite group $\Phi$, with
$\varphi(\bft_{3})=\cals$ in {\rm Theorem \ref{abstract-kernel}} ($k=2$, $\Sol$ case),
or $\varphi(\bft_{4})=\cals$ in {\rm Section \ref{hithere2}} ($k=3$, $\Solmn$ case),
we have an
isomorphism $H_{\varphi}^2(\bbzphi, \bbz^k) \cong H^1(\Phi;\coker(I-\cals))$
where $\coker(I-\cals)\cong (I-\cals)\inv\bbz^k/\bbz^k\subset T^k$
is a finite abelian group.
Thus the set of all
equivalence classes of extensions $Q$,
\[
1\lra \bbz^k\lra Q\lra \bbzphi\lra 1,
\]
%of $\bbz^2$ by $\bbzphi$
is in one-one correspondence
with $H^1(\Phi;\coker(I-\cals))$.
\end{theorem}


\begin{proof}
Since $\det(I-\cals)\not=0$,  $H^1(\Phi;\coker(I-\cals))$ is finite, as $\coker(I-\cals)$
is finite.

First we verify that $\varphi(\bbzphi)\subset \on{GL}(k,\bbz)=\aut(\bbz^k)$
leaves the group  $(I-\cals)\inv\bbz^k\subset \bbr^k$
containing $\bbz^k$ invariant.
Suppose $\bfa\in\bbr^k$ such that $(I-\cals)\bfa=\bfz\in\bbz^k$. Then,
\begin{align*}
(I-\cals)\big(\varphi(\bft_{k+1})\bfa\big)
&=(I-\cals)\big(\cals\bfa\big)\\
&=\cals\big((I-\cals)(\bfa)\big)\\
&=\cals(\bfz)\in\bbz^k.\\
\intertext{Now for $\varphi(\alphabar)$, if $\abar=+1$, then}
(I-\cals)\big(\varphi(\bar \alpha)\bfa\big)
&=\varphi(\bar \alpha)(I-\cals^{\abar})\bfa\\
&=\varphi(\bar \alpha)(I-\cals)\bfa\\
&=\varphi(\bar \alpha)\bfz\in\bbz^k;\\
\intertext{and if $\abar=-1$, then}
(I-\cals)\big(\varphi(\bar \alpha)\bfa\big)
&=\varphi(\bar \alpha)(I-\cals^{\abar})\bfa\\
&=\varphi(\bar \alpha)(-\cals^{-1})(I-\cals)\bfa\\
&=\varphi(\bar \alpha)(-\cals^{-1})\bfz\in\bbz^k.
\end{align*}
These show that, if $\bfa\in (I-\cals)\inv\bbz^k$, then
so are $\varphi(\bft_{k+1})\bfa$ and $\varphi(\bar \alpha)\bfa$.
Consequently,  $(I-\cals)\inv\bbz^k$ is $\varphi(\bbzphi)$-invariant.
Since $\bfa-\varphi(\bft_{k+1})\bfa=(I -\cals)\bfa \in \bbz^k$, $\bft_{k+1}$ acts trivially on
$\coker(I-\cals)$. We obtain an induced action of
$\bbzphi/\angles{\bft_{k+1}} \cong \Phi$ on $\coker(I-\cals)$, and so
$H^1(\Phi;\coker(I-\cals))$ is defined.

Suppose we have a class in $H^2(\bbzphi; \bbz^k)$, which defines an extension $Q$.
Since $\bbz^k\subset \bbr^k$ has the unique automorphism extension
property, there exists a push-out $\wt Q$
\cite[(5.3.4)]{seif-book}
fitting the commuting diagram
$$
\CD
1 @>>> \bbz^k @>>> Q      @>>> \bbzphi @>>> 1\\
@. @VVV @VVV @| @.\\
1 @>>> \bbr^k @>>> \wt Q      @>>> \bbzphi @>>> 1.\\
\endCD
$$
We claim that $\wt Q$ is the split extension $\bbr^k \rx \bbzphi$. Note that $\bbzphi$ contains $\bbz=\angles{\bft_{k+1}}$ as a finite index subgroup, and $H^2(\bbz; \bbr^k)$ vanishes. Then $H^2(\bbzphi; \bbr^k)$ is annihilated by the index of $\bbz=\angles{\bft_{k+1}}$ in $\bbzphi$ \cite[Proposition 10.1]{Brown}. Therefore, $H^2(\bbzphi; \bbr^k)$ vanishes. Thus we can always assume that such a $Q$ is embedded in $\bbr^k \rx \bbzphi$.
Since $\bbz\subset\bbzphi$ lifts back to $\gamsig$, it lifts back to
$\wt Q$
so that  $\wt Q$ contains $(\bzero,\bft_{k+1})\in \bbr^k\rx\bbzphi$.
For any element $\bft_{k+1}^n \alphabar\in\bbzphi$, pick a preimage
$\alpha= (a, \alphabar)\in \bbr^k\rx\bbzphi$, taking care that
$a=\bzero$ if $\alphabar=\id$.
Clearly, $\bft_{k+1}^n \alphabar\mapsto a$ defines a map
$\eta: \bbzphi\ra \bbr^k/\bbz^k= T^k$. Evidently, $\eta$ maps into
$\coker(I-\cals) \subset T^k$, which in turn, yields
\[
\eta: \Phi\ra \coker(I-\cals).
\]
We claim that $\eta$ is a crossed homomorphism.
Let $\alphabar, \betabar\in\Phi$, and let
$\eta(\alphabar)=\bfa, \eta(\betabar)=\bfb$.
Then
$\eta(\bft_{k+1}^m\alphabar)=\bfa, \eta(\bft_{k+1}^n\betabar)=\bfb$, and we have
\begin{align*}
(\bfa,\bft_{k+1}^m\alphabar)(\bfb,\bft_{k+1}^n\betabar)
&=(\bfa+\varphi(\bft_{k+1}^m\alphabar)(\bfb),\bft_{k+1}^m\alphabar \bft_{k+1}^n\betabar)\\
&=(\bfa+\varphi(\bft_{k+1}^m)(\varphi(\alphabar)(\bfb)),
\bft_{k+1}^m(\alphabar \bft_{k+1}^n\alphabar\inv)\alphabar\betabar).
\end{align*}
Since $\alphabar \bft_{k+1}^n\alphabar\inv=\bft_{k+1}^l$ for some $l$,
\begin{align*}
\eta(\alphabar \betabar)
&=\bfa+\varphi(\bft_{k+1}^m)(\varphi(\alphabar)(\bfb))\\
&=\eta(\alphabar)+\cals^m(\varphi(\alphabar)(\eta(\betabar)))\\
&=\eta(\alphabar)+\varphi(\alphabar)(\eta(\betabar)),
\end{align*}
where the last equality holds because
$\phi(\alphabar)(\eta(\betabar))\in\coker(I-\cals)$, and
the action of $\cals$ on $\coker(I-\cals)$ is trivial
(if $\bfa\in\coker(I-\cals)$, then
$(I-\cals)\bfa\in\bbz^k$, and hence $\bfa=\cals\bfa$ modulo $\bbz^k$).
Thus $\eta$ is a crossed homomorphism. Conversely, such a crossed homomorphism $\eta$ clearly gives rise to an
extension $Q$. Thus, we obtain a surjective map
\[Z^1(\Phi; \coker(I-S)) \ra H^2(\bbzphi; \bbz^k),\]
which we claim is a homomorphism. To see this, given \[\eta: \Phi \ra \coker(I-\cals),\] we find a $2$-cocycle
$f:\bbzphi \x \bbzphi \ra \bbz^k$ representing the extension $Q$ corresponding to $\eta$. Fix a lift $\wt \eta: \Phi \ra (I-S)^{-1}(\bbz^k)$ (not a homomorphism in general) of $\eta$. Then we can write any element of $Q$ as 
\[(\bfn +  \wt \eta(\alphabar), \bft_{k+1}^{m} \alphabar),\]
where $\bfn \in \bbz^k$ and $m \in \bbz$.
Now, for \[(\bfn_1 + \wt \eta(\alphabar), \bft_{k+1}^{m_1} \alphabar) \text{ and }  (\bfn_2 + \wt \eta(\betabar), \bft_{k+1}^{m_2} \betabar) \in Q,\]
\begin{align*}
&(\bfn_1 + \wt \eta(\alphabar), \bft_{k+1}^{m_1} \alphabar) (\bfn_2 + \wt \eta(\betabar), \bft_{k+1}^{m_2} \betabar)=\\ &(\bfn_1 + \cals^m \varphi(\alphabar)(\bfn_2) + \wt \eta(\alphabar) +  \cals^m \varphi(\alphabar)( \wt \eta(\betabar)), \bft_{k+1}^{m_1} \alphabar \bft_{k+1}^{m_2} \betabar ).
\end{align*}
Therefore, $Q$ is represented by the $2$-cocycle $f:\bbzphi \x \bbzphi \ra \bbz^k$ defined by
\begin{align*}
f( \bft_{k+1}^{m_1} \alphabar, \bft_{k+1}^{m_2} \betabar  ) = \wt \eta(\alphabar) +  \cals^m \varphi(\alphabar)( \wt \eta(\betabar))- \wt \eta(\alphabar \betabar).
\end{align*}
It is now clear that addition of crossed homomorphisms in $Z^1(\Phi; \coker(I-S))$
corresponds to addition of $2$-cocycles in $Z^2(\bbzphi; \bbz^k)$.


Now we shall prove that $Q$ splits if and only if the corresponding $\eta$
is a coboundary, i.e. $\eta\in B^1(\Phi;\coker(I-\cals))$. Note that this will imply that $Z^1(\Phi; \coker(I-S)) \ra H^2(\bbzphi; \bbz^k)$ induces an isomorphism \[H^1(\Phi; \coker(I-S)) \cong H^2(\bbzphi; \bbz^k).\]

A splitting $\bbzphi\ra Q$ induces a homomorphism
\[
s: \bbzphi\ra \wt Q.
\]
Suppose $s(\bft_{k+1})=(z,\bft_{k+1})$ with $z\in\bbz^k$.
(Even in this case, our definition of $\eta$ shows that, we will pick
$(\bzero,\bft_{k+1})$ as our preimage of $\bft_{k+1}$
so that $\eta(\bft_{k+1})=\bzero$, and
$\eta(\alphabar)=\bfa$ if $s(\alphabar)=(\bfa,\alphabar)$ for others).
Let
$y=(\cals-I)\inv z$. Then
\begin{align*}
(y,I) (z,\bft_{k+1}) (-y,I)
&=(y+z-\phi(\bft_{k+1})(y), \bft_{k+1})\\
&=(z+(I-\cals)(y), \bft_{k+1})\\
&=(\bzero, \bft_{k+1}),
\intertext{and}
(y,I) (\bfa,\alphabar) (-y,I)
&=(y+\bfa-\phi(\alphabar)y,\alphabar)\\
&=(\bfa+(I-\phi(\alphabar))y,\alphabar),\\
\intertext{by letting $\bfa+(I-\phi(\alphabar))y=\bfv$,}
&=(\bfv,\alphabar).
\end{align*}
Now,
\begin{align*}
(\bfv,\alphabar) (\bzero,\bft_{k+1})(\bfv,\alphabar)\inv
&=(\bfv-(\alphabar\bft_{k+1} \alphabar\inv) \bfv,\alphabar\bft_{k+1} \alphabar\inv)\\
&=(\bfv-\bft_{k+1}^{\abar} \bfv,\bft_{k+1}^{\abar})\\
&=((I-\cals^{\abar})\bfv,\bft_{k+1}^{\abar}).
\end{align*}
Since $\bbz$ is normal in $\bbzphi$, for $s$ to be a homomorphism,
we must have $(I-\cals^{\abar})\bfv=\bzero \text{ mod } \bbz$. This happens if and only if
$\bfv=\bzero \text{ mod } \bbz$, which holds if and only if
\[
\eta(\alphabar)=\bfa=(\phi(\alphabar)-I)(-y)=(\delta y) (\alphabar),
\]
so that $\eta$ is a coboundary.
\end{proof}



\begin{remark}[Cocycles and Coboundaries]
\label{cocycleConditions}
For each subgroup $\Phi$ of $D_4$, we describe both $Z^1(\Phi;\coker(I-\cals))$
and $B^1(\Phi;\coker(I-\cals))$.
For $\Phi \cong \bbz_2 \x \bbz_2$, we need to check that the commutator of
$(\bfa, \alphabar)$ and $(\bfb, \betabar)$ is in $\bbz^2$. For $\bbz_4$,
there is no cocycle condition to check (since $I+ \varphi(\alphabar)  + \varphi(\alphabar)^2 +
\varphi(\alphabar) ^3 = 0$). Likewise for $\bbz_4 \rx \bbz_2$, there is no cocycle
condition for the order 4 element.

(1)
$\Phi=\bbz_2=\angles{\alphabar}$. Then
\begin{align*}
Z^1(\Phi;\coker(I-\cals))
&=\{\bfa\in\coker(I-\cals):\
(I+\varphi(\alphabar))\bfa=\bzero\},\\
B^1(\Phi;\coker(I-\cals))
&=\{\bfa=(I-\varphi(\alphabar))\bfv:\ \bfv\in\coker(I-\cals)\}.
\end{align*}
\vskip4pt
(2)
$\Phi=\bbz_4=\angles{\alphabar}$. Then
\begin{align*}
Z^1(\Phi;\coker(I-\cals))
&=\{\bfa\in\coker(I-\cals)\},\\
B^1(\Phi;\coker(I-\cals))
&=\{\bfa=(I-\varphi(\alphabar))\bfv:\ \bfv\in\coker(I-\cals)\}.
\end{align*}
\vskip4pt
(3)
$\Phi=\bbz_2\x\bbz_2=\angles{\alphabar,\betabar}$. Then
\begin{align*}
Z^1(\Phi;\coker(I-\cals))&=\{(\bfa,\bfb)\in\coker(I-\cals):
(I+\varphi(\alphabar))\bfa=\bzero,\\
&(I+\varphi(\betabar))\bfb=\bzero,
(I-\varphi(\alphabar))\bfb=(I-\varphi(\betabar))\bfa\},
\end{align*}
\begin{align*}
B^1(\Phi;\coker(I-\cals))
&=\{(\bfa,\bfb)=((I-\varphi(\alphabar))\bfv,(I-\varphi(\betabar))\bfv):\\
&\bfv\in\coker(I-\cals)\}.
\end{align*}
\vskip4pt
(4)
$\Phi=\bbz_4\rx\bbz_2=\angles{\alphabar,\betabar |\
\alphabar^2, \betabar^2, (\betabar\alphabar)^4}$. Then
\begin{align*}
Z^1(\Phi;\coker(I-\cals))&=\{(\bfa,\bfb)\in\coker(I-\cals):
(I+\varphi(\alphabar))\bfa=\bzero,\\
&(I+\varphi(\betabar))\bfb=\bzero\},
\end{align*}
\begin{align*}
B^1(\Phi;\coker(I-\cals))
&=\{(\bfa,\bfb)=((I-\varphi(\alphabar))\bfv,(I-\varphi(\betabar))\bfv):\\
&\bfv\in\coker(I-\cals)\}.
\end{align*}
\vskip4pt
\end{remark}

\bigskip

%\begin{remark}
%There are nontrivial classes in
%$H_{\phi}^2(\bbzphi, \bbz^2) \cong H^1(\Phi;\coker(I-\cals))$
%which correspond to groups $Q$ with torsion. See Example \ref{maxhol}.
%\boxt{Mention exact statement of JB's where they have a mistake? - it
%may be better to mention in example section }
%\end{remark}

Suppose we found
an extension $Q$; that is, $\eta\in H^1(\Phi;\coker(I-\cals))$
with $\eta(\alphabar)=\bfa$. Then
\[
Q=\angles{\bft_1,\bft_2,\bft_3,
\alpha=(\bft_1^{a_1} \bft_2^{a_2} \bft_3^{a_3},A)}
\]
has the following presentation:
%
\begin{align}
\label{Solpresentation}
\hspace{0.5in}
\nonumber
&\bft_3(\bft_1^{n_1}\bft_2^{n_2})\bft_3\inv=\bft_1^{m_1}\bft_2^{m_2},
\text{ where } {\xy{m_1}{m_2}=\cals\xy{n_1}{n_2}},\\
\nonumber
&\alpha(\bft_1^{n_1}\bft_2^{n_2})\alpha\inv=\bft_1^{m_1}\bft_2^{m_2},
\text{ where } {\xy{m_1}{m_2}=\varphi(\alphabar)\xy{n_1}{n_2}},\\
&\alpha\bft_3\alpha\inv=\bft_1^{w_1}\bft_2^{w_2}\bft_3^{\abar},
\text{ where } {\xy{w_1}{w_2}=
\left(I-\cals^{\abar}\right)\xy{a_1}{a_2}},\\
\nonumber
&\alpha^2=\bft_1^{v_1}\bft_2^{v_2}\bft_3^{(1+\abar)a_3},
\text{ where } {\xy{v_1}{v_2}=\left(I+\varphi(\alphabar)\right)\xy{a_1}{a_2}},
\text{ if }\ A^2=I,\\
\nonumber
&\alpha^4=\id,
\text{ if }\ \ord(A)=4.
\end{align}


\begin{remark}
\label{abstractKernelSol}
An element of $H^1(\Phi;\coker(I-\cals))$ determines
a unique (up to isomorphism) extension of $\bbz^2$ by $\bbzphi$, which we can also
view as 
\[1 \ra \gamsig \ra Q \ra \Phi \ra 1.\]
This also follows from group cohomology. For an element of $H^1(\Phi;\coker(I-\cals))$ fixes the exponents $a_1, a_2, a_3$ in  equation (\ref{Solpresentation}), and defines an abstract kernel $\Phi \ra \out(\gamsig)$. As $\gamsig$ has trivial center, $H^2(\Phi; \calz(\gamsig))$
is trivial.
\end{remark}

\begin{corollary}
\label{a-conj-zero3}
Let $Q = \angles{\Gamma_\cals,
  (\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3},A)}$ be
an extension in {\rm Theorem \ref{H1-cokernel-theorem}}  with lattice $\Gamma_\cals =
\angles{\bft_1,
\bft_2, \bft_3}$. Suppose $\varphi(\alphabar)=-K$ and $\cals=nK\pm I$
(in {\rm Theorem \ref{abstract-kernel}},
$A$ has order 2, $\abar=1$, and $a_3=\frac 12$).
Then $H^1(\Phi;\coker(I-\cals))=0$.
Therefore, there exists $\bft_1^{v_1}\bft_2^{v_2}$ which conjugates
$(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{\frac12},A)$ to $(\bft_3^{\frac12},A)$
and leaves $\Gamma_\cals$ invariant.
\end{corollary}

\begin{proof}
We have $\det(I - \varphi(\alphabar)) = 1 + \det(\varphi(\alphabar)) -\tr(\varphi(\alphabar))$.
If $\det(\varphi(\alphabar))=-1$, this quantity is nonzero since
$\tr(\varphi(\alphabar)) \neq 0$.
Else, if $\det(\varphi(\alphabar))=1$, we had that $\lvert
\tr(\varphi(\alphabar)) \rvert > 2$. 
%To see this, note that
%$\varphi(\alphabar) = P \inv \sqrt{\Delta} A P$.
Consequently, $I - \varphi(\alphabar)$ is non-singular and we may take
% \[
% \xy{v_1}{v_2}=(I-\varphi(\alphabar)) \inv \xy{a_1}{a_2}.
% \]
$\bfv=(I-\varphi(\alphabar))\inv\bfa$. Then our $\bfv$ conjugates
$(\bft_3^{\frac 1 2}, A)$ to $(\bft_1^{a_1} \bft_2^{a_2}\bft_3^{\frac 1 2},
A)$.
It remains to show $\bfv\in (I-\cals)\inv\bbz^2$.
First observe that $\varphi(\alphabar)$ is a square root of $\cals$.
\begin{align*}
\varphi(\alphabar)^2
&=P\inv\sqrt{\Delta} A \sqrt{\Delta} A P =P\inv\Delta  A^2 P\ \text{(since
  $\abar=+1$, $A$ is diagonal)}\\
&=P\inv\Delta P=\cals.
\end{align*}
Now
\[
(I-\cals) \bfv = (I+\varphi(\alphabar))(I-\varphi(\alphabar)) \bfv =
(I+\varphi(\alphabar))\bfa \in \bbz^2.\qedhere
\]
\end{proof}

\begin{lemma}
\label{EasyTor}
When $\alpha =(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3}, A) \in Q$ satisfies
$a_3 = \frac{1}{2}$ and $\abar=1$, $\gamma \alpha$ is not torsion for any $\gamma \in
\gamsig$.
\end{lemma}
\begin{proof}
Note that $A$ is necessarily of order 2. Let $\pr: \Sol \ra \bbr$ denote the
quotient homomorphism of $\Sol$ by its nil-radical $\bbr^2$. Write $\gamma \in \gamsig$
as $\bft_1^{n_1}\bft_2^{n_2}\bft_3^{n_3}$. Application of $\pr$ to $(\gamma \alpha)^2$
yields
\[
\pr(\gamma \alpha)^2 = 2 n_3+1,
\]
from which we infer $\gamma \alpha$ is not torsion.
\qedhere
\end{proof}

\begin{theorem}[Classification of $\Sol$-geometry]
\label{ClassificationSol-geometry}
The following is a complete classification
crystallographic groups of $\Sol$. In each case,
we give $\Phi$, $\bbzphi$, the abstract kernel $\varphi:\bbzphi\ra\gltz$ (which
gives the action of $\bbzphi$ on $\angles{\bft_1,\bft_2}$), equations defining $\bfa=\xy{a_1}{a_2}$, $\bfb=\xy{b_1}{b_2}$
from $H^1(\Phi;\coker(I-\cals))$, and conditions for $Q$ to be torsion-free.
Note that $\varphi(\bft_3) = \cals$ by definition.
\end{theorem}

\noindent
For each abstract kernel $\bbzphi \ra \gltz$ give in Theorem \ref{abstract-kernel},
we give

\noindent
{\rm (1) } The holonomy $\Phi$, the conditions on $\cals$, and the abstract kernel \[\phi: \bbzphi \ra \gltz\] 

(this fixes the exponents on $\bft_3$).

\noindent
{\rm (2) } A computation of $H^1(\Phi;\coker(I-\cals))$ (this fixes the exponents on $\bft_1, \bft_2$).

\noindent
{\rm (3) } When the group is torsion-free.


\begin{enumerate}
\bigskip
\item[\gr{0}]
$\Phi=\text{trivial}$,
$\cals\in\sltz$
with $\tr(\cals)>2$;\ $Q=\gamsig$.
\par\noindent
$\bullet$ Torsion-free
%
\item[\gr{1}]
$\Phi=\bbz_2$: $A=
\left[\begin{matrix}
1 &0\\ 0& -1
\end{matrix}\right]$,
$\bbzphi=\bbz=\angles{\bft_3,\bar\alpha=(\bft_3^{\frac12},A)}$.\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$ with $\det(K)=-1$, $\tr(K)=n>0$.\parn
$\bullet$
$\vara=-K$\parn
$\bullet$ $H^1(\Phi;\coker(I-\cals))$ is trivial so that $\bfa=\bzero$. \parn
%
$\bullet$
$Q=\angles{\bft_1,\bft_2,\bft_3,
\alpha =(\bft_3^{\frac12},A)}$.\parn
$\bullet$
Torsion-free
%
\item[\gr{2a}]
$\Phi=\bbz_2$: $A=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,\
$\bbzphi=\bbz\x\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A)}$.\parn
$\bullet$
$\cals\in\sltz$ with $\tr(\cals)>2$.\parn
$\bullet$
$\vara=A$\parn
$\bullet$
$H^1(\Phi;\coker(I-\cals)) = \frac {\coker(I-\cals)}{2\coker(I-\cals)} \subseteq \bbz_2 \x \bbz_2 $.
That is, all $\bfa$ with $(I-\cals)\bfa\equiv \bfzero$, taken modulo $2\bfa$.\parn
$\bullet$
$Q=\angles{\bft_1,\bft_2,\bft_3,
\alpha =(\bft_1^{a_1} \bft_2^{a_2},A)}$.\parn
$\bullet$
Always has torsion
%
\item[\gr{2b}]
$\Phi=\bbz_2$: $A=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,
$\bbzphi=\bbz=\angles{\bft_3,\bar\alpha=(\bft_3^{\frac12},A)}$.\parn
$\bullet$
$\cals=n K-I$, $K\in\gltz$ with $\det(K)=+1$,  $\tr(K)=n>2$.\parn
$\bullet$ $\vara=-K$\parn
$\bullet$ $H^1(\Phi;\coker(I-\cals))$ is trivial so that $\bfa=\bzero$. \parn
$\bullet$
$Q=\angles{\bft_1,\bft_2,\bft_3,
\alpha =(\bft_3^{\frac12},A)}$.\parn
$\bullet$
Torsion-free
%
\item[\gr{3}]
$\Phi=\bbz_2$: $A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,\
$\bbzphi=\bbz\rx\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A)}$.\parn
$\bullet$
$\cals\in\sltz$, $\tr(\cals)>2$,
with $\sigma_{12}=-\sigma_{21}$.\parn
$\bullet$ $\vara=A$\parn
$\bullet$
$H^1(\Phi;\coker(I-\cals))\subseteq \bbz_2$ can be described as all $\bfa$ with $(I-\cals)\bfa\equiv \bfzero$,
$a_2\equiv -a_1$ taken modulo
$(v_1-v_2){\scriptstyle{\xy{1}{-1}}}$ where $(I-\cals)\bfv\equiv \bfzero$.\parn
$\bullet$
$Q=\angles{\bft_1,\bft_2,\bft_3,
\alpha =(\bft_1^{a_1} \bft_2^{a_2},A)}$.\parn
$\bullet$
Always has torsion
%
\item[\gr{3\ii}]
$\Phi=\bbz_2$: $A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,\
$\bbzphi=\bbz\rx\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A)}$.\parn
$\bullet$
$\cals\in\sltz$, $\tr(\cals)>2$,
with $\sigma_{11}=\sigma_{22}$.\parn
$\bullet$
$\vara=\left[\begin{matrix}
1&0\\ 0&-1
\end{matrix}\right]
$\parn
$\bullet$
$H^1(\Phi;\coker(I-\cals))\subseteq \bbz_2 \x \bbz_2$ can be described as all $\bfa$ with $(I-\cals)\bfa\equiv\bfzero$, $2 a_1\equiv 0$ (so $a_1=0$ or $\tfrac12$)
modulo $\xy{0}{2v_2}$ where $(I-\cals)\bfv\equiv \bfzero$.\parn
$\bullet$
$Q=\angles{\bft_1,\bft_2,\bft_3,
\alpha =(\bft_1^{a_1} \bft_2^{a_2},A)}.$\parn
$\bullet$
Torsion-free if and only if $a_1=\tfrac12$
and $a_2\not\equiv \frac{(\sigma_{11}+1)(2 n+1)}{2\sigma_{12}}$
for any $n\in\bbz$.
%
\par\noindent
\item[\gr{4}]
$\Phi=\bbz_4$: $A=
\left[\begin{matrix}
0 &1\\ -1& 0
\end{matrix}\right]$,
$\bbzphi=\bbz\rx\bbz_4=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A)}$.\parn
$\bullet$
$\cals\in\sltz$, symmetric, and $\tr(\cals)>2$.\parn
$\bullet$
$\vara=A$\parn
$\bullet$
$H^1(\Phi;\coker(I-\cals)) \cong \frac {\coker(I-\cals)}{\im({I-A})} \subseteq \bbz_2$. That is,
all $\bfa$ with $(I-\cals)\bfa\equiv \bfzero$, taken modulo $(I-A)\bfa$.
\parn
$\bullet$
$Q=\angles{\bft_1,\bft_2,\bft_3,
\alpha =(\bft_1^{a_1} \bft_2^{a_2},A)}$.\parn
$\bullet$
Always has torsion
%
\item[\gr{5}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
1 &0\\ 0& -1
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,\parn
$\bbzphi=\bbz\x\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{\frac12},A),
\bar\beta=(\bft_3^0,B)}$.\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$ with $\det(K)=-1$ and $\tr(K)=n>0$.\parn
$\bullet$
$\vara=-K$, $\varb=B$
\hfill \gr{1}+\gr{2a}\parn
$\bullet$
$H^1(\Phi;\coker(I-\cals)) \cong \frac {\coker(I+K)}{2\coker({I+K})} \subseteq \bbz_2 \x \bbz_2$. That is,
all $\bfb$ with $(I+K)\bfb\equiv \bfzero$, taken modulo $2\bfb$.\parn
$\bullet$
$Q=\angles{\bft_1,\bft_2,\bft_3,
\alpha=(\bft_3^{\frac12},A),
\beta=(\bft_1^{b_1} \bft_2^{b_2},B)}$\parn
$\bullet$
Always has torsion
%
\item[\gr{6a}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,\parn
$\bbzphi=(\bbz\x\bbz_2)\rx\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A),
\bar\beta=(\bft_3^0,B)}$.\parn
$\bullet$
$\cals\in\sltz$ with $\sigma_{12}=-\sigma_{21}$ and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=A$, $\varb=B$ \hfill \gr{3}+\gr{2a}\parn
%
$\bullet$
$H^1(\Phi;\coker(I-\cals))$ can be described as all
$\bfa$, $\bfb$ with $(I-\cals)\bfa\equiv \bfzero$ and $(I-\cals)\bfb\equiv \bfzero$,
$a_2\equiv -a_1$, $b_1-b_2-2 a_1 \equiv 0$, $\bfa$ and $\bfb$ modulo $(v_1-v_2)
{\scriptstyle{\xy{1}{-1}}}$ and  $2\bfv$ for $\bfv \in \coker(I-\cals)$, respectively.\parn
%
$\bullet$
$Q=\angles{\bft_1,\bft_2,\bft_3,
\alpha=(\bft_1^{a_1} \bft_2^{a_2},A),
\beta=(\bft_1^{b_1} \bft_2^{b_2},B)}$.\parn
$\bullet$
Always has torsion
%
\item[\gr{6a\ii}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,\parn
$\bbzphi=(\bbz\x\bbz_2)\rx\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A),
\bar\beta=(\bft_3^0,B)}$.\parn
$\bullet$
$\cals\in\sltz$
with $\sigma_{11}=\sigma_{22}$ and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=
\left[\begin{matrix}
1 &0\\ 0& -1
\end{matrix}\right]$,\
$\varb=B$
\hfill \gr{3\ii}+\gr{2a}\parn
$\bullet$
$H^1(\Phi;\coker(I-\cals))$ can be described as all $\bfa$, $\bfb$ with
$(I-\cals)\bfa\equiv \bfzero$ and $(I-\cals)\bfb\equiv \bfzero$,
 $2 a_1\equiv 0$, $2b_2-2a_2 \equiv 0$,
$\bfa$ and $\bfb$ taken modulo $\xy{0}{2v_2}$ and $2\bfv$ for $\bfv \in \coker(I-\cals)$,
respectively.
\parn
%
$\bullet$
$Q=\angles{\bft_1,\bft_2,\bft_3,
\alpha=(\bft_1^{a_1} \bft_2^{a_2},A),
\beta=(\bft_1^{b_1} \bft_2^{b_2},B)}$.\parn
$\bullet$
Always has torsion
%none of the following 4 equations have integer solutions for $n_1,n_2$:
%\begin{alignat*}{2}
%2 (a_1+n_1)=&0,
%&(\sigma_{11}+1)(a_1+n_1)+\sigma_{12}(a_2-n_2)&=0,\\
%2 (a_2-b_2+n_2)&=0,\qquad
%&(\sigma_{11}-1)(a_1+b_1+n_1)+\sigma_{12}(a_2-b_2-n_2)&=0.
%\end{alignat*}
%
%
\item[\gr{6b}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,\parn
$\bbzphi=\bbz\rx\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A),
\bar\beta=(\bft_3^{\frac12},B)}$.\parn
$\bullet$
$\cals=n K-I$, $K\in\sltz$ with $\tr(K)= n>2$; and
$k_{12}=-k_{21}$.\parn
$\bullet$ $\vara=A$, $\varb=-K$ \hfill \gr{3}+\gr{2b}\parn
$\bullet$
$H^1(\Phi;\coker(I-\cals)) \cong H^1(\angles{A};\coker(I+K)) \subseteq \bbz_2$, that is,
all $\bfa $ with $(I+K)\bfa \equiv \bfzero$ and $a_2\equiv -a_1$ modulo
$(v_1-v_2){\scriptstyle{\xy{1}{-1}}}$, where $(I+K)\bfv \equiv \bfzero$. \parn
$\bullet$
$Q=\angles{\bft_1,\bft_2,\bft_3,
\alpha=(\bft_1^{a_1} \bft_2^{a_2},A),
\beta=(\bft_3^{\frac12},B)}$.\parn
$\bullet$
Always has torsion
%
\item[\gr{6b\ii}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,\parn
$\bbzphi=\bbz\rx\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A),
\bar\beta=(\bft_3^{\frac12},B)}$.\parn
$\bullet$
$\cals=n K-I$, $K\in\sltz$ with $\tr(K)= n>2$;\
$k_{11}=k_{22}$.\parn
$\bullet$ $\vara=\matPpm$, $\varb=-K$ \hfill \gr{3\ii}+\gr{2b}\parn
$\bullet$ $H^1(\Phi;\coker(I-\cals)) \cong H^1(\angles{A};\coker(I+K)) \subseteq \bbz_2 \x \bbz_2$, that is,
all $\bfa $ with $(I+K)\bfa \equiv \bfzero$ and $2a_1\equiv 0$ taken modulo
$\xy{0}{2v_2}$, where $(I+K)\bfv \equiv \bfzero$. \parn
$\bullet$
$Q=\angles{\bft_1,\bft_2,\bft_3,
\alpha=(\bft_1^{a_1} \bft_2^{a_2},A),
\beta=(\bft_3^{\frac12},B)}$.\parn
$\bullet$
Torsion-free if and only if
$a_1=\tfrac12$
and $a_2\not\equiv \frac{(k_{11}-1)(2 n+1)}{2k_{12}}$ for any $n \in \bbz$.
%
%
\item[\gr{7}]
$\Phi=\bbz_4\rx\bbz_2$:
$A=\matNpp, B=\matPpm$,\parn
$\bbzphi=(\bbz\x\bbz_2)\rx\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A),
\bar\beta=(\bft_3^{\frac12},B)}$.\parn
$\bullet$
$\cals=n K+I$,$K\in\gltz$, $\det(K)=-1$, $\tr(K)>0$;
$k_{12}=-k_{21}$.\parn
$\bullet$
$\vara=A$, $\varb=-K$ \hfill (includes \gr{6a})\qquad  \gr{3}+\gr{1}\parn
$\bullet$ $H^1(\Phi;\coker(I-\cals))$ is all $\bfa$ with $(I-\cals)\bfa \equiv \bfzero$ and $a_2 \equiv -a_1$, taken modulo
$(v_1-v_2){\scriptstyle{\xy{1}{-1}}}$, where $(I+K)\bfv \equiv \bfzero$.  \parn
$\bullet$
$Q=\angles{\bft_1,\bft_2,\bft_3,
\alpha=(\bft_1^{a_1} \bft_2^{a_2},A),
\beta=(\bft_3^{\frac12},B)}$.\parn
$\bullet$
Always has torsion
%

\item[\gr{7\ii}]
$\Phi=\bbz_4\rx\bbz_2$:
$A=\matNpp, B=\matPpm$,\parn
$\bbzphi=(\bbz\x\bbz_2)\rx\bbz_2=\angles{\bft_3,\bar\alpha=(\bft_3^{0},A),
\bar\beta=(\bft_3^{\frac12},B)}$\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$, $\det(K)=-1$, $\tr(K)>0$;
$k_{11}=k_{22}$.\parn
$\bullet$
$\vara=\matPpm$, $\varb =-K$
\hfill (includes \gr{6a\ii})\qquad \gr{3\ii}+\gr{1}\parn
$\bullet$ $H^1(\Phi;\coker(I-\cals))$ is all $\bfa$ with $(I-\cals)\bfa \equiv \bfzero$ and $2a_1 \equiv 0$, taken modulo
$\xy{0}{2v_2}$, where $(I+K)\bfv \equiv \bfzero$.  \parn
$\bullet$
$Q=\angles{\bft_1,\bft_2,\bft_3,
\alpha=(\bft_1^{a_1} \bft_2^{a_2},A),
\beta=(\bft_3^{\frac12},B)}$.\parn
$\bullet$
Always has torsion
\par\noindent
\end{enumerate}
\bigskip

See equation (\ref{Solpresentation}) for a description of the relations.
Of course, when the holonomy has two generators $\alpha$, $\beta$,
we also have:
\begin{align*}
&[\alpha,\beta]=\bft \text{ for } \bft \in \gamsig,\
\text{ if $\Phi=\bbz_2\x\bbz_2$};\\
&(\alpha\beta)^4= \bft \text{ for } \bft \in \gamsig\
\text{ if $\Phi=\bbz_4\rx\bbz_2$}.
\end{align*}
%\end{theorem}
%\vfill\eject

\begin{proof}
Theorem \ref{abstract-kernel} classifies all the $\bbzphi$ and abstract kernels
$\phi:\bbzphi \ra \gltz$ up to conjugation in $\gltz$ (which is
essentially up to a change of generating set for the lattice).
This fixes $a_3, b_3$.
Theorem \ref{H1-cokernel-theorem} allows us to find $a_1, a_2, b_1, b_2$
for each abstract kernel $\varphi$. This is all we need for a $\Sol$ crystallographic
group $Q$. The proof of torsion-free criteria for cases \gr{3i}, \gr{6bi} is postponed
until Corollary \ref{SolTor}.
\bigskip

\item[\gr{0}]
See Theorem \ref{lattice-part-construction}.
\bigskip

\item[\gr{1}]
By Proposition \ref{simplifyAtilde} (2)\textcircled{b},
$\cals=n K+I$, where $\det(K)=-1$, $\tr(K)=n>0$. Then
$\varphi(\bar\alpha)=\cals^{a_3}\circ P\inv A P=-K$. Corollary \ref{a-conj-zero3} shows $H^1(\Phi;\coker(I-\cals))$ is trivial,
and thus we can take $a_1=a_2=0$.
By Lemma \ref{EasyTor}, $Q$ is torsion-free.

\bigskip

\item[\gr{2a}]
Let
$
A=\left[\begin{matrix}
-1&0\\ 0&-1
\end{matrix}\right].
$
There is no condition on $\cals$, as
$\varphi(\bar\alpha)=\cals^{0}\wa= A\in\gltz$ automatically.
\noindent
Now $\bfa$ must satisfy $(I-\cals)\bfa\in\bbz^2$
modulo $(I-\varphi(\bar\alpha))\bfa=2\bfa$,
since $(I+\varphi(\bar\alpha))\bfa=\bfzero\in\bbz^2$ already. Note that all elements of 
$H^1(\Phi; \coker(I-S))$ are of order 2, and is generated by at most 2 elements. Therefore,
$H^1(\Phi; \coker(I-S))$ is at most $\bbz_2 \x \bbz_2$.
The square of $\alpha$ is always the identity. Therefore, $Q$ has torsion.
\bigskip

\item[\gr{2b}]
Let
$$
A=\left[\begin{matrix}
-1&0\\ 0&-1
\end{matrix}\right],\ a_3=\tfrac12.
$$
\noindent
By Proposition \ref{simplifyAtilde} (2)\textcircled{b},
$\cals=n K-I$ with $\det(K)=+1$, $\tr(K)=n>2$. Then
$\vara =\cals^{a_3}\circ P\inv A P=-K$.
By Corollary \ref{a-conj-zero3}, we can take $a_1=a_2=0$ so that
$\alpha=(\bft_3^{\frac12},A)$. By Lemma \ref{EasyTor}, $Q$ is torsion-free.

\bigskip

\item[\gr{3}]
Let
\[
A=\left[\begin{matrix}
0&1\\ 1&0
\end{matrix}\right],\
\text{ and }\
\wa = A.
\]
The conditions on $\cals$ follow from
\ref{simplifyAtilde} (1)\textcircled{c}.
Let $\alpha=(\bft_1^{a_1} \bft_2^{a_2},A)$.
Then $\bfa$ must satisfy $(I-\cals)\bfa\in\bbz^2$,
\[
(I+\vara)\bfa=
\left[\begin{matrix}
1&1\\ 1&1
\end{matrix}\right] \bfa\equiv\bfzero
\text{ modulo }
(I-\vara)\bfv=
\left[\begin{matrix}
1&-1\\ -1&1
\end{matrix}\right] \bfv
\text{, for }
(I-\cals)\bfv\in\bbz^2,
\]
computing, we obtain the claimed conditions. Applying the coboundary operator
to the cocycles yields:
\[(I-\vara) \xy{a_1}{-a_1} = \xy{2a_1}{-2a_1}.\]
This computation implies that $H^1(\Phi;\coker(I-\cals))$ is at most $\bbz_2$.

\noindent
To show that $Q$ always has torsion, note that $a_2 \equiv -a_1$, and
\begin{align*}
\alpha^2
&=(\bft_1^{a_1}\bft_2^{-a_1},A)^2
=(\bft_1^{a_1}\bft_2^{-a_1}\cdot
A(\bft_1^{a_1}\bft_2^{-a_1}),I) \\
&=(\bft_1^{a_1}\bft_2^{-a_1}\cdot \bft_2^{a_1}\bft_1^{-a_1},I)\\
&=(e,I).
\end{align*}
\bigskip

\item[\gr{3\ii}]
Let
\[
A=\left[\begin{matrix}
0&1\\ 1&0
\end{matrix}\right],\
\text{ and }\
\wa =
\left[\begin{matrix}
1&0\\ 0&-1
\end{matrix}\right]\not=A.
\]
Conditions on $\cals$ follow from
\ref{simplifyAtilde} (1)\textcircled{c}.
Then $\bfa$ must satisfy $(I-\cals)\bfa\in\bbz^2$,
\[
(I+\vara)\bfa=
\left[\begin{matrix}
2&0\\ 0&0
\end{matrix}\right] \bfa\equiv\bfzero
\text{ modulo }
(I-\vara)\bfv=
\left[\begin{matrix}
0&0\\ 0&2
\end{matrix}\right] \bfv
\text{, for }
(I-\cals)\bfv\in\bbz^2.
\]
This shows that $a_1=0$ or $\frac12$. Note that $H^1(\Phi; \coker(I-\cals))$ can be at most
$\bbz_2 \x \bbz_2$.
The torsion criteria follow from Corollary \ref{SolTor}.
\bigskip

\item[\gr{4}]
Let
$$
A=\left[\begin{matrix}
0 &1\\ -1& 0
\end{matrix}\right],\
\text{ and } \wa = A.
$$
The conditions on $\cals$ follow from
\ref{simplifyAtilde} (1)\textcircled{b}.
By Remark \ref{cocycleConditions}, we just need
$\bfa \in \coker(I-\cals)$, taken modulo
$\im(I-\phi(\alphabar))$. Note that $\det(I-\phi(\alphabar))=2$, which implies
that $H^1(\Phi; \coker(I-S))$ is at most $\bbz_2$.
As $\alpha^4=e$, $Q$ always has torsion.
%
% There are 4 choices for $a_4$ which make $\alpha^4={\bft_4}^{\frac{i}{q}},
% \quad (i=0,1,2,3)$.
%  $\varPi$ is torsion-free only when $i>0$.

\bigskip


\item[\gr{5}]
Let $\Phi=\bbz_2\x\bbz_2=\angles{A,B}$, where
$$
A=\matPpm,\
B=\matPmm, \text{ and } a_3=\tfrac12.
$$
This is a combination of {\tt (1)+(2a)}.
From case {\tt(1)},
$\cals=I+n K \in\gltz$
with $\det(K)=-1$. By Corollary \ref{a-conj-zero3}, we can take $a_1=a_2=0$.
We need $\bfb$ to satisfy $(I-\cals)\bfb \equiv \bfzero$.
However, the cocycle conditions for
$\bbz_2 \x \bbz_2$ in Remark \ref{cocycleConditions}
show that we must have $(I - \phi(\alphabar))\bfb$ = $(I+K)\bfb \equiv \bfzero$.
In fact, since $(I-\cals) = (I-K)(I+K)$,
this condition implies $(I-\cals)\bfb \equiv \bfzero$. Since
we have already fixed $a_1=a_2=0$, for the coboundary in Theorem \ref{H1-cokernel-theorem}, we take $\bfb$ modulo $(I - \phi(\betabar))\bfv = 2 \bfv$ only when
$\bfv$ satisfies $(I - \phi(\alphabar))\bfv$ = $(I+K)\bfv \equiv \bfzero$.
Hence
\[
H^1(\Phi;\coker(I-\cals)) \cong \frac {\coker(I+K)}{2\coker({I+K})} \subseteq \bbz_2.
\]
As $\beta^2=e$, $Q$ is never torsion-free.

\bigskip

\item[\gr{6a}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,\ \text{ and } $\wa=A$.\parn
This is a combination of  \gr{3}+\gr{2a}.
We have $(I-\cals)\bfa\equiv \bfzero$ and $(I-\cals)\bfb\equiv \bfzero$.
Also, $\bfa$ and $\bfb$ also must satisfy the cocycle conditions
in Remark \ref{cocycleConditions}. Note that $(I+\phi(\alphabar))\bfa \equiv \bfzero$
forces $a_2\equiv -a_1$, whereas $(I-\ka) \bfb-(I-\kb) \bfa \equiv \bfzero$ forces
$b_1-b_2-2 a_1 \equiv 0$, $-b_1 + b_2 - 2 a_2 \equiv 0$. Since $a_2\equiv -a_1$, the second equation is redundant. We take $\bfa$ and $\bfb$
modulo $(I-\ka) \bfv$ and $(I-\kb)\bfv$, respectively, where $(I-\cals)\bfv \equiv \bfzero$.
As $Q$ contains a subgroup of type \gr{3}, $Q$ has torsion.
\bigskip


\item[\gr{6a\ii}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,\ \text{ and  }$\wa =
\left[\begin{matrix}
1&0\\ 0&-1
\end{matrix}\right]\not=A.$\parn
Similar to case \gr{6a}, this is a combination of  \gr{3\ii}+\gr{2a}.
From the cocycle condition $(I+\ka)\bfa \equiv \bfzero$, $a_1=0$ or $\tfrac12$.
The description of $H^1(\Phi, \coker(I-\cals))$ follows just like
in case \gr{6a}. As $\beta^2 = e$, $Q$ has torsion.



%Torsion cut 2
%
\bigskip

\item[\gr{6b}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,\ $\wa=A$, \text{ and } $b_3=\frac 12$.\parn
This is a combination of  \gr{3}+\gr{2b}.
To have $\wa = A$,
Proposition \ref{simplifyAtilde} (1)\textcircled{c}, (2)\textcircled{a} force
$\sigma_{12}=-\sigma_{21}$ and $\cals=n K-I$,
where $K\in\sltz$ with $\tr(K)= n>2$, which taken together yield the desired
conditions on $K$.
\noindent
By Corollary \ref{a-conj-zero3}, we can take $b_1=b_2=0$.
Hence,
\noindent
\begin{align*}
\alpha&=(\bft_1^{a_1}\bft_2^{a_2},A),\\
 \beta&=(\bft_3^{\frac12},B).
\end{align*}
\noindent
The cocycle conditions in Remark \ref{cocycleConditions} force
$(I + \ka) \bfa \equiv \bfzero$ as well as
$(I - \kb) \bfa = (I+K)\bfa \equiv \bfzero$. This shows
$\bfa \in \coker(I+K)$.
Since $b_1, b_2 =0$ is fixed, take $\bfa$ modulo $(I - \ka)\bfv$ only when
$(I-\kb)\bfv = (I+K)\bfv \equiv \bfzero$. Hence
\[
H^1(\Phi;\coker(I-\cals)) \cong H^1(\angles{A};\coker(I+K)) \subseteq \bbz_2.
\]
\noindent
There is always torsion in $Q$, as it contains a subgroup of type \gr{3}.

\bigskip
%
\item[\gr{6b\ii}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$,\ $\wa=\matPpm$, $b_3=\frac 12$.\parn
This is a combination of  \gr{3\ii}+\gr{2b};
Similar to case \gr{6b}, we apply Proposition \ref{simplifyAtilde} (1)\textcircled{c}, (2)\textcircled{a}
to obtain $\cals=n K-I$, where $K\in\sltz$ with $\tr(K)= n>2$,
$k_{11}=k_{22}$.
\noindent
By Corollary \ref{a-conj-zero3}, we can take $b_1=b_2=0$.
Hence,
\begin{align*}
\alpha&=(\bft_1^{a_1}\bft_2^{a_2},A),\\
 \beta&=(\bft_3^{\frac12},B).
\end{align*}
\noindent
The argument to establish
\[
H^1(\Phi;\coker(I-\cals)) \cong H^1(\angles{A};\coker(I+K)) \subseteq \bbz_2 \x \bbz_2
\]
is identical to that of \gr{6b}. In this case, we use
$\ka = \matPpm$ rather than $\ka = A$ and obtain
the desired conditions on $a_1, a_2$.
The torsion criteria follow from Corollary \ref{SolTor}.
\bigskip
%
%
\item[\gr{7}]
$\Phi=\bbz_4\rx\bbz_2:$
$
A=\matNpp, B=\matPpm, \wa=A, \text{ and } b_3 = \frac 1 2.
$
\parn
We apply Proposition \ref{simplifyAtilde} (1)\textcircled{c}, (2)\textcircled{b}
to obtain the conditions $\cals=n K+I$, where $K\in\gltz$
with $\det(K)=-1$ and  $\tr(K)>0$;
and $k_{12}=-k_{21}$.
\noindent
This is a combination \gr{3}+\gr{1} which includes \gr{6a}.
By Corollary \ref{a-conj-zero3}, we can take $b_1=b_2=0$.
Therefore,
\[
\alpha=(\bft_1^{a_1} \bft_2^{a_2},A),\
\beta=(\bft_3^{\frac12},B).
\]

For $(I-\cals)\bfa \equiv \bfzero$, the only cocycle
condition that $\bfa$ must satisfy is $(I+\ka)\bfa \equiv \bfzero$, which
forces $a_2\equiv -a_1$ (Remark \ref{cocycleConditions}).
\noindent
However, we have fixed $b_1=b_2=0$. Therefore, when computing the coboundaries,
we can take $\bfa$ modulo $(I-\ka)\bfv$ only for
$(I-\cals)\bfv$ where $\bfv$ also satisfies
$(I-\kb)\bfv=(I+K)\bfv \equiv \bfzero$. Note that $(I+K)\bfv \equiv \bfzero$
actually implies  $(I-\cals)\bfv \equiv \bfzero$ since $(I-\cals) = (I-K)(I+K)$.
There is always torsion in $Q$, as it contains a subgroup of type \gr{4}.
\bigskip
%
\item[\gr{7\ii}]
$\Phi=\bbz_4\rx\bbz_2:$
$
A=\matNpp, B=\matPpm, \wa=\matPpm, b_3 = \frac 1 2.
$
\parn
We apply Proposition \ref{simplifyAtilde} (1)\textcircled{c}, (2)\textcircled{b} to obtain
$\cals=n K+I$, where $K\in\gltz$ with $\det(K)=-1$ and  $\tr(K)=n>0$;
and $k_{11}=k_{22}$.\parn
This is a combination of \gr{3\ii}+\gr{1}, which includes \gr{6a\ii}.
By Corollary \ref{a-conj-zero3}, we can take $b_1=b_2=0$. Therefore,
\[
\alpha=(\bft_1^{a_1} \bft_2^{a_2},A),
\beta=(\bft_3^{\frac12},B).
\]
\noindent
The description for $H^1(\Phi; \coker(I-\cals))$ follows as in case \gr{7},
using $\ka = \matPpm$ rather than $\ka =A$.
There is always torsion in $Q$, as it contains a subgroup of type \gr{4}.


\noindent
This finishes the proof of Theorem \ref{ClassificationSol-geometry}.
\qedhere
\end{proof}




\step[{\bf Detecting Torsion in $\Sol$ Crystallographic Groups}]

A crystallographic group $Q$ of $\Sol$ is torsion-free precisely if it acts
freely on $\Sol$. Using this fact, we will give proofs of the torsion criteria in
Theorem \ref{ClassificationSol-geometry}.

\begin{proposition}[Compare {\cite[Theorem 5.3]{scott}}]
\label{topologySolManifolds}
Every
$\Sol$-geometry manifold $Q \backslash \Sol$ has the structure of
\begin{itemize}
\item The mapping torus of a hyperbolic linear self-diffeomorphism of $T^2$
(Q is of type \gr{0}, \gr{1}, or \gr{2b}).
\item The union of two twisted I-bundles over Klein bottles (Q is of type \gr{3\ii} or \gr{6b\ii}).
\end{itemize}
\bigskip

When $Q$ is of type \gr{3\ii}, $Q \backslash \Sol$ can be described as $T^2 \x I$ with
$T^2 \x \{0 \}$ identified to itself by the affine involution $\left( \xy{a_1} {a_2}, \matPpm \right)$, and $T^2 \x \{1 \}$ identified to itself by the affine involution
$\left( \xy{a_1} {a_2}, \left[ \begin{matrix} \sigma_{11} & -\sigma_{12}\\
\sigma_{21} & -\sigma_{11} \end{matrix} \right] \right)$.

If $-\ka = \left[\begin{matrix} -1 & 0 \\ 0 & 1\end{matrix} \right]$ is used instead of $\ka$,
then  $Q \backslash \Sol$ can be described as $T^2 \x I$ with
$T^2 \x \{0 \}$ identified to itself by the affine involution $\left( \xy{a_1} {a_2},
\left[\begin{matrix} -1 & 0 \\ 0 & 1\end{matrix} \right] \right)$, and
 $T^2 \x \{1 \}$ identified to itself by the affine involution
$\left( \xy{a_1} {a_2}, \left[ \begin{matrix} -\sigma_{11} & \sigma_{12}\\
-\sigma_{21} & \sigma_{11} \end{matrix} \right] \right)$.

When $Q$ is of type \gr{6b\ii}, $Q \backslash \Sol$ can be described as $T^2 \x I$ with
$T^2 \x \{0 \}$ identified to itself by the affine involution $\left( \xy{a_1} {a_2}, \matPpm \right)$, and $T^2 \x \{1 \}$ identified to itself by the affine involution
$\left( \xy{a_1} {a_2}, \left[ \begin{matrix} -k_{11} & k_{12}\\
-k_{21} & k_{11} \end{matrix} \right] \right)$.

\end{proposition}

\begin{proof}
If $Q = \Gamma_\cals$ is a $\Sol$ crystallographic group of type
\gr{0}, then $Q \cong \bbz^2 \rx_\cals \bbz$. The action of
$\Gamma_\cals$ on $\Sol$ is equivalent to an action of $\bbz^2 \rx_{\cals}
\bbz$ on $\bbr^2 \rx_\cals \bbr$.  A fundamental domain for this action is
given by the unit cube $I^3$ and evidently $Q \backslash \Sol$ is
given by $T^2 \x I$ with $T^2 \x \{0\}$ identified to $T^2 \x \{1\}$ via
$\cals$, which we view as a hyperbolic self-diffeomorphism of $T^2$.  Note
that $\bbr^2 \ra \bbr^2 \rx_\cals \bbr \ra \bbr$ induces the fiber bundle
with infinite cyclic structure group generated by $\cals$
\[
T^2 \ra {\Gamma_\cals} \backslash \Sol \ra S^1.
\]

If $Q$ is of type \gr{1}, then $Q \backslash \Sol$ is the quotient of
${\Gamma_\cals} \backslash \Sol$ by the involution defined by
$\alpha = \left(\bft_3^{\frac 1 2}, \matPpm\right)$. On the base of $T^2 \ra {\Gamma_\cals} \backslash \Sol \ra S^1$,
$\alpha$ acts as a translation. Thus a fundamental domain for the action of $\alpha$ is
given by $T^2 \x \left[0, \frac 1 2 \right]$. Note that $\alpha$ identifies
$T^2 \x \{0\}$ with $T^2 \x \{\frac 1 2\}$ via $\varphi(\alphabar) = -K$, which is
a square root of $\cals$. This $\varphi(\alphabar)$ is a hyperbolic matrix with
one eigenvalue positive and the other negative, and $Q \backslash \Sol$
is the mapping torus of $\varphi(\alphabar)$. The argument for type \gr{2b}
is identical, in this case we obtain the mapping torus of a hyperbolic
matrix with both eigenvalues negative.

Now suppose $Q$ is of type \gr{3\ii}. Then $Q$ is the quotient of ${\Gamma_\cals} \backslash \Sol$ by the involution defined by
$\alpha = \left(\bft_1^{a_1} \bft_2^{a_2}, \matNpp\right)$. Here $\alpha$ acts as a reflection
on the base $S^1$. As in the preceding case, a fundamental domain for this action
is given by $T^2 \x \left[0, \frac 1 2 \right]$. Now $\alpha$ identifies $T^2 \x \{0\}$ to
itself and $T^2 \x \{\frac 1 2\}$ to itself. We seek criteria for $\alpha$ to act freely on these boundary
tori so that they are identified to Klein bottles. Then
cutting $Q \backslash \Sol$ along $T^2 \x \{\frac 1 4\}$ yields two twisted $I$-bundles over Klein bottles.
Indeed, $(\bft_1^{a_1}\bft_2^{a_2},A).\bft_1^{x_1} \bft_2^{x_2} = \bft_1^{a_1}\bft_2^{a_2}A(\bft_1^{x_1} \bft_2^{x_2})$
shows that $\alpha$ acts on $T^2$ as the affine transformation $(\bfa,\varphi(\alphabar))$. For an arbitrary
element $\bft_1^{x_1} \bft_2^{x_2} \bft_3^{\frac 1 2} \in T^2 \x \{\frac 1 2\}$,
\begin{align*}
\bft_3 (\bft_1^{a_1}\bft_2^{a_2},A).\bft_1^{x_1} \bft_2^{x_2}\bft_3^{\frac 1 2}& =\bft_3 \bft_1^{a_1}\bft_2^{a_2} A(\bft_1^{x_1} \bft_2^{x_2}) A(\bft_3^{\frac 1 2})\\
&=\bft_3 \bft_1^{a_1}\bft_2^{a_2} A(\bft_1^{x_1} \bft_2^{x_2}) \bft_3^{-\frac 1 2}\\
&=\left( \bft_3 \bft_1^{a_1}\bft_2^{a_2} \bft_3^{-1} \right)
\left( \bft_3 A(\bft_1^{x_1} \bft_2^{x_2}) \bft_3^{-1} \right) \bft_3^{\frac 1 2} \in T^2 \x \{\frac 1 2\}.
\end{align*}

Since conjugation by $\bft_3$ is the action of $\cals$, we see that $\alpha$ acts on
$T^2$ as the affine transformation $(\cals \bfa, \cals \varphi(\alphabar))$. But
since $\bfa \in \coker(I-\cals)$, this simplifies to $(\bfa, \cals \varphi(\alphabar))$.
The condition that $\sigma_{11}=\sigma_{22}$ ensures that $\cals \varphi(\alphabar)$
has order 2. Note that cutting $Q \backslash \Sol$ along $T^2 \x \{\frac 1 4\}$ results in
two components, each a twisted I-bundle over a Klein bottle.

The argument in case \gr{6b\ii} is nearly identical. In this case, note that
$Q$ contains a group of type \gr{2b}, say $Q'$, as an index 2 subgroup.
Therefore, $Q \backslash \Sol$ is the quotient of $Q' \backslash \Sol$ by
$\alpha = \left(\bft_1^{a_1} \bft_2^{a_2}, \matNpp\right)$. But because $Q' \backslash \Sol$
admits the structure of a $T^2$ bundle over $S^1$, the construction for \gr{3\ii}
applies. A fundamental domain for the action of $\alpha$ on $Q' \backslash \Sol$
is given by $T^2 \x \{\frac 1 4\}$. As in case \gr{3\ii}, $\alpha$ acts on
$T^2 \x \{0\}$ affinely as $(\bfa,\varphi(\alphabar))$. For an arbitrary
element $\bft_1^{x_1} \bft_2^{x_2} \bft_3^{\frac 1 4} \in T^2 \x \{\frac 1 4\}$, we have
\begin{align*}
(\bft_3^{\frac12}, B) (\bft_1^{a_1}\bft_2^{a_2},A).&\bft_1^{x_1} \bft_2^{x_2}\bft_3^{\frac 1 4}\\ &=
\bft_3^{\frac12} B(\bft_1^{a_1}\bft_2^{a_2}) BA(\bft_1^{x_1} \bft_2^{x_2}) BA(\bft_3^{\frac 1 4})\\
&=\bft_3^{\frac12} B(\bft_1^{a_1}\bft_2^{a_2})
\bft_3^{-\frac12} \bft_3^{\frac12} BA(\bft_1^{x_1} \bft_2^{x_2}) \bft_3^{-\frac 1 4}\\
&=\left(\bft_3^{\frac12} B(\bft_1^{a_1}\bft_2^{a_2})
\bft_3^{-\frac12} \right) \left( \bft_3^{\frac12} BA(\bft_1^{x_1} \bft_2^{x_2})  \bft_3^{-\frac 1 2} \right) \bft_3^{\frac 1 4}
\in T^2 \x \{\frac 1 4\}.
\end{align*}

Now conjugation
by $(\bft_3^{\frac 12}, B)$ is the action of $\varphi(\betabar)=-K$ on
$T^2$. Hence $\alpha$ acts affinely on $T^2 \x \{\frac 1 4\}$ as
$(\varphi(\betabar)\bfa, \varphi(\betabar) \varphi(\alphabar))$.
But we had the condition $(I-\varphi(\betabar))\bfa = (I+K)\bfa \in \bbz^2$, so
this simplifies to $(\bfa, \varphi(\betabar) \varphi(\alphabar))$.
As in case \gr{3\ii}, we obtain a union of two twisted I-bundles over Klein bottles.
\qedhere
\end{proof}


\begin{corollary}[Torsion in $\Sol$ Crystallographic Groups]
\label{SolTor}

\noindent
Consider the crystallographic groups of $\Sol$ in {\rm Theorem \ref{ClassificationSol-geometry}}.

A group of type \gr{3\ii} is torsion-free if
and only if
\begin{equation}
\label{tor1}
a_1=\frac12\ \text{ and }
a_2\not\equiv \frac{(\sigma_{11}+1)(2 n+1)}{2\sigma_{12}} \text{ for any } n \in \bbz.
\end{equation}

A group of type \gr{6b\ii} is torsion-free if
and only if
\begin{equation}
\label{tor2}
a_1=\frac12 \ \text{ and }
a_2\not\equiv \frac{(k_{11}-1)(2 n+1)}{2k_{12}}
\text{ for any } n \in \bbz.
\end{equation}
\qedhere
\end{corollary}
\begin{proof}
By Proposition \ref{topologySolManifolds}, for type \gr{3\ii} groups, we need the affine involutions $\left( \xy{a_1} {a_2},
\left[\begin{matrix} 1 & 0 \\ 0 & -1\end{matrix} \right] \right)$ and $\left( \xy{a_1} {a_2}, \left[ \begin{matrix} \sigma_{11} & -\sigma_{12}\\
\sigma_{21} & -\sigma_{11} \end{matrix} \right] \right)$ to act freely on $T^2$, which occurs precisely when $a_1$ and $a_2$ satisfy (\ref{tor1}).

For type \gr{6\ii} groups, we need the affine involutions $\left( \xy{a_1} {a_2},
\left[\begin{matrix} 1 & 0 \\ 0 & -1\end{matrix} \right] \right)$ and $\left( \xy{a_1} {a_2}, \left[ \begin{matrix} -k_{11} & k_{12}\\
-k_{21} & k_{11} \end{matrix} \right] \right)$ to act freely on $T^2$, which occurs precisely when $a_1$ and $a_2$ satisfy (\ref{tor2})
\qedhere
\end{proof}


\section{Crystallographic Groups of $\Solof$}

Here we provide a classification of the crystallographic groups of $\Solof$.
%This is joint work with K.B. Lee \cite{solof-geo}. 
Given a crystallographic group
$\varPi \subset \Solof \rx D_4,$ $\wt \Gamma := \varPi \cap \Solof$ is a lattice of $\Solof$
with finite quotient group $\Phi \subseteq D_4$. We have the short exact sequence
\[1 \ra \wt \Gamma \ra \varPi \ra \Phi \ra 1.\]
The first step will be to classify the lattices $\wt \Gamma$ of $\Solof$.

\step[{\bf Lattices of $\Solof$}]
\label{latsolof}

Let $G$ denote either $\Solof = \Nil \rx \bbr$ or $\Solr$.
Given a lattice $\tgamsig$ of $G$, $\wt\Gamma \cap \calz(G) \cong \bbz$ is a lattice of $\calz(G) \cong \bbr$ and
the projection map \[G \ra G / \calz(G) \cong \Sol\] maps $\tgamsig$ to a lattice of $\Sol$, which is isomorphic to $\Gamma_{\cals}$, for some $\cals\in\sltz$ with $\rm{trace}(\cals)>2$. Thus, $\tgamsig$ is the central extension
\[ 1 \lra \bbz \lra \tgamsig \lra \Gamma_{\cals} \lra 1.\]

Such central extensions of $\bbz$ by $\gamsig$ are
classified by the second cohomology group $H^2(\gamsig;\bbz)$.

\begin{theorem}
\label{lattice-part-construction}
Let $\cals\in\sltz$ with $\rm{trace}(\cals)>2$.
There is a one-one correspondence between the equivalence classes of all central extensions
\[1\lra \bbz\lra \wt\Gamma\ \lra\gamsig\lra 1\]
and the set $\bbz \oplus \text{Coker}(\cals-I)$. 
\end{theorem}

\begin{proof}
Recall $\gamsig=\bbz^2 \rx_\cals \bbz$.
\begin{align*}
H^2(\bbz^2\rx_{\cals}\bbz;\bbz)
&=\text{Free} \left(H_2(\bbz^2\rx_{\cals}\bbz;\bbz)\right)\oplus
  \text{Torsion} \left(H_1(\bbz^2\rx_{\cals}\bbz;\bbz)\right)\\
&=\bbz \oplus (\bbz^2/(\cals-I)\bbz^2)\\
&=\bbz \oplus \text{Coker}(\cals-I).
\end{align*}
The first term $\bbz$ comes from the central direction.
The second term $\text{Coker}(\cals-I)$ is finite, because
$\det(\cals-I)=2-\tr(\cals)\not=0$. For $\{q,(m_1,m_2)\}\in \bbz \oplus \text{Coker}(\cals-I)$, we shall
denote the corresponding $\wt\Gamma$ by $\lgamsig$.
\qedhere
\end{proof}


Here we shall show that $\lgamsig$ with $q \neq 0$ embeds as a lattice in $\Solof$.
An $\cals\in\sltz$ with $\tr(\cals)>2$ produces $P$ and $\Delta$. Then
we had the map (\ref{def-phi}) in Subsection \ref{latsol}:
\begin{align}
\phi: \bbz^2\rx_\cals\bbz &\lra \kern24pt\Sol\\
\notag \left(\xy{x}{y},u\right)
&\longmapsto
\notag \left(P\xy{x}{y},u\ln(\lambda)\right).
\end{align}
We combine this with $\Sol\ra\Solof$
$$
\left(\left[\begin{matrix} x\\ y \end{matrix}\right],u\right)
\longmapsto
\left[
\begin{matrix} 1 &e^u x &*\\
               0 &e^u &y\\
               0 &0 &1 \end{matrix}
\right].
$$
The only ambiguities are the central slots. So let
\begin{align}
\label{embedSolofLattice}
\begin{split}
\bfe_1&=\left(\left[\begin{matrix} 1\\ 0 \end{matrix} \right],0\right)
\longmapsto
(P\bfe_1,0)\longmapsto
\bft_1=
\left[\begin{matrix}
1 & p_{11} &c_1\\
0 & 1 &p_{21}\\
0 & 0 &1 \end{matrix}
\right]
,\\
\bfe_2&=\left(\left[\begin{matrix} 0\\ 1 \end{matrix} \right],0\right)
\longmapsto
(P\bfe_2,0)\longmapsto
\bft_2=
\left[\begin{matrix}
1 & p_{12} &c_2\\
0 & 1 &p_{22}\\
0 & 0 &1 \end{matrix}
\right],\\
\bfe_3&
=\left(\left[\begin{matrix} 0\\ 0 \end{matrix} \right],1\right)
\longmapsto
(0,\ln(\lambda))\longmapsto
\bft_3=
\left[\begin{matrix}
1 & 0 &c_3\\
0 & {\lambda} &0\\
0 & 0 &1 \end{matrix}
\right],\\
&\kern6pt\phantom{=\left(\left[\begin{matrix} 0\\
1 \end{matrix} \right],0\right)
\longmapsto}
\bft_4=
\left[\begin{matrix}
1 & 0&1\\
0 & 1 &0\\
0 & 0 &1 \end{matrix}
\right].
\end{split}
\end{align}
where $c_i$'s are to be determined. Then $[\bft_1,\bft_2]=\bft_4$
(regardless the values of $c_i$'s).

\begin{lemma}
\label{determine-TT}
%Let $\cals\in\sltz$ with $\tr(\cals)>2$, eigenvalue $\lambda$;
%let $P$ be a matrix diagonalizing $\cals$.
For any integers $q, m_1,m_2$,
there exist unique $c_1, c_2$ for which $\{\bft_1,\bft_2,\bft_3,\bft_4\}$
forms a group $\lgamsig$ with the presentation
\begin{align*}
\lgamsig=\langle
\bft_1,\bft_2,\bft_3,\bft_4^{\frac 1q}\ |\
[\bft_1,\bft_2]&=\bft_4,\ \text{$\bft_4$ is central,}\\
\bft_3 \bft_1 \bft_3\inv
&= \bft_1^{\sigma_{11}} \bft_2^{\sigma_{21}} \bft_4^{\frac{m_1}{q}},\\
\bft_3 \bft_2 \bft_3\inv&= \bft_1^{\sigma_{12}} \bft_2^{\sigma_{22}}
\bft_4^{\frac{m_2}{q}}\rangle.
\end{align*}

Consequently, $\lgamsig$ is solvable and contains $\Gamma_q =\angles{\bft_1, \bft_2, \bft^{\frac 1q}_4}$ as its discrete nil-radical, where $\Gamma_q$ is a lattice of $\nil$.
\end{lemma}

\begin{proof}
We only need to verify the last two equalities. But they
become a system of  equations on $c_i$'s
\begin{align}
\label{c1c2-equation}
\begin{split}
(1-\sigma_{11})c_1 -\sigma_{21} c_2  &={\frac{m_1}{q}}
-\frac{\sigma_{21} (\sigma_{12}+1 - \sigma_{11} + \sigma_{11} \sqrt{T^2-4})}{2 \sqrt{T^2-4}},\\
-\sigma_{12} c_1+ (1-\sigma_{22})c_2 &={\frac{m_2}{q}}
+\frac{\sigma_{12} (\sigma_{21} +1 - \sigma_{22} - \sigma_{22} \sqrt{T^2-4})}{2 \sqrt{T^2-4}},
\end{split}
\end{align}
where $T=\sigma_{11}+\sigma_{22}$.
Since $I-\cals$ is non-singular, there exists a unique solution for
$c_1, c_2$.
\end{proof}

\begin{remark}
(1) There are 1-parameter family of $P$'s diagonalizing $\cals$ with
  $\det(P)=1$.
In the above lemma, the $c_i$'s are independent of choice of $P$ because
equation (\ref{c1c2-equation}) has coefficients only from the matrix $\cals$.

(2) Note that a lattice $\lgamsig$ of $\Solof$ projects to the lattice
$\gamsig$ of $\Sol$.
%In place of $\bft_4$, we can take $\bft'_4 = {\bft_4}^{\frac1q}$ so that
%$[\bft_1,\bft_2]={\bft'_4}^{q}$ for any integer $q$. Then $\lgamsig=\langle
%\bft_1,\bft_2,\bft_3,{\bft_4}^{\frac1q}\rangle$ is a lattice of $\Solof$.

(3) The equation (\ref{c1c2-equation}) also shows the cohomology
classification.
Suppose $\{c_1,c_2\}$ and $\{c'_1,c'_2\}$ are solutions for the equations
with $\{m_1,m_2\}$ and $\{m'_1,m'_2\}$, respectively.
Then $(c'_1-c_1,c'_2-c_2)\in\big(\frac1q\bbz\big)^2$ if and only if
$(m'_1-m_1,m'_2-m_2)\in \text{Coker}(\cals^T-I) \cong \coker(\cals-I)$.
This happens if and only if
$\lgamsig=\wt\Gamma_{(\cals;q,m'_1,m'_2)}$.

(4) Notice that $c_3$ does not show up in the presentation of the lattice
$\lgamsig$. This implies that $c_3$ can be changed without affecting the
isomorphism type of the lattice.
\end{remark}

\begin{notation}[Standard lattice]
\label{standard-lattice}
The lattice generated by
\[
\bft_1=
\left[\begin{matrix}
1 & p_{11} &c_1\\
0 & 1 &p_{21}\\
0 & 0 &1 \end{matrix}
\right]
,\\
\bft_2=
\left[\begin{matrix}
1 & p_{12} &c_2\\
0 & 1 &p_{22}\\
0 & 0 &1 \end{matrix}
\right],\\
\bft_3=
\left[\begin{matrix}
1 & 0 &c_3\\
0 & {\lambda} &0\\
0 & 0 &1 \end{matrix}
\right],\\
\bft_4^{\frac 1q}=
\left[\begin{matrix}
1 & 0&\frac 1q\\
0 & 1 &0\\
0 & 0 &1 \end{matrix}
\right]
\]
{\bf with $c_3=0$}, is called a \emph{standard lattice} of $\Solof$.
\end{notation}
Therefore, any lattice of $\Solof$ is isomorphic to a standard lattice.
However, a non-standard lattice (i.e., $c_3\not=0$) will be needed when we consider finite extensions of $\wt \Gamma_\cals$ (in the holonomy $\bbz_4$ case).

%\begin{remark}
%For a lattice $\lgamsig$ of $\Solof$, embedded as in assignment (\ref{embedSolofLattice}), the real powers of $\bft_3$ and $\bft_4$ are well defined elements of $\Solof$, that is, $\bft_3^{s_3}$ and $\bft_4^{s_4}$
%for $s_3, s_4 \in \bbr$ are well defined.

%Rational powers of $\bft_i$ for $i=1,2$ are also well defined elements of $\Solof$. To see this, note that it suffices to define $\bft_i^\frac{1}{n}$ for $i=1,2$ and $n=1,2,3, \dots$. 

%\noindent
%We have
%\[\bft_i^{\frac{1}{n}}=
%\left[\begin{matrix}
%1 & \frac{p_{1i}}{n} &c\\
%0 & 1 & \frac{p_{2i}}{n}\\
%0 & 0 &1 \end{matrix}
%\right],\]
%where $c$ is a real number to be determined. Raising $\bft_i^\frac{1}{n}$ to the $n$-th power,
%\begin{align*}
%(\bft_i^{\frac{1}{n}})^n&=
%\left[\begin{matrix}
%1 & p_{1i} & n c + v\\
%0 & 1 & p_{2i}\\
%0 & 0 &1 \end{matrix}
%\right] \\&= \bft_i
%= \left[\begin{matrix}
%1 & p_{1i} & c_i\\
%0 & 1 & p_{2i}\\
%0 & 0 &1 \end{matrix}
%\right] ,
%\end{align*}
%where $v$ is an expression independent of $c$. Now we can solve $n c + v=c_i$, and obtain a unique value for $c$.
%\[(\bft_i^{\frac{1}{n}})^n=
%\left[\begin{matrix}
%1 & p_{1i} & n c + v\\
%0 & 1 & p_{2i}\\
%0 & 0 &1 \end{matrix}
%\right] = \bft_i = \left[\begin{matrix}
%1 & p_{1i} & c_i\\
%0 & 1 & p_{2i}\\
%0 & 0 &1 \end{matrix}
%\right] ,\]

%\end{remark}

\noindent
The following  lemmas on lattices of $\Solof$ will be needed in the next section.

\begin{lemma}
\label{rational1}
Let $\lgamsig$ be a lattice of $\Solof$, embedded as in assignment (\ref{embedSolofLattice}). Let $r_1, r_2 \in \bbq$. Then
\[\bft_1^{r_1} \bft_2^{r_2} = \bft_2^{r_2} \bft_1^{r_1} \bft_4^{r_1 r_2}.\]
\end{lemma}
\begin{proof}
We compute that
\[ \left[ \bft_1^{r_1}, \bft_2^{r_2} \right] = \bft_4^{r_1 r_2 \det(P)} = \bft_4^{r_1 r_2}.  \qedhere\]
\end{proof}

\begin{lemma}
\label{rational2}
Let $\lgamsig$ be a lattice of $\Solof$, embedded as in assignment (\ref{embedSolofLattice}). Let $a_1, a_2 \in \bbq$. Then, for $\abar=\pm 1$,
\begin{align*}
\bft_3^{\abar} \bft_1^{a_1} \bft_2^{a_2} \bft_3^{-\abar} = \bft_1^{l_1} \bft_2^{l_2} \bft_4^{v}, 
\text{ where } 
{\xy{l_1}{l_2}=
\cals^{\abar}\xy{a_1}{a_2}}, \text{ and } v \in \bbq.
\end{align*} 
\end{lemma}
\begin{proof}
The definition of $\lgamsig$ shows that ${\xy{l_1}{l_2}=
\cals^{\abar}\xy{a_1}{a_2}}$. It remains to show that $v$ rational.
Because $a_1$ and $a_2$ are rational, there is a positive integer $n$
so that $n a_1, n a_2 \in \bbz$. By Lemma \ref{rational1},
\begin{align*}
(\bft_1^{a_1} \bft_2^{a_2})^n &= \bft_1^{a_1} \bft_2^{a_2} \cdots \bft_1^{a_1} \bft_2^{a_2} (n \text{ times})\\
&= \bft_1^{n a_1} \bft_2^{n a_2} \bft_4^{u'}, \text{ for } u' \in \bbq.
\end{align*}
Therefore, 
\begin{align*}
\bft_3^{\abar} (\bft_1^{a_1} \bft_2^{a_2})^n \bft_3^{-\abar}  &= \bft_3^{\abar} \bft_1^{n a_1} \bft_2^{n a_2} \bft_4^{u'}  \bft_3^{-\abar}\\ & =\bft_3^{\abar} \bft_1^{n a_1} \bft_2^{n a_2}   \bft_3^{-\abar} \bft_4^{u'}\\
&= \bft_1^{n_1} \bft_2^{n_2} \bft_4^{u} \text{ for some } n_1,n_2 \in \bbz, \text{ and } u \in \bbq,
\end{align*}
where the last equality follows from that $n a_1$  and $n a_2$ are integers, together with the relations in Lemma \ref{determine-TT}.

On the other hand, we have that
\begin{align*}
\bft_3^{\abar} (\bft_1^{a_1} \bft_2^{a_2})^n \bft_3^{-\abar}  &= 
\bft_3^{\abar} \bft_1^{a_1} \bft_2^{a_2}\bft_3^{-\abar} \cdots \bft_3^{\abar} \bft_1^{a_1} \bft_2^{a_2}\bft_3^{-\abar} (n \text{ times})\\
&= \bft_1^{l_1} \bft_2^{l_2} \bft_4^{v} \cdots \bft_1^{l_1} \bft_2^{l_2} \bft_4^{v}  (n \text{ times}) \\
&=  \bft_1^{n l_1} \bft_2^{n l_2}  \bft_4^{nv + w} \text{ for some } w \in \bbq,
\end{align*}
where the last equality follows from Lemma \ref{rational1}.

\noindent
Consequently, we have 
\[ \bft_1^{n_1} \bft_2^{n_2} \bft_4^{u} = \bft_1^{n l_1} \bft_2^{n l_2}  \bft_4^{nv + w}.\]
This forces $n_1= n l_1$ and $n_2= {n l_2}$. Therefore, 
\[ nv + w = u.\]
Since $n \in \bbz$, $u, w \in \bbq$, it follows that $v \in \bbq$.
\qedhere
\end{proof}


\begin{example}
For
$\cals=
\left[
\begin{matrix} 2 &1\\
               5 &3 \end{matrix}
\right]$.
We have $\lambda=\frac12 \left(5+\sqrt{21}\right)$.
Take
\[
P=
\left[
\begin{array}{cc}
 -\frac{5 d}{\sqrt{21}} & \left(\frac{1}{2}-\frac{1}{2 \sqrt{21}}\right) d \\
 -\frac{1}{d} & \frac{-1-\sqrt{21}}{10 d}
\end{array}
\right],
\]
($d\not=0$ arbitrary) and $q=1$ to get
\begin{align*}
c_1&=\tfrac{1}{42}\left(5(7+\sqrt{21})+ 28 m_1-70 m_2\right)\\
c_2&=\tfrac{1}{42}\left(-(49+\sqrt{21})-14 m_1+14 m_2\right).
\end{align*}
Therefore, $\wt\Gamma_{(\cals;1,m_1,m_2)}$ is represented as follows:
\[
\begin{array}{ll}
\bft_1=\left[
\begin{array}{ccc}
 1 & -\frac{5 d}{\sqrt{21}} & \frac{1}{42} \left(28 {m_1}-70 {m_2}+5 \sqrt{21}+35\right) \\
 0 & 1 & -\frac{1}{d} \\
 0 & 0 & 1 \\
\end{array}
\right], &\\
\bft_2=\left[
\begin{array}{ccc}
 1 & -\frac{1}{42} \left(-21+\sqrt{21}\right) d & \
\frac{1}{42} \left(-14 {m_1}+14 {m_2}-\sqrt{21}-49\right) \\
 0 & 1 & \frac{-1-\sqrt{21}}{10 d} \\
 0 & 0 & 1 \\
\end{array}
\right],&\\
\bft_3=\left[
\begin{array}{ccc}
 1 & 0 & c_3 \\
 0 & \frac{1}{2} \left(5+\sqrt{21}\right) & 0 \\
 0 & 0 & 1 \\
\end{array}
\right],\qquad
\bft_4=\left[
\begin{array}{ccc}
 1 & 0 & 1 \\
 0 & 1 & 0 \\
 0 & 0 & 1 \\
\end{array}
\right].
\end{array}
\]
These have relations
\[
\bft_3 \bft_1 \bft_3\inv= \bft_1^{2}\bft_2^{5} \bft_4^{m_1},\
\bft_3 \bft_2 \bft_3\inv= \bft_1^{} \bft_2^{3} \bft_4^{m_2}.
\]
Notice that
\[
\cals-I=
\left[\begin{matrix}
1&1\\ 5&2
\end{matrix}\right],
\]
and $\text{Coker}(\cals-I)=\bbz_3$.
Therefore, depending on $(m_1,m_2)$, we have 3 non-equivalent
$\wt\Gamma_{(\cals;1,m_1,m_2)}$'s.
Of course, one can throw in a finer generator
$\bft_4^{\frac{1}{q}}$, in place of $\bft_4^{}$.
\end{example}
\bigskip

\step[{\bf Classification of $\Solof$ Crystallographic Groups}]

Let $\varPi \subset \Solof \rx K$ be a crystallographic group of $\Solof$, where $K$ is a maximal compact
subgroup of $\aut(\Solof)$. As all maximal compact subgroups of $\Solof$ are conjugate, 
we can assume that $K$ is the maximal compact subgroup \[D_4 = \left\langle \left[\begin{matrix} 0 & -1 \\ 1 & 0 
\end{matrix} \right], \left[\begin{matrix} 1 & 0 \\ 0 & -1 
\end{matrix} \right] \right\rangle\] of $\aut(\Solof)$, whose action on $\Solof$ is described in Proposition \ref{autosolof}.
Furthermore, we can conjugate $\varPi$ in $\aff(\Solof)$ so that the lattice of $\varPi$ is some
$\lgamsig$, where $\lgamsig$ is embedded in $\Solof$ as in assignment (\ref{embedSolofLattice}) in Subsection \ref{latsolof}.

\begin{proposition}
\label{solofShape}
\noindent
{\rm{(1)}} Any crystallographic group $\varPi$ of $\Solof$ can be conjugated into
$\Solof\rx D_4$ in such a way that
\[
\varPi \cap \Solof=\angles{\bft_1,\bft_2,\bft_3,\bft_4^{\frac1q}},
\]
where
\[
\bft_1=
\left[\begin{matrix}
1 & p_{11} &c_1\\
0 & 1 &p_{21}\\
0 & 0 &1 \end{matrix}
\right],\
\bft_2=
\left[\begin{matrix}
1 & p_{12} &c_2\\
0 & 1 &p_{22}\\
0 & 0 &1 \end{matrix}
\right],\
\bft_3=
\left[\begin{matrix}
1 & 0 &c_3\\
0 & \lambda &0\\
0 & 0 &1 \end{matrix}
\right],\
\bft_4=
\left[\begin{matrix}
1 & 0 &1\\
0 & 1 &0\\
0 & 0 &1 \end{matrix}
\right].
\]

\noindent
{\rm{(2)}} The holonomy group $\Phi$ is generated by at most two elements of $D_4$, and thus $\varPi$ is generated by $\angles{\bft_1, \bft_2, \bft_3, \bft_4^{\frac1q}}$ and at most two isometries of the form $(\bft_1^{a_1} \bft_2^{a_2} \bft_3^{a_3} \bft_4^{a_4} , A)$, for $A \in D_4$ and real numbers $a_i$.
\end{proposition}


\begin{proof}
Let $\wt\Gamma=\varPi\cap\Solof$. This lattice must meet the center of $\Solof$ in a lattice:
$\wt\Gamma\cap \calz(\Solof)$ is a lattice of
$\calz(\Solof)$, say generated by $\bft_4^{\frac1q}$. Also
$\wt\Gamma\cap\Nil$ is a lattice of the nilradical $\Nil$, so we can find
generators $\angles{\bft_1,\bft_2,\bft_4^{\frac1q}}$ of this lattice as
given in the statement.
The remaining one generator for the lattice $\wt\Gamma$ must project
down to a generator of the quotient $\bbz$. It must be of the form
\[
\bft''_3=
\left[\begin{matrix}
1 & a &c_3\\
0 & {\lambda} &b\\
0 & 0 &1 \end{matrix}
\right]
\]
Conjugation by
$
\left[\begin{matrix}
1 & \frac{a}{1-\lambda} & 0\\
0 & {\lambda} &-\frac{b \lambda}{1-\lambda}\\
0 & 0 &1 \end{matrix}
\right]
$
maps $\bft''_3$ to the form of $\bft_3$. Recall that $\wt\Gamma/\calz(\wt\Gamma)$ gives rise to $\cals$, and we can find $P = (p_{i j})$ diagonalizing $\cals$.
Assuming $\det(P)=1$ (this can always be achieved, as in the case of $\Sol$), we have
$[\bft_1,\bft_2]=\bft_4$.
Therefore, any lattice is conjugate to a lattice
$\angles{\bft_1,\bft_2,\bft_3,\bft_4^{\frac1q}}$.
\end{proof}

However, we will see that we can always take $c_3=0$, except possibly in the case when
the holonomy $\Phi$ is $\bbz_4$.
Because lattices of $\Solof$ project to lattices of $\Sol$,
the projections $\Solof \ra \Sol$ and $\aut(\Solof) \ra \aut(\Sol)$ induce
a projection $\Solof \rx D_4 \ra \Sol \rx D_4$ which carries a $\Solof$ crystallographic group 
to a $\Sol$ crystallographic group. Furthermore, when $\varPi$ is embedded in $\Solof \rx D_4$
as in Proposition \ref{solofShape}, the lattice $\wt \Gamma_\cals = \lgamsig$ projects to a standard
lattice $\gamsig$ of $\Sol$. That is, we have the following commuting diagram:


\[
\CD
@. 1 @. 1 @.  @.\\
@. @VVV @VVV @. @.\\
@. \frac1q\bbz @=  \frac1q\bbz @. @.\\
@. @VVV @VVV @. @.\\
1 @>>> \tgamsig @>>> \varPi @>>> \Phi @>>> 1\\
@. @VVV @VVV @| @.\\
1 @>>> \gamsig @>>> Q @>>> \Phi @>>> 1\\
@. @VVV @VVV @. @.\\
@. 1 @. 1 @.  @.\\
\endCD
\]

Here is an outline of our procedure for obtaining the $\Solof$ crystallographic groups.
\bigskip


\noindent
{\rm (1)} For each $\Sol$ crystallographic group $Q$ from Theorem \ref{ClassificationSol-geometry},
classify lifts of $\gamsig $ to lattices of $\wt \Gamma_\cals$ of $\Solof$ (Theorem \ref{lattice-part-construction}).\\

\noindent
{\rm (2)} Show that the abstract kernel $\Phi \ra \out(\gamsig)$ will induce an abstract kernel $\Phi \ra \out(\wt \Gamma_\cals)$, after possibly making the central generator of $\wt \Gamma_\cals$ finer (i.e., taking a larger value of $q$).\\

\noindent
{\rm (3)} Once an abstract kernel $\Phi \ra \out(\wt \Gamma_\cals)$ is fixed, show that
after possibly taking a larger value of $q$, the obstruction in $H^3(\Phi; \bbz)$ to the existence of $\varPi$ vanishes. Then
 the equivalence classes of extensions
\[ 1 \ra \wt \Gamma_\cals \ra \varPi \ra \Phi \ra 1 \] are classified by $H^2(\Phi; \calz(\wt \Gamma_\cals)) = H^2(\Phi; \bbz)$. The details follow.\\



Our goal is finding all crystallographic groups $\varPi$ of $\Solof$ which
project down to $Q$. In general, it is \emph{not} true that there exists
$\varPi$ fitting the commutative diagram of exact sequences
\[
\CD
@. \bbz @= \bbz @.\\
@. @VVV @VVV @.\\
1 @>>> \tgamsig @>>> \varPi @>>> \Phi @>>> 1\\
@. @VVV @VVV @| @.\\
1 @>>> \Gamma_\cals @>>> Q @>>> \Phi @>>> 1\\
\endCD
\]
Even though $\wt\Gamma_{\cals}$ always exists, for $\varPi$ to exist, sometimes the
kernel $\bbz$ needs to be ``inflated''. It turns out that, after appropriate
inflation, an extension $\varPi$ always exists. We will need the following lemma.

%Suppose in Proposition \ref{solofShape}, we have fixed the $c_i$, as well
%as set $q=1$, thus
%fixing the lattice $\wt \Gamma_{(\cals;1,n_1,n_2)} = \angles{\bft_1, \bft_2, \bft_3, \bft_4}$.
%For a generator $A \in \Phi$, let
%\[\alpha = (\bft_1^{a_1} \bft_2^{a_2} \bft_3^{a_3} \bft_4^{{a_4}}, A) = %(a,A).\]
%The $a_1, a_2$, and $a_3$ are determined by the abstract kernel
%$ %\Phi \ra \out(\gamsig)$ (Remark \ref{abstractKernelSol}). We consider the 
%effect that conjugation by $\alpha$ has on $\wt \Gamma_{(\cals;1,n_1,n_2)}$.
%Note that conjugation by $\alpha$ is independent of $a_4$. 

%\begin{lemma}
%\label{discrete}
%Let $S$ be a subset of a Lie group $G$ inheriting the discrete topology. Then the image of $S$ under either left or right translation is discrete. That is, for $g \in G$,
%$\ell_{g}(S)=gS$ and $r_{g}(S)=Sg$ are discrete.
%\end{lemma}
%\begin{proof}
%Note that discreteness is a topological property.
%Any Lie group admits a left invariant metric, so that $\ell_g$ is an %isometry, and thus $\ell_g$ maps discrete subspaces to discrete subspaces.
%With a right invariant metric,  $r_g$ is an isometry, and thus maps %discrete subspaces to discrete subspaces.
%\qedhere
%\end{proof}

\begin{proposition}
\label{relations}
Let $Q$ be a $\Sol$ crystallographic group with standard lattice $\gamsig$.
Suppose in {\rm Proposition \ref{solofShape}}, we have fixed the $c_i$, as well
as set $q=1$, thus
fixing the lattice $\wt \Gamma_{(\cals;1,n_1,n_2)} = \angles{\bft_1, \bft_2, \bft_3, \bft_4}$.
For any generator $A \in \Phi$, let
\[\alpha = (\bft_1^{a_1} \bft_2^{a_2} \bft_3^{a_3} \bft_4^{{a_4}}, A) = (a,A).\]
The powers $a_1, a_2$, and $a_3$ are determined by the abstract kernel
$\Phi \ra \out(\gamsig)$ {\rm(Remark \ref{abstractKernelSol})}. Consider the 
effect that conjugation by $\alpha$ has on $\wt \Gamma_{(\cals;1,n_1,n_2)}$. Note that conjugation by $\alpha$ is independent of $a_4$.
We have the relations:
\begin{align*}
&\alpha \bft_1 \alpha\inv=\bft_1^{m_1}\bft_2^{m_2}  \bft_4^{{v_1}},
\text{ where } {\xy{m_1}{m_2}=\varphi(\alphabar)\xy{1}{0}},\\
&\alpha \bft_2 \alpha\inv=\bft_1^{n_1}\bft_2^{n_2} \bft_4^{{v_2}},
\text{ where } {\xy{n_1}{n_2}=\varphi(\alphabar)\xy{0}{1}},\\
&\alpha\bft_3\alpha\inv=\bft_1^{w_1}\bft_2^{w_2}\bft_3^{\abar} \bft_4^{{v_3}},
\text{ where } {\xy{w_1}{w_2}=
\left(I-\cals^{\abar}\right)\xy{a_1}{a_2}},\\
&\alpha\bft_4 \alpha\inv={\bft_4^{\ahat}},
\end{align*}
where $v_1$ and $v_2$ are rational numbers. We can adjust $c_3$ so that $v_3$ is rational, as well.
\end{proposition}
\begin{proof}

First we claim that the image of $\wt \Gamma_{(\cals;1,n_1,n_2)}$ under conjugation by $\alpha$,
\[\mu(\alpha)(\wt \Gamma_{(\cals;1,n_1,n_2)} ) = \alpha \wt \Gamma_{(\cals;1,n_1,n_2)} \alpha \inv ,\]
 is a lattice
of $\Solof$ lifting the standard lattice $\gamsig$ of $\Sol$.

As conjugation by $\alpha$ defines an automorphism of $\Solof$,
$\mu(\alpha)(\wt \Gamma_{(\cals;1,n_1,n_2)} )$ is again a subgroup of $\Solof$. Since conjugation by $\alpha$ is an affine diffeomorphism of $\Solof$, $\mu(\alpha)(\wt \Gamma_{(\cals;1,n_1,n_2)} )$ is also discrete. Because $\mu(\alpha)(\wt \Gamma_{(\cals;1,n_1,n_2)} )$ lifts the 
standard lattice $\gamsig$ of $\Sol$, and maps the generator $\bft_4$ of 
\[\calz(\Solof) \cap \wt \Gamma_{(\cals;1,n_1,n_2)}\] to ${\bft_4^{\ahat}}$, it follows that
$\mu(\alpha)(\wt \Gamma_{(\cals;1,n_1,n_2)} )$ is cocompact.
Therefore $\mu(\alpha)(\wt \Gamma_{(\cals;1,n_1,n_2)} )$ is a lattice of $\Solof$ lifting $\gamsig$.


All such lifts are given in Lemma \ref{determine-TT}. From equation (\ref{c1c2-equation}) in Subsection \ref{latsolof}, we see that for any two solutions $c_1, c_2$ and $c'_1, c'_2$, both $c'_1 - c_1$ and $c'_2 - c_2$ must be rational. Thus $v_1$ and $v_2$ are rational numbers.

For 
\begin{align}
\label{teh}
\alpha\bft_3\alpha\inv=\bft_1^{w_1}\bft_2^{w_2}\bft_3^{\abar} \bft_4^{{v_3}},
\text{ where } {\xy{w_1}{w_2}=
\left(I-\cals^{\abar}\right)\xy{a_1}{a_2}},
\end{align}
 we need only consider two cases by Theorem \ref{ClassificationSol-geometry}. First, consider the case when $a_3=\frac{1}{2}$. Then $A$ must be diagonal, so that $\abar=+1$:
\begin{align*}
\alpha &= (\bft_3^{\frac{1}{2}} \bft_4^{a_4}, A)\\
\abar&=+1.
\end{align*}
In the computation below, let $\hat \bft_3$ denote $\bft_3$ with the $(1,3)$-slot set to be zero, so that $\bft_3 = \hat \bft_3 \bft_4^{c_3}$.
\begin{align*}
\alpha\bft_3\alpha\inv &= \bft_3^{\frac{1}{2}} A(\bft_3 ) \bft_3^{-\frac{1}{2}} \\
&=\bft_3^{\frac{1}{2}} A( \hat \bft_3 \bft_4^{c_3} ) \bft_3^{-\frac{1}{2}}\\
&=\bft_3^{\frac{1}{2}} {\hat \bft_3} \bft_4^{\ahat c_3}  \bft_3^{-\frac{1}{2}}\\
&=\bft_3^{\frac{1}{2}} {\hat \bft_3} \bft_4^{ c_3} \bft_4^{- c_3}  \bft_4^{\ahat c_3}  \bft_3^{-\frac{1}{2}}\\
&= \bft_3^{\frac{1}{2}}  \bft_3  \bft_3^{-\frac{1}{2}} \bft_4^{(\ahat-1) c_3}= \bft_3 \bft_4^{(\ahat-1)c_3}.
\end{align*}
Since $\ahat=\pm 1$, there is always a choice of $c_3$ which makes $(\ahat-1)c_3 \in \bbq$.

\noindent
Now consider the case when $a_3=0$,
$
\alpha = (\bft_1^{a_1} \bft_2^{a_2}  \bft_4^{a_4}, A)
$.
We compute:
\begin{align*}
\alpha\bft_3\alpha\inv &=\bft_1^{a_1} \bft_2^{a_2} A(\bft_3 ) \bft_2^{-a_2} \bft_1^{-a_1} \\
&= \bft_1^{a_1} \bft_2^{a_2} A( \hat \bft_3 \bft_4^{c_3} ) \bft_2^{-a_2} \bft_1^{-a_1}\\
&=\bft_1^{a_1} \bft_2^{a_2} {\hat \bft_3}^{\abar} \bft_4^{\ahat c_3}  \bft_2^{-a_2} \bft_1^{-a_1}\\
&= \bft_1^{a_1} \bft_2^{a_2} {\hat \bft_3}^{\abar} \bft_4^{\abar c_3} \bft_4^{-\abar c_3} \bft_4^{\ahat c_3}  \bft_2^{-a_2} \bft_1^{-a_1}\\
&= \bft_1^{a_1} \bft_2^{a_2} \bft_3^{\abar} \bft_4^{(\ahat-\abar) c_3}  \bft_2^{-a_2} \bft_1^{-a_1}\\
&= \left(\bft_1^{a_1} \bft_2^{a_2} \bft_3^{\abar} \bft_2^{-a_2} \bft_1^{-a_1} \bft_3^{-\abar} \right) {\bft_3}^{\abar} \bft_4^{(\ahat-\abar) c_3}.
\end{align*}
Now by Lemmas \ref{rational1} and \ref{rational2}, and using that $a_1$, $a_2$ are rational, we have 
\begin{align*}
\left( \bft_1^{a_1} \bft_2^{a_2} \bft_3^{\abar} \bft_2^{-a_2} \bft_1^{-a_1} \bft_3^{-\abar} \right) \bft_3^{\abar} \bft_4^{(\ahat-\abar) c_3} &= 
 \left( \bft_1^{b_1}  \bft_2^{b_2} \bft_4^{u}  \right) \bft_3^{\abar}  \bft_4^{(\ahat-\abar) c_3} \\
&= \bft_1^{b_1}  \bft_2^{b_2} \bft_3^{\abar}  \bft_4^{u+(\ahat-\abar) c_3},
\end{align*}
for a rational number $u$. Equating this with equation (\ref{teh}), we obtain
\begin{align*}
\bft_1^{w_1}\bft_2^{w_2}\bft_3^{\abar} \bft_4^{{v_3}} =  \bft_1^{b_1}  \bft_2^{b_2} \bft_3^{\abar}  \bft_4^{u+(\ahat-\abar) c_3}.
\end{align*}
Now $w_1=b_1$ and $w_2=b_2$ is forced. Therefore,
\[v_3 = u+(\ahat-\abar) c_3.\]
Because $\ahat=\pm 1$, $\abar= \pm 1$, and $u$ is rational,
$c_3$ can always be chosen so that $v_3$ is rational.\qedhere
\end{proof}

\begin{proposition}
Let $Q$ be a crystallographic group of $\Sol$, from {\rm Theorem \ref{ClassificationSol-geometry}}.  Then there exists $c_i$ and $q$ so that
the abstract kernel $\Phi \ra \out(\gamsig)$
induces $\Phi \ra \out(\lgamsig)$, where $\lgamsig = \angles{\bft_1, \bft_2, \bft_3, \bft_4^{\frac{1}{q}}}$ is a lattice of $\Solof$ lifting $\gamsig$.
\end{proposition}
\begin{proof}
The abstract kernel of $\Phi \ra \out(\gamsig)$ is given by, for $A \in \Phi$, 
\[\mu(\alpha): \gamsig \ra \gamsig, \text{ where } \alpha = (\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3},A) \in Q.\]
For any integer $q>0$, we can form the lattice of $\Solof$
\[\angles{\wt \Gamma_{(\cals;1,n_1,n_2)}, \bft_4^{\frac {1}{q}}} = \wt \Gamma_{(\cals;q,q n_1,q n_2)}.\] That is, we add a finer generator of the central direction to  $\wt \Gamma_{(\cals;1,n_1,n_2)}$. 

Now, for each generator of $A \in \Phi$, the $v_i$ in Proposition \ref{relations} are rational. Therefore, for $q$ large enough, $\wt \Gamma_{(\cals;q,q n_1,q n_2)}$ is invariant under conjugation by $(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3} \bft_4^{a_4},A)$, for each $A \in \Phi$. As this conjugation is independent of lift of $(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3},A)$ to  $(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3} \bft_4^{a_4},A)$, with $m_1 = q n_1$ and $m_2 = q n_2$,
we obtain an abstract kernel $\Phi \ra \out(\lgamsig)$.
\qedhere
\end{proof}


\begin{proposition}
\label{inflation}
Let $Q\hookrightarrow \Sol\rx D_4$ be a crystallographic group for $\Sol$
containing $\gamsig$. Assume that the abstract kernel $\Phi \ra \out(\gamsig)$
induces $\Phi \ra \out(\lgamsig)$. Then for some $p>0$, there exists $\varPi$ which fits the following commuting diagram
\[
\CD
@. 1 @. 1 @.  @.\\
@. @VVV @VVV @. @.\\
@. \frac1{pq}\bbz @=  \frac1{pq}\bbz @. @.\\
@. @VVV @VVV @. @.\\
1 @>>> \wt \Gamma_{(\cals;pq,p m_1,p m_2)} @>>> \varPi @>>> \Phi @>>> 1\\
@. @VVV @VVV @| @.\\
1 @>>> \gamsig @>>> Q @>>> \Phi @>>> 1\\
@. @VVV @VVV @. @.\\
@. 1 @. 1 @.  @.\\
\endCD
\]
\end{proposition}

\begin{proof}
Since
the center of $\wt \Gamma_{(\cals;q, m_1, m_2)}$ is $\frac1{q} \bbz$ and $\Phi$ is finite, $H^3(\Phi;\tfrac1{q} \bbz)$
is finite. This means the obstruction class to the existence of the
extension vanishes if we use $\frac1{pq} \bbz$ for the coefficients, for some
$p>0$. That is, it vanishes inside $H^3(\Phi;\tfrac1{pq} \bbz)$. Thus, with such
$pq$, the center of $\wt \Gamma_{(\cals;pq,p m_1,p m_2)}$ is $\frac1{pq} \bbz$, and an extension $\varPi$
exists.
\qedhere
\end{proof}

So we can assume that after appropriate inflation, there exists
an extension $\varPi$ with lattice $\lgamsig$, for some $q>0$. The Seifert
Construction will show that such an abstract extension actually embeds in $\Solof \rx D_4$ as a crystallographic group.

\begin{theorem}%[Seifert Construction]
\label{seifert-const}
Let $\gammt=\lgamsig$ be a lattice of $\Solof$,
and
\[
1\lra \gammt \lra \varPi \lra \Phi \lra 1
\]
be an extension of $\gammt$ by a finite group $\Phi$.
Then there exists a homomorphism
\[
\theta: \varPi\ra \Solof\rx D_4 \subset \Solof\rx\aut(\Solof)
\]
carrying $\gammt$ onto a standard lattice.
This $\theta$ is injective if and only if
%the kernel of $\varphi: \varPi\ra\out(\gammt)\ra\out(\Solof)$
%is torsion-free (in particular, if $\Phi\subset D_4$).
$\Phi\subset D_4$.
Such $\theta$ is unique up to conjugation by an element of
$\Solof\rx\aut(\Solof)$.
\end{theorem}

\begin{proof}
$\Solof$ is completely solvable, and we can apply the Seifert construction
\cite[Theorem 7.3.2]{seif-book} with $G=\Solof$ and $W=\text{\{point\}}$. Since
$\Phi$ is finite,
the homomorphism $\varphi: \varPi\ra\out(\gammt)\ra\out(\Solof)$ has
finite image in $\out(\Solof)$, and it lifts back to a finite subgroup $C$ of
$\aut(\Solof)$. But this $C$ can be conjugated into
$D_4\subset \aut(\Solof)$, a maximal compact subgroup.
Consequently, we have a commuting diagram
\[
\CD
1 @>>> \gammt @>>> \varPi @>>> \Phi @>>> 1\\
@. @VVV @VVV @VVV @.\\
1 @>>> \Solof   @>>> \Solof\rx D_4 @>>> D_4 @>>> 1
\endCD
\]
The essence of the argument is showing that the cohomology set
$H^2(\Phi;\Solof)$ is trivial for any finite group $\Phi$.
The uniqueness is a result of \cite[Corollary 7.7.4]{seif-book}. It also
comes from $H^1(\Phi;\Solof)=0$.
\qedhere
\end{proof}
After inflation, the Seifert Construction produces a crystallographic group of $\Solof$.
Often we can assume that $c_3 = 0$, that is, $\lgamsig$ is
a standard lattice of $\Solof$.
Recall that $\aut(\Solof)=\bbr\rx\aut(\Sol)$ (Proposition \ref{autosolof}),
where $\hat k \in \bbr$ acts by
\[
\left[
\begin{matrix} 1 &e^u x &z\\
               0 &e^u &y\\
               0 &0 &1 \end{matrix}
\right]
\longmapsto
\left[
\begin{matrix} 1 &e^u x &z + k u \\
               0 &e^u &y\\
               0 &0 &1 \end{matrix}
\right].
\]
We have the following:

\begin{theorem}
\label{abstract-kernel-theorem}
For all holonomy groups,
except $\bbz_4$, a crystallographic group $\varPi$ of $\Solof$ embeds into
$\Solof\rx D_4$ in such a way that $\varPi\cap\Solof$ is a standard lattice {\rm ($c_3=0$)}.

\end{theorem}
\begin{proof}
Let $e$ denote the identity element of $\Solof$.
For the statement concerning $c_3$,
conjugation by $(e,\hat k)$ with $k=-\frac {c_3} {{\lambda}}$ sets $c_3=0$ in $\bft_3$.
However, this conjugation moves  $D_4$ to $\hat k D_4 \hat k \inv$.

Suppose every $A\in \Phi$ satisfies $\abar\hat A=+1$.
Since such $A$ commute with $\hat k$,
conjugation by $(e,\hat k)$ leaves the holonomy group $\Phi$ inside $D_4$ while
moving $\bft_3$ so that $c_3=0$.
This applies to, from the list of Theorem \ref{abstract-kernel},
all the groups lifting $\Sol$ crystallographic groups of type \gr{2a}, \gr{2b}, \gr{3}, \gr{3i}, \gr{6a}, \gr{6ai}, \gr{6b}, and
\gr{6bi}.

Suppose $\Phi$ contains $A=\matPpm$.  Then Corollary \ref{a-conj-zero3} and
Lemma \ref{conj-in-t4}
show that a generator $\alpha$ of $\varPi$ projecting to $A \in \Phi$ can be conjugated to $\alpha=(\bft_3^{\frac12},A)$
(so that $a_1=a_2=a_4=0$).
Then, we shall show that
$\bft_3=\hat\bft_3\bft_4^{c_3}$ can be replaced by $\hat\bft_3$
(where $\hat\bft_3$ is $\bft_3$ with $c_3=0$).
\begin{align*}
\alpha^2
&=(\bft_3^{\frac12},A)^2
=((\hat\bft_3\bft_4^{c_3})^{\frac12},A)^2
=(\hat\bft_3\bft_4^{c_3})^{\frac12} A((\hat\bft_3\bft_4^{c_3})^{\frac12})\\
&=\hat\bft_3^{\frac12} \bft_4^{\frac{c_3}2} \cdot \hat\bft_3^{\frac12} \bft_4^{-\frac{c_3}2}
=\hat\bft_3.
\end{align*}
Thus $\hat\bft_3=\alpha^2\in\varPi$, and we can take $\hat\bft_3$ instead of
$\bft_3$ as a generator for the same group (which is
apparently redundant since $\alpha$ is in the set already).
From the list in Theorem \ref{abstract-kernel},
the groups \gr{1}, \gr{5}, \gr{7} and \gr{7i}
contain such an $A$ in the holonomy.

The only case that is not covered by these two cases is when $\Phi=\bbz_4$
(type \gr{4} in the list), which is discussed below in our main classification (Theorem 
\ref{ClassificationSolof-geometry}).\qedhere
\end{proof}

\begin{lemma}
\label{conj-in-t4}
If $\det(A)=-1$, by conjugation, $a_4$ can be made 0.
\end{lemma}

\begin{proof}
Suppose $\det(A)=-1$. Conjugation by
$\bft_4^{-\frac{a_4}{2}}$ fixes the lattice $\lgamsig$,
and moves $(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3}\bft_4^{{a_4}},A)$ to
$(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3},A)$.
\end{proof}

\begin{proposition}[Fixing $a_4, b_4$]
Consider the commuting diagram in {\rm Proposition \ref{inflation}}.
Given $Q$ and integers $q, m_1, m_2$, we had $\lgamsig$. The only thing
that remains for the construction of $\varPi$ is fixing $a_4,b_4$. As is
well known, all the extensions $\varPi$ in the short exact sequence
\[
1\ra\lgamsig\ra\varPi\ra \Phi\ra 1
\]
are classified by $H^2(\Phi;\calz(\lgamsig))=H^2(\Phi;\bbz)$.
When $\Phi=\angles{A}$,
\[
H^2(\bbz_p;\bbz)=
\begin{cases}
0,       &\text{if $\ahat=-1$};\\
\bbz_p,  &\text{if $\ahat=1$},
\end{cases}
\]
see \textup{\cite[Theorem 7.1, p.122]{maclane}}
.

In actual calculation, this becomes an equation
\[
\alpha^p=\bft_1^{n_1} \bft_2^{n_2} \bft_3^{n_3} \bft_4^{k_4}
\]
for integers $n_i$ and  $k_4=\tfrac{i}{q}$, $i=0,1,\cdots,p-1$.
\qed
\end{proposition}

\begin{remark}
When $\Phi=\angles{A,B}$ is not cyclic, $\ahat=\bhat=+1$ never happens, so we can set
one of $a_4$, $b_4$ to zero. Thus, $H^2(\Phi; \calz(\tgamsig))$ is cyclic for all $\Phi$.
\end{remark}

\step[{\bf Detecting Torsion in $\Solof$ Crystallographic Groups}]

Given a lattice $\wt \Gamma_\cals$ of $\Solof$ (which projects to a lattice $\gamsig$ of $\Sol$), the short exact sequence
\[1 \ra \calz(\wt \Gamma_\cals) \ra \wt \Gamma_\cals \ra \gamsig \ra 1\]
induces an $S^1$-bundle over the solvmanifold $\gamsig \backslash \Sol$,
\[S^1 \ra \wt \Gamma_\cals \backslash (\Solof) \ra \gamsig \backslash \Sol.\]
The following lemma will be useful for determining when a $\Solof$ crystallographic group has torsion.

\begin{lemma}
\label{torSolof}
Let $\wt \Gamma_\cals$ be a lattice of $\Solof$, projecting
to a standard lattice $\gamsig$ of $\Sol$, and suppose that
for $\alpha \in  \Solof \rx D_4$, the group
$\varPi = \angles{\wt \Gamma_\cals, \alpha}$ is crystallographic.
Let $\bar \alpha$ denote the projection of $\alpha$ to $\Sol \rx D_4$.
When the automorphism part of $\alpha$ 
acts as a reflection on the center of $\Solof$, $\varPi$ is torsion-free if and only if $\angles{\gamsig,\bar \alpha}$ is torsion-free.
\end{lemma}
\begin{proof}
Obviously, if $\angles{\gamsig,\bar \alpha}$ is torsion-free, then $\varPi$
must be torsion-free. For the converse, suppose that $\angles{\gamsig,\bar \alpha}$
has torsion. In this case, the action of $\bar \alpha$ on the solvmanifold
$\gamsig \bs \Sol$ must fix a point. Observe that the action of $\alpha$
on the solvmanifold $\wt \Gamma_\cals \bs \Solof$ is $S^1$ fiber preserving.
Therefore, a circle fiber is left invariant under the action of $\alpha$. Since
$\alpha$ acts as reflection on the fiber, $\alpha$ must fix a point. Therefore,
$\varPi$ has torsion.
\qedhere
\end{proof}


%\subsubsection{Main Classification}

\begin{theorem}[Classification of $\Solof$-geometry]
\label{ClassificationSolof-geometry}
The following is a complete list of
crystallographic groups of $\Solof$, viewed as lifts of
$\Sol$ crystallographic groups, see {\rm Theorem \ref{ClassificationSol-geometry}}.
\end{theorem}

\noindent
We assume that in each case $q, c_1$, $c_2$, and $c_3$ are fixed so as to induce 
an abstract kernel $\Phi \ra \out(\lgamsig)$ with vanishing obstruction
in $H^3(\Phi; \tgamsig)$. Recall that this was possible by 
Proposition \ref{inflation} and Theorem \ref{seifert-const}.
In all cases, except, $\Phi = \bbz_4$, we can take $c_3=0$ (Theorem \ref{abstract-kernel-theorem}).
In the $\bbz_4$ holonomy case, we have two different (up to isomorphism) choices for $c_3$.
Whenever the holonomy group contains an automorphism which is
off-diagonal, the orbifold is non-orientable. To avoid repetition, we do not give
the classification of the exponents on $\bft_1, \bft_2$ by $H_{\varphi}^2(\bbzphi; \bbz^2) \cong H^1(\Phi; \coker(I-\cals))$, as these exponents are fixed by $Q$ (see Theorem \ref{ClassificationSol-geometry}).
\begin{enumerate}
\item[]
\bigskip
\item[\gr{0}]
$\Phi=\text{trivial}$ \parn
$\bullet$ $c_3=0$ \parn
$\bullet$ $\varPi=\wt\Gamma_{(\cals;q,m_1,m_2)}$.\parn
$\bullet$ Torsion-free
%
\item[\gr{1}]
$\Phi=\bbz_2$: $A=
\left[\begin{matrix}
1 &0\\ 0& -1
\end{matrix}\right]$\parn
$\bullet$ $c_3=0$\parn
$\bullet$
$\varPi=\angles{\bft_1,\bft_2,\bft_3,\bft_4^{\frac{1}{q}},
\alpha =(\bft_3^{\frac12},A)}$.\parn
$\bullet$ $H^2(\Phi; \calz(\tgamsig))$ is trivial.\parn
$\bullet$ Torsion-free
%
\item[\gr{2a}]
$\Phi=\bbz_2$: $A=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$\parn
$\bullet$ $c_3=0$ \parn
$\bullet$ $\varPi=\angles{\bft_1,\bft_2,\bft_3,\bft_4^{\frac{1}{q}},
\alpha =(\bft_1^{a_1} \bft_2^{a_2}\bft_4^{a_4},A)}$. \parn
$\bullet$ $H^2(\Phi; \calz(\tgamsig)) = \bbz_2$.  There are two choices for $a_4$, the solutions of 
$\alpha^2 = \bft_4^{\frac {i} {q}}$ ($i=0,1$). \parn
$\bullet$
Torsion-free for $i=1$
%
\item[\gr{2b}]
$\Phi=\bbz_2$: $A=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$\parn
$\bullet$ $c_3=0$ \parn
$\bullet$ $\varPi=\angles{\bft_1,\bft_2,\bft_3,\bft_4^{\frac{1}{q}},
\alpha =(\bft_3^{\frac12}\bft_4^{a_4},A)}$.\parn
$\bullet$ $H^2(\Phi; \calz(\tgamsig))=\bbz_2$, $a_4=0$ or $\frac{1}{2q}$.\parn
$\bullet$ Torsion-free
%
\item[\gr{3}]
$\Phi=\bbz_2$: $A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$\parn
$\bullet$ $c_3=0$\parn
$\bullet$
$\varPi=\angles{\bft_1,\bft_2,\bft_3,\bft_4^{\frac{1}{q}},
\alpha =(\bft_1^{a_1} \bft_2^{a_2},A)}$.\parn
$\bullet$ $H^2(\Phi; \calz(\tgamsig))$ is trivial.\parn
$\bullet$
Always has torsion
%
\item[\gr{3\ii}]
$\Phi=\bbz_2$: $A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$\parn
$\bullet$ $c_3=0$\parn
$\bullet$
$\varPi=\angles{\bft_1,\bft_2,\bft_3,\bft_4^{\frac{1}{q}},
\alpha =(\bft_1^{a_1} \bft_2^{a_2},A)}$.\parn
$\bullet$ $H^2(\Phi; \calz(\tgamsig))$ is trivial.\parn
$\bullet$
$\varPi$ is torsion-free if and only if $a_1=\tfrac12$
and $a_2\not\equiv \frac{(\sigma_{11}+1)(2 n+1)}{2\sigma_{12}}$
for any $n\in\bbz$.
%
\par\noindent
\item[\gr{4}]
$\Phi=\bbz_4$: $A=
\left[\begin{matrix}
0 &1\\ -1& 0
\end{matrix}\right]$\parn
$\bullet$
There are two choices for $c_3$ in $\bft_3$. They are solutions of
$d=0$ or $d=\tfrac1q$ for $c_3$, where
$\alpha\bft_3 \alpha\inv
=\bft_1^{(1-\sigma_{22})a_1+\sigma_{12}a_2}\bft_2^{
\sigma_{21}a_1+(1-\sigma_{11})a_2}\bft_3^{-1}\bft_4^{d}$. Each corresponds to 
a different abstract kernel $\Phi \ra \out(\tgamsig)$.\parn
$\bullet$
$\varPi=\angles{\bft_1,\bft_2,\bft_3,\bft_4^{\frac{1}{q}},
\alpha =(\bft_1^{a_1} \bft_2^{a_2}\bft_4^{a_4},A)}$.\parn
$\bullet$
$H^2(\Phi; \calz(\tgamsig)) = \bbz_4$. There are 4 choices for $a_4$ which make $\alpha^4=\bft_4^{\frac{i}{q}}$,
$\quad (i=0,1,2,3)$\parn
$\bullet$
$\varPi$ is torsion-free only when $i=1,3$.
%
\item[\gr{5}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
1 &0\\ 0& -1
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$\parn
$\bullet$ $c_3=0$\parn
$\bullet$ $\varPi=\angles{\bft_1,\bft_2,\bft_3,\bft_4^{\frac{1}{q}},
\alpha=(\bft_3^{\frac12},A)
\beta=(\bft_1^{b_1} \bft_2^{b_2} \bft_4^{b_4},B)}$.\parn
$\bullet$ $H^2(\Phi; \calz(\tgamsig))=\bbz_2$.
There are two choices for $b_4$, the solutions of
$\beta^2=\bft_4^{\frac{i}{q}}, (i=0,1)$.\parn
$\bullet$
$\varPi$ is torsion-free only when
$i=1$.
%
\item[\gr{6a}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$\parn
$\bullet$ $c_3=0$\parn
$\bullet$
$\varPi=\angles{\bft_1,\bft_2,\bft_3,\bft_4^{\frac{1}{q}},
\alpha=(\bft_1^{a_1} \bft_2^{a_2},A),
\beta=(\bft_1^{b_1} \bft_2^{b_2}\bft_4^{b_4},B)}$.\parn
$\bullet$ $H^2(\Phi; \calz(\tgamsig))=\bbz_2$.
There are two choices for $b_4$, the solutions of
$\beta^2=\bft_4^{\frac{i}{q}}, (i=0,1)$.\parn
$\bullet$
Always has torsion
%
\item[\gr{6a\ii}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$\parn	
$\bullet$ $c_3=0$\parn
$\bullet$ $\varPi=\angles{\bft_1,\bft_2,\bft_3,\bft_4^{\frac{1}{q}},
\alpha=(\bft_1^{a_1} \bft_2^{a_2},A),
\beta=(\bft_1^{b_1} \bft_2^{b_2}\bft_4^{b_4},B)}$.\parn
$\bullet$ $H^2(\Phi; \calz(\tgamsig))=\bbz_2$.
There are two choices for $b_4$, the solutions of
$\beta^2=\bft_4^{\frac{i}{q}}, (i=0,1)$.\parn
$\bullet$
$\varPi$ is torsion-free if and only if
$i=1$,
$a_1=\tfrac12$, $a_2 = b_2 + \frac 12$,
$b_1\not\equiv \frac {\sigma_{12}(2 n+1)}{2(\sigma_{11}-1)} + \frac 1 2$,
$b_2\not\equiv \frac{(\sigma_{11}+1)(2 m+1)}{2\sigma_{12}} + \frac 1 2$ for any $m,n \in \bbz$.
%none of the following 4 equations have integer solutions for $n_1,n_2$:
%\begin{alignat*}{2}
%2 (a_1+n_1)=&0,
%&(\sigma_{11}+1)(a_1+n_1)+\sigma_{12}(a_2-n_2)&=0,\\
%2 (a_2-b_2+n_2)&=0,\qquad
%&(\sigma_{11}-1)(a_1+b_1+n_1)+\sigma_{12}(a_2-b_2-n_2)&=0.
%\end{alignat*}
%
%
\item[\gr{6b}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$\parn
$\bullet$ $c_3=0$ \parn
$\bullet$
$\varPi=\angles{\bft_1,\bft_2,\bft_3,\bft_4^{\frac{1}{q}},
\alpha=(\bft_1^{a_1} \bft_2^{a_2},A),
\beta=(\bft_3^{\frac12}\bft_4^{b_4},B)}$.\parn
$\bullet$ $H^2(\Phi; \calz(\tgamsig))=\bbz_2$.
There are two choices for $b_4$, the solutions of
$\beta^2=\bft_3 \bft_4^{\frac{i}{q}}, (i=0,1)$.\parn
$\bullet$
Always has torsion
%
\item[\gr{6b\ii}]
$\Phi=\bbz_2\x\bbz_2$:
$A=
\left[\begin{matrix}
0 &1\\ 1& 0
\end{matrix}\right]$,
$B=
\left[\begin{matrix}
-1 &0\\ 0& -1
\end{matrix}\right]$\parn
$\bullet$ $c_3=0$\parn
$\bullet$
$\varPi=\angles{\bft_1,\bft_2,\bft_3,\bft_4^{\frac{1}{q}},
\alpha=(\bft_1^{a_1} \bft_2^{a_2},A),
\beta=(\bft_3^{\frac12}\bft_4^{b_4},B)}$.\parn
$\bullet$
$H^2(\Phi; \calz(\tgamsig))=\bbz_2$.
There are two choices for $b_4$, the solutions of
$\beta^2=\bft_3\bft_4^{\frac{i}{q}}, (i=0,1)$.\parn
$\bullet$
$\varPi$ is torsion-free if and only if
$a_1=\tfrac12$
and $a_2\not\equiv \frac{(k_{11}-1)(2 n+1)}{2k_{12}}$ for any $n \in \bbz$.
%
%
\item[\gr{7}]
$\Phi=\bbz_4\rx\bbz_2$:
$A=\matNpp, B=\matPpm$\parn
$\bullet$ $c_3=0$ \parn
$\bullet$ $\varPi=\angles{\bft_1,\bft_2,\bft_3,\bft_4^{\frac{1}{q}},
\alpha=(\bft_1^{a_1} \bft_2^{a_2},A),
\beta=(\bft_3^{\frac12}\bft_4^{b_4},B)}$.\parn
$\bullet$
$H^2(\Phi; \calz(\tgamsig))=\bbz_4$.
There are $4$ choices for $b_4$, the solutions of
$(\beta\alpha)^4=\bft_4^{\frac{j}{q}}, (j=0,1,2,3)$.\parn
$\bullet$
Always has torsion
%

\item[\gr{7\ii}]
$\Phi=\bbz_4\rx\bbz_2$:
$A=\matNpp, B=\matPpm$ \parn
$\bullet$ $c_3=0$ \parn
$\bullet$ $\varPi=\angles{\bft_1,\bft_2,\bft_3,\bft_4^{\frac{1}{q}},
\alpha=(\bft_1^{a_1} \bft_2^{a_2},A),
\beta=(\bft_3^{\frac12}\bft_4^{b_4},B)}$.\parn
$\bullet$
$H^2(\Phi; \calz(\tgamsig))=\bbz_4$.
There are $4$ choices for $b_4$, the solutions of
$(\beta\alpha)^4=\bft_4^{\frac{j}{q}}, (j=0,1,2,3)$.\parn
$\bullet$
%
% The conditions that $a_1, a_2$ satisfy
% \begin{align*}
% \begin{array}{l}
% a_1 = \frac {1 - k_{12} } {2 k_{11}} + \frac i k_{11} \text{ for }
% i=0,1,\dots k_{11}-1 \\
% a_2 = \frac 1 2
% \end{array}
% \end{align*}
% are necessary for $\varPi$ to be torsion-free. When these conditions are
% satisfied,
$\varPi$ is torsion-free if and only if $j=1,3$ and
$a_1 = \frac 12$ and $a_2= -\frac {k_{21}+1}{2k_{11}} + \frac i {k_{11}}$ for $i=0, \ldots, k_{11}-1$.
\par\noindent
\end{enumerate}
\bigskip

%The relations are given as follows.

%\begin{align*}
%&[\bft_1\bft_2]=\bft_4,\
%\bft_4 \text{ central},\\
%&\begin{matrix}
%\bft_3 \bft_1 \bft_3\inv= \bft_1^{\sigma_{11}} \bft_2^{\sigma_{21}}\bft_4^{m_1},\\
%\bft_3 \bft_2 \bft_3\inv= \bft_1^{\sigma_{12}}
%\bft_2^{\sigma_{22}}\bft_4^{m_2 },\end{matrix}\\
%&\begin{matrix}
%\alpha\bft_1\alpha\inv=\bft_1^{\tau_{11}} \bft_2^{\tau_{21}} \bft_4^{d_1}, \\
%\alpha\bft_2\alpha\inv=\bft_1^{\tau_{12}} \bft_2^{\tau_{22}} \bft_4^{d_2}, \end{matrix}
%\text{ where }
%\cals^{a_3}\wa=
%\left[\begin{matrix}
%\tau_{11} & \tau_{12}\\ \tau_{21} & \tau_{22}
%\end{matrix}\right],\\
%&\alpha\bft_3\alpha\inv=\bft_1^{w_1}\bft_2^{w_2}\bft_3^{\abar} \bft_4^{d_3},
%\text{ where } {\xy{w_1}{w_2}=\left(I-\cals^{\abar}\right)\xy{a_1}{a_2}},\\
%&\alpha\bft_4\alpha\inv=\bft_4^{\ahat},\\
%&\alpha^2=\bft_1^{v_1}\bft_2^{v_2}\bft_3^{(1+\abar)a_3}\bft_4^{d_4},\
%\text{ where } {\xy{v_1}{v_2}=\left(I+\cals^{a_3}\wa\right)\xy{a_1}{a_2}},
%\text{ if }\ A^2=I,\\
%&\alpha^4=\bft_4^{4 a_4}, \text{ if } A^4=I \text{ (and $a_3=0$);}\\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%&\begin{matrix}
%\beta\bft_1\beta\inv=\bft_1^{\tau'_{11}} \bft_2^{\tau'_{21}} \bft_4^{d'_1}, \\
%\beta\bft_2\beta\inv=\bft_1^{\tau'_{12}} \bft_2^{\tau'_{22}} \bft_4^{d'_2}, \end{matrix}
%\text{ where }
%\cals^{b_3}\wb=
%\left[\begin{matrix}
%\tau'_{11} & \tau'_{12}\\ \tau'_{21} & \tau'_{22}
%\end{matrix}\right],\\
%&\beta\bft_3\beta\inv=\bft_1^{w'_1}\bft_2^{w'_2}\bft_3^{\bbar} \bft_4^{d'_3},
%\text{ where } {\xy{w'_1}{w'_2}=\left(I-\cals^{\bbar}\right)\xy{b_1}{b_2}},\\
%&\beta\bft_4\beta\inv=\bft_4^{\bhat},\\
%&\beta^2=\bft_1^{v'_1}\bft_2^{v'_2}\bft_3^{(1+\bbar)b_3}\bft_4^{d'_4},\
%\text{ where } {\xy{v'_1}{v'_2}=\left(I+\cals^{b_3}\wa\right)\xy{b_1}{b_2}},
%\text{ if }\ B^2=I,\\
%%&\beta^4=\bft_4^{2 b_4}, \text{ if } B^4=I \text{ (and $b_3=0$).}\\
%&[\alpha,\beta]=\bft \text{ for } \bft \in \tgamsig,\
%\text{ if $\Phi=\bbz_2\x\bbz_2$};\\
%&(\alpha\beta)^4= \bft \text{ for } \bft \in \tgamsig\ 
%\text{ if $\Phi=\bbz_4\rx\bbz_2$}.
%\end{align*}

\begin{proof}

%Theorem \ref{abstract-kernel} classifies all the $\bbzphi$ and abstract kernels
%$\phi:\bbzphi \ra \gltz$ up to conjugation in $\gltz$ (which is
%essentially up to a change of generating set for the lattice).
%This fixes $a_3, b_3$.
%Theorem \ref{H1-cokernel-theorem} allows us to find $a_1, a_2, b_1, b_2$
%for each abstract kernel $\phi$. This is all we need for a $\Sol$ crystallographic
%group $Q$.

%We have shown that every $\Sol$ crystallographic group $Q$ lifts to
%a $\Solof$ crystallographic group.
%However, here is a procedure of actually finding the lifts $\varPi$ for a given $Q$
%\emph{without relying on the Seifert fibering construction.}
%The group $Q$ has $a_1,a_2,a_3$ (and $b_1,b_2,b_3$) already.
%All we need now is fixing the central component of each generator,
%$\bft_1, \bft_2, \bft_3, \alpha, \beta$. They are
%\[
%c_1, c_2, c_3, a_4, b_4.
%\]
%Recall that $c_1, c_2$ are uniquely determined by the parameters $m_1, m_2$
%in
%\[
%\bft_3 \bft_1 \bft_3\inv
%= \bft_1^{\sigma_{11}} \bft_2^{\sigma_{21}} \bft_4^{\frac{m_1}{q}},
%\qquad
%\bft_3 \bft_2 \bft_3\inv= \bft_1^{\sigma_{12}} \bft_2^{\sigma_{22}}
%\bft_4^{\frac{m_2}{q}},
%\]
%See Lemma \ref{determine-TT}.

In cases \gr{2a}, \gr{2b} and \gr{4}, $\Phi=\bbz_p$, $p=2$ or $4$.
Let
$\bft_4^{\frac1q}$ be a free generator of $\tgamsig\cap\calz(\Solof).$
Since $\det(A)=+1$, $\alpha^p$ has $\bft_4$ component ${\bft_4}^{p\cdot
a_4+\ell}$, where $\ell$ is independent of $a_4$.
The equation
\[
p\cdot a_4+\ell=0
\]
has solutions
\[
a_4, a_4+\tfrac{1}{p\cdot q},\cdots,a_4+\tfrac{p-1}{p\cdot q},
\]
each corresponding to a different class in $H^2(\Phi; \calz(\tgamsig))$. In fact, the number $\ell$ is always a rational number, and hence so is
$a_4$ (or $b_4$).
The remaining cases when $\Phi=\bbz_2\x\bbz_2$ or $D_4$ are similar. We set one of exponents on
$\bft_4=0$ by Lemma \ref{conj-in-t4}, and apply the above technique to find the remaining exponent on $\bft_4$.

%The last case is $\Phi=D_4$ (\gr{7} and  \gr{7\ii}).
%We will see that 
%We take $\gamma= \alpha \beta=(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{-\frac12}
%\bft_4^{-b_4},AB)$. Now we proceed as the previous cases. Since $\det(AB)=+1$,
%$\gamma^4$ has $\bft_4$ component ${\bft_4}^{-4\cdot
%b_4+\ell}$, where $\ell$ is independent of $b_4$.
%The equation
%\[
%-4\cdot b_4+\ell=0
%\]
%has a unique solution for $b_4$. This determines $b_4$.
%Once this is done, then a complete list of $b_4$ is
%\[
%b_4, b_4+\tfrac{1}{4\cdot q}, b_4+\tfrac{2}{4\cdot q}, b_4+\tfrac{3}{4\cdot q}.
%\]
%This yields all representations of $\varPi$ coming from $Q$ into $\aff(\Solof)$.

\bigskip
\noindent
\gr{0}
See Theorem \ref{lattice-part-construction}.
\bigskip

\noindent
\gr{1} Since $\ahat=-1$, Lemma \ref{conj-in-t4} implies
$a_4$ can be conjugated to zero.
So, $\varPi=\angles{\bft_1,\bft_2,\bft_3,\bft_4,
\alpha =(\bft_3^{\frac12},A)}$. As $Q$ is torsion-free, so is $\varPi$.
\bigskip

\noindent
\gr{2a}
There are two choices for $a_4$, the solutions of
$\alpha^2=\bft_4^{\frac{i}{q}}, (i=0,1)$.
Recall that $Q$ is never torsion-free, and thus $\varPi$ is torsion-free only when
$\alpha^2=\bft_4^{\frac{1}{q}}$. That is, when $i=1$, the class representing $\varPi$
in $H^2(\bbz_2; \calz(\tgamsig))$ corresponds to
the non-split (torsion-free) extension  
\[1 \ra \tgamsig \ra \varPi \ra \bbz_2\ra 1.\]

\noindent
\gr{2b}
By the $\Sol$ case, we can take $a_1=a_2=0$ so that
$\alpha=(\bft_3^{\frac12}\bft_4^{a_4},A)$. Then
$\alpha^2=\bft_3\bft_4^{2 a_4}$.
Therefore, $a_4=0$ or $\frac{1}{2q}$.
As $Q$ is torsion-free, so is $\varPi$.
\bigskip

\noindent
\gr{3}
By Lemma \ref{conj-in-t4}, we may assume $a_4=0$, equivalently,
$H^2(\Phi; \calz(\tgamsig))$ vanishes.
Let $\alpha=(\bft_1^{a_1} \bft_2^{a_2},A)$.
As $Q$ always has torsion, Lemma \ref{torSolof} applies to show that $\varPi$ has torsion.
Alternatively, explicit computation, like in the $\Sol$ case, also shows that $\varPi$ has torsion.
For $a_2 \equiv -a_1$, and one can compute that
\begin{align*}
\alpha^2
=(\bft_1^{a_1}\bft_2^{-a_1},A)^2
&=(e,I).
\end{align*}

\noindent
\gr{3\ii}
By Lemma \ref{conj-in-t4}, we may assume $a_4=0$, that is $H^2(\Phi; \calz(\tgamsig))$ vanishes.
Therefore, $\alpha=(\bft_1^{a_1} \bft_2^{a_2},A)$.
Lemma \ref{torSolof} applies to show that $\varPi$ is torsion-free precisely when $Q$ is torsion-free, and hence precisely when the criteria from Theorem \ref{ClassificationSol-geometry} \gr{3i} are satisfied.
\bigskip

\noindent
\gr{4}
This is the only case where a non-standard lattice is present, that is
$c_3 \neq 0$.
%There are two choices for $c_3$ in $\bft_3$. They are solutions of
%$d=0$ or $d=\tfrac1q$ for $c_3$, where
%\[
%\alpha\bft_3 \alpha\inv
%=\bft_1^{(1-\sigma_{22})a_1+\sigma_{12}a_2}\bft_2^{
%\sigma_{21}a_1+(1-\sigma_{11})a_2}\bft_3^{-1}\bft_4^{d}.
%\]
%If $c_3=0$, then the lattice is standard, otherwise non-standard. 
We have the relations
{\small
\begin{align*}
[\bft_1\bft_2]&=\bft_4,\quad
\bft_4 \text{ central},\\
\bft_3 \bft_1 \bft_3\inv
&= \bft_1^{\sigma_{11}} \bft_2^{\sigma_{21}}\bft_4^{m_1},\\
\bft_3 \bft_2 \bft_3\inv&= \bft_1^{\sigma_{12}}
\bft_2^{\sigma_{22}}\bft_4^{m_2 },\\
\alpha\bft_1 \alpha\inv
&=\bft_2\inv \bft_4^{v_1}\\
\alpha\bft_2 \alpha\inv
&=\bft_1 \bft_4^{v_2}\\
\alpha\bft_3 \alpha\inv
&=\bft_1^{(1-\sigma_{22})a_1+\sigma_{12}a_2}\bft_2^{
\sigma_{21}a_1+(1-\sigma_{11})a_2}\bft_3^{-1}\bft_4^{u_4+2 c_3},\\
\alpha^4
&=\bft_4^{4 a_4-(a_1-a_2)^2+v_4}, \text{ where}\\
\end{align*}
\begin{align*}
%
(\sigma_{11} + \sigma_{22} -2)v_1
&=
(-\sigma_{21}+\sigma_{22}-1) m_1+(\sigma_{11}-\sigma_{21}-1) m_2\\
&+\tfrac12
(\sigma_{11}-\sigma_{11} \sigma_{21}+\sigma_{22}-(\sigma_{11}^2+\sigma_{21}-2 \sigma_{11} \sigma_{21}) \sigma_{22}-\sigma_{11} \sigma_{22}^2)\\
&- (\sigma_{11}+\sigma_{22}-2)\cdot a_1\\
%
(\sigma_{11} + \sigma_{22} -2)v_2
&=
(1-\sigma_{21}-\sigma_{22}) m_1+(\sigma_{11}+\sigma_{21}-1) m_2\\
&-\tfrac12 (\sigma_{11}-\sigma_{22}) (-1-\sigma_{21}+\sigma_{11} \sigma_{22}) - (\sigma_{11}+\sigma_{22}-2)\cdot a_2\\
%
{2 \sigma_{12} (\sigma_{11} + \sigma_{22} -2)}u_4
&=\left((\phm\sigma_{22} m_1 -\sigma_{12} m_2)
+ \tfrac12(\phm\sigma_{11}-\sigma_{12})\sigma_{12} \sigma_{22}\right)\cdot a_1\\
&+\left((-\sigma_{12} m_1 + \sigma_{11} m_2)
+ \tfrac12 (-\sigma_{12}+\sigma_{22})\sigma_{11}\sigma_{12}\right) \cdot a_2\\
&+ \tfrac12\sigma_{12} \sigma_{22}\cdot a_1^2+ (\sigma_{22}-\sigma_{11}
\sigma_{22})\cdot a_1 a_2 + \tfrac12 (\sigma_{11}-2) \sigma_{12}\cdot a_2^2,\\
%
(\sigma_{11} + \sigma_{22} -2)v_4
&=
2 a_1 (\sigma_{22}+\sigma_{11} (\sigma_{21}\sigma_{22}-\sigma_{21}-\sigma_{22}^2)+2 (\sigma_{22}-1) m_1-2\sigma_{21} m_2)\\
&+ 2 a_2 (\sigma_{11}-\sigma_{22}(\sigma_{11} \sigma_{21} -\sigma_{21} +\sigma_{11}^2)-2 \sigma_{21} m_1+2 (\sigma_{11}-1) m_2)\\
\end{align*}
}
%This takes some serious calculations.
Recall that $m_1,m_2$ and all $\sigma_{ij}$'s are integers.
Since
\[
\xy{a_1}{a_2}=
\tfrac{1}{\sigma_{11}+\sigma_{22}-2}
\left[\begin{matrix}
\sigma_{11}-1& \sigma_{12}\\ \sigma_{21}&\sigma_{22-1}
\end{matrix}\right]
\xy{u_1}{u_2},\quad u_1, u_2\in\bbz,
\]
$v_1, v_2, v_4$ and $u_4$ are all \emph{rational numbers}.
Note that $c_3$ and $a_4$ appear only once in $\alpha\bft_3 \alpha\inv$ and
$\alpha^4$, respectively.
The above equalities show that both $c_3$ and $a_4$ must also be
rational numbers.
With a pre-determined $q$ in $\varPi\cap\calz(\Solof)=\angles{\bft_4^{\frac1q}}$,
% suppose $a_4=a,c_3=c$ is one solution. Then
% $a_4=a+\frac{i}{4q}, c_3=c+\frac{j}{2q}$,\
% ($0\leq i<4,\ 0\leq j<2$)  can be another solution.
% They may have co-relations.
suppose $c_3=-\tfrac{u_4}{2}$ (a solution for $u_4+2 c_3=0$). Then
$c_3'=c_3+\frac{1}{2q}$ is another solution so that
\[
\alpha\bft_3 \alpha\inv
=\bft_1^{(1-\sigma_{22})a_1+\sigma_{12}a_2}\bft_2^{
\sigma_{21}a_1+(1-\sigma_{11})a_2}\bft_3^{-1}\bft_4^{d},
\]
where $d=0$ or $\tfrac1q$. Unless $c_3$ is a multiple of $\tfrac1q$, the
corresponding lattice is non-standard. In case $c_3=(\tfrac1q)^p$, then
$\bft_3=\hat\bft_3\cdot (\bft_4^{\frac1q})^p$, where $\hat\bft_3$ has
no $\bft_4$-component. Since  $(\bft_4^{\frac1q})^p\in\varPi$, we
can replace the generator $\bft_3$ by $\hat\bft_3$ so that the lattice
becomes standard.


For $a_4$, we have
\[
\alpha^4=\bft_4^{4a_4-(a_1-a_2)^2+v_4},
\]
Then there are 4 choices for $a_4$,
$a_4=\frac{(a_1-a_2)^2-v_4+i}{4q}, (i=0,1,2,3)$.
These are the solutions of
\[
\alpha^4=\bft_4^{\frac{i}{q}}, \quad (i=0,1,2,3).
\]

\noindent
For $i=0, 2$, $\varPi$ has torsion. To see this when $i=2$, note that
\[  (\bft_4^{-\frac{1}{q}} \alpha^2)^2 = \bft_4^{-\frac{2}{q}}\bft_4^{\frac{2}{q}}=e.\]
For $i=1,3$, $\varPi$ is torsion-free.
\bigskip

\noindent
\gr{5}
By Corollary \ref{a-conj-zero3}, we can take $a_1=a_2=0$.
Since $\det(A)=-1$, we may assume $a_4=0$ by Lemma \ref{conj-in-t4}.
There are two choices for $b_4$, the solutions of
$\beta^2=\bft_4^{\frac{i}{q}}, (i=0,1)$. Thus $H^2(\Phi, \calz(\tgamsig)) = \bbz_2$.
% The equality
% \[
% (\bft\cdot \beta)^2=\bft_4^{2 (b_4+n_4)+ *}
% \]
% shows
Like case \gr{2a}, $\varPi$ is torsion-free only when
$\beta^2=\bft_4^{\frac{1}{q}}$.
\bigskip

\noindent
\gr{6a}
There are two choices for $b_4$, the solutions of
$\beta^2=\bft_4^{\frac{i}{q}}, (i=0,1)$. That is, $H^2(\Phi, \calz(\tgamsig)) = \bbz_2$.
As $\varPi$ contains a group of type \gr{3a}, it is never torsion-free.
\bigskip
%
\noindent
\gr{6a\ii}
\noindent
There are two choices for $b_4$, the solutions of
$\beta^2=\bft_4^{\frac{i}{q}}, (i=0,1)$. That is, $H^2(\Phi, \calz(\tgamsig)) = \bbz_2$.
For $\varPi$ to be torsion-free,
the subgroups
$\angles{\tgamsig,\alpha}$,
$\angles{\tgamsig,\beta}$, and
$\angles{\tgamsig,\alpha \beta}$, where
\[
\alpha \beta = \left(\bft_1^{a_1 + b_1} \bft_2^{a_2 - b_2} \bft_4^{b'_4}, \left[ \begin{matrix}0 & -1 \\ -1 & 0 \end{matrix} \right] \right)
\]
must all be torsion-free.
\noindent
The group $\angles{\tgamsig,\beta}$ is torsion-free precisely when $b_4$ satisfies
$\beta^2=\bft_4^{\frac{1}{q}}$.

We need criteria for
$(\bft_1^{a_1} \bft_2^{a_2},A)$ and $(\bft_1^{a_1+b_1} \bft_2^{a_2-b_2},AB)$ to
be of infinite order.
By Lemma \ref{torSolof}, these they are of infinite order precisely when
they are infinite order, projected to $\Sol$.
By computing when
the appropriate affine maps in Proposition \ref{topologySolManifolds}
are fixed point free, we obtain the conditions
$a_1=\frac 12$, $a_2 = b_2 + \frac 12$,
$b_1\not\equiv \frac {\sigma_{12}(2 n+1)}{2(\sigma_{11}-1)} + \frac 1 2$
$b_2\not\equiv \frac{(\sigma_{11}+1)(2 m+1)}{2\sigma_{12}} + \frac 1 2$ for any $m,n \in \bbz$.
\bigskip

\noindent
\item[\gr{6b}]
Note that, $a_4=0$ by Lemma \ref{conj-in-t4}, and there are two choices for $b_4$, the solutions of
$\beta^2=\bft_3\bft_4^{\frac{i}{q}}, (i=0,1)$. Hence $H^2(\Phi, \calz(\tgamsig)) = \bbz_2$. 
As $\varPi$ contains a subgroup of type \gr{3},
which has torsion, it must have torsion as well.
\bigskip
%
\noindent
\item[\gr{6b\ii}]
Note $a_4=0$ by Lemma \ref{conj-in-t4}, and
there are two choices for $b_4$, the solutions of
$\beta^2=\bft_3\bft_4^{\frac{i}{q}}, (i=0,1)$. Thus $H^2(\Phi, \calz(\tgamsig)) = \bbz_2$. 
\noindent
Recall from Theorem \ref{ClassificationSol-geometry} that
$Q$ (and hence $\varPi$, by Lemma \ref{torSolof}) is torsion-free precisely when the 
affine maps
\begin{align*}
\begin{array}{l}
\left( \xy{a_1} {a_2}, \matPpm \right) \\
\left( \xy{a_1} {a_2}, \left[ \begin{matrix} -k_{11} & k_{12}\\
-k_{21} & k_{11} \end{matrix} \right] \right)
\end{array}
\end{align*}
are fixed point free on $T^2$, which occurs precisely when
$a_1=\tfrac12$
and $a_2\not\equiv \frac{(k_{11}-1)(2 n+1)}{2k_{12}}$ for any $n \in \bbz$.
\bigskip

\noindent
\item[\gr{7}]

We may take $a_4=0$ without loss of generality by Lemma \ref{conj-in-t4}.
$(\beta\alpha)^4=\bft_4^{4b_4+\ell}$ shows that
there are $4$ choices for $b_4$, the solutions of
$(\beta\alpha)^4={\bft_4}^{\frac{j}{q}}, (j=0,1,2,3)$. Hence $H^2(D_4; \calz(\tgamsig)) = \bbz_4$.
As $\varPi$ contains a subgroup of type \gr{3}, it has torsion.
\bigskip
%
\noindent
\item[\gr{7\ii}]
Like case \gr{7}, we take $a_4=0$ and there are $4$ choices for $b_4$, the solutions of
$(\beta\alpha)^4={\bft_4}^{\frac{j}{q}}, (j=0,1,2,3)$, so that $H^2(D_4; \calz(\tgamsig)) = \bbz_4$.
\noindent
For $\varPi$ to be
torsion-free, $\angles{\tgamsig, \beta\alpha}$ is necessarily
torsion-free. This forces $b_4$ to satisfy $(\beta\alpha)^4=\bft_4^{\frac{j}{q}}, 
(j =1,3)$. Note that $\beta$ and the holonomy element corresponding to $-B$,
$\alpha \beta \alpha$, are of infinite order when projected to $\Sol$, and hence also on $\Solof$. 
Thus the only remaining holonomy elements to consider are $\alpha$ and
\[
\beta \alpha \beta =\left(\bft_1^{-k_{11}a_1-k_{12}a_2}\bft_2^{-k_{21}a_1-k_{11}a_2}\bft_4^{2b_4+v} ,\left[ \begin{matrix} 0 & -1 \\ -1 & 0\end{matrix} \right] \right).
\]
We need criteria for
$(\bft_1^{a_1} \bft_2^{a_2},A)$ and $(\bft_1^{-k_{11}a_1 - k_{12}a_2}
\bft_2^{ -k_{21}a_1 - k_{11}a_2} A,B A B)$ to
be infinite order, when projected to $\Sol$. 

By Proposition \ref{topologySolManifolds},
we just need to ensure that the appropriate affine maps are fixed point
free on $T^2$, and this occurs precisely when
\begin{align*}
\begin{array}{l}
a_1 = \frac 1 2 ,\ \
a_2\not\equiv \frac{(\sigma_{11}+1)(2 n+1)}{2\sigma_{12}},\\
\frac{-k_{21}}{2} - k_{11}a_2 = \frac 12,\\
\frac{-k_{11}}{2} - k_{12}a_2 \not\equiv \frac{\sigma_{12}(2 n+1)}{2(\sigma_{11}-1)}.
\end{array}
\end{align*}

With $a_1 = \frac 12$ and $a_2= -\frac {k_{21}+1}{2k_{11}} + \frac i {k_{11}}$ for $i=0, \ldots, k_{11}-1$,
using that $\det(K)=-1$ and $K^2=\cals$, one can compute that the remaining criteria are satisfied. In fact, we compute
\begin{align*}
\frac{(\sigma_{11}+1)(2 n+1)}{2\sigma_{12}} &= \frac{(k_{11}^2+k_{12}k_{21}+1)(2n+1)}{4k_{11}k_{12}}\\
&=\frac{2k_{12}k_{21}(2n+1)}{4k_{11}k_{12}}=\frac{k_{21}(2n+1)}{2k_{11}}.
\end{align*}
Now, for $m \in \bbz$,
\begin{align*}
 a_2= -\frac {k_{21}+1}{2k_{11}} + \frac i {k_{11}} &= \frac{k_{21}(2n+1)}{2k_{11}} + m \implies \\
 \frac {-k_{21}-1+2i}{2k_{11}} &=  \frac{k_{21}(2n+1)+2m k_{11}}{2k_{11}} \implies\\
 \frac {-1+2i}{2k_{11}} &= \frac{2k_{21}(n+1)+2m k_{11}}{2k_{11}},
\end{align*}
a contradiction for any integers $i,n,m$, as the numerators are 
of different parity. For the remaining criterion, we compute
\begin{align*}
\frac{-k_{11}}{2} - k_{12}a_2 &= \frac{-k_{11}}{2} - k_{12} \left( -\frac {k_{21}+1}{2k_{11}} + \frac i {k_{11}}   \right) \\
&= \frac{ -k_{11}^2+ k_{12}k_{21} + k_{12} - 2k_{12}i}{2k_{11}} \\
&= \frac{1 + k_{12} - 2k_{12}i} {2k_{11}}.
\end{align*}
Now, for $m \in \bbz$,
\begin{align*}
\frac{1 + k_{12} - 2k_{12}i} {2k_{11}} &= \frac{\sigma_{12}(2 n+1)}{2(\sigma_{11}-1)}+m \implies \\
\frac{1 + k_{12} - 2k_{12}i} {2k_{11}} &= \frac{2k_{11}k_{12}(2 n+1)}{2(k_{11}^2 + k_{12}k_{21}-1)}+m \implies\\
\frac{1 + k_{12} - 2k_{12}i} {2k_{11}} &= \frac{2k_{11}k_{12}(2 n+1)}{4k_{11}^2}+m = 
 \frac{k_{12}(2 n+1)+2m k_{11}}{2k_{11}} \implies \\
  \frac{1 - 2k_{12}i} {2k_{11}} &=  \frac{2 (n k_{12} +m k_{11}) }{2k_{11}},
\end{align*}
a contradiction for any integers $i,n,m$, as the numerators are 
of different parity.
Therefore, with $a_1 = \frac 12$ and $a_2= -\frac {k_{21}+1}{2k_{11}} + \frac i {k_{11}}$ for $i=0, \ldots, k_{11}-1$, and $(\beta\alpha)^4 = \bft_4^{\frac{j}{q}},  (j=1,3)$,
$\varPi$ is torsion-free.

%torsion cut 4
\noindent
This completes the proof of Theorem \ref{ClassificationSolof-geometry}.
\end{proof}

We have shown that every $\Solof$ crystallographic group $\varPi$
can be conjugated so that the holonomy $\Phi$ is a subgroup of our
fixed $D_4 \subset \aut(\Solof)$. We take $c_3=0$ (standard lattice)
for all holonomy groups except $\bbz_4$.
In this case, we have two values for $c_3$. This corresponds to two 
distinct abstract kernels, which give rise to different isomorphism classes.
Alternatively, in the $\bbz_4$ case, one may assume $c_3=0$. Then, however, $\Phi$ is not a subgroup of $D_4$. Rather,
$\Phi \subset \hat k D_4 \hat k \inv$, where $\hat k \in \bbr$ acts by
\[
\left[
\begin{matrix} 1 &e^u x &z\\
               0 &e^u &y\\
               0 &0 &1 \end{matrix}
\right]
\longmapsto
\left[
\begin{matrix} 1 &e^u x &z + k u \\
               0 &e^u &y\\
               0 &0 &1 \end{matrix}
\right].
\]

\begin{remark}
Let $\varPi$ and $\varPi'$ be two crystallographic groups of $\Solof$, with projections to $\Sol$ crystallographic groups, $Q$ and $Q'$ , respectively. That is, 
\begin{align*}
\varPi / (\varPi \cap \calz(\Solof))&\cong Q,\\
\varPi' / (\varPi' \cap \calz(\Solof)) &\cong Q'.
\end{align*}
Our classification is up to equivalence of extensions, a finer notion of
equivalence than isomorphism. 
However, we observe that $Q$ and $Q'$ being isomorphic is necessary for $\varPi$
and $\varPi'$ to be isomorphic. This is easy to see from the rigidity property of
these crystallographic groups (Theorem \ref{Rigidity}). Any isomorphism
between $\varPi$ and $\varPi'$ is conjugation by an element of $\aff(\Solof) = \Solof \rx \aut(\Solof)$. Such a conjugation must carry $\varPi \cap \calz(\Solof)$ to $\varPi' \cap \calz(\Solof)$, and hence must induce an isomorphism between $Q$ and $Q'$. 
\end{remark}

%\end{theorem}
%\vfill\eject

\step[{\bf Examples}]

First we show that $\Solof \rx D_4$ embeds into the affine group $\aff(4) \hookrightarrow \gl(5,\bbr)$.
% J. Milnor \cite{miln77-1} asked if every torsion-free
% polycyclic-by-finite group $\varPi$ occurs as the fundamental group of
% a compact, complete affinely flat manifold.   This is equivalent to
% asking if $\varPi$ can act on $\bbr^{K}$ properly
% as affine motions with $\varPi\bs\bbr^{K}$ compact.
% This has been disproved even for nilpotent groups, and no
% complete solutions are known.
% We show our $\Solof$-crystallographic groups have such affine structures.
% %This idea comes from \cite{HL} where an affine structure of
% %$\Sol$ was described.
The following correspondence is an injective homomorphism
of Lie groups, $\Solof\hookrightarrow \aff(4)$,
\begin{equation}
\label{affine-rep}
\left[
\begin{array}{ccc}
 1 & e^u x & z \\
 0 & e^u & y \\
 0 & 0 & 1
\end{array}
\right]
\lra
\left[
\begin{array}{ccccc}
 1 & -\frac{1}{2} e^{-u} y & \frac{e^u x}{2} & 0 & z-\frac{x y}{2} \\
 0 & e^{-u} & 0 & 0 & x \\
 0 & 0 & e^u & 0 & y \\
 0 & 0 & 0 & 1 & u \\
 0 & 0 & 0 & 0 & 1
\end{array}
\right].
\end{equation}
Moreover, the automorphisms
\[
\left[\begin{matrix}
a&0\\0&d
\end{matrix}\right],\quad
\left[\begin{matrix}
0&b\\ c&0
\end{matrix}\right]\in\aut(\Solof)
\]
can also be embedded as
\[
\left[\begin{matrix}
a d &0 &0 &0 &0\\
0 &a &0 &0 &0\\
0 &0 &d &0 &0\\
0 &0 &0 &1 &0\\
0 &0 &0 &0 &1
\end{matrix}\right],\quad
\left[\begin{matrix}
-b c &0 &0 &0 &0\\
0 &0 &b &0 &0\\
0 &c &0 &0 &0\\
0 &0 &0 &-1 &0\\
0 &0 &0 &0 &1
\end{matrix}\right],
\]
respectively, where $a,b,c,d$ are $\pm 1$.
Note that, if we remove the first row and the first column
from $\aff(4)$, we get a representation of $\Sol$ into $\aff(3)$.

If we write the element $(\bfa,A)\in \Solof\rx D_4$
as the product $\bfa\cdot A$, then the group operation of $\Solof\rx D_4$
is compatible with the matrix product in this affine group.
The action of $A$ on $\bfa$ is by conjugation. That is,
\begin{align*}
(\bfa\cdot A)(\bfb\cdot B)&=\bfa A \bfb B\\
&=\bfa (A \bfb A \inv) \cdot AB\\
&=(\bfa,A) \cdot (\bfb,B).
\end{align*}

\begin{example}[\gr{4} Non-standard lattice]
\label{z4-nonzeroc3-example}
This is an example where $c_3$ is non-zero
%(Theorem \ref{structure-theorem})
(Theorem \ref{ClassificationSolof-geometry}). Let
$
A=\left[\begin{matrix}
0&1\\ -1&0
\end{matrix}\right]
$
and
$\alpha=(\bft_1^{a_1} \bft_2^{a_2} \bft_4^{a_4},A)$.
Let
$
\cals=
\left[\begin{matrix}
1&2\\ 2&5
\end{matrix}\right].
$
Then $\lambda=3+2 \sqrt{2}$, and with
\[
P=
\left[
\begin{array}{cc}
 -\frac{1}{2} \sqrt{2+\sqrt{2}} & \frac{\sqrt{2-\sqrt{2}}}{2} \\
 -\frac{1}{\sqrt{2 \left(2+\sqrt{2}\right)}} & -\frac{1}{2} \sqrt{2+\sqrt{2}}
\end{array}
\right],
\]
$\varPi=\angles{\bft_1,\bft_2,\bft_3, \bft_4^{},\alpha}$, where
$\alpha=(\bft_1^{a_1}\bft_2^{a_2}\bft_4^{a_4},A)\in \Solof\rx\aut(\Solof)$,
has a representation into $\aff(4)$:
\begin{align*}
\bft_1&=\left[
\begin{array}{ccccc}
 1 & 0 & \frac{1}{2} & 0 & \frac{1}{2}(2{m_1}-{m_2}-3) \\
 0 & 1 & 0 & 0 & 1 \\
 0 & 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 0 & 1
\end{array}
\right],\\
\bft_2&=\left[
\begin{array}{ccccc}
 1 & -\frac{1}{2} & 0 & 0 & \frac{1}{2} (-{m_1}-1) \\
 0 & 1 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 & 1 \\
 0 & 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 0 & 1
\end{array}
\right],\\
\bft_3&=\left[
\begin{array}{ccccc}
 1 & 0 & 0 & 0 & {c_3} \\
 0 & 1 & 2 & 0 & 0 \\
 0 & 2 & 5 & 0 & 0 \\
 0 & 0 & 0 & 1 & \ln \left(3+2 \sqrt{2}\right) \\
 0 & 0 & 0 & 0 & 1
\end{array}
\right],\\
\bft_4&=\left[
\begin{array}{ccccc}
 1 & 0 & 0 & 0 & 1 \\
 0 & 1 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 0 & 1
\end{array}
\right],\\
(a,A)&=\left[
\begin{array}{ccccc}
 1 & -\frac{{a_1}}{2} & -\frac{{a_2}}{2} & 0 & \frac{1}{2} (2 {a_4}-{a_2} ({m_1}+1)+{a_1} ({a_2}+2 {m_1}-{m_2}-3)) \\
 0 & 0 & 1 & 0 & {a_1} \\
 0 & -1 & 0 & 0 & {a_2} \\
 0 & 0 & 0 & -1 & 0 \\
 0 & 0 & 0 & 0 & 1
\end{array}
\right].
\end{align*}
$\varPi$ has a presentation
\begin{align*}
&[\bft_1,\bft_2]=\bft_4,\ \text{ and $\bft_4$ is central}\\
&\bft_3\bft_1\bft_3\inv=\bft_1^{} \bft_2^{2} \bft_4^{m_1}, \
\bft_3\bft_2\bft_3\inv=\bft_1^{2} \bft_2^{5} \bft_4^{m_2},\\
&\alpha\bft_1\alpha\inv=\bft_2^{-1} \bft_4^{\frac12(-4-2 a_1+m_1-m_2)},\
\alpha\bft_2\alpha\inv=\bft_1^{}\bft_4^{\frac12(2-2 a_2-3 m_1+m_2)},\\
&\alpha\bft_3\alpha\inv=\bft_1^{-4 a_1+2 a_2}
\bft_2^{2 a_1}
\bft_3^{-1}
\bft_4^{5 a_1^2+2 c_3+a_1 (-5+5 m_1-2 m_2)+a_2 (3-a_2-2 m_1+m_2)},\\
&\alpha\bft_4\alpha\inv=\bft_4^{},\
\alpha^4=\bft_4^{-a_1^2+4 a_4-a_2 (2+a_2+2 m_1)+2 a_1 (-3+a_2+2 m_1-m_2)}.
\end{align*}
From
$
(I-\cals)\inv=
\left[\begin{matrix}
1 &-\tfrac12\\ -\tfrac12& 0
\end{matrix}\right],
$
we have $\coker(I-\cals)=\bbz_2\x\bbz_2$ generated by
\[
\xy{\tfrac12}{0},\quad \xy{0}{\tfrac12}.
\]
%Now the system of equations
%\begin{align*}
%\xy{w_1}{w_2}&=
%(I-\cals\inv)\bfa
%=\left[
%\begin{array}{cc}
% -4 & 2 \\
%  2 & 0
%\end{array}
%\right]
%\xy{a_1}{a_2}\in\bbz^2\\
%\xy{v_1}{v_2}&=
%(I+A+A^2+A^3)\bfa
%=\left[
%\begin{array}{cc}
%  0 & 0 \\
%  0 & 0
%\end{array}
%\right]
%\xy{a_1}{a_2}\in\bbz^2
%\end{align*}
%has 4 solutions modulo $\bbz^2$;
%\[
%\xy{a_1}{a_2}
%=
%\tfrac{1}{2}\xy{0}{0},\
%\tfrac{1}{2}\xy{1}{0},\
%\tfrac{1}{2}\xy{0}{1},\
%\tfrac{1}{2}\xy{1}{1}.
%\]
The coboundary is
\[
\im(I-\vara)=
\im
\left[\begin{matrix}
1&-1\\ 1&1
\end{matrix}\right]
=
\left\{\xy{0}{0},\xy{\tfrac12}{\tfrac12}\right\}.
\]
Thus we have only two cases
\[
\xy{a_1}{a_2}=
\xy{0}{0},\ \
\xy{\tfrac12}{0}.
\]
For simplicity, we shall assume $m_1=m_2=0$.

\noindent
With $\xy{a_1}{a_2}=\xy{0}{0}$,
$\varPi=\angles{\bft_1,\bft_2,\bft_3, \bft_4^{},\alpha}$, 
$\alpha=(\bft_4^{a_4},A)
\in \Solof\rx\aut(\Solof)$,
has a presentation
\begin{align*}
&[\bft_1,\bft_2]=\bft_4,\ \text{ and $\bft_4$ is central}\\
&\bft_3\bft_1\bft_3\inv=\bft_1^{} \bft_2^{2}, \
\bft_3\bft_2\bft_3\inv=\bft_1^{2} \bft_2^{5},\\
&\alpha\bft_1\alpha\inv=\bft_2^{-1} \bft_4^{-2},\
\alpha\bft_2\alpha\inv=\bft_1^{}\bft_4^{},\\
&\alpha\bft_3\alpha\inv=\bft_3\inv\bft_4^{2c_3}, \
\alpha\bft_4\alpha\inv=\bft_4^{},\\
&\alpha^4=\bft_4^{4 a_4}.
\end{align*}
%
The minimum $q$ for $\tgamsig$ is $q=1$, and we have choices
$a_4=0,\frac14,\frac12,\frac34$ and $c_3=0,\frac12$
(any combination of $a_4$ and $c_3$), with the same center.
So, there are 8 distinct groups.
Half of them (with $c_3=0$) have standard lattices, and the rest
(with $c_3=\tfrac12$) have non-standard lattices.

With $\xy{a_1}{a_2}=\xy{\frac12}{0}$,
$\varPi=\angles{\bft_1,\bft_2,\bft_3, \bft_4^{},\alpha}$, where
$\alpha=(\bft_1^{\frac12}\bft_4^{a_4},A)
\in \Solof\rx\aut(\Solof)$,
has a presentation
\begin{align*}
&[\bft_1,\bft_2]=\bft_4,\ \text{ and $\bft_4$ is central}\\
&\bft_3\bft_1\bft_3\inv=\bft_1^{} \bft_2^{2}, \
\bft_3\bft_2\bft_3\inv=\bft_1^{2} \bft_2^{5},\\
&\alpha\bft_1\alpha\inv=\bft_2^{-1} \bft_4^{-\frac52},\
\alpha\bft_2\alpha\inv=\bft_1^{}\bft_4^{},\\
&\alpha\bft_3\alpha\inv=\bft_1^{-2}\bft_2^{}\bft_3^{-1}
\bft_4^{-\frac54+2c_3},
\alpha\bft_4\alpha\inv=\bft_4^{},\\
&\alpha^4=\bft_4^{-\frac{13}{4}+4a_4}.
\end{align*}
%
The minimum $q$ for $\tgamsig$ is $q=2$ (which comes out of
$\alpha\bft_1\alpha\inv=\bft_2^{-1} \bft_4^{-\frac52}$),
and we have choices
$a_4=
\frac{1}{16},\frac{3}{16},\frac{5}{16},\frac{7}{16}$ and
$c_3=\frac{1}{8}$ or $\frac{3}{8}$
(any combination of $a_4$ and $c_3$), with the same center.
So, there are 8 distinct groups.
%Note that all these groups are not-isomorphic to each other.
All these groups have non-standard lattices, because no $c_3$ is an
integer multiple of $\frac1q$ for $q=2$.
\end{example}
\begin{example}[\gr{6b\ii}]
Let $\Phi=\bbz_2\x\bbz_2=\angles{A,B}$, where
\[
A=\matNpp,\ B=\matPmm \text{ with } \wa=\matPpm,\ \wb=B.
\]
Then $\cals$ must be of the form $n K-I$. Let us take
\[
K=
\left[\begin{matrix}
3&2\\ 4&3
\end{matrix}\right].
\text{ Then, } 
\cals=
\left[\begin{matrix}
17&12\\ 24&17
\end{matrix}\right].
\]
Note that $k_{11}=k_{22}$ and $\tr(K)=n=6>2$.
Therefore, $\cals$ satisfies the conditions for case \gr{6b\ii}
and
\begin{align*}
\alpha&=(\bft_1^{a_1}\bft_2^{a_2},A),\\
\beta&=(\bft_3^{\frac12}\bft_4^{b_4},B).
\end{align*}
Then $\lambda=17+ 12\sqrt{2}$, and with
\[
P=
\left[
\begin{array}{cc}
 -\frac{1}{\sqrt[4]{2}} & \frac{1}{2^{3/4}} \\
 -\frac{1}{\sqrt[4]{2}} & -\frac{1}{2^{3/4}} \\
\end{array}
\right],
\]
the equation in Lemma \ref{determine-TT} yields
\begin{align*}
c_1&=\tfrac{1}{4} (102+\sqrt{2}+2 m_1-3 m_2)\\
c_2&=\tfrac{1}{8} (-204-\sqrt{2}-3 m_1+4 m_2).
\end{align*}
%Our crystallographic group
%$\varPi=\angles{\bft_1,\bft_2,\bft_3, \bft_4^{},\alpha, \beta}$
%has a representation into $\aff(4)$:
%\begin{align*}
%\bft_1&=\left[
%\begin{array}{ccccc}
% 1 & 0 & \frac{1}{2} & 0 & \frac{1}{4} (102+2 m_1-3 m_2) \\
% 0 & 1 & 0 & 0 & 1 \\
%  0 & 0 & 1 & 0 & 0 \\
% 0 & 0 & 0 & 1 & 0 \\
 %0 & 0 & 0 & 0 & 1
%\end{array}
%\right],\\
%\bft_2&=\left[
%\begin{array}{ccccc}
% 1 & -\frac{1}{2} & 0 & 0 & -\frac{3}{8} (68+-m_1)+\frac{1}{2}m_2 \\
% 0 & 1 & 0 & 0 & 0 \\
% 0 & 0 & 1 & 0 & 1 \\
% 0 & 0 & 0 & 1 & 0 \\
%  0 & 0 & 0 & 0 & 1
%\end{array}
%\right],\\
%\bft_3&=\left[
%\begin{array}{ccccc}
% 1 & 0 & 0 & 0 & c_3 \\
% 0 & 17 & 12 & 0 & 0 \\
% 0 & 24 & 17 & 0 & 0 \\
% 0 & 0 & 0 & 1 & \ln \left(17+12\sqrt{2}\right) \\
% 0 & 0 & 0 & 0 & 1
%\end{array}
% \right],\\
%\bft_4&=\left[
%\begin{array}{ccccc}
% 1 & 0 & 0 & 0 & 1 \\
% 0 & 1 & 0 & 0 & 0 \\
% 0 & 0 & 1 & 0 & 0 \\
% 0 & 0 & 0 & 1 & 0 \\
% 0 & 0 & 0 & 0 & 1
%\end{array}
%\right],\\
%(a,A)&=\left[
%\begin{array}{ccccc}
% -1 & -\frac{a_2}{2} & 
%\\
 %0 & 1 & 0 & 0 & a_1 \\
% 0 & 0 & -1 & 0 & a_2 \\
% 0 & 0 & 0 & -1 & 0 \\
% 0 & 0 & 0 & 0 & 1
%\end{array}
%\right],\\
%(b,B)&=\left[
%\begin{array}{ccccc}
% 1 & 0 & 0 & 0 & {b_4}+\frac{c_3}{2} \\
% 0 & -3 & -2 & 0 & 0 \\
% 0 & -4 & -3 & 0 & 0 \\
% 0 & 0 & 0 & 1 & \frac{1}{2} \ln \left(17+12
%   \sqrt{2}\right) \\
% 0 & 0 & 0 & 0 & 1 \\
%\end{array}
%\right].
%====\end{align*}

\noindent
$\coker(I+K)$ is $\bbz_4\x\bbz_2$, and we have
\[
\xy{v_1}{v_2}
=
\xy{\frac{i}{4}}{\frac{j}{2}},\quad i=0,1,2,3;\ j=0,1.
\]
From $(I+\varphi(\alphabar))\bfa=\bzero$, we have
\begin{equation}
\label{example6bi-a}
\bfa=\xy{a_1}{a_2}
=\xy{\frac{i}{2}}{0}, \xy{0}{\frac{j}{2}},\quad i,j=0,1.
\end{equation}
For any $\bfv\in\coker(I+K)$,
\begin{align*}
(\delta\bfv)(A)&=(I-\wa)\bfv=\xy{0}{2 v_2}\equiv\xy{0}{0}.
\end{align*}
Therefore, $H^1(\angles{A};\coker(I+K))=\bbz_2\x\bbz_2$
(quotient of  $\bbz_4\x\bbz_2$)
is generated by $\bfa$ in (\ref{example6bi-a}) above.
Our group $\varPi$ has presentation
\begin{align*}
&[\bft_1,\bft_2]=\bft_4,\
[\bft_i,\bft_4]=1,\text{ for } i=1,2,3,\\% [\bft_2,\bft_4]=1,\ [\bft_3,\bft_4]=1,\\
% \text{ and\ $\bft_4$ is central},
&\bft_3\bft_1\bft_3\inv=\bft_1^{17} \bft_2^{24} \bft_4^{m_1}, \
\bft_3\bft_2\bft_3\inv=\bft_1^{12} \bft_2^{17}\bft_4^{m_2},\\
&\alpha\bft_1\alpha\inv=\bft_1^{}\bft_4^{-51-a_2-m_1+\frac32 m_2},\
\alpha\bft_2\alpha\inv=\bft_2^{-1}\bft_4^{-a_1},\\
&\alpha\bft_3\alpha\inv=\bft_1^{-16 a_1+12 a_2}\bft_2^{8(3 a_1-2 a_2)}\bft_3^{-1}
\\
&\hskip38pt\cdot\bft_4^{
204 a_1^2+a_1 (-272 a_2+17 (60+m_1)-24 m_2)+a_2 (-714+90 a_2-12 m_1+17 m_2)},\\
&\alpha\bft_4\alpha\inv=\bft_4^{-1},\\
&\beta\bft_1\beta\inv=\bft_1^{-3}\bft_2^{-4}\bft_4^{-6+\frac12 m_1-m_2},\
\beta\bft_2\beta\inv=\bft_1^{-2}\bft_2^{-3}\bft_4^{\frac12(-108-m_1+m_2)},\\
&\beta\bft_3\beta\inv=\bft_3^{},\
\beta\bft_4\beta\inv=\bft_4^{},\\
&\alpha^2=\bft_1^{2 a_1} \bft_4^{-a_1 (51+a_2+m_1)+\frac32  a_1 m_2}\\
&\beta^2=\bft_3^{}\bft_4^{2 b_4},\\
&[\alpha,\beta]
=\bft_1^{4 a_1-2 a_2}\bft_2^{-4(a_1-a_2)}\bft_3^{-1}
\bft_4^{
6 a_1^2-2 b_4+\frac12 a_2 (306+10 a_2+5 m_1-7 m_2)
+a_1 (-204-12 a_2-\frac72 m_1+5 m_2)}.
\end{align*}

Of the 8 groups, seven groups have torsion, and one is torsion-free.
\begin{alignat*}{2}
\xy{a_1}{a_2}&=\xy{0}{0},\xy{0}{\tfrac{1}{2}}:\ &&\quad
\alpha^2=\id,\\
\xy{a_1}{a_2}&=\xy{\tfrac12}{0}:
&&\hskip12pt \text{Both $\varPi$ and $Q$ are torsion-free},\\
\xy{a_1}{a_2}&=\xy{\tfrac12}{\tfrac14}:
&&\quad(\bft_2^{}\bft_3^{}\alpha)^2=\id,\\
\xy{a_1}{a_2}&=\xy{\tfrac12}{\frac12}:
&&\quad(\bft_1^{-1}\bft_3^{}\alpha)^2=\id.\\
\end{alignat*}
Note that $\xy{a_1}{a_2}=\xy{\tfrac12}{0}$ is the only case
that satisfies $a_1=\frac12$ and 
$a_2\not\equiv \frac{(k_{11}-1)(2 m+1)}{2 k_{12}}=\frac{2 m+1}{2}\equiv\frac12$,
as stated in Theorem \ref{ClassificationSolof-geometry}.
\end{example}
\bigskip

\begin{example}[\gr{7\ii}]
\label{maxhol}
This example has maximal holonomy group $D_4$, but it does not contain
all the possible holonomy actions. For example,
groups of type \gr{6b} or \gr{6b\ii} are not subgroups of this group.
Let $\Phi=\bbz_4\rx\bbz_2=\angles{A,B}$, where
$$
A=\matNpp,\ B=\matPpm,
$$
and
\begin{align*}
\alpha&=(\bft_1^{a_1}\bft_2^{a_2},A),\\
\beta&=(\bft_3^{\frac12}\bft_4^{b_4},B).
\end{align*}
Our $\cals$ is of the form
\[
\cals=n K+I,
\]
where
\[
K=\left[\begin{matrix}
k_{11} &k_{12}\\ k_{21} & k_{22}
\end{matrix}\right]
\]
with $\det(K)=-1$ and  $\tr(K)=n\not=0$.
Now for $\varphi(\alphabar)$, we have two choices. We can take $\wa=A$
or $\wa=\matPpm$.
Since the former yields always a torsion group, we take the latter case.
Therefore we consider case \gr{7\ii}, $k_{11}=k_{22}$.

As an example, we take
\[
K=
\left[\begin{matrix}
1&2\\ 1&1
\end{matrix}\right],\quad
\text{therefore\ }
n=k_{11}+k_{22}=2,\quad
\cals=n K+I=
\left[\begin{matrix}
3&4\\ 2&3
\end{matrix}\right].
\]
Then $\lambda=3+2 \sqrt{2}$, and with
\[
P=
\left[
\begin{array}{rr}
 -\frac{1}{\sqrt[4]{2^3}} &\frac{1}{\sqrt[4]{2}}\\
 -\frac{1}{\sqrt[4]{2^3}} &-\frac{1}{\sqrt[4]{2}}
\end{array}
\right],
\]
the equation in Lemma \ref{determine-TT} yields
\begin{align*}
c_1&=\frac18 (-12+\sqrt{2}+4 m_1-4 m_2)\\
c_2&=\frac14 (-\sqrt{2}-4 m_1+2 m_2).
\end{align*}
Recall we  can take $c_3=0$ by Theorem \ref{ClassificationSolof-geometry}.
Our crystallographic group
$\varPi=\angles{\bft_1,\bft_2,\bft_3, \bft_4^{},\alpha, \beta}$
has a representation into $\aff(4)$:
\begin{align*}
\bft_1&=\left[
\begin{array}{ccccc}
 1 & 0 & \frac{1}{2} & 0 & \frac{1}{2} ({m_1}-{m_2}-3) \\
 0 & 1 & 0 & 0 & 1 \\
 0 & 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 0 & 1
\end{array}
\right],\\
\bft_2&=\left[
\begin{array}{ccccc}
 1 & -\frac{1}{2} & 0 & 0 & \frac{1}{2} ({m_2}-2 {m_1}) \\
 0 & 1 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 & 1 \\
 0 & 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 0 & 1
\end{array}
\right],\\
\bft_3&=\left[
\begin{array}{ccccc}
 1 & 0 & 0 & 0 & 0 \\
 0 & 3 & 4 & 0 & 0 \\
 0 & 2 & 3 & 0 & 0 \\
 0 & 0 & 0 & 1 & \ln \left(3+2 \sqrt{2}\right)\\
 0 & 0 & 0 & 0 & 1
\end{array}
\right]\  \text{(because $c_3=0$)},\\
\bft_4&=\left[
\begin{array}{ccccc}
 1 & 0 & 0 & 0 & 1 \\
 0 & 1 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 0 & 1
\end{array}
\right],\\
(a,A)&=\left[
\begin{array}{ccccc}
 -1 & -\frac{a_2}{2} & -\frac{a_1}{2} &
   0 & \frac{1}{2} (a_1
   (a_2+m_1-m_2-3)+a_2
   (m_2-2 m_1)) \\
 0 & 1 & 0 & 0 & a_1 \\
 0 & 0 & -1 & 0 & a_2 \\
 0 & 0 & 0 & -1 & 0 \\
 0 & 0 & 0 & 0 & 1 \\
\end{array}
\right],\\
(b,B)&=\left[
\begin{array}{ccccc}
 -1 & 0&0&0&b_4\\
 0 & -1 & -2 & 0 & 0 \\
 0 & -1 & -1 & 0 & 0 \\
 0 & 0 & 0 & 1 & \frac{1}{2} \log \left(3+2 \sqrt{2}\right) \\
 0 & 0 & 0 & 0 & 1 \\
\end{array}
\right].
\end{align*}

\noindent
We have
\[
\coker(I-\cals)=(\bbz_2)^2
=
\left\{\tfrac12\xy{0}{0},
\tfrac12\xy{0}{1},
\tfrac12\xy{1}{0},
\tfrac12\xy{1}{1}
\right\}.
\]
Now
\[
\varphi(\alphabar)=\matPpm,\quad
\varphi(\betabar)=-K,
\]
yields
\[
I+\ka=
\left[\begin{matrix}
2&0\\0&0
\end{matrix}\right],\
I+\kb=
\left[\begin{matrix}
0&-2\\-1&0
\end{matrix}\right].
\]
Then
$(I+\varphi(\alphabar))\bfa\equiv\bfzero$ yields $2 a_1\equiv 0$, which is not
a new condition.
We have 4 choices for $\bfa$,
\[
\xy{a_1}{a_2}
=\tfrac12\xy{0}{0},
\tfrac12\xy{0}{1},
\tfrac12\xy{1}{0},
\tfrac12\xy{1}{1}.
\]
The coboundary $\im(I-\vara)$ yields the trivial group,
and hence there are 4 non-equivalent groups $\varPi$.
The group $\varPi$ has a presentation
\begin{align*}
\hskip24pt
&[\bft_1,\bft_2]=\bft_4,\
[\bft_i,\bft_4]=1\ (i=1,2,3),\\
% \text{ and\ $\bft_4$ is central},
&\bft_3\bft_1\bft_3\inv=\bft_1^{3} \bft_2^{2} \bft_4^{m_1}, \
\bft_3\bft_2\bft_3\inv=\bft_1^{4} \bft_2^{3}\bft_4^{m_2},\\
&\alpha\bft_1\alpha\inv=\bft_1^{}\bft_4^{3-a_2-m_1+m_2},\
\alpha\bft_2\alpha\inv=\bft_2^{-1}\bft_4^{-a_1},\\
&\alpha\bft_3\alpha\inv=\bft_1^{-2 a_1+4 a_2}\bft_2^{2 (a_1-a_2)}\bft_3^{-1}
\bft_4^{3 a_1^2-a_1 (3+6 a_2-3 m_1+2 m_2)+a_2 (6+2 a_2-4 m_1+3 m_2)},\\
&\alpha\bft_4\alpha\inv=\bft_4^{-1},\\
&\beta\bft_1\beta\inv=\bft_1^{-1}\bft_2^{-1}\bft_4^{\frac12(-1-2 m_1+m_2)},\
\beta\bft_2\beta\inv=\bft_1^{-2}\bft_2^{-1}\bft_4^{-4+m_1-m_2},\\
&\beta\bft_3\beta\inv=\bft_3^{},\
\beta\bft_4\beta\inv=\bft_4^{-1},\\
&\alpha^2=\bft_1^{2 a_1}\bft_4^{-a_1 (-3+a_2+m_1-m_2)},\\
&\beta^2=\bft_3^{},\\
&(\alpha \beta)^4=\bft_4^{
-4 b_4+a_1^2+ 4 a_1 a_2 +2 a_2^2 -2 a_1(3-m_1 +m_2)-2 a_2(2m_1-m_2)}.
%&(\alpha \beta)(\beta \alpha)=\bft_1^{4(a_1+a_2)} \bft_2^{-2(a_1+a_2)}
%\bft_3^{-1} \bft_4^{(a_1 + a_2) (6 + 3 a_1 + 2 a_2 - 4 m_1 + 3 m_2)}
\end{align*}
Out of these 4 groups (by the choice of $a_i$'s),
only one group can be torsion-free, and 3 groups have torsion.
%Thus out of the 4 classes in $H^1(\Phi; \coker(I-\cals))$,
%two of the non-trivial classes always have torsion.
\begin{alignat*}{2}
\xy{a_1}{a_2}&=\xy{0}{0},\xy{0}{\tfrac12}:\ &&\quad
\alpha^2=\id.\\
\xy{a_1}{a_2}&=\xy{\tfrac12}{\tfrac12}:
&&\quad\Big(\bft_2^{-1}(\alpha\beta)^2\alpha\Big)^2=\id\\
\xy{a_1}{a_2}&=\xy{\frac12}{0}: &&a_2 = - \frac{k_{21}+1}{2k_{11}} \equiv 0\quad
\end{alignat*}
When $m_i=0$, $a_1=\frac 1 2$, $a_2=0$, $q=4$ (minimum), 
$b_4$ takes values $\frac{j}{16}$, $0\leq j\le 4$.
When $b_4=0$ or $\frac{2}{16}$, $\varPi$ is torsion-free.
When $b_4=\frac{1}{16}$ or $\frac{3}{16}$, $\varPi$ has torsion.
\end{example}

\section{Crystallographic Groups of $\Solr$}
\label{hithere}

Here we provide a classification of the crystallographic groups of $\Solr$.
Given a crystallographic group
$\varPi \subset (\Solr) \rx (D_4 \x \bbz_2),$ $\wt \Gamma = \varPi \cap (\Solr)$ is a lattice of $\Solr$
with finite quotient group $\Psi \subseteq D_4 \x \bbz_2$. We have the short exact sequence
\[1 \ra \wt \Gamma \ra \varPi \ra \Psi \ra 1.\]


\step[{\bf Lattices of $\Solr$}]

In the cohomology classification of lattices in Theorem \ref{lattice-part-construction}, the $\Solr$ case
corresponds precisely to when $q=0$. We will see that $\lgamsigz$ embeds as a lattice of $\Solr$.

An $\cals\in\sltz$ with $\tr(\cals)>2$ produces $P$ and $\Delta$. Using $P$ and $\Delta$, we embed $\bbz^2 \rx \bbz$ into $\Sol$ as in assignment (\ref{def-phi}) in Subsection \ref{latsol}, which we now lift
to $\Solr$.
Recall the definition of $\Solr$ as $\bbr^3 \rx_{\phi(u)} \bbr$, where 
\[\phi(u)=
\left[ \begin{matrix} e^{-u} & 0 & 0\\ 0 & e^{u} & 0\\ 0 & 0 & 1 \end{matrix}\right].\]
The only ambiguities are the central slots. So let
\begin{align}
\begin{split}
\bfe_1&=\left(\left[\begin{matrix} 1\\ 0 \end{matrix} \right],0\right)
\longmapsto
(P\bfe_1,0)\longmapsto
\bft_1=
\left(  \left[\begin{matrix}
p_{11} \\
p_{21}\\
c_1\end{matrix}
\right], 0 \right)
,\\
\bfe_2&=\left(\left[\begin{matrix} 0\\ 1 \end{matrix} \right],0\right)
\longmapsto
(P\bfe_2,0)\longmapsto
\bft_2=
\left(  \left[\begin{matrix}
p_{12} \\
p_{22}\\
c_2\end{matrix}
\right], 0 \right)
,\\
\bfe_3&
=\left(\left[\begin{matrix} 0\\ 0 \end{matrix} \right],1\right)
\longmapsto
(0,\ln(\lambda))\longmapsto
\bft_3=
\left(  \left[\begin{matrix}
0 \\
0\\
c_3 \end{matrix}
\right], \ln(\lambda) \right),\\
&\kern6pt\phantom{=\left(\left[\begin{matrix} 0\\
1 \end{matrix} \right],0\right)
\longmapsto}
\bft_4=
\left(  \left[\begin{matrix}
0 \\
0\\
1 \end{matrix}
\right], 0 \right),
\end{split}
\end{align}
where $c_i$'s will be determined later. Then $[\bft_1,\bft_2]=e$
(regardless the values of $c_i$'s).

\step[{\bf Warning}] In this notation, the central factor of $\Solr$ goes to the 3rd slot of $\bbr^3 \subset \bbr^3 \rx \bbr.$ That is,

\[\Solr \ni (e, s) \mapsto \left(  \left[\begin{matrix}
0 \\
0\\
s \end{matrix}
\right], 0 \right) \in \bbr^3 \rx \bbr.\]

\begin{lemma}
\label{determine-TT2}
%Let $\cals\in\sltz$ with $\tr(\cals)>2$, eigenvalue $\lambda$;
%let $P$ be a matrix diagonalizing $\cals$.
For any integers $m_1,m_2$,
there exist unique $c_1, c_2$ for which $\{\bft_1,\bft_2,\bft_3,\bft_4\}$
forms a group $\lgamsigz$ with the presentation
\begin{align*}
\lgamsigz=\langle
\bft_1,\bft_2,\bft_3,\bft_4 |\
[\bft_1,\bft_2]&=\bft^{0}_4 =e,\ \text{$\bft_4$ is central,}\\
\bft_3 \bft_1 \bft_3\inv
&= \bft_1^{\sigma_{11}} \bft_2^{\sigma_{21}} \bft_4^{m_1},\\
\bft_3 \bft_2 \bft_3\inv&= \bft_1^{\sigma_{12}} \bft_2^{\sigma_{22}}
\bft_4^{m_2}\rangle.
\end{align*}

Consequently, $\lgamsigz$ is solvable and contains $\bbz^3 = \angles{\bft_1, \bft_2, \bft_4}$ as its discrete nil-radical, a lattice of $\bbr^3$ in $\bbr^3 \subset \bbr^3 \rx \bbr$.
\end{lemma}

\begin{proof}
We only need to verify the last two equalities. But they
become a system of  equations on $c_i$'s
\begin{align}
\label{c1c2-equation2}
\begin{split}
(1-\sigma_{11})c_1 -\sigma_{21} c_2  &=m_1\\
-\sigma_{12} c_1+ (1-\sigma_{22})c_2 &=m_2
\end{split}
\end{align}
Since $I-\cals$ is non-singular, there exists a unique solution for
$c_1, c_2$.
\end{proof}

\begin{remark}
(1) There are 1-parameter family of $P$'s diagonalizing $\cals$ with
  $\det(P)=1$.
In the above lemma, the $c_i$'s are independent of choice of $P$  because
equation (\ref{c1c2-equation2}) has coefficients only from the matrix $\cals$..

(2) The equation (\ref{c1c2-equation2}) also shows the cohomology
classification.
Suppose $\{c_1,c_2\}$ and $\{c'_1,c'_2\}$ are solutions for the equations
with $\{m_1,m_2\}$ and $\{m'_1,m'_2\}$, respectively.
Then $(c'_1-c_1,c'_2-c_2)\in \bbz^2$ if and only if
$(m'_1-m_1,m'_2-m_2)\in \text{Coker}(\cals^T-I) \cong \coker(\cals-I) $.
This happens if and only if
$\lgamsigz=\wt\Gamma_{(\cals;0,m'_1,m'_2)}$.

(3) Notice that $c_3$ does not show up in the presentation of the lattice
$\lgamsigz$. This implies that $c_3$ can be changed without affecting the
isomorphism type of the lattice.
\end{remark}

\begin{notation}[Standard lattice]
\label{standard-lattice2}
The lattice generated by
\[
\bft_1=
\left(  \left[\begin{matrix}
p_{11} \\
p_{21}\\
c_1\end{matrix}
\right], 0 \right)
,\\
\bft_2=
\left(  \left[\begin{matrix}
p_{12} \\
p_{22}\\
c_2\end{matrix}
\right], 0 \right),\\
\bft_3=
\left(  \left[\begin{matrix}
0 \\
0\\
c_3 \end{matrix}
\right], \ln(\lambda) \right),\\
\bft_4=
\left(  \left[\begin{matrix}
0 \\
0\\
1 \end{matrix}
\right], 0 \right)
\]
{\bf with $c_3=0$}, is called a \emph{standard lattice} of $\Solr$.
\end{notation}

\noindent
Any lattice of $\Solr$ is isomorphic to a standard lattice.

\step[{\bf Classification of $\Solr$ Crystallographic Groups}]

Given a crystallographic group $\varPi \subset \isom(\Solr) = (\Solr) \rx (D_4 \x \bbz_2)$, we can conjugate $\varPi$ in $\aff(\Solr)$ so that
$\lgamsigz$ satisfies criteria of Notation \ref{standard-lattice2}, \emph{except possibly} for the condition that $c_3=0$. Indeed, like the case of $\Solof$,
 we will see that $c_3=0$ cannot always be achieved when we have fixed our maximal compact subgroup of $\aut(\Solr)$ to be $D_4 \x \bbz_2$. Note that
 $\lgamsigz$ projects to a standard lattice of $\Sol$.
 
 \begin{notation}
 We will let $\tau$ denote the automorphism of $\Solr$
  given by a reflection in the central direction. In the description of $\Solr$ as $\bbr^3 \rx \bbr$, $(\bfx, t) \in \bbr^3 \rx \bbr$ is mapped to $(T\bfx, t) \in \bbr^3 \rx \bbr$ under $\tau$, where
  \[ T = \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix} \right]. \]
 \end{notation}
 
 \begin{proposition}
 \label{solrShape}
 \noindent
 {\rm{(1)}} Any crystallographic group $\varPi$ of $\Solr$ can be conjugated into
 $(\Solr) \rx (D_4 \x \bbz_2)$ in such way that
 \[
 \tgamsig = \varPi \cap (\Solr)=\angles{\bft_1,\bft_2,\bft_3,\bft_4}
 \]
 where
 \[
 \bft_1=
 \left(  \left[\begin{matrix}
 p_{11} \\
 p_{21}\\
 c_1\end{matrix}
 \right], 0 \right),\
 \bft_2=
\left(  \left[\begin{matrix}
p_{12} \\
p_{22}\\
c_2\end{matrix}
\right], 0 \right),\
 \bft_3=
 \left(  \left[\begin{matrix}
 0 \\
 0\\
 c_3 \end{matrix}
 \right], \ln(\lambda) \right),\
 \bft_4=
\left(  \left[\begin{matrix}
0 \\
0\\
1 \end{matrix}
\right], 0 \right).
 \]
 
 \noindent
 {\rm{(2)}} The holonomy group $\Psi$ is generated by at most three elements of $ D_4 \x \bbz_2$, and therefore $\varPi$ is generated by $ \angles{\bft_1, \bft_2, \bft_3, \bft_4}$ and at most three isometries of the form $(\bft_1^{b_1} \bft_2^{b_2} \bft_3^{b_3} \bft_4^{{b_4}} , B)$ for $B \in D_4 \x \bbz_2$ and real numbers $b_i$.
 \end{proposition}
 \begin{proof}
 Let $\rho: \Solr \ra \Sol$ denote projection. Given a lattice $\tgamsig$ of 
 $\Solr$, $\tgamsig$ must meet the center of $\Solr$ (the $\bbr$ factor of $\Solr$) in a lattice
 $\tgamsig \cap \calz(\Solr) \cong \bbz$. Let $\bft_4$ denote a generator
 of $\tgamsig \cap \bbr$. Then $\bft_4$ must be of the form
 \[\bft_4=
 \left(  \left[\begin{matrix}
 0 \\
 0\\
 \ell \end{matrix}
 \right], 0 \right).\]
Let $\hat {\ell}$ denote the automorphism
 of $\Solr$ defined by multiplication by ${\ell}$ on the center:
  \[\hat {\ell}:
  \left(  \left[\begin{matrix}
  x \\
  y\\
  t \end{matrix}
  \right], u \right) \lra \left(  \left[\begin{matrix}
    x \\
    y\\
    \ell t \end{matrix}
    \right], u \right).\]
After conjugating $\tgamsig$ by $(e, \hat {\ell})^{-1}$ in $\aff(\Solr)$, $\bft_4$ has the desired form. For the remaining generators, note that $\rho(\tgamsig)$ must project to a lattice of $\Sol$. The conjugations by elements of $\aff(\Sol)$ in Proposition \ref{lattice-a=0} also lift to conjugations by elements of $\aff(\Solr)$ (see Proposition \ref{autosolr}), so that $\bft_1, \bft_2$, and $\bft_3$ have the desired form. Therefore, we can assume that $\rho(\tgamsig)$ is a standard $\Sol$ lattice $\gamsig$.
Finally, observe that the conjugations in $\aff(\Solr)$ do not change the maximal compact subgroup $D_4 \x \bbz_2$.
 \qedhere
 \end{proof}
 
 
Similar to the $\Solof$ case, the quotient of $\varPi$ by
 $\calz(\tgamsig) = \angles{\bft_4}$, \[Q= \varPi / \angles{\bft_4} = \varPi / \bbz,\] is
 a finite extension of the $\Sol$ lattice $\gamsig$ by $\Psi$. In fact, we have the following commuting diagram.
\[
\CD
@. 1 @. 1 @.  @.\\
@. @VVV @VVV @. @.\\
@. \bbz @= \bbz @. @.\\
@. @VVV @VVV @. @.\\
1 @>>> {\wt \Gamma_\cals} @>>> \varPi @>>> \Psi @>>> 1\\
@. @VVV @VVV @| @.\\
1 @>>> \gamsig @>>> Q @>>> \Psi @>>> 1\\
@. @VVV @VVV @. @.\\
@. 1 @. 1 @.  @.\\
\endCD
\]

Recall that we have let $\tau$ denote the generator of the $\bbz_2$ factor in $D_4 \x \bbz_2$. The inclusion $\Psi \hookrightarrow D_4 \x \bbz_2$ lifts an inclusion of $\Phi \hookrightarrow D_4$,
where $\Phi$ is one of the holonomy groups in List (\ref{bbzphi-list}) in Subsection \ref{classSol}. That is, there is a $\Phi \subseteq D_4$ for which the diagram commutes:
\[
\CD
\Psi @>>> D_4 \x \bbz_2\\
@VVV @VV{/ \bbz_2 = \angles{\tau}}V\\
\Phi @>>> D_4\\
\endCD
\]

Crystallographic groups of $\Solr$ do not in general project to crystallographic groups of $\Sol$, as $\tau$ projects
to the trivial automorphism of $\Sol$. Using the notation of our
$\Sol$ crystallographic group classification, $Q$ fits the 
short exact sequence
\[1 \ra \bbz^2 = \angles{\bft_1, \bft_2} \ra Q \ra \bbzpsi \ra 1,\] to which
Theorem \ref{H1-cokernel-theorem} applies. The group $\bbzpsi$ fits the short
exact sequence
\[1 \ra \bbz = \angles{\bft_3} \ra \bbzpsi \ra \Psi \ra 1.\]
Similar to the $\Sol$ case, $\bbzpsi$ is a subgroup of  $\frac{1}{2}\bbz \rx (D_4 \x \bbz_2)$, where $\frac{1}{2}\bbz = \angles{\bft_3^{\frac12}}$. That is, any element of $\bbzpsi$ can be expressed as $(\bft_3^{a_3},A)$, where $a_3=0$ or $\frac12$, and $A\in  (D_4 \x \bbz_2)$.
We need to classify all such $\bbzpsi$ and abstract kernels $\varphi: \bbzpsi \ra \gltz$.
The inclusion $\bbzpsi \hookrightarrow \frac{1}{2}\bbz \rx (D_4 \x \bbz_2)$ lifts an inclusion of $\bbzphi \hookrightarrow \frac{1}{2}\bbz \rx D_4$, where $\bbzphi$ is a group of one of the types from list (\ref{bbzphi-list}) in Subsection \ref{classSol}.
That is, the following diagram commutes (note that  $\bbz_2 = \angles{\tau}$ is central in $\frac{1}{2}\bbz \rx (D_4 \x \bbz_2)$):
\[
\CD
\bbzpsi @>>> \frac{1}{2}\bbz \rx (D_4 \x \bbz_2)\\
@VVV @VV{/ \bbz_2 = \angles{\tau}}V\\
\bbzphi @>>> \frac{1}{2}\bbz \rx D_4\\
\endCD
\]


There are many more such groups in the $\Solr$ case than
in the $\Sol$ case.
For example, notice that $\bbzpsi$ may contain an element of the form $\bar \gamma = (\bft_3^{0}, \tau)$ with $\varphi(\bar \gamma) = I$. Such a $\bbzpsi$ did not arise in the $\Sol$ case.

Here is our methodology 
for classifying the $\Solr$ crystallographic groups. Essentially,
we classify all $Q$, then examine how to lift them to $\Solr$.
 
\noindent
{\rm(1)} Classify the possible abstract kernels $\varphi: \bbzpsi \ra \gltz$ with $\varphi(\bft_3) = \cals$. This
fixes the holonomy group and the exponents on $\bft_3$. 

\noindent
{\rm(2)} Apply Theorem \ref{H1-cokernel-theorem} to obtain all $Q$. This fixes
the exponents of $\bft_1, \bft_2$, as elements of $H^2(\bbzpsi; \bbz^2) \cong H^1(\Psi;\coker(I-\cals))$.

\noindent
{\rm(3)} Classify lifts of $\gamsig$ to lattices of $\wt \Gamma_\cals$ of $\Solr$ (Theorem \ref{lattice-part-construction}).

\noindent
{\rm(4)} Determine which $\Solr$ lattices  $\wt \Gamma_\cals$ admit
an abstract kernel $\Psi \ra \out(\wt \Gamma_\cals)$, induced by conjugation by lifts of elements of $Q$.

\noindent
{\rm(5)} Once an abstract kernel $\Psi \ra \out(\wt \Gamma_\cals)$ is fixed, the equivalence classes of extensions
\[ 1 \ra \wt \Gamma_\cals \ra \varPi \ra \Psi \ra 1 \] are classified by $H^2(\Psi; \calz(\wt \Gamma_\cals)) = H^2(\Psi; \bbz)$, fixing the exponents on $\bft_4$.



\begin{lemma}[cf. Lemma \ref{conj-in-t4}]
\label{conj-in-t4Solr}
If $(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3}\bft_4^{a_4},A\tau) \in \varPi$ ($A \in D_4$, $\tau$ is reflection on center of $\Solr$), then by conjugation, $a_4$ can be made 0.
\end{lemma}

\begin{proof}
Conjugation by
$\bft_4^{-\frac{a_4}{2}}$ fixes the lattice $\lgamsigz$,
and moves $(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3}\bft_4^{{a_4}},A)$ to
$(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3},A)$.
\end{proof}


\begin{lemma}[cf. Lemma \ref{PPM-A30}]
\label{PPM-A302}
Let \[A=
\left[\begin{matrix}
1 &0 \\ 0& -1
\end{matrix}\right].\]
If $\bbzpsi$ contains an element of the form $\alphabar = (\bft_3^{0}, A)$ or $(\bft_3^{0}, A\tau)$, then $\bbzpsi$ cannot arise as the quotient of a $\Solr$ crystallographic group.
\end{lemma}
\begin{proof}
The matrix
\[
\varphi(\bft_3)\varphi(\alphabar)= \cals \varphi(\alphabar) = P \inv \Delta P {P \inv} A P = P \inv \Delta A P
\]
must be integral. The sum of its eigenvalues, $\frac{1}{\lambda} - \lambda$, must
be an integer. On the other hand, since $\frac{1}{\lambda} + \lambda$ is an integer,
we infer that $2{\lambda}$ is an integer, so that $\lambda$ is rational, which is a contradiction. A similar argument works for $-A$ as well.
\qedhere
\end{proof}

\step[{\bf Existence of $\Psi \ra \out(\wt\Gamma_\cals)$}]
\label{existence}

Given a $\Sol$ crystallographic group $Q$ with abstract kernel
$\Psi \ra \gamsig$, we showed the \emph{existence} of a $\Solof$ lattice $\lgamsig$, 
lifting $\gamsig$,
together with induced abstract kernel $\Psi \ra \out(\lgamsig)$ (Proposition \ref{inflation}).

Given $\beta = (\bft_1^{b_1} \bft_2^{b_2} \bft_3^{b_3} \bft_4^{b_4}, B) \in (\Solr) \rx (D_4 \x \bbz_2)$, here we provide precise conditions for a $\Solr$ lattice $\tgamsig = \angles{\bft_1,\bft_2, \bft_3, \bft_4}$ to be invariant under conjugation by $\beta$, so that we obtain an abstract kernel $\Psi \ra  \out(\lgamsig)$.
The computations below are independent of $b_4$ and so we set $b_4=0$. The conditions on $c_i$ in the following tables are of course modulo the integers.

\noindent

\begin{table}[H]
\caption{Conjugation on $\bft_1$ and $\bft_2$ when holonomy action is trivial on center and $a_3=0$}
\label{t1}
\begin{center}
\begin{tabular}{|l| p{3.8cm} | p{3cm}|}
\hline
$\alpha = (\bft_1^{a_1}\bft_2^{a_2}\bft_3^{0}, A)$   & $\mu(\alpha)$ on $\bft_1, \bft_2$ & Conditions on $c_i$\\
\hline
$A = \matPmm$, $\varphi(\bar \alpha) = A$ & $\alpha \bft_1 \alpha \inv = \bft_1^{-1} \bft_4^{2c_1}$ \newline $\alpha \bft_2 \alpha \inv = \bft_2^{-1} \bft_4^{2c_2}$ & $c_1=0$ or $\frac12$ \newline $c_2=0$ or $\frac12$  \\
\hline
$A = \matNpm$, $\varphi(\bar \alpha) = A$ & $\alpha \bft_1 \alpha \inv = \bft_2^{-1} \bft_4^{c_1+c_2}$ \newline $\alpha \bft_2 \alpha \inv = \bft_1 \bft_4^{-c_1+c_2}$ & $(c_1,c_2)=(0,0)$ \newline $(c_1,c_2)=(\frac12,\frac12)$\\
\hline
$A = \matNpp$, $\varphi(\bar \alpha) = A$ & $\alpha \bft_1 \alpha \inv = \bft_2 \bft_4^{c_1-c_2}$ \newline $\alpha \bft_2 \alpha \inv = \bft_1 \bft_4^{-c_1+c_2}$ & $c_1-c_2 =0$\\
\hline
$A = \matNpp$, $\varphi(\bar \alpha) = \matPpm$ & $\alpha \bft_1 \alpha \inv = \bft_1$ \newline $\alpha \bft_2 \alpha \inv = \bft_2^{-1} \bft_4^{2 c_2}$ & $c_2 = 0$ or $\frac12$\\
\hline
\end{tabular}
\end{center}
\end{table}


\noindent

\begin{table}[H]
\caption{Conjugation on $\bft_1$ and $\bft_2$ when holonomy action is trivial on center and $a_3=\frac{1}{2}$}
\label{t3}
\begin{center}
\begin{tabular}{|p{2.3cm}| p{6cm} | p{4.3cm}|}
\hline
$\alpha = (\bft_3^{\frac 1 2}, A)$   & $\mu(\alpha)$ on $\bft_1, \bft_2$ & Conditions on $c_i$\\
\hline
 $\varphi(\bar \alpha) = -K$ & $\alpha \bft_1 \alpha \inv =  \bft_1^{-k_{11}} \bft_2^{-k_{21}} \bft_4^{c_1+k_{11}c_1+k_{21}c_2}$ \newline $\alpha \bft_2 \alpha \inv = \bft_1^{-k_{12}} \bft_2^{-k_{22}} \bft_4^{c_2+k_{12}c_1+k_{22}c_2}$ & $(c_1,c_2) \in  \coker(I+K^T)$  \\
\hline
\end{tabular}
\end{center}
\end{table}


\noindent

\begin{table}[H]
\caption{Conjugation on $\bft_1$ and $\bft_2$ when holonomy action is reflection on center and $a_3=0$}
\label{t2}
\begin{center}
\begin{tabular}{|l| p{3.7cm} | p{3cm}|}
\hline
$\alpha = (\bft_1^{a_1}\bft_2^{a_2}\bft_3^{0}, A\tau)$   & $\mu(\alpha)$ on $\bft_1, \bft_2$ & Conditions on $c_i$\\
\hline
$A = \matPmm$, $\varphi(\bar \alpha) = A$ & $\alpha \bft_1 \alpha \inv = \bft_1^{-1}$ \newline $\alpha \bft_2 \alpha \inv = \bft_2^{-1}$ & None \\
\hline
$A = \matNpm$, $\varphi(\bar \alpha) = A$ & $\alpha \bft_1 \alpha \inv = \bft_2^{-1} \bft_4^{-c_1+c_2}$ \newline $\alpha \bft_2 \alpha \inv = \bft_1 \bft_4^{-c_1-c_2}$ & $(c_1,c_2)=(0,0)$ \newline $(c_1,c_2)=(\frac12,\frac12)$\\
\hline
$A = \matNpp$, $\varphi(\bar \alpha) = A$ & $\alpha \bft_1 \alpha \inv = \bft_2 \bft_4^{-c_1-c_2}$ \newline $\alpha \bft_2 \alpha \inv = \bft_1 \bft_4^{-c_1-c_2}$ & $c_1+c_2 =0$\\
\hline
$A = \matNpp$, $\varphi(\bar \alpha) = \matPpm$ & $\alpha \bft_1 \alpha \inv = \bft_1  \bft_4^{-2 c_1}$ \newline $\alpha \bft_2 \alpha \inv = \bft_2^{-1}$ & $c_1 = 0$ or $\frac12$\\
\hline
$A = I$ & $\alpha \bft_1 \alpha \inv = \bft_1  \bft_4^{-2 c_1}$ \newline $\alpha \bft_2 \alpha \inv = \bft_2 \bft_4^{-2 c_2} $ & $c_1=0$ or $\frac12$ \newline $c_2=0$ or $\frac12$  \\
\hline
\end{tabular}
\end{center}
\end{table}


\noindent

\begin{table}[H]
\caption{Conjugation on $\bft_1$ and $\bft_2$ when holonomy action is reflection on center and $a_3=\frac{1}{2}$}
\label{t4}
\begin{center}
\begin{tabular}{|p{2.3cm}| p{6.2cm} | p{4.3cm}|}
\hline
$\alpha = (\bft_3^{\frac 1 2}, A\tau)$   & $\mu(\alpha)$ on $\bft_1, \bft_2$ & Conditions on $c_i$\\
\hline
$\varphi(\bar \alpha) = -K$ & $\alpha \bft_1 \alpha \inv = \bft_1^{-k_{11}} \bft_2^{-k_{21}} \bft_4^{-c_1+k_{11}c_1+k_{21}c_2}$ \newline $\alpha \bft_2 \alpha \inv = \bft_1^{-k_{12}} \bft_2^{-k_{22}} \bft_4^{-c_2+k_{12}c_1+k_{22}c_2}$ & $(c_1,c_2) \in \coker(K^T- I) $  \\
\hline
$\varphi(\bar \alpha) = K$ & $\alpha \bft_1 \alpha \inv = \bft_1^{k_{11}} \bft_2^{k_{21}} \bft_4^{-c_1-k_{11}c_1-k_{21}c_2}$ \newline $\alpha \bft_2 \alpha \inv = \bft_1^{k_{12}} \bft_2^{k_{22}} \bft_4^{-c_2-k_{12}c_1-k_{22}c_2}$ & $(c_1,c_2) \in \coker(I + K^T) $  \\
\hline
\end{tabular}
\end{center}
\end{table}


We now study the holonomy action on the $\bft_3$ generator of a $\Solr$ lattice; that is, we study $\alpha \bft_3\alpha\inv$. From the $\Sol$ relations, equation (\ref{Solpresentation}) in Subsection \ref{classSol}:
\[
\alpha\bft_3\alpha\inv=\bft_1^{w_1}\bft_2^{w_2}\bft_3^{\abar} \bft_4^{w_3},
\text{ where } {\xy{w_1}{w_2}=
\left(I-\cals^{\abar}\right)\xy{a_1}{a_2}}.
\]
Of course, $w_1$ and $w_2$ are integers; we need $w_3$ to be an integer.

\noindent

\begin{table}[H]
\caption{Conjugation on $\bft_3$ when holonomy action is trivial on center and $a_3=0$}
\label{t5}
\begin{center}
\begin{tabular}{|p{3.1cm}| p{6cm} | p{3.5cm}|}
\hline
$\alpha = (\bft_1^{a_1}\bft_2^{a_2}\bft_3^{0}, A)$   & $\mu(\alpha)$ on $\bft_3$ & Conditions on $c_i$\\
\hline
$\abar=+1$ & $\alpha \bft_3 \alpha \inv = \bft_1^{w_1} \bft_2^{w_2} \bft_3 \bft_4^{-c_1 w_1 - c_2 w_2}$ & $c_1 w_1 + c_2 w_2 = 0$ \\
\hline
$\abar=-1$ & $\alpha \bft_3 \alpha \inv =\bft_1^{w_1} \bft_2^{w_2} \bft_3^{-1} \bft_4^{2c_3 -c_1 w_1 - c_2 w_2}$ & $c_3 = \frac{c_1 w_1 + c_2 w_2}{2}$ or \newline  $\frac{c_1 w_1 + c_2 w_2}{2} + \frac12$\\
\hline
\end{tabular}
\end{center}
\end{table}


\noindent

\begin{table}[H]
\caption{Conjugation on $\bft_3$ when holonomy action is trivial on center and $a_3=\frac{1}{2}$}
\label{t7}
\begin{center}
\begin{tabular}{|p{4cm}| p{4cm} | p{3cm}|}
\hline
$\alpha = (\bft_3^{\frac 1 2}, A)$   & $\mu(\alpha)$ on $\bft_3$ & Conditions on $c_i$\\
\hline
 $\varphi(\bar \alpha) = -K, (\abar=+1)$ & $\alpha \bft_3 \alpha \inv = \bft_3$   & None  \\
\hline
\end{tabular}
\end{center}
\end{table}


\noindent

\begin{table}[H]
\caption{Conjugation on $\bft_3$ when holonomy action is reflection on center and $a_3=0$}
\label{t6}
\begin{center}
\begin{tabular}{|p{3.5cm}| p{6cm} | p{3.5cm}|}
\hline
$\alpha = (\bft_1^{a_1}\bft_2^{a_2}\bft_3^{0}, A\tau)$   & $\mu(\alpha)$ on $\bft_3$ & Conditions on $c_i$\\
\hline
$\abar=+1$ & $\alpha \bft_3 \alpha \inv =\bft_1^{w_1} \bft_2^{w_2} \bft_3 \bft_4^{-2c_3 -c_1 w_1 - c_2 w_2}$ & $c_3 = -\frac{c_1 w_1 + c_2 w_2}{2}$ or \newline  $-\frac{c_1 w_1 + c_2 w_2}{2} + \frac12$ \\
\hline
$\abar=-1$ & $\alpha \bft_3 \alpha \inv = \bft_1^{w_1} \bft_2^{w_2} \bft_3^{-1} \bft_4^{-c_1 w_1 - c_2 w_2}$ & $c_1 w_1 + c_2 w_2=0$\\
\hline
\end{tabular}
\end{center}
\end{table}



\noindent

\begin{table}[H]
\caption{Conjugation on $\bft_3$ when holonomy action is reflection on center and $a_3=\frac{1}{2}$}
\label{t8}
\begin{center}
\begin{tabular}{|p{4cm}| p{4cm} | p{3cm}|}
\hline
$\alpha = (\bft_3^{\frac 1 2}, A\tau)$   & $\mu(\alpha)$ on $\bft_3$ & Conditions on $c_i$\\
\hline
 $\varphi(\bar \alpha) = \pm K, (\abar=+1)$ & $\alpha \bft_3 \alpha \inv = \bft_3 \bft_4^{-2 c_3}$ & $c_3 = 0$ or $\frac12$  \\
\hline
\end{tabular}
\end{center}
\end{table}


\begin{remark}
When $c_1=c_2=c_3=0$, so that $\tgamsig =  \gamsig \x \bbz$, the above criteria on the $c_i$
are immediately satisfied.
\end{remark}

\begin{lemma}[cf. Theorem \ref{abstract-kernel-theorem}]
\label{solrc3}
Let $\varPi \subset (\Solr) \rx (D_4 \x \bbz_2)$ be a $\Solr$ crystallographic group,
with lattice embedded as in {\rm Proposition \ref{solrShape}}. If all elements of $\varPi$
are of the form $(\bfx, A) \in \varPi$ with $\abar=1$, or  $(\bfx, A\tau) \in \varPi$
with $\abar=-1$, then we can assume that $c_3=0$ (This is analogous to the $\Solof$ case when
$\abar \ahat=+1$ for all holonomy elements $A$).
\end{lemma}
\begin{proof}
Let $\hat {k}$ denote the automorphism
 of $\Solr$ defined as:
  \[\hat {k}:
  \left(  \left[\begin{matrix}
  x \\
  y\\
  t \end{matrix}
  \right], u \right) \lra \left(  \left[\begin{matrix}
    x \\
    y\\
     t + ku \end{matrix}
    \right], u \right).\]
Setting $k=\frac{c_3}{\ln(\lambda)}$, we see that conjugation by
$(e, \hat {k}) \inv \in \aff(\Solr)$ sets $c_3=0$, while fixing the
maximal compact subgroup $D_4 \x \bbz_2$.
\end{proof}


\step[{\bf Computation of Cyclic Powers}]

Before our main classification result, we provide computations of the cyclic powers of all possible $\alpha$. This will be useful for determining the exponents on $\bft_4$, as well as checking whether or not a crystallographic group has torsion.

\noindent

\begin{table}[H]
\caption{Cyclic powers when holonomy acts trivially on center and $a_3=0$}
\label{t9}
\begin{center}
\begin{tabular}{|p{6cm}| p{7cm} |}
\hline
$\alpha = (\bft_1^{a_1}\bft_2^{a_2}\bft_3^{0}\bft_4^{a_4}, A)$ & $\alpha^n \hspace{1cm} (n=\ord(A))$\\
\hline
$A = \matPmm$, $\varphi(\bar \alpha) = A$ & $\alpha^2 = \bft_4^{2(a_4+{a_1}{c_1}+{a_2}{c_2})}$ \\
\hline
$A = \matNpm$, $\varphi(\bar \alpha) = A$ & $\alpha^4 = \bft_4^{4(a_4+{a_1}{c_1}+{a_2}{c_2})}$\\
\hline
$A = \matNpp$, $\varphi(\bar \alpha) = A$ & $\alpha^2 = \bft_1^{a_1+a_2} \bft_2^{a_1+a_2} \bft_4^{2a_4+{a_1}{c_1}+{a_2}{c_2}-{a_2}{c_1}-{a_1}{c_2}}$ \\
\hline
$A = \matNpp$, $\varphi(\bar \alpha) = \matPpm$ & $\alpha^2 = \bft_1^{2a_1} \bft_4^{2(a_4+{a_2}{c_2})}$ \\
\hline
\end{tabular}
\end{center}
\end{table}


\noindent

\begin{table}[H]
\caption{Cyclic powers when holonomy acts trivially on center and $a_3=\frac{1}{2}$}
\label{t11}
\begin{center}
\begin{tabular}{|p{5cm}| p{7cm} |}
\hline
$\alpha = (\bft_3^{\frac 1 2} \bft_4^{a_4}, A)$   & $\alpha^n \hspace{1cm} (n=\ord(A))$\\
\hline
 $\varphi(\bar \alpha) = -K, (\bar A = +1)$ & $\alpha^2 = \bft_3 \bft_4^{2a_4}$ \\
\hline
\end{tabular}
\end{center}
\end{table}


\noindent

\begin{table}[H]
\caption{Cyclic powers when holonomy acts as reflection on center and $a_3=0$}
\label{t10}
\begin{center}
\begin{tabular}{|p{6cm}| p{7cm} |}
\hline
$\alpha = (\bft_1^{a_1}\bft_2^{a_2}\bft_3^{0}\bft_4^{a_4}, A\tau)$ & $\alpha^n \hspace{1cm} (n=\ord(A))$\\
\hline
$A = \matPmm$, $\varphi(\bar \alpha) = A$ & $\alpha^2 = e$ \\
\hline
$A = \matNpm$, $\varphi(\bar \alpha) = A$ & $\alpha^4 = e$\\
\hline
$A = \matNpp$, $\varphi(\bar \alpha) = A$ & $\alpha^2 = \bft_1^{a_1+a_2} \bft_2^{a_1+a_2} \bft_4^{-({a_1}{c_1}+{a_2}{c_2}+{a_2}{c_1}+{a_1}{c_2})}$ \\
\hline
$A = \matNpp$, $\varphi(\bar \alpha) = \matPpm$ & $\alpha^2 = \bft_1^{2a_1} \bft_4^{-2 a_1 c_1}$ \\
\hline
$A = I$, & $\alpha^2 = \bft_1^{2a_1} \bft_2^{2a_2} \bft_4^{-2 ({a_1}{c_1}+{a_2}{c_2})}$ \\
\hline
\end{tabular}
\end{center}
\end{table}


\noindent

\begin{table}[H]
\caption{Cyclic powers when holonomy acts as reflection on center and $a_3=\frac{1}{2}$}
\label{t111}
\begin{center}
\begin{tabular}{|p{5cm}| p{7cm} |}
\hline
$\alpha = (\bft_3^{\frac 1 2} \bft_4^{a_4}, A\tau)$   & $\alpha^n \hspace{1cm} (n=\ord(A))$\\
\hline
 $\varphi(\bar \alpha) = \pm K, (\bar A = +1)$ & $\alpha^2 = \bft_3 \bft_4^{-c_3}$ \\
\hline
\end{tabular}
\end{center}
\end{table}



%It will be useful to record the following facts
%about \[\coker(\cals - I)=(\cals-I)\inv / \bbz^2\] in the table below. We list
%the maximal subgroup of $\left\langle \xy{\frac12}{0},\xy{0}{\frac12} \right\rangle$ that 
%$\coker(\cals - I)$ contains.

%\noindent
%\begin{table}[H]
%\label{t12}
%\begin{tabular}{|p{5cm}| p{7cm} |}
%\hline
%Image of $\cals$ in $\gl(2,\bbz_2)$  & \\
%\hline
%$\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]$ & $\left\langle %\xy{\frac12}{0},\xy{0}{\frac12} \right\rangle \in \coker(\cals - I)$\\
%\hline
%$\left[\begin{matrix}1 & 0 \\ 1 & 1\end{matrix}\right]$ & $\left\langle \xy{0}{\frac12} %\right\rangle \in \coker(\cals - I)$\\
%\hline
%$\left[\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}\right]$ & $\left\langle \xy{\frac12}{0} %\right\rangle \in \coker(\cals - I)$\\
%\hline
%$\left[\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right]$ & %$\left\langle\xy{\frac12}{\frac12} \right\rangle \in \coker(\cals - I)$\\
%\hline
%$\left[\begin{matrix}0 & 1 \\ 1 & 1\end{matrix}\right]$ & $\left\langle \xy{0}{0} %\right\rangle \in \coker(\cals - I)$\\
%\hline
%$\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]$ & $\left\langle \xy{0}{0} %\right\rangle \in \coker(\cals - I)$\\
%\hline
%\end{tabular}
%\end{table}

\step[{\bf Detecting Torsion in $\Solr$ Crystallographic Groups}]

Given a lattice $\wt \Gamma_\cals$ of $\Solr$ (which projects to a lattice $\gamsig$ of $\Sol$), the short exact sequence
\[1 \ra \calz(\wt \Gamma_\cals) \ra \wt \Gamma_\cals \ra \gamsig \ra 1\]
induces an $S^1$-bundle over the solvmanifold $\gamsig \backslash \Sol$,
\[S^1 \ra \wt \Gamma_\cals \backslash (\Solr) \ra \gamsig \backslash \Sol.\]

\begin{lemma}[cf. Lemma \ref{torSolof}]
\label{torSolr}
Let $\wt \Gamma_\cals$ be a lattice of $\Solr$, projecting
to a standard lattice $\gamsig$ of $\Sol$, and suppose that
for $\alpha \in  (\Solr) \rx (D_4 \x \bbz_2)$, the group
$\varPi = \angles{\wt \Gamma_\cals, \alpha}$ is crystallographic.
Let $\bar \alpha$ denote the projection of $\alpha$ to $\Sol \rx D_4$.
When the automorphism part of $\alpha$ 
acts as a reflection on the center of $\Solr$, $\varPi$ is torsion-free if and only if $\angles{\gamsig,\bar \alpha}$ is torsion-free.
\end{lemma}
\begin{proof}
Obviously, if $\angles{\gamsig,\bar \alpha}$ is torsion-free, then $\varPi$
must be torsion-free. For the converse, suppose that $\angles{\gamsig,\bar \alpha}$
has torsion. In this case, the action of $\bar \alpha$ on the solvmanifold
$\gamsig \bs \Sol$ must fix a point. Observe that the action of $\alpha$
on the solvmanifold $\wt \Gamma_\cals \bs (\Solr)$ is $S^1$ fiber preserving.
Therefore, a circle fiber is left invariant under the action of $\alpha$. Since
$\alpha$ acts as reflection on the fiber, $\alpha$ must fix a point. Therefore,
$\varPi$ has torsion.
\qedhere
\end{proof}

\begin{lemma}[cf. Lemma \ref{EasyTor}]
\label{EasyTorSolr}
When $\beta=(\bft_1^{b_1}\bft_2^{b_2}\bft_3^{b_3}\bft_4^{b_4}, B) \in \varPi$ satisfies
$b_3 = \frac{1}{2}$ and $\bbar=1$, $\gamma \beta$ is not torsion for any $\gamma \in
\tgamsig$.
\end{lemma}
\begin{proof}
Note that $B$ is necessarily of order 2. Let $\pr: \Solr \ra \bbr$ denote the
quotient homomorphism of $\Solr$ by its nil-radical $\bbr^3$. Write $\gamma \in \tgamsig$
as $\bft_1^{n_1}\bft_2^{n_2}\bft_3^{n_3}\bft_4^{n_4}$. Application of $\pr$ to $(\gamma \beta)^2$
yields
\[
\pr(\gamma \beta)^2 = 2 n_3+1,
\]
from which we infer $\gamma \beta$ is not torsion.
\qedhere
\end{proof}


%\subsubsection{Main Classification}

We are now ready the classify the $\Solr$ crystallographic groups. We will enumerate the $\Solr$ crystallographic groups associated to each abstract kernel
$\varphi: \bbzpsi \ra \gltz$. The $\Solr$ crystallographic groups will be ordered according to the abstract kernel in Theorem \ref{abstract-kernel} that $\varphi$ lifts.
 We start our classification with the trivial case (\gr{0} Family). In all cases, we give:

\noindent
{\rm(1)} The holonomy $\Psi$, the conditions on $\cals$, and the abstract kernel \[\varphi: \bbzpsi \ra \gltz\] (this fixes the exponents on $\bft_3$).

\noindent
{\rm(2)} A computation of $H^1(\Psi;\coker(I-\cals))$ (this fixes the exponents on $\bft_1, \bft_2$). As a matter of notation, we set
\[\bfa=\xy{a_1}{a_2}, \bfb=\xy{b_1}{b_2}, \text{ and } \bfr=\xy{t_1}{t_2}, \]
the exponents on $\bft_1, \bft_2$ corresponding to holonomy elements $A$ or $A\tau$, $B$ or $B\tau$, and $\tau$, respectively.

%We will always let $\bfr$ denote the exponents on $\bft_1$ and $\bft_2$ corresponding
%to the holonomy element $\tau$.

\noindent
{\rm(3)} Conditions on $c_1, c_2, c_3$ (this fixes the lattice $\tgamsig$, and induces an abstract kernel $\Psi \ra \out(\tgamsig)$).

\noindent
{\rm(4)} Exponents on $\bft_4$, which are classified by $H^2(\Psi; \calz(\tgamsig))$.

\noindent
{\rm(5)} When the group is torsion-free.


\begin{theorem}[\gr{0} Family]
\label{0family}
The crystallographic groups of $\Solr$, lifting an abstract kernel
$\bbzphi \ra \gltz$ of case \gr{0} of {\rm Theorem \ref{abstract-kernel}} are:
\end{theorem}
\begin{enumerate}
\item[\gr{0a}]
$\Psi=\{ e \}$, $\bbzpsi=\angles{\bft_3}$\parn
$\bullet$ $\cals\in\sltz$ with $\tr(\cals)>2$\parn
$\bullet$ $c_3=0$\parn
$\bullet$ $\varPi=\angles{\tgamsig}$ \parn
$\bullet$ Torsion-free \parn

\item[\gr{0b}]
$\Psi=\bbz_2$, $\bbzpsi=\angles{\bft_3, \bar\gamma=(\bft^{0}_3, \tau)}$\parn
$\bullet$ $\cals\in\sltz$ with $\tr(\cals)>2$\parn
$\bullet$ $\varphi(\bar \gamma) = I$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfr$ with $(I-\cals)\bfr \equiv \bfzero$ and  $2 \bfr \equiv \bfzero$ \parn
$\bullet$ $c_1=0$ or $\frac12$, and $c_2=0$ or $\frac12$\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Table \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \gamma=(\bft^{t_1}_1\bft^{t_2}_2 \bft^{0}_3, \tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))$ is trivial.\parn
$\bullet$ Torsion-free when $\bfr\neq\bfzero$. \parn

\item[\gr{0c}]
$\Psi=\bbz_2$, $\bbzpsi=\angles{\bft_3, \bar\gamma=(\bft^{\frac12}_3, \tau)}$\parn
$\cals=n K-I$, where $K\in\gltz$ with $\det(K)=+1$,  $\tr(K)=n>2$.\parn
$\bullet$ $\varphi(\bar \gamma) = K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is trivial, so that we can take $\bfr=\bfzero$.\parn
$\bullet$ $(c_1, c_2) \in \coker(I+K^T)$\parn
$\bullet$ $c_3=0$\parn
$\bullet$ $\varPi=\angles{\tgamsig, \gamma=(\bft^{\frac12}_3, \tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))$ is trivial.\parn
$\bullet$ Torsion-free \parn
\end{enumerate}
\begin{proof}
The arguments in Theorem \ref{abstract-kernel} establish the three different possibilities
for $\bbzpsi$. In particular, \gr{2b} from Theorem \ref{abstract-kernel} gives the conditions on $\cals$ in case \gr{0c}. The difference is that $\varphi(\bar \gamma)$ is given by $K$, not $-K$, as in  Theorem \ref{abstract-kernel} \gr{2b}.

\bigskip
In case \gr{0b}, the cocycle condition for $\bfr \in \coker(I-\cals)$ is that $(I+\varphi(\bar\gamma))\bfr=2\bfr \equiv \bfzero$, whereas the coboudary vanishes.
This shows our description of $H^1(\Psi;\coker(I-\cals))$.
In case \gr{0c}, $H^1(\Psi;\coker(I-\cals))$ vanishes by Corollary \ref{a-conj-zero3}.

\bigskip
In all cases, the conditions on $c_1, c_2$ follow from the Tables \ref{t2} and \ref{t4}. We can take $c_3=0$ in case \gr{0a} by Lemma \ref{solrc3}, and $c_3=0$ is forced
in case \gr{0c} by Table \ref{t111}. Table \ref{t6} shows that we have two choices for $c_3$ in case \gr{0b}.

\bigskip
Evidently, groups of type \gr{0b} are torsion-free when $\bfr \neq \bfzero$.
 The groups of type \gr{0c} are torsion-free Lemma \ref {EasyTorSolr}.
In all three cases, we take $a_4=0$ by Lemma \ref{conj-in-t4Solr}, and
$H^2(\Psi; \calz(\tgamsig))$ vanishes. 
\qedhere
\end{proof}

\begin{theorem}[\gr{1} Family]
\label{1family}
The crystallographic groups of $\Solr$, lifting an abstract kernel
$\bbzphi \ra \gltz$ of case \gr{1} of {\rm Theorem \ref{abstract-kernel}},
\[A=\left[\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right], \text{ are}\]
\end{theorem}
\begin{enumerate}
\item[\gr{1a}]
$\Psi=\bbz_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha=(\bft^{\frac 12}_3,A)}$\parn
$\bullet$ $\cals=n K+I$, where $K\in\gltz$ with $\det(K)=-1$,  $\tr(K)=n>0$.\parn
$\bullet$ $\vara=-K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is trivial, so that we can take $\bfa=\bfzero$.\parn
$\bullet$ $(c_1, c_2) \in \coker(I+K^T)$\parn
$\bullet$ $c_3=0$\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha=(\bft^{\frac 12}_3 \bft^{a_4}_4,A)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig)) = \bbz_2$, $a_4=0$ or $\frac12$\parn
$\bullet$ Torsion-free \parn

\item[\gr{1b}]
$\Psi=\bbz_2$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{\frac 12}_3,A \tau)}$\parn
$\bullet$ $\cals=n K+I$, where $K\in\gltz$ with $\det(K)=-1$,  $\tr(K)=n>0$.\parn
$\bullet$ $\vara=-K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is trivial, so that we can take $\bfa=\bfzero$.\parn
$\bullet$ $(c_1, c_2) \in \coker(K^T-I)$\parn
$\bullet$ $c_3=0$\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{\frac 12}_3,A \tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))$ is trivial.\parn
$\bullet$ Torsion-free \parn

\item[\gr{1c}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{\frac 12}_3,A), \bar\gamma=(\bft^{0}_3,\tau)}$\parn
$\bullet$ $\cals=n K+I$, where $K\in\gltz$ with $\det(K)=-1$,  $\tr(K)=n>0$.\parn
$\bullet$ $\vara=-K, \varphi(\bar \gamma) = I$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfr$ with $(I+K)\bfr \equiv \bfzero$ and  $2 \bfr \equiv \bfzero$ \parn
$\bullet$ $(c_1, c_2) \in \coker(I+K^T)$ and $c_1=0$ or $\frac12$, $c_2=0$ or $\frac12$\parn
$\bullet$ One choice for $c_3$, Table \ref{t6} gives two different values for $c_3$,
but only one satisfies $(\alpha\gamma)^2 \in \tgamsig$.\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{\frac 12}_3 \bft^{a_4}_4,A), \gamma=(\bft^{t_1}_1\bft^{t_2}_2 \bft^{0}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $a_4=0$ or $\frac12$\parn
$\bullet$ Torsion-free when $\bfr\neq\bfzero$. \parn
\end{enumerate}
\begin{proof}
First, observe that
\begin{align*}
\bbzpsi&=\angles{\bft_3,  \bar\alpha=(\bft^{\frac 12}_3,A), \bar\gamma=(\bft^{\frac12}_3,\tau)}\\
&=\angles{\bft_3,  (\bft^{0}_3,A\tau), (\bft^{\frac12}_3,\tau)}
\end{align*}
cannot be lifted to a $\Solr$ crystallographic group by Lemma \ref{PPM-A302}, and 
we therefore disregard this case.

\bigskip
In cases \gr{1a} and \gr{1b}, $H^1(\Psi;\coker(I-\cals))$ vanishes by Corollary \ref{a-conj-zero3}.
In case \gr{1c}, Corollary \ref{a-conj-zero3} implies we may take $\bfa=\bfzero$.
Then the cocycle conditions imply that $\bfr$ must satisfy \[(I+K)\bfr \equiv \bfzero\]
and \[(I+\varphi(\bar\gamma))\bfr=2\bfr \equiv \bfzero.\]

\bigskip
In all cases, the conditions on $c_1, c_2$ follow from the Tables \ref{t2}, \ref{t3}, and \ref{t4}. We can take $c_3=0$ in case \gr{1a} by Lemma \ref{solrc3}, and $c_3=0$ is forced
in case \gr{1b} by Table \ref{t111}. Table \ref{t6} shows that we have two choices for $c_3$ in case \gr{1c}. However, 
\[(\alpha \gamma)^2 = \bft_1^{n_1}\bft_2^{n_2}\bft_3^{} \bft_4^{-c_3+v},\]
for integers $n_i$, and a rational number $v$. Only one choice for $c_3$ (modulo the integers) will ensure
$-c_3+v$ is integral.

\bigskip
The two choices for $a_4$ in cases \gr{1a} and \gr{1c} arise from Table \ref{t11}:
\[\alpha^2 = \bft_3 \bft_4^{2a_4}.\qedhere\]
\end{proof}


\begin{theorem}[\gr{2a} Family]
\label{2afamily}
The crystallographic groups of $\Solr$, lifting an abstract kernel
$\bbzphi \ra \gltz$ of case \gr{2a} of {\rm Theorem \ref{abstract-kernel}},
\[A=\left[\begin{matrix} -1 & 0\\ 0 & -1 \end{matrix} \right], \text{ are}\]
\end{theorem}
\begin{enumerate}
\item[\gr{2aa}]
$\Psi=\bbz_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A)}$\parn
$\bullet$ $\cals\in\sltz$ with $\tr(\cals)>2$\parn
$\bullet$ $\vara=A$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals)) = \frac {\coker(I-\cals)}{2\coker(I-\cals)}$.
That is, all $\bfa$ with $(I-\cals)\bfa\equiv \bfzero$, taken modulo $2\bfa$.\parn
$\bullet$ $c_1=0$ or $\frac12$, $c_2=0$ or $\frac12$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $a_4=-({a_1}{c_1}+{a_2}{c_2})+ \frac{i}{2}, i=0,1$\parn
$\bullet$ Torsion-free when $i=1$\parn

\item[\gr{2ab}]
$\Psi=\bbz_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A \tau)}$\parn
$\bullet$ $\cals\in\sltz$ with $\tr(\cals)>2$\parn
$\bullet$ $\vara=A$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals)) = \frac {\coker(I-\cals)}{2\coker(I-\cals)}$.
That is, all $\bfa$ with $(I-\cals)\bfa\equiv \bfzero$, taken modulo $2\bfa$.\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Table \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3,A \tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))$ is trivial.\parn
$\bullet$ Always has torsion \parn

\item[\gr{2ac}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A), \bar\gamma=(\bft^{0}_3,\tau)}$\parn
$\bullet$ $\cals\in\sltz$ with $\tr(\cals)>2$\parn
$\bullet$ $\vara=A, \varphi(\bar \gamma) = I$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfa, \bfr$ with $(I-\cals)\bfa \equiv (I-\cals)\bfr\equiv\bfzero$ satisfying $2\bfr\equiv\bfzero$ \parn
$\bullet$ $c_1=0$ or $\frac12$, $c_2=0$ or $\frac12$\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Tables \ref{t5} and \ref{t6} \parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \gamma=(\bft^{t_1}_1\bft^{t_2}_2 \bft^{0}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $a_4=-({a_1}{c_1}+{a_2}{c_2})+ \frac{i}{2} (i=0,1)$\parn
$\bullet$ Always has torsion \parn

\item[\gr{2ad}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A), \bar\gamma=(\bft^{\frac 12}_3,\tau)}$\parn
$\bullet$ $\cals=n K-I$, where $K\in\gltz$ with $\det(K)=+1$,  $\tr(K)=n>2$.\parn
$\bullet$ $\vara=A, \varphi(\bar \gamma) = K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals)) = \frac {\coker(I-K)}{2\coker(I-K)}$.
That is, all $\bfa$ with $(I-K)\bfa\equiv \bfzero$, taken modulo $2\bfa$.\parn
$\bullet$ $(c_1, c_2) \in \coker(I+K^T)$ and $c_1=0$ or $\frac12$, $c_2=0$ or $\frac12$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \gamma=(\bft^{\frac 12}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $a_4=-({a_1}{c_1}+{a_2}{c_2})+ \frac{i}{2} (i=0,1)$\parn
$\bullet$Torsion-free when $i=1$\parn

\item[\gr{2ae}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A\tau),\bar \gamma=(\bft^{\frac 12}_3,\tau)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\gltz$ with $\det(K)=+1$,  $\tr(K)=n>2$.\parn
$\bullet$ $\vara=A, \varphi(\bar \gamma) = K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals)) = \frac {\coker(I-K)}{2\coker(I-K)}$.
That is, all $\bfa$ with $(I-K)\bfa\equiv \bfzero$, taken modulo $2\bfa$.\parn
$\bullet$ $(c_1, c_2) \in \coker(I+K^T)$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A\tau), \gamma=(\bft^{\frac 12}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, two choices for $a_4$. We have
$(\alpha \gamma)^2 = \bft_1^{n_1} \bft_2^{n_2}\bft_3 \bft_4^{2a_4 + v}$ for integers $n_1$, and some rational number $v$, and therefore two solutions for $a_4$.\parn
$\bullet$ Always has torsion \parn
\end{enumerate}
\begin{proof}

\bigskip

The computation for $H^1(\Psi;\coker(I-\cals))$ for cases \gr{2aa} and \gr{2ab} follows 
immediately from that of case \gr{2a} in Theorem \ref{ClassificationSol-geometry}. That
for cases \gr{2ac}, \gr{2ad}, and \gr{2ae} follows from computing the appropriate cocycle conditions
in Remark \ref{cocycleConditions}, together with application of Corollary \ref{a-conj-zero3}. In particular, in cases \gr{2ad} and \gr{2ae}, we take $\bfr=\bfzero$ by Corollary \ref{a-conj-zero3}; the commutator cocycle condition in cases \gr{2ad} and \gr{2ae} then reduces to
\begin{align*}
(I-\phi(\bar \gamma))\bfa &=  (I-\phi(\bar \alpha))\bfr\\
(I-K)\bfa &\equiv \bfzero.
\end{align*}

\bigskip
In all cases, the conditions on $c_1, c_2$ follow from the Tables \ref{t1}, \ref{t2}, and \ref{t4}. We can take $c_3=0$ in case \gr{2aa} by Lemma \ref{solrc3}. We have two choices for $c_3$ in cases \gr{2ab} and \gr{2ac}. For cases \gr{2ad} and \gr{2ae}, notice that from Table \ref{t111}, $\gamma^2 = \bft_3 \bft_4^{-c_3}$, so that $c_3=0$ is forced.

\bigskip
The two choices for $a_4$ in cases \gr{2aa}, \gr{2ac}, and \gr{2ad} arise from Table \ref{t11}:
\[\alpha^2 = \bft_4^{2(a_4+{a_1}{c_1}+{a_2}{c_2})}.\]

\bigskip
The torsion criteria are clear, given that $\Sol$ groups of type \gr{2a}
in Theorem \ref{ClassificationSol-geometry} always have torsion. Notice
that $\gamma$ in case \gr{2ad} cannot contribute torsion by Lemma \ref{EasyTorSolr}.
Lemma \ref{torSolr} implies that \gr{2ab} always has torsion, and that
\gr{2ac} also has torsion as it contains case \gr{2ab}. Case \gr{2ae} is always torsion by Lemma \ref{torSolr}.
\qedhere
\end{proof}

\begin{theorem}[\gr{2b} Family]
\label{2bfamily}
The crystallographic groups of $\Solr$, lifting an abstract kernel
$\bbzphi \ra \gltz$ of case \gr{2b} of {\rm Theorem \ref{abstract-kernel}},
\[A=\left[\begin{matrix} -1 & 0\\ 0 & -1 \end{matrix} \right], \text{ are}\]
\end{theorem}
\begin{enumerate}
\item[\gr{2ba}]
$\Psi=\bbz_2$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{\frac 12}_3,A)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\gltz$ with $\det(K)=+1$,  $\tr(K)=n>2$.\parn
$\bullet$ $\vara=-K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is trivial so that $\bfa=\bzero$. \parn
$\bullet$ $(c_1, c_2) \in \coker(I+K^T)$\parn
$\bullet$ $c_3=0$\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{\frac 12}_3 \bft^{a_4}_4,A)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $a_4=0$, $\frac12$\parn
$\bullet$ Torsion-free\parn

\item[\gr{2bb}]
$\Psi=\bbz_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{\frac 12}_3,A \tau)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\gltz$ with $\det(K)=+1$,  $\tr(K)=n>2$.\parn
$\bullet$ $\vara=-K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is trivial so that $\bfa=\bzero$. \parn
$\bullet$ $(c_1, c_2) \in \coker(K^T-I)$\parn
$\bullet$ $c_3=0$\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{\frac 12}_3,A \tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))$ is trivial.\parn
$\bullet$ Torsion-free \parn

\item[\gr{2bc}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{\frac 12}_3,A), \bar\gamma=(\bft^{0}_3,\tau)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\gltz$ with $\det(K)=+1$,  $\tr(K)=n>2$.\parn
$\bullet$ $\vara=-K, \varphi(\bar \gamma) = I$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ all $\bfr$ with $(I+K)\bfr \equiv \bfzero$, 
$2\bfr \equiv\bfzero$.\parn
$\bullet$ $(c_1, c_2) \in \coker(I+K^T)$ and $c_1=0$ or $\frac12$, $c_2=0$ or $\frac12$\parn
$\bullet$ One choice for $c_3$, Table \ref{t6} gives two different values for $c_3$,
but only one satisfies $(\alpha\gamma)^2 \in \tgamsig$.\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{\frac 12}_3 \bft^{a_4}_4,A), \gamma=(\bft^{t_1}_1\bft^{t_2}_2 \bft^{0}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig)) = \bbz_2$, $a_4=0$, $\frac12$\parn
$\bullet$ Torsion-free when $\bfr \neq \bfzero$. \parn

\end{enumerate}
\begin{proof}

First, note that
\begin{align*}
\bbzpsi&=\angles{\bft_3, \bar\alpha= (\bft^{\frac 12}_3,A), \bar\gamma=(\bft^{\frac{1}{2}}_3,\tau)}\\
&=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A\tau), \bar\gamma=(\bft^{\frac{1}{2}}_3,\tau)}
\end{align*}
is a lift of type \gr{2a} from the $\Sol$ classification, and it corresponds to type \gr{2ae} in Theorem \ref{2afamily}.

\bigskip
The computation for $H^1(\Psi;\coker(I-\cals))$ for cases \gr{2ba} and \gr{2bb} follows 
immediately from that of case \gr{2b} in Theorem \ref{ClassificationSol-geometry}. That
for case \gr{2bc} follows from computing the appropriate cocycle conditions
in Remark \ref{cocycleConditions}, together with application of Corollary \ref{a-conj-zero3}. 

\bigskip
In all cases, the conditions on $c_1, c_2$ follow from the Tables \ref{t2}, \ref{t3}, and \ref{t4}. We can take $c_3=0$ in case \gr{2ba} by Lemma \ref{solrc3}. From Table \ref{t111}, $c_3=0$ is forced in case \gr{2bb}. Table \ref{t6} shows that we have two choices for $c_3$ in case \gr{2bc}. However, 
\[(\alpha \gamma)^2 = \bft_1^{n_1}\bft_2^{n_2}\bft_3^{} \bft_4^{-c_3+v},\]
for integers $n_i$, and a rational number $v$. Only one choice for $c_3$ (modulo the integers) will ensure
$-c_3+v$ is integral.

\bigskip
For the torsion criteria, notice that $\Sol$ groups of type \gr{2b} were always torsion-free.
\qedhere
\end{proof}

\begin{theorem}[\gr{3} Family]
\label{3family}
The crystallographic groups of $\Solr$, lifting an abstract kernel
$\bbzphi \ra \gltz$ of case \gr{3} of {\rm Theorem \ref{abstract-kernel}},
\[A=\left[\begin{matrix} 0 & 1\\ 1 & 0 \end{matrix} \right], \text{ are}\]
\end{theorem}
\begin{enumerate}
\item[\gr{3a}]
$\Psi=\bbz_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A)}$\parn
$\bullet$
$\cals\in\sltz$
with $\sigma_{12}=-\sigma_{21}$ and
$\tr(\cals)>2$.\parn
$\bullet$ $\vara=A$\parn
$\bullet$
$H^1(\Psi;\coker(I-\cals))$ can be described as all $\bfa$ with $(I-\cals)\bfa\equiv \bfzero$,
$a_2\equiv -a_1$ taken modulo
$(v_1-v_2){\scriptstyle{\xy{1}{-1}}}$ where $(I-\cals)\bfv\equiv \bfzero$.\parn
$\bullet$ $c_1-c_2=0$\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $a_4 = 0$ or $\frac12$\parn
$\bullet$ Torsion-free when $a_4=\frac12$

\item[\gr{3b}]
$\Psi=\bbz_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A \tau)}$\parn
$\bullet$
$\cals\in\sltz$
with $\sigma_{12}=-\sigma_{21}$ and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=A$\parn
$\bullet$
$H^1(\Psi;\coker(I-\cals))$ can be described as all $\bfa$ with $(I-\cals)\bfa\equiv \bfzero$,
$a_2\equiv -a_1$ taken modulo
$(v_1-v_2){\scriptstyle{\xy{1}{-1}}}$ where $(I-\cals)\bfv\equiv \bfzero$.\parn
$\bullet$ $c_1+c_2=0$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3,A \tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))$ is trivial.\parn
$\bullet$ Always has torsion \parn

\item[\gr{3c}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3, \bar \alpha=(\bft^{0}_3,A), \bar\gamma=(\bft^{0}_3,\tau)}$\parn
$\bullet$
$\cals\in\sltz$
with $\sigma_{12}=-\sigma_{21}$ and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=A, \varphi(\bar \gamma) = I$\parn
$\bullet$
$H^1(\Psi;\coker(I-\cals))$ can be described as all $\bfa$, $\bfr$ with $(I-\cals)\bfa\equiv (I-\cals)\bfr \equiv \bfzero$, $\bfr=(0,0)$ or $(\frac12,\frac12)$,\newline
and $a_2\equiv -a_1$ taken modulo
$(v_1-v_2){\scriptstyle{\xy{1}{-1}}}$ where $(I-\cals)\bfv\equiv \bfzero$.\parn
$\bullet$ $(c_1,c_2)=(0,0)$ or $(c_1,c_2)=(\frac12, \frac12)$\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Tables \ref{t5} and \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \gamma=(\bft^{t_1}_1\bft^{t_2}_2 \bft^{0}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $a_4 = 0$ or $\frac12$\parn
$\bullet$ Always has torsion \parn

\item[\gr{3d}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3,\bar \alpha= (\bft^{0}_3,A), \bar\gamma=(\bft^{\frac 12}_3,\tau)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\sltz$ with $\tr(K)= n>2$ and
$k_{12}=-k_{21}$.\parn
$\bullet$ $\vara=A, \varphi(\bar \gamma) =K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ can be described as all $\bfa$ with $(I-K)\bfa\equiv \bfzero$,
$a_2\equiv -a_1$ taken modulo
$(v_1-v_2){\scriptstyle{\xy{1}{-1}}}$ where $(I-K)\bfv\equiv \bfzero$.\parn
$\bullet$ $(c_1, c_2) \in \coker(I+K^T)$, $c_1-c_2=0$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \gamma=(\bft^{\frac 12}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))= \bbz_2$, $a_4 = 0$ or $\frac12$\parn
$\bullet$ Always has torsion \parn
\end{enumerate}
\begin{proof}

\bigskip
The computation for $H^1(\Psi;\coker(I-\cals))$ for cases \gr{3a} and \gr{3b} follows 
immediately from that of case \gr{3} in Theorem \ref{ClassificationSol-geometry}. That
for cases \gr{3c} and \gr{3d} follows from computing the appropriate cocycle conditions
in Remark \ref{cocycleConditions}, together with application of Corollary \ref{a-conj-zero3}. Let us examine \gr{3c} in greater detail. Notice that the cyclic power cocycle condition forces $2\bfr \equiv 0$. Furthermore, consider the commutator cocycle condition in case \gr{3c}:
\begin{align*}
(I-\varphi(\alpha))\bfr &=(I-\varphi(\gamma))\bfa\\
\left[ \begin{matrix} 1 & -1 \\ -1 & 1\end{matrix} \right] \bfr &= \bfzero
\end{align*}
This forces $\bfr$ to be either $(0,0)$ or $(\frac12,\frac12)$.


\bigskip
In all cases, the conditions on $c_1, c_2$ follow from the Tables \ref{t1}, \ref{t2}, and \ref{t4}. We can take $c_3=0$ in case \gr{3b} by Lemma \ref{solrc3}. From Table \ref{t111}, $c_3=0$ is forced in case \gr{3d}. For cases \gr{3a} and \gr{3c}, we have two choices for $c_3$.

\bigskip
For the choices we have for $a_4$ in cases \gr{3a}, \gr{3c}, and \gr{3d}, note that $c_1-c_2=0$ and by Table \ref{t9},
\[
\alpha^2 = \bft_1^{a_1+a_2} \bft_2^{a_1+a_2} \bft_4^{2a_4+{a_1}{c_1}+{a_2}{c_2}-{a_2}{c_1}-{a_1}{c_2}}=\bft_4^{2a_4+{a_1}(c_1-c_2)-{a_2}(c_1-c_2)}=\bft_4^{2a_4}.
\]

\bigskip
For the torsion criteria, notice that $\Sol$ groups of type \gr{3} always have torsion. Thus
Lemma \ref{torSolr} applies to show that $\alpha$ is always a torsion element in case \gr{3b}, $\alpha\gamma$ is a torsion element in case \gr{3c}, and $\alpha\gamma$ is a torsion element in case \gr{3d}. Only in case \gr{3a} can we resolve torsion in the central direction of $\Solr$.

\qedhere
\end{proof}



\begin{theorem}[\gr{3\ii} Family]
\label{3ifamily}
The crystallographic groups of $\Solr$, lifting an abstract kernel
$\bbzphi \ra \gltz$ of case \gr{3\ii} of {\rm Theorem \ref{abstract-kernel}},
\[A=\left[\begin{matrix} 0 & 1\\ 1 & 0 \end{matrix} \right], \text{ are}\]
\end{theorem}
\begin{enumerate}
\item[\gr{3\ii a}]
$\Psi=\bbz_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A)}$\parn
$\bullet$
$\cals\in\sltz$
with $\sigma_{11}=\sigma_{22}$ and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=\matPpm$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ can be described as all $\bfa$ with $(I-\cals)\bfa\equiv\bfzero$, $2 a_1\equiv 0$ (so $a_1=0$ or $\tfrac12$)
modulo $\xy{0}{2v_2}$ where $(I-\cals)\bfv\equiv \bfzero$.\parn
$\bullet$ $c_2=0$ or $\frac12$\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $a_4 = -{a_2}{c_2}+\frac{i}{2} (i=0,1)$\parn
\begin{itemize}
\item When $i=0$, torsion-free when $a_1=\tfrac12$
and $a_2\not\equiv \frac{(\sigma_{11}+1)(2 n+1)}{2\sigma_{12}}$
for any $n\in\bbz$
\item When $i=1$, always torsion-free
\end{itemize}

\item[\gr{3\ii b}]
$\Psi=\bbz_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A \tau)}$\parn
$\bullet$
$\cals\in\sltz$
with $\sigma_{11}=\sigma_{22}$ and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=\matPpm$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ can be described as all $\bfa$ with $(I-\cals)\bfa\equiv\bfzero$, $2 a_1\equiv 0$ (so $a_1=0$ or $\tfrac12$)
modulo $\xy{0}{2v_2}$ where $(I-\cals)\bfv\equiv \bfzero$.\parn
$\bullet$ $c_1=0$ or $\frac12$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3,A \tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))$ is trivial.\parn
$\bullet$ Torsion-free when $a_1=\tfrac12$
and $a_2\not\equiv \frac{(\sigma_{11}+1)(2 n+1)}{2\sigma_{12}}$
for any $n\in\bbz$.

\item[\gr{3\ii c}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3,\bar \alpha= (\bft^{0}_3,A),\bar \gamma=(\bft^{0}_3,\tau)}$\parn
$\bullet$
$\cals\in\sltz$
with $\sigma_{11}=\sigma_{22}$ and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=\matPpm, \varphi(\bar \gamma) =I$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ can be described as all $\bfa$, $\bfr$ with $(I-\cals)\bfa \equiv (I-\cals)\bfr\equiv\bfzero$, $2 \bfr \equiv 0$, $2 a_1\equiv 0$
with $\bfa$ taken modulo $\xy{0}{2v_2}$ where $(I-\cals)\bfv\equiv \bfzero$.\parn
$\bullet$ $c_1=0$ or $\frac12$, and $c_2=0$ or $\frac12$\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Tables \ref{t5} and \ref{t6} \parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \gamma=(\bft^{t_1}_1\bft^{t_2}_2 \bft^{0}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $a_4 = -{a_2}{c_2}+\frac{i}{2} (i=0,1)$\parn
$\bullet$ Torsion-free when $\bfr \neq \bfzero$, $t_1 + a_1 = \frac12$, 
$t_2 + a_2\not\equiv \frac{(\sigma_{11}+1)(2 n+1)}{2\sigma_{12}}$ for any $n\in\bbz$, and...
\begin{itemize}
\item when $i=0$, in addition $a_1=\tfrac12$
and $a_2\not\equiv \frac{(\sigma_{11}+1)(2 n+1)}{2\sigma_{12}}$ for any $n\in\bbz$
\item when $i=1$, no additional condition
\end{itemize}

\item[\gr{3\ii d}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A), \bar\gamma=(\bft^{\frac 12}_3,\tau)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\sltz$ with $\tr(K)= n>2$ and
$k_{11}=k_{22}$.\parn
$\bullet$ $\vara=\matPpm, \varphi(\bar \gamma) = K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ can be described as all $\bfa$ with $(I-K)\bfa\equiv\bfzero$, $2 a_1\equiv 0$
modulo $\xy{0}{2v_2}$ where $(I-K)\bfv\equiv \bfzero$.\parn
$\bullet$ $c_2=0$ or $\frac12$, and $(c_1,c_2) \in \coker(I+K^T)$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \gamma=(\bft^{\frac 12}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $a_4 = -{a_2}{c_2}+\frac{i}{2} (i=0,1)$\parn
$\bullet$ Torsion-free ...
\begin{itemize}
\item when $i=0$, $a_1=\tfrac12$
and $a_2\not\equiv \frac{(k_{11}+1)(2 n+1)}{2k_{12}}$ for any $n \in \bbz$.
\item when $i=1$, the $\angles{\tgamsig,(\alpha\gamma)}$ must be torsion-free, projected to $\Sol$.
\end{itemize}
\end{enumerate}
\begin{proof}

\bigskip
The computation for $H^1(\Psi;\coker(I-\cals))$ for cases \gr{3\ii a} and \gr{3\ii b} follows 
immediately from that of case \gr{3\ii} in Theorem \ref{ClassificationSol-geometry}. That
for cases \gr{3\ii c} and \gr{3\ii d} follows from computing the appropriate cocycle conditions
in Remark \ref{cocycleConditions}, together with application of Corollary \ref{a-conj-zero3}. Let us examine \gr{3\ii c} in greater detail. Notice that the cyclic power cocycle condition forces $2\bfr \equiv 0$. We do not obtain an additional condition on $\bfr$ from commutator cocycle condition in case \gr{3\ii c}:
\begin{align*}
(I-\varphi(\alpha))\bfr &=(I-\varphi(\gamma))\bfa\\
\left[ \begin{matrix} 0 & 0 \\ 0 & 2\end{matrix} \right] \bfr &= \bfzero.
\end{align*}

\bigskip
In all cases, the conditions on $c_1, c_2$ follow from the Tables \ref{t1}, \ref{t2}, and \ref{t4}. We can take $c_3=0$ in case \gr{3\ii b} by Lemma \ref{solrc3}. From Table \ref{t111}, $c_3=0$ is forced in case \gr{3\ii d}. For cases \gr{3\ii a} and \gr{3\ii c}, we have two choices for $c_3$.

\bigskip
For the choices we have for $a_4$ in cases \gr{3\ii a}, \gr{3\ii c}, and \gr{3\ii d}, note that
\[
\alpha^2 = \bft_1^{2a_1} \bft_4^{2(a_4+{a_2}{c_2})}.
\]

\bigskip
For case \gr{3\ii a}, when $i=1$, $\varPi$ is always torsion-free. But when $i=0$,
the projection of $\varPi$ to a $\Sol$ crystallographic group of type \gr{3\ii} must
be torsion-free. The torsion criteria now follow from Theorem \ref{ClassificationSol-geometry}, case \gr{3\ii}.

\bigskip
For case \gr{3\ii b}, by Lemma \ref{torSolr}, the projection of $\varPi$ to a $\Sol$
crystallographic group of type \gr{3\ii} must be torsion-free, and hence the desired criteria.

\bigskip
For case \gr{3\ii c}, we first require that $\bfr \neq \bzero$ for $\gamma$ to be of infinite order. In addition,
\[
\gamma \alpha = (\bft^{a_1+t_1}_1\bft^{a_2+t_2}_2 \bft^{a_4}_4, A \tau)
\]
must be infinite order when projected to $\Sol$ by Lemma \ref{torSolr}. Thus we obtain
the stated conditions on $a_1 + t_1$ and $a_2 + t_2$.

Now, when  $i=1$, $\alpha$ is never torsion. But when $i=0$, the projection of $\alpha$
to $\Sol$ must be of infinite order, and we obtain additional conditions on $a_1$ and $a_2$.

\bigskip
For case \gr{3\ii d}, note that when $i=0$, both 
\begin{align*}
\alpha &= (\bft^{a_1}_1\bft^{a_2}_2, A)\\
\alpha \gamma &= (\bft^{a_1}_1\bft^{a_2}_2 \bft^{\frac12}_3, A \tau)\\
\end{align*}
must be torsion-free, when projected to $\Sol$. This 
occurs precisely when the $\Sol$ crystallographic group
of type \gr{3\ii},
\[
Q'=\angles{\bft_1, \bft_2, \bft_3^{\frac12} , \alpha}
\]
is torsion-free. Note that the lattice of $Q'$ 
is 
$\angles{\bft_1, \bft_2, \bft^{\frac12}_3} = \bbz^2 \rx_K \bbz$.
The desired criteria on $a_1$ and $a_2$ now follow by applying the result of
Theorem \ref{ClassificationSol-geometry}, case \gr{3\ii} with $K$ in place of $\cals$.

When $i=1$, $\alpha$ and $\gamma$ are always of infinite order. But Lemma \ref{torSolr} implies that
\begin{align*}
\alpha \gamma &= (\bft^{a_1}_1\bft^{a_2}_2 \bft^{\frac12}_3, A \tau)\\
\end{align*}
must be of infinite order, when projected to $\Sol$.
This 
occurs precisely when the $\Sol$ crystallographic group
of type \gr{3\ii},
\[
Q''=\angles{\bft_1, \bft_2, \bft_3 , \alpha \gamma}
\]
is torsion-free. We can conjugate $\alpha\gamma$ by $\bft_3^{-\frac{1}{4}}$ to obtain
\[
\bft_3^{-\frac{1}{4}} (\bft^{a_1}_1\bft^{a_2}_2 \bft^{\frac12}_3, A \tau) \bft_3^{\frac{1}{4}} = (\bft^{a'_1}_1\bft^{a'_2}_2 \bft^{0}_3 \bft^{a_4}_4, A \tau)
\]
Now one could apply  Theorem \ref{ClassificationSol-geometry}, case \gr{3\ii} to obtain criteria on $a'_1$, $a'_2$.
\qedhere
\end{proof}

\begin{theorem}[\gr{4} Family]
\label{4family}
The crystallographic groups of $\Solr$, lifting an abstract kernel
$\bbzphi \ra \gltz$ of case \gr{4} of {\rm Theorem \ref{abstract-kernel}},
\[A=\left[\begin{matrix} 0 & 1\\ -1 & 0 \end{matrix} \right], \text{ are}\]
\end{theorem}
\begin{enumerate}
\item[\gr{4a}]
$\Psi=\bbz_4$, $\bbzpsi=\angles{\bft_3,\bar\alpha=  (\bft^{0}_3,A)}$\parn
$\bullet$
$\cals\in\sltz$, symmetric and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=A$\parn
$\bullet$
$H^1(\Psi;\coker(I-\cals)) \cong \frac {\coker(I-\cals)}{\im({I-A})}$.\parn
$\bullet$ $(c_1,c_2)=(0,0)$ or $(c_1,c_2)=(\frac12,\frac12)$\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig,\alpha=  (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_4$, $a_4=-({a_1}{c_1}+{a_2}{c_2}) + \frac{i}{4} (i=0,1,2,3)$\parn
$\bullet$ Torsion-free for $i=1,3$

\item[\gr{4b}]
$\Psi=\bbz_4$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A \tau)}$\parn
$\bullet$
$\cals\in\sltz$, symmetric and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=A$\parn
$\bullet$
$H^1(\Psi;\coker(I-\cals)) \cong \frac {\coker(I-\cals)}{\im({I-A})}$.\parn
$\bullet$ $(c_1,c_2)=(0,0)$ or $(c_1,c_2)=(\frac12,\frac12)$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3,A \tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))$ is trivial.\parn
$\bullet$ Always has torsion \parn

\item[\gr{4c}]
$\Psi=\bbz_4 \x \bbz_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A), \bar\gamma= (\bft^{0}_3,\tau)}$\parn
$\bullet$
$\cals\in\sltz$, symmetric and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=A$\parn
$\bullet$
$H^1(\Psi;\coker(I-\cals))$ is all $\bfa$, $\bfr$ with $(I-\cals)\bfa\equiv(I-\cals)\bfr\equiv\bfzero$, $\bfr=(0,0)$ or $(\frac12,\frac12)$,
and $\bfa$ taken modulo $(I-A)\bfv$, for $(I-\cals)\bfv \equiv \bfzero$.\parn
$\bullet$ $(c_1,c_2)=(0,0)$ or $(c_1,c_2)=(\frac12,\frac12)$\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Tables \ref{t5} and \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \gamma= (\bft^{t_1}_1\bft^{t_2}_2 \bft^{0}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_4$, $a_4=-({a_1}{c_1}+{a_2}{c_2}) + \frac{i}{4} (i=0,1,2,3)$\parn
$\bullet$ Always has torsion \parn

\item[\gr{4d}]
$\Psi=\bbz_4 \x \bbz_2$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A), \bar\gamma=(\bft^{\frac 12}_3,\tau)}$\parn
$\bullet$ $\cals=n K-I$, where $K\in\sltz$, symmetric with $\tr(K)= n>2$.\parn
$\bullet$ $\vara=A, \varphi(\bar \gamma) = K$\parn
$\bullet$ $H^1(\Psi;\coker(I-S)) \cong \frac {\coker(I-K)}{\im({I-A})}$.\parn
$\bullet$ $(c_1, c_2) \in \coker(I+K^T)$, $(c_1,c_2)=(0,0)$ or $(c_1,c_2)=(\frac12,\frac12)$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \gamma=(\bft^{\frac 12}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_4$, $a_4=-({a_1}{c_1}+{a_2}{c_2}) + \frac{i}{4} (i=0,1,2,3)$\parn
$\bullet$ Always has torsion \parn
\end{enumerate}
\begin{proof}

\bigskip
The computation for $H^1(\Psi;\coker(I-\cals))$ for cases \gr{4a} and \gr{4b} follows 
from that of case \gr{4} in Theorem \ref{ClassificationSol-geometry}. That
for cases \gr{4c} and \gr{4d} follows from computing the appropriate cocycle conditions
in Remark \ref{cocycleConditions}, together with application of Corollary \ref{a-conj-zero3}. For case \gr{4c}, notice that the cyclic power cocycle condition for $\gamma$ forces $2\bfr \equiv 0$. We obtain an additional condition on $\bfr$ from the commutator cocycle condition in case \gr{4c}:
\begin{align*}
(I-\varphi(\alpha))\bfr &=(I-\varphi(\gamma))\bfa\\
\left[ \begin{matrix} 1 & -1 \\ 1 & 1\end{matrix} \right] \bfr &= \bfzero,
\end{align*}
which forces $(t_1,t_2) = (0,0)$ or $(t_1,t_2) = (\frac12,\frac12).$

\bigskip
In all cases, the conditions on $c_1, c_2$ follow from the Tables \ref{t1}, \ref{t2}, and \ref{t4}. We can take $c_3=0$ in case \gr{4b} by Lemma \ref{solrc3}. From Table \ref{t111}, $c_3=0$ is forced in case \gr{4d}. For cases \gr{4a} and \gr{4c}, we have two choices for $c_3$.

\bigskip
For the choices we have for $a_4$ in cases \gr{4a}, \gr{4c}, and \gr{4d}, note that from Table \ref{t9}
\[
\alpha^4 = \bft_4^{4(a_4+{a_1}{c_1}+{a_2}{c_2})}.
\]


\bigskip
Recall that $\Sol$ groups of type \gr{4} always have torsion. Lemma \ref{torSolr} shows  that groups of type \gr{4b}, \gr{4c}, and \gr{4d} all have torsion. Only in case \gr{4a} 
can we resolve torsion in the central direction, just like the case of 
$\Solof$ crystallographic groups of type \gr{4} (Theorem \ref{ClassificationSolof-geometry}).
\qedhere
\end{proof}


\begin{theorem}[\gr{5} Family]
\label{5family}
The crystallographic groups of $\Solr$, lifting an abstract kernel
$\bbzphi \ra \gltz$ of case \gr{5} of {\rm Theorem \ref{abstract-kernel}},
\[A=\left[\begin{matrix} 1 & 0\\ 0 & -1 \end{matrix} \right],
B=\left[\begin{matrix} -1 & 0\\ 0 & -1\end{matrix} \right], \text{ are}\]
\end{theorem}
\begin{enumerate}
\item[\gr{5a}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3,  \bar\alpha= (\bft^{\frac 12}_3,A), \bar\beta=(\bft^{0}_3,B)}$\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$, $\det(K)=-1$, and  $\tr(K)=n>0$.\parn
$\bullet$ $\vara=-K, \varb = B$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals)) \cong \frac {\coker(I+K)}{2\coker({I+K})}$\parn
$\bullet$ $(c_1,c_2) \in \coker(I+K^T)$, $c_1=0$ or $\frac12$, $c_2=0$ or $\frac12$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha= (\bft^{\frac 12}_3 \bft^{a_4}_4,A), \beta=(\bft^{b_1}_1\bft^{b_2}_2\bft^{0}_3 \bft^{b_4}_4,B)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2 \x \bbz_2$, $a_4=0$ or $\frac12$, $b_4=-({a_1}{c_1}+{a_2}{c_2})+ \frac{i}{2} (i=0,1)$\parn
$\bullet$ Torsion-free when $i=1$


\item[\gr{5b}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{\frac 12}_3,A \tau),\bar\beta= (\bft^{0}_3,B)}$\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$, $\det(K)=-1$, $\tr(K)=n>0$.\parn
$\bullet$ $\vara=-K, \varb = B$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals)) \cong \frac {\coker(I+K)}{2\coker({I+K})}$\parn
$\bullet$ $(c_1,c_2) \in \coker({K^T}-I)$, $c_1=0$ or $\frac12$, $c_2=0$ or $\frac12$\par
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{\frac 12}_3,A \tau),\beta= (\bft^{b_1}_1\bft^{b_2}_2\bft^{0}_3 \bft^{b_4}_4,B)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $b_4=-({a_1}{c_1}+{a_2}{c_2})+ \frac{i}{2} (i=0,1)$\parn
$\bullet$ Torsion-free when $i=1$

\item[\gr{5c}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{\frac 12}_3,A), \bar\beta=(\bft^{0}_3,B \tau)}$\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$, $\det(K)=-1$, $\tr(K)=n>0$.\parn
$\bullet$ $\vara=-K, \varb = B$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals)) \cong \frac {\coker(I+K)}{2\coker({I+K})}$\parn
$\bullet$ $(c_1,c_2) \in \coker(I+K^T)$\parn
$\bullet$ Unique choice for $c_3$ (up to isomorphism), found by solving $(\alpha \beta)^2 \in \tgamsig$, this
$c_3$ must satisfy the conditions in Table \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{\frac 12}_3 \bft^{a_4}_4,A), \beta=(\bft^{b_1}_1\bft^{b_2}_2\bft^{0}_3,B \tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $a_4=0$ or $\frac12$\parn
$\bullet$ Always has torsion \parn

\item[\gr{5d}]
$\Psi=\bbz^3_2$, $\bbzpsi=\angles{\bft_3,  \bar\alpha= (\bft^{\frac 12}_3,A), \bar\beta=(\bft^{0}_3,B), \bar\gamma=(\bft^{0}_3,\tau)}$\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$, $\det(K)=-1$, and $\tr(K)=n>0$.\parn
$\bullet$ $\vara=-K, \varb = B, \varphi(\bar \gamma) = I$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfb$, $\bfr$ with
$(I+K)\bfb \equiv (I+K)\bfr \equiv\bfzero$, $2\bfr\equiv\bfzero$, and $\bfb$ taken modulo $2\bfb$.\parn
$\bullet$ $(c_1,c_2) \in \coker(I+K^T)$, $c_1=0$ or $\frac12$, $c_2=0$ or $\frac12$\parn
$\bullet$ Unique choice for $c_3$ (up to isomorphism), found by solving $(\alpha \gamma)^2 \in \tgamsig$, this
$c_3$ must satisfy the conditions in Tables \ref{t5} and  \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha= (\bft^{\frac 12}_3 \bft^{a_4}_4,A), \beta=(\bft^{b_1}_1\bft^{b_2}_2\bft^{0}_3 \bft^{b_4}_4,B),\gamma=(\bft^{t_1}_1\bft^{t_2}_2\bft^{0}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2 \x \bbz_2$, $a_4=0$ or $\frac12$, $b_4=-({a_1}{c_1}+{a_2}{c_2})+ \frac{i}{2} (i=0,1)$\parn\parn
$\bullet$ Always has torsion \parn

\end{enumerate}
\begin{proof}
First, observe that
\begin{align*}
\bbzpsi&=\angles{\bft_3, \bar\alpha=(\bft^{\frac 12}_3,A), \bar\beta=(\bft^{0}_3,B), \bar\gamma=(\bft^{\frac12}_3,\tau)}\\
&=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A\tau), \bar\beta=(\bft^{0}_3,B), \bar\gamma=(\bft^{\frac12}_3,\tau)}
\end{align*}
cannot be lifted to a $\Solr$ crystallographic group by Lemma \ref{PPM-A302}, and 
we therefore disregard this case.

Also, note that
\begin{align*}
\bbzpsi&=\angles{\bft_3, \bar\alpha=(\bft^{\frac 12}_3,A\tau), \bar\beta=(\bft^{0}_3,B\tau)}\\
&=\angles{\bft_3,  \bar\alpha=(\bft^{\frac 12}_3,-A),  \bar\beta=(\bft^{0}_3,B\tau)}
\end{align*}
is conjugate to a $\Solr$ crystallographic group of type \gr{5c}, as $A$ and $-A$ are conjugate in $D_4 \x \bbz_2$.

\bigskip
The statements regarding $H^1(\Psi;\coker(I-\cals))$ for cases \gr{5a}, \gr{5b}, and \gr{5c} follow 
from that of case \gr{5} in Theorem \ref{ClassificationSol-geometry}. The statement for Case
\gr{5d} follows from the cocycle conditions
in Remark \ref{cocycleConditions}, together with application of Corollary \ref{a-conj-zero3}.

\bigskip
The conditions on the $c_i$, as well as the choices for the exponents on $\bft_4$, follow from our proceeding classification of $\Solr$ groups
with cyclic holonomy. 

\bigskip
For the torsion criteria, note that in cases \gr{5a} and \gr{5b}, both
$\angles{\tgamsig, \alpha}$ and $\angles{\tgamsig, \alpha\beta}$ are torsion-free
by Lemma \ref{EasyTorSolr}. In these cases we must take $i=1$ to get
that $\angles{\tgamsig, \beta}$ (and hence $\varPi$) is torsion-free. Groups of
type \gr{5c} have torsion by Lemma \ref{torSolr}. As a group of type \gr{5d} contains
a group of type \gr{5c}, it likewise has torsion.
\qedhere
\end{proof}

\begin{theorem}[\gr{6a} Family]
\label{6afamily}
The crystallographic groups of $\Solr$, lifting an abstract kernel
$\bbzphi \ra \gltz$ of case \gr{6a} of {\rm Theorem \ref{abstract-kernel}},
\[A=\left[\begin{matrix} 0 & 1\\ 1 & 0 \end{matrix} \right],
B=\left[\begin{matrix} -1 & 0\\ 0 & -1\end{matrix} \right], \text{ are}\]
\end{theorem}
\begin{enumerate}
\item[\gr{6aa}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A), \bar\beta=(\bft^{0}_3,B)}$\parn
$\bullet$
$\cals\in\sltz$ with $\sigma_{12}=-\sigma_{21}$ and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=A, \varb = B$\parn
$\bullet$
$H^1(\Psi;\coker(I-\cals))$ is
$\bfa$, $\bfb$ with $(I-\cals)\bfa\equiv(I-\cals)\bfb\equiv\bfzero$,
$a_2\equiv -a_1$, $b_1-b_2-2 a_1 \equiv 0$, $\bfa$ and $\bfb$ modulo $(v_1-v_2)
{\scriptstyle{\xy{1}{-1}}}$ and  $2\bfv$ for $\bfv \in \coker(I-\cals)$, respectively.\parn
$\bullet$ $(c_1,c_2)=(0,0)$ or $(\frac12,\frac12)$\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{b_1}_1\bft^{b_2}_2 \bft^{0}_3 \bft^{b_4}_4,B)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2 \x \bbz_2$, $a_4 = 0$ or $\frac12$, $b_4=-({a_1}{c_1}+{a_2}{c_2})+ \frac{i}{2} (i=0,1)$\parn
$\bullet$ Always has torsion \parn

\item[\gr{6ab}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A \tau), \bar\beta=(\bft^{0}_3,B)}$\parn
$\bullet$
$\cals\in\sltz$ with $\sigma_{12}=-\sigma_{21}$ and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=A, \varb = B$\parn
$\bullet$
$H^1(\Psi;\coker(I-\cals))$ is
$\bfa$, $\bfb$ with $(I-\cals)\bfa\equiv(I-\cals)\bfb\equiv\bfzero$,
$a_2\equiv -a_1$, $b_1-b_2-2 a_1 \equiv 0$, $\bfa$ and $\bfb$ modulo $(v_1-v_2)
{\scriptstyle{\xy{1}{-1}}}$ and  $2\bfv$ for $\bfv \in \coker(I-\cals)$, respectively.\parn
$\bullet$ $(c_1,c_2)=(0,0)$ or $(\frac12,\frac12)$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3,A \tau), \beta=(\bft^{b_1}_1\bft^{b_2}_2 \bft^{0}_3 \bft^{b_4}_4 ,B)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $b_4=-({a_1}{c_1}+{a_2}{c_2})+ \frac{i}{2} (i=0,1)$\parn
$\bullet$ Always has torsion \parn

\item[\gr{6ac}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A), \bar\beta=(\bft^{0}_3,B \tau)}$\parn
$\bullet$
$\cals\in\sltz$ with $\sigma_{12}=-\sigma_{21}$ and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=A, \varb = B$\parn
$\bullet$
$H^1(\Psi;\coker(I-\cals))$ is
$\bfa$, $\bfb$ with $(I-\cals)\bfa\equiv(I-\cals)\bfb\equiv\bfzero$,
$a_2\equiv -a_1$, $b_1-b_2-2 a_1 \equiv 0$, $\bfa$ and $\bfb$ modulo $(v_1-v_2)
{\scriptstyle{\xy{1}{-1}}}$ and  $2\bfv$ for $\bfv \in \coker(I-\cals)$, respectively.\parn
$\bullet$ $c_1-c_2=0$\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Tables \ref{t5} and \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{b_1}_1\bft^{b_2}_2 \bft^{0}_3,B \tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $a_4 = 0$ or $\frac12$\parn
$\bullet$ Always has torsion \parn

\item[\gr{6ad}]
$\Psi=\bbz^3_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A), \bar\beta=(\bft^{0}_3,B),\bar\gamma= (\bft^{0}_3,\tau)}$\parn
$\bullet$
$\cals\in\sltz$ with $\sigma_{12}=-\sigma_{21}$ and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=A, \varb = B, \varphi(\bar \gamma) = I$\parn
$\bullet$
$H^1(\Psi;\coker(I-\cals))$ is
$\bfa$, $\bfb$, $\bfr$ with $(I-\cals)\bfa\equiv(I-\cals)\bfb\equiv(I-\cals)\bfr\equiv\bfzero$, $\bfr=(0,0)$ or $(\frac12,\frac12)$,
$a_2\equiv -a_1$, $b_1-b_2-2 a_1 \equiv 0$, \newline$\bfa$ and $\bfb$ modulo $(v_1-v_2)
{\scriptstyle{\xy{1}{-1}}}$ and  $2\bfv$ for $\bfv \in \coker(I-\cals)$, respectively.\parn
$\bullet$ $(c_1,c_2)=(0,0)$ or $(\frac12,\frac12)$\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Tables \ref{t5} and \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{b_1}_1\bft^{b_2}_2 \bft^{0}_3 \bft^{b_4}_4,B),\gamma= (\bft^{t_1}_1\bft^{t_2}_2 \bft^{0}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2 \x \bbz_2$, $a_4 = 0$ or $\frac12$, $b_4=-({a_1}{c_1}+{a_2}{c_2})+ \frac{i}{2} (i=0,1)$\parn
$\bullet$ Always has torsion \parn

\item[\gr{6ae}]
$\Psi=\bbz^3_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A), \bar\beta=(\bft^{0}_3,B), \bar\gamma=(\bft^{\frac 12}_3,\tau)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\sltz$ with $\tr(K)= n>2$ and
$k_{12}=-k_{21}$.\parn
$\bullet$ $\vara=A, \varb = B, \varphi(\bar \gamma) = K$\parn
$\bullet$
$H^1(\Psi;\coker(I-\cals))$ is
$\bfa$, $\bfb$ with $(I-K)\bfa\equiv(I-K)\bfb\equiv\bfzero$,
$a_2\equiv -a_1$, $b_1-b_2-2 a_1 \equiv 0$, $\bfa$ and $\bfb$ modulo $(v_1-v_2)
{\scriptstyle{\xy{1}{-1}}}$ and  $2\bfv$ for $\bfv \in \coker(I-K)$, respectively.\parn
$\bullet$ $(c_1,c_2) \in \coker(I+K^T)$, and $(c_1,c_2)=(0,0)$ or $(\frac12,\frac12)$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{b_1}_1\bft^{b_2}_2 \bft^{0}_3 \bft^{b_4}_4,B), \gamma=(\bft^{\frac 12}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2 \x \bbz_2$, $a_4 = 0$ or $\frac12$, $b_4=-({a_1}{c_1}+{a_2}{c_2})+ \frac{i}{2} (i=0,1)$\parn
$\bullet$ Always has torsion \parn

\item[\gr{6af}]
$\Psi=\bbz^3_2$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A), \bar\beta=(\bft^{0}_3,B \tau), \bar\gamma=(\bft^{\frac 12}_3,\tau)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\sltz$ with $\tr(K)= n>2$ and
$k_{12}=-k_{21}$.\parn
$\bullet$ $\vara=A, \varb = B, \varphi(\bar \gamma) = K$\parn
$\bullet$
$H^1(\Psi;\coker(I-\cals))$ is
$\bfa$, $\bfb$ with $(I-K)\bfa\equiv(I-K)\bfb\equiv\bfzero$,
$a_2\equiv -a_1$, $b_1-b_2-2 a_1 \equiv 0$, $\bfa$ and $\bfb$ modulo $(v_1-v_2)
{\scriptstyle{\xy{1}{-1}}}$ and  $2\bfv$ for $\bfv \in \coker(I-K)$, respectively.\parn
$\bullet$ $c_1-c_2=0$, $(c_1,c_2)\in\coker(I+K^T)$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t5} and Table \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{b_1}_1\bft^{b_2}_2 \bft^{0}_3 \bft^{b_4}_4,B \tau), \gamma=(\bft^{\frac 12}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2 \x \bbz_2$,  $a_4 = 0$ or $\frac12$,
two choices for $b_4$. We have
$(\beta \gamma)^2 = \bft_1^{n_1} \bft_2^{n_2}\bft_3 \bft_4^{2b_4 + v}$ for integers $n_1$, and some rational number $v$, and therefore two solutions for $b_4$.\parn
$\bullet$ Always has torsion \parn
\end{enumerate}
\begin{proof}
First, observe that if a $\Solr$ crystallographic group $\varPi$ has
\begin{align*}
\bbzpsi&=\angles{\bft_3, \bar\alpha=(\bft^{0}_3,A\tau), \bar\beta=(\bft^{0}_3,B\tau)}\\
&=\angles{\bft_3, (\bft^{0}_3,-A), (\bft^{0}_3,B\tau)},
\end{align*}
then it is conjugate to a $\Solr$ crystallographic group of type \gr{6ac}, as
$-A$ and $A$ are conjugate in $D_4 \x \bbz_2$ (and $B\tau$ remains fixed under this conjugation).

\bigskip
The statements regarding $H^1(\Psi;\coker(I-\cals))$ for cases \gr{6aa}, \gr{6ab}, and \gr{6ac} follow 
from that of case \gr{6a} in Theorem \ref{ClassificationSol-geometry}. The statements for cases \gr{6ad}, \gr{6ae}, and \gr{6af} follow from the cocycle conditions
in Remark \ref{cocycleConditions}, together with application of Corollary \ref{a-conj-zero3}.

\bigskip
The conditions on the $c_i$, as well as the choices for the exponents on $\bft_4$, follow from our proceeding classification of $\Solr$ groups
with cyclic holonomy. In particular, note that in case \gr{6aa}, from the cyclic 
cases, we obtain two conditions on $(c_1, c_2)$
\begin{align*}
&c_1-c_2=0\\
&c_1=0 \text{ or } \frac12, c_2=0 \text{ or } \frac12,
\end{align*}
and these two conditions force $(c_1,c_2) = (0,0)$ or $(\frac12,\frac12)$.

\bigskip
All $\Solr$ crystallographic groups of this type have torsion, just like their
corresponding $\Sol$ crystallographic groups.
In case \gr{6aa}, we need the groups 
\begin{align*}
&\angles{\tgamsig, \alpha}\\
&\angles{\tgamsig, \beta}\\
&\angles{\tgamsig, \alpha\beta}
\end{align*}
to all be torsion-free. All these groups
are torsion, when projected to $\Sol$ crystallographic groups, and so we must resolve the
torsion using the $\bbr$ direction of $\Solr$. We have:
\begin{align*}
a_4=0, i=0 &\implies \text{ all have torsion}\\
a_4=\frac12, i=0 &\implies \angles{\tgamsig, \beta} \text{ has torsion}\\
a_4=0, i=1 &\implies \angles{\tgamsig, \alpha} \text{ has torsion}\\
a_4=\frac12, i=1 &\implies \angles{\tgamsig, \alpha\beta} \text{ has torsion}
\end{align*}
Now groups of type \gr{6ad} and \gr{6ae} have torsion, as they contain a group of type
\gr{6aa} as a subgroup. To see that groups of type  \gr{6ab}, \gr{6ac}, and \gr{6af} have torsion,
we may appeal to Lemma \ref{torSolr}.\qedhere
\end{proof}

\begin{theorem}[\gr{6a\ii} Family]
\label{6aifamily}
The crystallographic groups of $\Solr$ lifting an abstract kernel
$\bbzphi \ra \gltz$ of case \gr{6a\ii} of {\rm Theorem \ref{abstract-kernel}},
\[A=\left[\begin{matrix} 0 & 1\\ 1 & 0 \end{matrix} \right],
B=\left[\begin{matrix} -1 & 0\\ 0 & -1\end{matrix} \right], \text{ are}\]
\end{theorem}
\begin{enumerate}
\item[\gr{6a\ii a}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A), \bar\beta=(\bft^{0}_3,B)}$\parn
$\bullet$
$\cals\in\sltz$
with $\sigma_{11}=\sigma_{22}$ and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=\matPpm, \varb = B$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfa$, $\bfb$ with
$(I-\cals)\bfa \equiv (I-\cals)\bfb \equiv \bfzero$,
 $2 a_1\equiv 0$, $2b_2-2a_2 \equiv \bfzero$,
$\bfa$ and $\bfb$ taken modulo $\xy{0}{2v_2}$ and $2\bfv$ for $\bfv \in \coker(I-\cals)$,
respectively.\parn
$\bullet$ $c_1=0$ or $\frac12$, $c_2=0$ or $\frac12$\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{b_1}_1\bft^{b_2}_2\bft^{0}_3 \bft^{b_4}_4,B)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2 \x \bbz_2$, $a_4 = -{a_2}{c_2}+\frac{i}{2} (i=0,1)$, $b_4=-({a_1}{c_1}+{a_2}{c_2})+ \frac{j}{2} (j=0,1)$\parn
$\bullet$ Torsion-free when $j=1$, and...
\begin{itemize}
\item when $i=0$, $a_1=\tfrac12$
and $a_2\not\equiv \frac{(\sigma_{11}+1)(2 n+1)}{2\sigma_{12}}$ for any $n\in\bbz$
\item when $i=1$, $a_1+b_1 \neq  \frac{\sigma_{12}(2 n+1)}{2(\sigma_{11}-1)}$ for any $n\in\bbz$ and $a_2-b_2 =\frac 1 2$
\end{itemize}

\item[\gr{6a\ii b}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A \tau), \bar\beta=(\bft^{0}_3,B)}$\parn
$\bullet$
$\cals\in\sltz$
with $\sigma_{11}=\sigma_{22}$ and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=\matPpm, \varb = B$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfa$, $\bfb$ with
$(I-\cals)\bfa \equiv (I-\cals)\bfb \equiv \bfzero$,
 $2 a_1\equiv 0$, $2b_2-2a_2 \equiv \bfzero$,
$\bfa$ and $\bfb$ taken modulo $\xy{0}{2v_2}$ and $2\bfv$ for $\bfv \in \coker(I-\cals)$,
respectively.\parn
$\bullet$ $c_1=0$ or $\frac12$, $c_2=0$ or $\frac12$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3,A \tau), \beta=(\bft^{b_1}_1\bft^{b_2}_2 \bft^{0}_3 \bft^{b_4}_4,B)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $b_4=-({a_1}{c_1}+{a_2}{c_2})+ \frac{j}{2} (j=0,1)$\parn
$\bullet$ Torsion-free when $j=1$, $a_1=\tfrac12$, $a_2 = b_2 + \frac12$,
$b_1\not\equiv \frac {\sigma_{12}(2 n+1)}{2(\sigma_{11}-1)} + \frac 1 2$,
$b_2\not\equiv \frac{(\sigma_{11}+1)(2 m+1)}{2\sigma_{12}} + \frac 1 2$ for any $m,n \in \bbz$.


\item[\gr{6a\ii c}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A), \bar\beta=(\bft^{0}_3,B \tau)}$\parn
$\bullet$
$\cals\in\sltz$
with $\sigma_{11}=\sigma_{22}$ and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=\matPpm, \varb = B$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfa$, $\bfb$ with
$(I-\cals)\bfa \equiv (I-\cals)\bfb \equiv \bfzero$,
 $2 a_1\equiv 0$, $2b_2-2a_2 \equiv \bfzero$,
$\bfa$ and $\bfb$ taken modulo $\xy{0}{2v_2}$ and $2\bfv$ for $\bfv \in \coker(I-\cals)$,
respectively.\parn
$\bullet$ $c_2=0$ or $\frac12$\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Tables \ref{t5} and \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{b_1}_1\bft^{b_2}_2 \bft^{0}_3,B \tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))= \bbz_2$, $a_4 = -{a_2}{c_2}+\frac{i}{2} (i=0,1)$\parn
$\bullet$ Always has torsion \parn

\item[\gr{6a\ii d}]
$\Psi=\bbz^3_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A), \bar\beta=(\bft^{0}_3,B), \bar\gamma=(\bft^{0}_3,\tau)}$\parn
$\bullet$
$\cals\in\sltz$
with $\sigma_{11}=\sigma_{22}$ and $\tr(\cals)>2$.\parn
$\bullet$ $\vara=\matPpm, \varb = B, \varphi(\bar \gamma) = I$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfa$, $\bfb$, $\bfr$ with
$(I-\cals)\bfa \equiv (I-\cals)\bfb \equiv (I-\cals)\bfr\equiv\bfzero$,
 $2 a_1\equiv 0$, $2b_2-2a_2 \equiv \bfzero$, $2\bfr=\bfzero$,
$\bfa$ and $\bfb$ taken modulo $\xy{0}{2v_2}$ and $2\bfv$ for $\bfv \in \coker(I-\cals)$,
respectively.\parn
$\bullet$ $c_1=0$ or $\frac12$, $c_2=0$ or $\frac12$\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Tables \ref{t5} and \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{b_1}_1\bft^{b_2}_2 \bft^{0}_3 \bft^{b_4}_4,B), \gamma=(\bft^{t_1}_1\bft^{t_2}_2 \bft^{0}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2 \x \bbz_2$, $a_4 = -{a_2}{c_2}+\frac{i}{2} (i=0,1)$, $b_4=-({a_1}{c_1}+{a_2}{c_2})+ \frac{j}{2} (j=0,1)$\parn
$\bullet$ Always has torsion \parn

\item[\gr{6a\ii e}]
$\Psi=\bbz^3_2$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A), \bar\beta=(\bft^{0}_3,B),\bar\gamma= (\bft^{\frac 12}_3,\tau)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\sltz$ with $\tr(K)= n>2$ and
$k_{11}=k_{22}$.\parn
$\bullet$ $\vara=\matPpm, \varb = B, \varphi(\bar \gamma) = K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfa$, $\bfb$ with
$(I-K)\bfa \equiv (I-K)\bfb \equiv \bfzero$,
 $2 a_1\equiv 0$, $2b_2-2a_2 \equiv \bfzero$,
$\bfa$ and $\bfb$ taken modulo $\xy{0}{2v_2}$ and $2\bfv$ for $\bfv \in \coker(I-K)$,
respectively.\parn
$\bullet$ $c_1=0$ or $\frac12$, $c_2=0$ or $\frac12$, and $(c_1,c_2) \in \coker(I+K^T)$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{b_1}_1\bft^{b_2}_2 \bft^{0}_3 \bft^{b_4}_4,B),\gamma= (\bft^{\frac 12}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2 \x \bbz_2$, $a_4 = -{a_2}{c_2}+\frac{i}{2} (i=0,1)$, $b_4=-({a_1}{c_1}+{a_2}{c_2})+ \frac{j}{2} (j=0,1)$\parn
$\bullet$ Torsion-free when $j=1$, and when the torsion-free criteria for $\angles{\tgamsig, \alpha,\beta}$ \gr{6a\ii a}, and $\angles{\tgamsig, \gamma\alpha,\beta}$ \gr{6a\ii b} are satisfied.

\item[\gr{6a\ii f}]
$\Psi=\bbz^3_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A), \bar\beta=(\bft^{0}_3,B \tau), \bar\gamma=(\bft^{\frac 12}_3,\tau)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\sltz$ with $\tr(K)= n>2$ and
$k_{11}=k_{22}$.\parn
$\bullet$ $\vara=\matPpm, \varb = B, \varphi(\bar \gamma) = K$\parn
$\bullet$
$H^1(\Psi;\coker(I-\cals))$ can be described as all $\bfa$, $\bfb$ with
$(I-K)\bfa\equiv(I-K)\bfb \equiv$ $ \bfzero$,
 $2 a_1\equiv 0$, $2b_2-2a_2 \equiv \bfzero$,
$\bfa$ and $\bfb$ taken modulo $\xy{0}{2v_2}$ and $2\bfv$ for $\bfv \in \coker(I-K)$,
respectively.
\parn
$\bullet$ $c_2=0$ or $\frac12$, $(c_1,c_2) \in \coker(I+K^T)$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t5} and Table \ref{t6} \parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{b_1}_1\bft^{b_2}_2 \bft^{0}_3 \bft^{b_4}_4,B \tau), \gamma=(\bft^{\frac 12}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2 \x \bbz_2$, $a_4 = -{a_2}{c_2}+\frac{i}{2} (i=0,1)$, two choices for $b_4$. We have
$(\beta \gamma)^2 = \bft_1^{n_1} \bft_2^{n_2}\bft_3 \bft_4^{2b_4 + v}$ for integers $n_1$, and some rational number $v$, and therefore two solutions for $b_4$.\parn
$\bullet$ Always has torsion
\end{enumerate}
\begin{proof}
Observe that if a $\Solr$ crystallographic group $\varPi$ has
\begin{align*}
\bbzpsi&=\angles{\bft_3, \bar\alpha=(\bft^{0}_3,A\tau), \bar\beta=(\bft^{0}_3,B\tau)}\\
&=\angles{\bft_3,  (\bft^{0}_3,-A), (\bft^{0}_3,B\tau)},
\end{align*}
then it is conjugate to a $\Solr$ crystallographic group of type \gr{6a\ii c}, similar
to the situation in Theorem \ref{6afamily}.

\bigskip
See Theorem \ref{ClassificationSol-geometry}, case \gr{6a\ii}, for $H^1(\Psi; \coker(I-\cals))$ in cases  \gr{6a\ii a},  \gr{6a\ii b}, and \gr{6a\ii c}. For cases \gr{6a\ii d}, \gr{6a\ii e}, and \gr{6a\ii f} use the cocycle conditions in Remark \ref{cocycleConditions}, together with Corollary \ref{a-conj-zero3}. The conditions on the $c_i$ follow from previous cases.
The computation of $H^2(\Psi; \calz(\tgamsig))$ is similar to that in
Theorem \ref{6afamily}, and so we obtain the stated exponents on $\bft_4$.

\bigskip
The groups of type \gr{6a\ii c} (and hence also those of type \gr{6a\ii d} and \gr{6a\ii f}) always have torsion by Lemma \ref{torSolr}.

In case \gr{6a\ii a}, we must take $j=1$ to ensure that $\angles{\tgamsig, \beta}$ is torsion-free. When $i=0$, $\angles{\tgamsig, \alpha\beta}$ is torsion-free. 
When $i=1$, $\angles{\tgamsig, \alpha}$ is torsion-free. Thus,
\begin{align*}
i=0 &\implies \angles{\tgamsig, \alpha} \text{ must be torsion-free projected to } \Sol \\
i=1,&\implies  \angles{\tgamsig, \alpha\beta}\text{ must be torsion-free projected to } \Sol
\end{align*}
Both the groups $\angles{\tgamsig, \alpha}$ and $\angles{\tgamsig, \alpha\beta}$
are $\Sol$ crystallographic groups of type \gr{3\ii}. The projection of
$\alpha \beta$ to $\Sol$ is 
\[(\bft_1^{a_1+b_1} \bft_2^{a_2-b_2},AB).\] 
By computing when
the appropriate affine maps in Proposition \ref{topologySolManifolds}
are fixed point free, we obtain the desired conditions.
See also Theorem \ref{ClassificationSolof-geometry}, case \gr{6a\ii}.

In case \gr{6a\ii b}, we must take $j=1$. 
Lemma \ref{torSolr} then implies that both the groups $\angles{\tgamsig, \alpha}$
and $\angles{\tgamsig, \alpha\beta}$ must both be torsion-free when projected to $\Sol$.
The proof for this case now proceeds as in Theorem \ref{ClassificationSolof-geometry}, case \gr{6a\ii}.

For case \gr{6a\ii e}, we must take $j=1$, which is
is necessary for the group $\angles{\tgamsig, \gamma, \beta}$ to be torsion-free. Then, $\varPi$ will be torsion-free precisely when the groups $\angles{\tgamsig, \alpha, \beta}$ and $\angles{\tgamsig, \gamma\alpha, \beta}$ are torsion-free, as the union of these three $\bbz_2^2$ holonomy
$\Solr$ crystallographic groups exhausts $\varPi$. 
\qedhere
\end{proof}


\begin{theorem}[\gr{6b} Family]
\label{6bfamily}
The crystallographic groups of $\Solr$ lifting an abstract kernel
$\bbzphi \ra \gltz$ of case \gr{6b} of {\rm Theorem \ref{abstract-kernel}},
\[A=\left[\begin{matrix} 0 & 1\\ 1 & 0 \end{matrix} \right],
B=\left[\begin{matrix} -1 & 0\\ 0 & -1\end{matrix} \right], \text{ are}\]
\end{theorem}
\begin{enumerate}
\item[\gr{6ba}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A), \bar\beta=(\bft^{\frac 12}_3,B)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\sltz$ with $\tr(K)= n>2$ and
$k_{12}=-k_{21}$.\parn
$\bullet$ $\vara=A, \varb = -K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals)) \cong H^1(\angles{A};\coker(I+K))$, that is,
all $\bfa $ with $(I+K)\bfa \equiv \bfzero$ and $a_2\equiv -a_1$ modulo
$(v_1-v_2){\scriptstyle{\xy{1}{-1}}}$, for $(I+K)\bfv \equiv \bfzero$. \parn
$\bullet$ $c_1-c_2=0$, $(c_1,c_2) \in \coker(I+K^T)$\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{\frac 12}_3 \bft^{b_4}_4,B)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2 \x \bbz_2$, $a_4 = 0$ or $\frac12$, $b_4=0$, $\frac12$\parn
$\bullet$ Torsion-free when $a_4=\frac12$, $b_4=0$

\item[\gr{6bb}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A \tau), \bar\beta=(\bft^{\frac 12}_3,B)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\sltz$ with $\tr(K)= n>2$ and
$k_{12}=-k_{21}$.\parn
$\bullet$ $\vara=A, \varb = -K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals)) \cong H^1(\angles{A};\coker(I+K))$, that is,
all $\bfa $ with $(I+K)\bfa \equiv \bfzero$ and $a_2\equiv -a_1$ modulo
$(v_1-v_2){\scriptstyle{\xy{1}{-1}}}$, for $(I+K)\bfv \equiv \bfzero$. \parn
$\bullet$ $c_1+c_2=0$, $(c_1,c_2) \in \coker(I+K^T)$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3,A \tau), \beta=(\bft^{\frac 12}_3 \bft^{b_4}_4,B)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))= \bbz_2$, $b_4=0$, $\frac12$\parn
$\bullet$ Always has torsion \parn

\item[\gr{6bc}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A), \bar\beta=(\bft^{\frac 12}_3,B \tau)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\sltz$ with $\tr(K)= n>2$ and
$k_{12}=-k_{21}$.\parn
$\bullet$ $\vara=A, \varb = -K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals)) \cong H^1(\angles{A};\coker(I+K))$, that is,
all $\bfa $ with $(I+K)\bfa \equiv \bfzero$ and $a_2\equiv -a_1$ modulo
$(v_1-v_2){\scriptstyle{\xy{1}{-1}}}$, for $(I+K)\bfv \equiv \bfzero$. \parn
$\bullet$ $c_1-c_2=0$, $(c_1,c_2) \in \coker(K^{T}-I)$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{\frac 12}_3,B \tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $a_4 = 0$ or $\frac12$\parn
$\bullet$ Always has torsion \parn

\item[\gr{6bd}]
$\Psi=\bbz^3_2$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A), \bar\beta=(\bft^{\frac 12}_3,B),\bar\gamma= (\bft^{0}_3,\tau)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\sltz$ with $\tr(K)= n>2$ and
$k_{12}=-k_{21}$.\parn
$\bullet$ $\vara=A, \varb = -K, \varphi(\bar \gamma) = I$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ can be described as all $\bfa$, $\bfr$ with $(I+K)\bfa\equiv (I+K)\bfr \equiv \bfzero$, $\bfr=(0,0)$ or $(\frac12,\frac12)$
and $a_2\equiv -a_1$, \newline $\bfa$ taken modulo
$(v_1-v_2){\scriptstyle{\xy{1}{-1}}}$ where $(I+K)\bfv\equiv \bfzero$.\parn
$\bullet$ $(c_1,c_2)=(0,0)$ or $(c_1,c_2)=(\frac12,\frac12)$, and $(c_1,c_2) \in \coker(I+K^T)$\parn
$\bullet$ Unique choice for $c_3$ (up to isomorphism), found by solving $(\beta \gamma)^2 \in \tgamsig$, this
$c_3$ must satisfy the conditions in Tables \ref{t5} and  \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{\frac 12}_3 \bft^{b_4}_4,B),\gamma= (\bft^{t_1}_1\bft^{t_2}_2 \bft^{0}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2 \x \bbz_2$, $a_4 = 0$ or $\frac12$, $b_4=0$, $\frac12$\parn
$\bullet$ Always has torsion \parn

\end{enumerate}
\begin{proof}
First, observe that if a $\Solr$ crystallographic group $\varPi$ has
\begin{align*}
\bbzpsi&=\angles{\bft_3, \bar\alpha=(\bft^{0}_3,A\tau), \bar\beta=(\bft^{\frac12}_3,B\tau)}\\
&=\angles{\bft_3,  (\bft^{\frac12}_3,-A), (\bft^{\frac12}_3,B\tau)},
\end{align*}
then it is conjugate to a $\Solr$ crystallographic group of type \gr{6bc}, as
conjugation by $\bft^{-\frac14}_3$
sends $(\bft^{\frac12}_3,-A)$ to $(\bft^{0}_3,-A)$, and
$-A$ and $A$ are conjugate in $D_4 \x \bbz_2$.

Also, note that
\begin{align*}
\bbzpsi&=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A), \bar\beta=(\bft^{\frac 12}_3,B),\bar\gamma= (\bft^{\frac 12}_3,\tau)}\\
&=\angles{\bft_3, \bar\alpha=(\bft^{0}_3,A), \bar\beta=(\bft^{0}_3,B\tau),\bar\gamma= (\bft^{\frac 12}_3,\tau)}
\end{align*}
is a lift of type \gr{6a} from the $\Sol$ classification, and it corresponds to type \gr{6af} in Theorem \ref{6afamily}.

\bigskip
The statements regarding $H^1(\Psi;\coker(I-\cals))$ for cases \gr{6ba}, \gr{6bb}, and \gr{6bc} follow
from that of case \gr{6b} in Theorem \ref{ClassificationSol-geometry}. The statement for case \gr{6bd} follows from the cocycle conditions
in Remark \ref{cocycleConditions}, together with application of Corollary \ref{a-conj-zero3}.

\bigskip
The conditions on the $c_i$, as well as the choices for the exponents on $\bft_4$, follow from our proceeding classification of $\Solr$ groups
with cyclic holonomy. In case \gr{6bd}, from the cyclic 
cases, we obtain three conditions on $(c_1, c_2)$
\begin{align*}
&c_1-c_2=0\\
&c_1=0 \text{ or } \frac12, c_2=0 \text{ or } \frac12\\
&(c_1,c_2) \in \coker(I+K^T),
\end{align*}
and the first two conditions force $(c_1,c_2) = (0,0)$ or $(\frac12,\frac12)$.

\bigskip
For case \gr{6ba}, notice that both $\angles{\tgamsig, \alpha}$ and $\angles{\tgamsig, \alpha\beta}$ project to 
$\Sol$ groups with torsion, whereas $\angles{\tgamsig, \beta}$ is already torsion-free on
$\Solr$. These three groups are torsion-free only when $a_4=\frac12$ and $b_4=0$.
Lemma \ref{torSolr} implies the remaining groups of this type all have torsion.
\qedhere
\end{proof}


\begin{theorem}[\gr{6b\ii} Family]
\label{6bifamily}
The crystallographic groups of $\Solr$ lifting an abstract kernel
$\bbzphi \ra \gltz$ of case \gr{6b\ii} of {\rm Theorem \ref{abstract-kernel}},
\[A=\left[\begin{matrix} 0 & 1\\ 1 & 0 \end{matrix} \right],
B=\left[\begin{matrix} -1 & 0\\ 0 & -1\end{matrix} \right], \text{ are}\]
\end{theorem}
\begin{enumerate}
\item[\gr{6b\ii a}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A), \bar\beta=(\bft^{\frac 12}_3,B)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\sltz$ with $\tr(K)= n>2$ and
$k_{11}=k_{22}$.\parn
$\bullet$ $\vara=\matPpm, \varb = -K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals)) \cong H^1(\angles{A};\coker(I+K))$, that is,
all $\bfa $ with $(I+K)\bfa \equiv \bfzero$ and $2a_1\equiv 0$ taken modulo
$\xy{0}{2v_2}$, where $(I+K)\bfv \equiv \bfzero$.\parn
$\bullet$ $c_2=0$ or $\frac12$, $(c_1,c_2) \in \coker(I+K^T)$\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{\frac 12}_3 \bft^{b_4}_4,B)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2 \x \bbz_2$, $a_4 = -{a_2}{c_2}+\frac{i}{2} (i=0,1)$, $b_4=0$, $\frac12$\parn
$\bullet$ Torsion-free...
\begin{itemize}
\item when $i=0, b_4=0$, $a_1=\tfrac12$
and $a_2\not\equiv \frac{(k_{11}-1)(2 n+1)}{2k_{12}}$ for any $n \in \bbz$.
\item when $i=1, b_4=0$, always
\item when $i=0, b_4=\frac{1}{2}$, $a_1=\tfrac12$
and $a_2\not\equiv \frac{(\sigma_{11}+1)(2 n+1)}{2k_{12}}$ for any $n \in \bbz$.
\item when $i=1, b_4=\frac{1}{2}$, $\angles{\tgamsig, \alpha\beta}$ must be torsion-free, projected to $\Sol$.
\end{itemize}


\item[\gr{6b\ii b}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A \tau), \bar\beta=(\bft^{\frac 12}_3,B)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\sltz$ with $\tr(K)= n>2$ and
$k_{11}=k_{22}$.\parn
$\bullet$ $\vara=\matPpm, \varb = -K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals)) \cong H^1(\angles{A};\coker(I+K))$, that is,
all $\bfa $ with $(I+K)\bfa \equiv \bfzero$ and $2a_1\equiv 0$ taken modulo
$\xy{0}{2v_2}$, where $(I+K)\bfv \equiv \bfzero$.\parn
$\bullet$ $c_1=0$ or $\frac12$, $(c_1,c_2) \in \coker(I+K^T)$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3,A \tau), \beta=(\bft^{\frac 12}_3 \bft^{b_4}_4,B)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $b_4=0$, $\frac12$\parn
$\bullet$ Torsion-free if and only if
$a_1=\tfrac12$
and $a_2\not\equiv \frac{(k_{11}-1)(2 n+1)}{2k_{12}}$ for any $n \in \bbz$.


\item[\gr{6b\ii c}]
$\Psi=\bbz^2_2$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A), \bar\beta=(\bft^{\frac 12}_3,B \tau)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\sltz$ with $\tr(K)= n>2$ and
$k_{11}=k_{22}$.\parn
$\bullet$ $\vara=\matPpm, \varb = -K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals)) \cong H^1(\angles{A};\coker(I+K))$, that is,
all $\bfa $ with $(I+K)\bfa \equiv \bfzero$ and $2a_1\equiv 0$ taken modulo
$\xy{0}{2v_2}$, where $(I+K)\bfv \equiv \bfzero$.\parn
$\bullet$ $c_2=0$ or $\frac12$, $(c_1,c_2) \in \coker(K^{T}-I)$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3\bft^{a_4}_4,A), \beta=(\bft^{\frac 12}_3,B \tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $a_4 = -{a_2}{c_2}+\frac{i}{2} (i=0,1)$\parn
$\bullet$ Torsion-free...
\begin{itemize}
\item when $i=0$, $a_1=\tfrac12$
and $a_2\not\equiv \frac{(k_{11}-1)(2 n+1)}{2k_{12}}$ for any $n \in \bbz$.
\item when $i=1$, $\angles{\tgamsig, \alpha\beta}$ is torsion-free, projected to $\Sol$.
\end{itemize}

\item[\gr{6b\ii d}]
$\Psi=\bbz^3_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A), \bar\beta= (\bft^{\frac 12}_3,B), \bar\gamma=(\bft^{0}_3,\tau)}$\parn
$\bullet$
$\cals=n K-I$, where $K\in\sltz$ with $\tr(K)= n>2$ and
$k_{11}=k_{22}$.\parn
$\bullet$ $\vara=\matPpm, \varb = -K, \varphi(\bar \gamma) = I$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is
all $\bfa$, $\bfr$ with $(I+K)\bfa \equiv(I+K)\bfr\equiv\bfzero$, $2a_1\equiv 0$, and $2\bfr=0$,
$\bfa$ taken modulo
$\xy{0}{2v_2}$, where $(I+K)\bfv \equiv \bfzero$.\parn
$\bullet$ $c_1=0$ or $\frac12$, $c_2=0$ or $\frac12$, $(c_1,c_2) \in \coker(I+K^T)$\parn
$\bullet$ Unique choice for $c_3$ (up to isomorphism), found by solving $(\beta \gamma)^2 \in \tgamsig$, this
$c_3$ must satisfy the conditions in Tables \ref{t5} and  \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3\bft^{a_4}_4,A), \beta= (\bft^{\frac 12}_3\bft^{b_4}_4,B), \gamma=(\bft^{t_1}_1\bft^{t_2}_2 \bft^{0}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2 \x \bbz_2$, $a_4 = -{a_2}{c_2}+\frac{i}{2} (i=0,1)$, $b_4=0$, $\frac12$\parn
$\bullet$ Torsion-free when $\bfr \neq \bfzero$ and when the torsion-free criteria
for $\angles{\tgamsig, \alpha,\beta}$ \gr{6b\ii a}, and $\angles{\tgamsig, \gamma\alpha,\beta}$ \gr{6b\ii b} are satisfied.

\end{enumerate}
\begin{proof}
First, observe that if a $\Solr$ crystallographic group $\varPi$ has
\begin{align*}
\bbzpsi&=\angles{\bft_3, \bar\alpha=(\bft^{0}_3,A\tau), \bar\beta=(\bft^{\frac12}_3,B\tau)}\\
&=\angles{\bft_3,  (\bft^{\frac12}_3,-A), (\bft^{\frac12}_3,B\tau)},
\end{align*}
then it is conjugate to a $\Solr$ crystallographic group of type \gr{6b\ii c}, similar
to the situation in Theorem \ref{6bfamily}.

Also, note that
\begin{align*}
\bbzpsi&=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A), \bar\beta=(\bft^{\frac 12}_3,B),\bar\gamma= (\bft^{\frac 12}_3,\tau)}\\
&=\angles{\bft_3, \bar\alpha=(\bft^{0}_3,A), \bar\beta=(\bft^{0}_3,B\tau),\bar\gamma= (\bft^{\frac 12}_3,\tau)}
\end{align*}
is a lift of type \gr{6a\ii} from the $\Sol$ classification, and it corresponds to type \gr{6a\ii f} in Theorem \ref{6aifamily}.

\bigskip
See Theorem \ref{ClassificationSol-geometry}, case \gr{6b\ii}, for $H^1(\Psi; \coker(I-\cals))$ in cases  \gr{6b\ii a},  \gr{6b\ii b}, and \gr{6b\ii c}. For case \gr{6b\ii d}, use the cocycle conditions in Remark \ref{cocycleConditions}, together with Corollary \ref{a-conj-zero3}. The conditions on the $c_i$ follow from previous cases.
The computation of $H^2(\Psi; \calz(\tgamsig))$ is identical to that in
Theorem \ref{6bfamily}, and so we obtain the stated exponents on $\bft_4$.

In case \gr{6b\ii a}, the subgroup $\angles{\tgamsig, \beta}$ is torsion-free
by Lemma \ref{EasyTorSolr}. 
\begin{align*}
i=0, b_4=0 &\implies \angles{\tgamsig, \alpha}, \angles{\tgamsig, \alpha \beta} \text{ must be torsion-free projected to } \Sol\\
i=1, b_4=0 &\implies \text{ already torsion-free}\\
i=0, b_4=\frac12 &\implies \angles{\tgamsig, \alpha} \text{ must be torsion-free projected to } \Sol\\
i=1, b_4=\frac12 &\implies \angles{\tgamsig, \alpha\beta} \text{ must be torsion-free projected to } \Sol
\end{align*}
Thus, when $i=0, b_4=0$, we apply the criteria for case \gr{6b\ii} in
Theorem \ref{ClassificationSol-geometry}. When $i=1, b_4=0$, both $\angles{\tgamsig, \alpha}$ and $\angles{\tgamsig, \alpha \beta}$ are already torsion-free. When $i=0, b_4=\frac12$, $\angles{\tgamsig, \alpha \beta}$ is torsion-free, and we apply the criteria for case \gr{3\ii} in Theorem \ref{ClassificationSol-geometry} to ensure that $\angles{\tgamsig, \alpha}$ is torsion-free.

In case \gr{6b\ii b}, Lemma \ref{torSolr} applies to show that the projection of $\varPi$
to a $\Sol$ group of type \gr{6b\ii} must be torsion-free, so we apply the criteria for case \gr{6b\ii} in
Theorem \ref{ClassificationSol-geometry}.

In case \gr{6b\ii c}, when $i=0$, the projection of $\varPi$ to $\Sol$ must be torsion-free.
When $i=1$, $\angles{\tgamsig, \alpha}$ is already, torsion-free, but we must ensure that  $\angles{\tgamsig, \alpha\beta}$ is torsion-free when projected to $\Sol$.

When $\varPi$ is of type \gr{6b\ii d}, of course $\bfr\neq\bfzero$ is necessary and sufficient for the group $\angles{\tgamsig, \gamma, \beta}$ to be torsion-free. Then, $\varPi$ will be torsion-free precisely when the groups $\angles{\tgamsig, \alpha, \beta}$ and $\angles{\tgamsig, \gamma\alpha, \beta}$ are torsion-free, as the union of these three $\bbz_2^2$ holonomy
$\Solr$ crystallographic groups exhausts $\varPi$. See Example \ref{bbz23}.
\qedhere
\end{proof}



\begin{theorem}[\gr{7} Family]
\label{7family}
The crystallographic groups of $\Solr$ lifting an abstract kernel
$\bbzphi \ra \gltz$ of case \gr{7} of {\rm Theorem \ref{abstract-kernel}},
\[A=\left[\begin{matrix} 0 & 1\\ 1 & 0 \end{matrix} \right],
B=\left[\begin{matrix} 1 & 0\\ 0 & -1\end{matrix} \right], \text{ are}\]
\end{theorem}
\begin{enumerate}
\item[\gr{7a}]
$\Psi=D_4$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A),\bar\beta= (\bft^{\frac 12}_3,B)}$\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$, $\det(K)=-1$, $\tr(K)>0$, and
$k_{12}=-k_{21}$.\parn
$\bullet$ $\vara=A, \varb = -K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfa$ with $(I-\cals)\bfa \equiv \bfzero$ and $a_2 \equiv -a_1$, taken modulo
$(v_1-v_2){\scriptstyle{\xy{1}{-1}}}$, where $(I+K)\bfv \equiv \bfzero$.  \parn
$\bullet$ $c_1-c_2=0$, $(c_1,c_2)\in\coker(I+K^T)$\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3\bft^{a_4}_4,A),\beta= (\bft^{\frac 12}_3\bft^{b_4}_4,B)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig)) = \bbz_2 \x \bbz_2$, $a_4=0$ or $\frac12$, $b_4=0$ or $\frac12$\parn
$\bullet$ Always has torsion

\item[\gr{7b}]
$\Psi=D_4$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A\tau), \bar\beta=(\bft^{\frac 12}_3,B)}$\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$, $\det(K)=-1$, $\tr(K)>0$, and
$k_{12}=-k_{21}$.\parn
$\bullet$ $\vara=A, \varb = -K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfa$ with $(I-\cals)\bfa \equiv \bfzero$ and $a_2 \equiv -a_1$, taken modulo
$(v_1-v_2){\scriptstyle{\xy{1}{-1}}}$, where $(I+K)\bfv \equiv \bfzero$.  \parn
$\bullet$ $c_1+c_2=0$, $(c_1,c_2)\in\coker(I+K^T)$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3,A\tau), \beta=(\bft^{\frac 12}_3 \bft^{b_4}_4,B)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig)) =\bbz_2$, $b_4=0$ or $\frac12$\parn
$\bullet$ Always has torsion

\item[\gr{7c}]
$\Psi=D_4$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A), \bar\beta=(\bft^{\frac 12}_3,B \tau)}$\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$, $\det(K)=-1$, $\tr(K)>0$, and
$k_{12}=-k_{21}$.\parn
$\bullet$ $\vara=A, \varb = -K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfa$ with $(I-\cals)\bfa \equiv \bfzero$ and $a_2 \equiv -a_1$, taken modulo
$(v_1-v_2){\scriptstyle{\xy{1}{-1}}}$, where $(I+K)\bfv \equiv \bfzero$.  \parn
$\bullet$ $c_1-c_2=0$, $(c_1,c_2)\in\coker(K^{T}-I)$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{\frac 12}_3 ,B\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $a_4=0$ or $\frac12$\parn
$\bullet$ Always has torsion

\item[\gr{7d}]
$\Psi=D_4$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A \tau), \bar\beta=(\bft^{\frac 12}_3,B \tau)}$\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$, $\det(K)=-1$, $\tr(K)>0$, and
$k_{12}=-k_{21}$.\parn
$\bullet$ $\vara=A, \varb = -K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfa$ with $(I-\cals)\bfa \equiv \bfzero$ and $a_2 \equiv -a_1$, taken modulo
$(v_1-v_2){\scriptstyle{\xy{1}{-1}}}$, where $(I+K)\bfv \equiv \bfzero$.  \parn
$\bullet$ $c_1+c_2=0$, $(c_1,c_2)\in\coker(K^T-I)$\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A \tau), \beta=(\bft^{\frac 12}_3 ,B\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_4$, four choices for $a_4$, the solutions of
$(\alpha\beta)^4=\bft_4^{{j}}, (j=0,1,2,3)$.\parn
$\bullet$ Always has torsion

\item[\gr{7e}]
$\Psi=D_4 \x \bbz_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A), \bar\beta=(\bft^{\frac 12}_3,B),\bar\gamma= (\bft^{0}_3,\tau)}$\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$, $\det(K)=-1$, $\tr(K)>0$, and
$k_{12}=-k_{21}$.\parn
$\bullet$ $\vara=A, \varb = -K, \varphi(\bar \gamma) = I$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfa$, $\bfr$ with $(I-\cals)\bfa \equiv(I-\cals)\bfr\equiv \bfzero$, $a_2 \equiv -a_1$ and $(t_1,t_2)=(0,0)$ or $(\frac12,\frac12)$, \newline $\bfa$ taken modulo
$(v_1-v_2){\scriptstyle{\xy{1}{-1}}}$, where $(I+K)\bfv \equiv \bfzero$.  \parn
$\bullet$ $(c_1,c_2)=(0,0)$ or $(\frac12,\frac12)$, $(c_1,c_2)\in\coker(I+K^T)$\parn
$\bullet$ Unique choice for $c_3$ (up to isomorphism), found by solving $(\beta \gamma)^2 \in \tgamsig$, this
$c_3$ must satisfy the conditions in Tables \ref{t5} and  \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{\frac 12}_3 \bft^{b_4}_4,B),\gamma= (\bft^{t_1}_1\bft^{t_2}_2 \bft^{0}_3,\tau)}$ \parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2 \x \bbz_2$, $a_4=0$ or $\frac12$, $b_4=0$ or $\frac12$\parn
$\bullet$ Always has torsion
\end{enumerate}
\begin{proof}
Observe that
\begin{align*}
\bbzpsi&=\angles{\bft_3, \bar\alpha=(\bft^{0}_3,A), \bar\beta=(\bft^{\frac12}_3,B), \bar\gamma=(\bft^{\frac12}_3,\tau)}\\
&=\angles{\bft_3,  (\bft^{0}_3,A), (\bft^{0}_3,B\tau), (\bft^{\frac12}_3,\tau)},
\end{align*}
does not lift to a $\Solr$ crystallographic group by Lemma \ref{PPM-A302}.

\bigskip
The statements regarding $H^1(\Psi;\coker(I-\cals))$ for cases \gr{7a}, \gr{7b}, \gr{7c}, and \gr{7d} follow 
from that of case \gr{7} in Theorem \ref{ClassificationSol-geometry}. The statement for case \gr{7e} follows from the cocycle conditions
in Remark \ref{cocycleConditions}, together with  Corollary \ref{a-conj-zero3}. Note also that the conditions on the $c_i$ follow from previous cases.

\bigskip
However, the holonomy $\Psi$ is no longer a direct product of cyclic groups.
In case \gr{7a}, the holonomy $\Psi = D_4$ acts trivially on $\calz(\tgamsig)$, and with
this trivial action,
\[H^2(D_4; \bbz) = \bbz_2 \x \bbz_2.\]
Indeed, solving $\alpha^2 \in \tgamsig$ and $\beta^2 \in \tgamsig$
yield the choices $a_4=0$ or $a_4=\frac12$, and $b_4=0$ or $b_4=\frac12$.

In case \gr{7b}, we set $a_4=0$ by Lemma \ref{conj-in-t4Solr}, and obtain the two choices
for $b_4$ by solving $\beta^2 \in \tgamsig$. Case \gr{7c} is similar, except that we take
$b_4=0$. For \gr{7d}, $\angles{\tgamsig,\alpha\beta}$ is a $\Solr$ group of type \gr{4a}, so that we obtain 4 choices for $a_4$.

\bigskip
A $\Solr$ group of type \gr{7a} contains a subgroup of
type \gr{4a}, $\angles{\tgamsig, \alpha\beta}$.
In the notation of Theorem \ref{4family}, $\angles{\tgamsig, \alpha\beta}$ corresponds to the torsion case, that is, $i=0$ or $2$. Thus, all groups of type \gr{7a} (and hence of 
type \gr{7e}) have torsion.


The $\Solr$ groups of type \gr{7b} and \gr{7c} have a subgroup of type \gr{4b}, $\angles{\tgamsig, \alpha\beta}$, which has torsion. 

For case \gr{7d}, Lemma \ref{torSolr} applied
to the subgroup $\angles{\tgamsig,\alpha}$ shows that $\varPi$ has torsion.
\qedhere
\end{proof}



\begin{theorem}[\gr{7\ii} Family]
\label{7ifamily}
The crystallographic groups of $\Solr$ lifting an abstract kernel
$ \bbzphi \ra \gltz$ of case \gr{7\ii} of {\rm Theorem \ref{abstract-kernel}},
\[A=\left[\begin{matrix} 0 & 1\\ 1 & 0 \end{matrix} \right]
B=\left[\begin{matrix} 1 & 0\\ 0 & -1\end{matrix} \right], \text{ are}\]
\end{theorem}
\begin{enumerate}
\item[\gr{7\ii a}]
$\Psi=D_4$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A),\bar\beta= (\bft^{\frac 12}_3,B)}$\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$, $\det(K)=-1$, $\tr(K)=n>0$,
$k_{11}=k_{22}$.\parn
$\bullet$ $\vara=\matPpm, \varb = -K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfa$ with $(I-\cals)\bfa \equiv \bfzero$ and $2a_1 \equiv 0$, taken modulo
$\xy{0}{2v_2}$, where $(I+K)\bfv \equiv \bfzero$.  \parn
$\bullet$ $c_2=0$ or $\frac12$, $(c_1,c_2) \in \coker(I+K^T)$.\parn
$\bullet$ Two choices for $c_3$ (up to isomorphism), see Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A),\beta= (\bft^{\frac 12}_3 \bft^{b_4}_4,B)}$\parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2 \x \bbz_2$,  $a_4 = -{a_2}{c_2}+\frac{i}{2} (i=0,1)$, $b_4=0$ or $\frac12$\parn
$\bullet$ Always has torsion

\item[\gr{7\ii b}]
$\Psi=D_4$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A\tau),\bar\beta= (\bft^{\frac 12}_3,B)}$\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$, $\det(K)=-1$, $\tr(K)=n>0$,
$k_{11}=k_{22}$.\parn
$\bullet$ $\vara=\matPpm, \varb = -K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfa$ with $(I-\cals)\bfa \equiv \bfzero$ and $2a_1 \equiv 0$, taken modulo
$\xy{0}{2v_2}$, where $(I+K)\bfv \equiv \bfzero$.\parn
$\bullet$ $c_1=0$ or $\frac12$, $(c_1,c_2) \in \coker(I+K^T)$.\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3,A \tau),\beta= (\bft^{\frac 12}_3 \bft^{b_4}_4,B)}$\parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $b_4=0$ or $\frac12$\parn
$\bullet$ Always has torsion

\item[\gr{7\ii c}]
$\Psi=D_4$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A), \bar\beta=(\bft^{\frac 12}_3,B\tau)}$\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$, $\det(K)=-1$, $\tr(K)=n>0$,
$k_{11}=k_{22}$.\parn
$\bullet$ $\vara=\matPpm, \varb = -K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfa$ with $(I-\cals)\bfa \equiv \bfzero$ and $2a_1 \equiv 0$, taken modulo
$\xy{0}{2v_2}$, where $(I+K)\bfv \equiv \bfzero$.  \parn
$\bullet$ $c_2=0$ or $\frac12$, $(c_1,c_2) \in \coker(K^{T}-I)$.\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t5}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{\frac 12}_3 ,B \tau)}$\parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_2$, $a_4 = -{a_2}{c_2}+\frac{i}{2} (i=0,1)$\parn
$\bullet$ Always has torsion

\item[\gr{7\ii d}]
$\Psi=D_4$, $\bbzpsi=\angles{\bft_3,  \bar\alpha=(\bft^{0}_3,A \tau), \bar\beta=(\bft^{\frac 12}_3,B\tau)}$\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$, $\det(K)=-1$, $\tr(K)=n>0$,
$k_{11}=k_{22}$.\parn
$\bullet$ $\vara=\matPpm, \varb = -K$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfa$ with $(I-\cals)\bfa \equiv \bfzero$ and $2a_1 \equiv 0$, taken modulo
$\xy{0}{2v_2}$, where $(I+K)\bfv \equiv \bfzero$.  \parn
$\bullet$ $c_1=0$ or $\frac12$, $(c_1,c_2) \in \coker(K^T-I)$.\parn
$\bullet$ $c_3=0$, must satisfy conditions in Table \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig,  \alpha=(\bft^{a_1}_1\bft^{a_2}_2\bft^{0}_3 \bft^{a_4}_4,A\tau), \beta=(\bft^{\frac 12}_3 ,B \tau)}$\parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig))=\bbz_4$, four choices for $a_4$, the solutions of
$(\alpha\beta)^4=\bft_4^{j}, (j=0,1,2,3)$.\parn
$\bullet$ Torsion-free $j=1,3$ and  $a_1 = \frac 12$ and $a_2= -\frac {k_{21}+1}{2k_{11}} + \frac i {k_{11}}$ for $i=0, \ldots, k_{11}-1$

\item[\gr{7\ii e}]
$\Psi=D_4 \x \bbz_2$, $\bbzpsi=\angles{\bft_3, \bar\alpha= (\bft^{0}_3,A), \bar\beta=(\bft^{\frac 12}_3,B), \bar\gamma=(\bft^{0}_3,\tau)}$\parn
$\bullet$
$\cals=n K+I$, $K\in\gltz$, $\det(K)=-1$, $\tr(K)=n>0$,
$k_{11}=k_{22}$.\parn
$\bullet$ $\vara=\matPpm, \varb = -K, \varphi(\bar \gamma) = I$\parn
$\bullet$ $H^1(\Psi;\coker(I-\cals))$ is all $\bfa$, $\bfr$ with $(I-\cals)\bfa \equiv (I-\cals)\bfr\equiv\bfzero$, $2a_1 \equiv 0$, $2\bfr\equiv 0$,
$\bfa$ taken modulo
$\xy{0}{2v_2}$, where $(I+K)\bfv \equiv \bfzero$.  \parn
$\bullet$ $c_1=0$ or $\frac12$, $c_2=0$ or $\frac12$, $(c_1,c_2) \in \coker(I+K^T)$.\parn
$\bullet$ Unique choice for $c_3$ (up to isomorphism), found by solving $(\beta \gamma)^2 \in \tgamsig$, this
$c_3$ must satisfy the conditions in Tables \ref{t5} and  \ref{t6}\parn
$\bullet$ $\varPi=\angles{\tgamsig, \alpha= (\bft^{a_1}_1\bft^{a_2}_2 \bft^{0}_3 \bft^{a_4}_4,A), \beta=(\bft^{\frac 12}_3 \bft^{b_4}_4,B), \gamma=(\bft^{t_1}_1\bft^{t_2}_2 \bft^{0}_3,\tau)}$\parn
$\bullet$ $H^2(\Psi; \calz(\tgamsig)) = \bbz_2 \x \bbz_2$, $a_4 = -{a_2}{c_2}+\frac{i}{2} (i=0,1)$, $b_4=0$ or $\frac12$\parn
$\bullet$ Always has torsion
\end{enumerate}
\begin{proof}
The group
\begin{align*}
\bbzpsi&=\angles{\bft_3, \bar\alpha=(\bft^{0}_3,A), \bar\beta=(\bft^{\frac12}_3,B), \bar\gamma=(\bft^{\frac12}_3,\tau)}\\
&=\angles{\bft_3,  (\bft^{0}_3,A), (\bft^{0}_3,B\tau), (\bft^{\frac12}_3,\tau)},
\end{align*}
does not lift to a $\Solr$ crystallographic group by Lemma \ref{PPM-A302}.

\bigskip
See Theorem \ref{ClassificationSol-geometry}, case \gr{7\ii}, for $H^1(\Psi; \coker(I-\cals))$ in cases  \gr{7\ii a},  \gr{7\ii b},  \gr{7\ii c}, and \gr{7\ii d}. For case \gr{7\ii e} use the cocycle conditions in Remark \ref{cocycleConditions}, together with Corollary \ref{a-conj-zero3}. The conditions on the $c_i$ follow from previous cases.
The computation of $H^2(\Psi; \calz(\tgamsig))$ is identical to that in
Theorem \ref{7family}, and so we obtain the stated exponents on $\bft_4$.

\bigskip
A $\Solr$ group of type \gr{7\ii a} contains a subgroup of
type \gr{4a}, $\angles{\tgamsig, \alpha\beta}$, which has torsion, as
in Theorem \ref{7family}. Thus, all groups of type \gr{7\ii a} (and hence of 
type \gr{7\ii e}) have torsion.

Also, like in Theorem \ref{7family}, groups of type  \gr{7\ii b} and \gr{7\ii c},
have a torsion subgroup of type \gr{4b}, that is $\angles{\tgamsig, \alpha\beta}$.

Only case \gr{7\ii d} remains. Our criteria for $\varPi$ to be torsion-free
are identical to the $\Solof$ case (Theorem \ref{ClassificationSolof-geometry}, case \gr{7\ii}).
 First, 
the subgroup of type \gr{4a}, $\angles{\tgamsig, \alpha\beta}$, must be torsion-free, which
occurs when $j=1,3$ in $(\alpha \beta)^4 = \bft^{j}_4$. By Lemma \ref{EasyTorSolr}, the elements $\beta$ and $\alpha\beta\alpha$ cannot contribute torsion.
The only remaining elements are $\alpha$ and $\beta\alpha\beta$. That is, we need the groups $\angles{\tgamsig, \alpha}$ and $\angles{\tgamsig, \beta\alpha\beta}$ to be torsion-free. They are torsion-free precisely when their projection to $\Sol$ groups of type
\gr{3\ii} are torsion-free by Lemma \ref{torSolr}. The projections of $\alpha$ and $\beta \alpha \beta$
to $\Sol$ are:
\begin{align*}
&(\bft_1^{a_1} \bft_2^{a_2},A),\\ 
&(\bft_1^{-k_{11}a_1 - k_{12}a_2}
\bft_2^{ -k_{21}a_1 - k_{11}a_2} A,B A B)
\end{align*}
The proof now proceeds exactly as in the $\Solof$ case (Theorem \ref{ClassificationSolof-geometry}, case \gr{7\ii}). See Example \ref{D_4}.
\qedhere
\end{proof}


\begin{remark}
In each case, one must explicitly check that a group $\varPi$ exists with the lattice $\tgamsig$ and given exponents on the $\bft_i$. In particular, for generators $A$ and $B$ of $\Psi$, let $\alpha$ and $\beta$ be generators of $\varPi$, projecting to $A$ and $B$, respectively. For each relation $r(A,B)=e$ between $A$ and $B$ in $\Psi$, we need to check that $r(\alpha, \beta) \in \tgamsig$. 

In the special case when $c_1=c_2=c_3=0$, so that the lattice $\tgamsig$ is a direct product
$\gamsig \x \bbz$, once we have an abstract kernel $\Psi \ra \out(\tgamsig)$ these
conditions are immediately satisfied, and an extension exists.

For example, suppose that \[\Psi = \angles{A,B} = \bbz_2 \x \bbz_2.\]
Then for 
\begin{align*}
\alpha &= (\bft_1^{a_1} \bft_2^{a_2} \bft_3^{a_3} \bft_4^{a_4}, A)\\
\beta &= (\bft_1^{b_1} \bft_2^{b_2} \bft_3^{b_3} \bft_4^{b_4}, B)
\end{align*}
With $a_1, a_2, a_3$, and $a_4$ given in the classification, we also need to check that
\[\alpha\beta\alpha \inv \beta  \inv = \bft_1^{n_1}\bft_2^{n_2}\bft_3^{n_3}\bft_4^{n_4}.\]
for integers $n_i$. In fact, $n_1, n_2, n_3$ are always integral (Theorem \ref{H1-cokernel-theorem}). But it must be checked that $n_4$ is integral for an extension to exist.
\end{remark}

\step[{\bf Examples}]

\begin{example}[\gr{6b\ii d}]
\label{bbz23}
Here we give an example of a torsion-free $\Solr$ crystallographic group
with $\bbz_2^3$ holonomy. See Theorem \ref{6bifamily}.
Take
\[
K=
\left[\begin{matrix}
3&2\\ 4&3
\end{matrix}\right].
\text{  Then, } 
\cals=
\left[\begin{matrix}
17&12\\ 24&17
\end{matrix}\right].
\]
Note that $k_{11}=k_{22}$ and $\cals$ satisfy the conditions for case \gr{6b\ii d}
in Theorem \ref{6bifamily}. Form $\bbz^2 \rx_{\cals} \bbz$ and embed it in $\Sol$ as a
standard lattice, $\gamsig$. We lift $\gamsig$ to the lattice $\tgamsig=\gamsig \x \bbz$ of $\Solr$. Simply set $c_1=c_2=c_3=0$ in Notation \ref{standard-lattice2}. We have defined our lattice $\tgamsig$ of $\varPi$. Now we must define
\begin{align*}
\alpha&=(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{0}\bft_4^{a_4},A),\\
\beta&=(\bft_3^{\frac12}\bft_4^{b_4},B),\\
\gamma&=(\bft_1^{t_1}\bft_2^{t_2}\bft_3^{0},\tau),
\end{align*}
so that $\varPi$ is torsion-free.
Let us set
\begin{align*}
\bfa=\xy{a_1}{a_2} &= \xy{0}{0} \\
\bfr=\xy{t_1}{t_2} &= \xy{\frac12}{0} \\
a_4&=\frac12\\
b_4&=0.\\
\end{align*}
Notice that $\bfa$ and $\bfr$ satisfy all cocycle conditions. Checking
the torsion criteria of Theorem \ref{6bifamily}, we see that the group
$\angles{\tgamsig, \gamma, \beta}$ is torsion-free since $\bfr\neq\bfzero$.
The $\Solr$ group
of type \gr{6b\ii a}, $\angles{\tgamsig, \alpha, \beta}$, is torsion-free by our choice of $a_4$ and $b_4$. Finally, we check that the $\Solr$ group of type \gr{6b\ii b}, $\angles{\tgamsig, \gamma \alpha, \beta}$, is torsion-free. Note that
\begin{align*}
\gamma \alpha&=(\bft_1^{\frac12}\bft_2^{0}\bft_3^{0}\bft_4^{\frac12},A\tau).
\end{align*}
The exponent on $\bft_1$ is $\frac12$, and as for the exponent on $\bft_2$,
\begin{align*}
0 \not\equiv \frac{(k_{11}-1)(2 n+1)}{2k_{12}} = \frac{(2 n+1)}{2} = n+\frac12,
\end{align*}
for any $n\in \bbz$. Therefore the criteria for $\angles{\tgamsig, \gamma \alpha, \beta}$ to be torsion-free
are satisfied. From this it follows that $\varPi$ is torsion-free. 

\bigskip

\noindent
So $\varPi \bs (\Solr)$ defines an infra-solvmanifold of $\Solr$ with $\bbz_2^3$ holonomy.

\bigskip

\noindent
This example is analogous to the following flat $4$-manifold $\varPi' \bs \bbr^4$ with $\bbz_2^3$ holonomy. We define \[\varPi' \subset \bbr^4 \rx \on{O}(4,\bbr)\] by
$\varPi' = \angles{\bbz^4, \alpha, \beta, \gamma}$, where
\begin{align*}
\alpha &= \left(  \left[\begin{matrix} 0 \\ 0 \\ 0 \\ \frac12 \end{matrix}\right]  ,     \left[\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}\right]        \right) \\
\beta &= \left(  \left[\begin{matrix} 0 \\ 0 \\ \frac12 \\ 0 \end{matrix}\right]  ,     \left[\begin{matrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}\right]        \right) \\
\gamma &= \left(  \left[\begin{matrix} \frac12 \\ 0 \\0 \\ 0 \end{matrix}\right]  ,     \left[\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{matrix}\right]        \right). \\
\end{align*}
\end{example}

\begin{example}[\gr{7\ii d}]
\label{D_4}
Here we give an example of a torsion-free $\Solr$ crystallographic group
with $D_4$ holonomy. See Theorem \ref{7ifamily}.
We take
\[
K=
\left[\begin{matrix}
1&2\\ 1&1
\end{matrix}\right],\quad
\text{therefore\ }
n=k_{11}+k_{22}=2,\quad
\cals=n K+I=
\left[\begin{matrix}
3&4\\ 2&3
\end{matrix}\right].\]
As in Example \ref{bbz23}, $\bbz^2 \rx_{\cals} \bbz$ and embeds in $\Sol$ as a
standard lattice, $\gamsig$. We lift $\gamsig$ to the lattice $\tgamsig=\gamsig \x \bbz$ of $\Solr$ by setting $c_1=c_2=c_3=0$ in Notation \ref{standard-lattice2}. Now we must define
\begin{align*}
\alpha&=(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{0}\bft_4^{a_4},A\tau),\\
\beta&=(\bft_3^{\frac12},B\tau),\\
\end{align*}
so that $\varPi$ is torsion-free.
We compute that $(\alpha \beta)^4 = \bft_4^{4 a_4}$ with $c_i=0$, and we define
\begin{align*}
\bfa=\xy{a_1}{a_2} &= \xy{\frac12}{0} \\
a_4&=\frac14.\\ 
\end{align*}
Note that the cocycle conditions on $\bfa$ are satisfied; namely,
$\bfa \in \coker(I-\cals)$. Also the torsion-free criteria for case \gr{7\ii d} of 
Theorem \ref{7ifamily} are satisfied, as $(\alpha\beta)^4=\bft_4$, $a_1=\frac12$, and
$a_2=-\frac {k_{21}+1}{2k_{11}}\equiv 0$.

\bigskip

\noindent
Thus $\varPi \bs (\Solr)$ is an infra-solvmanifold of $\Solr$ with $D_4$ holonomy.
\end{example}

\section{Crystallographic Groups of $\Solmn$}
\label{hithere2}

We have the family of solvable Lie groups of the form $\bbr^3 \rx_{\phi} \bbr$, where
\[ \phi(u) = \left[ \begin{matrix} e^{\theta u} & 0 & 0\\ 0 & e^{u} & 0\\ 0 & 0 & e^{-(1+\theta)u} \end{matrix}\right], \theta > 1.\]
The parameter $\theta$ corresponds precisely to the isomorphism type of the Lie group.
Since we are concerned with crystallographic groups, we need to assume that $\theta$ is chosen so that our Lie group admits a lattice. This is equivalent to $\phi(u_0)$ being
conjugate to an element of $\on{SL}(3,\bbz)$, for some $u_0$. Then the characteristic polynomial of $\phi(u_0)$ is $x^3-mx^2+nx-1=(x- e^{\theta u})(x-e^{u})(x-e^{-(1+\theta)u})$ for $m,n \in \bbz$, and we will denote the group $\bbr^3 \rx_{\phi} \bbr$ by $\Solmn$. It is known that there are only countably many such $\theta$ \cite{unimodular-groups}. For the rest of this section, fix such a $\theta$.

Here we provide a classification of the crystallographic groups of $\Solmn$.
Given a crystallographic group
\[\varPi \subset \Solmn \rx \bbz_2^3,\] $\wt \Gamma = \varPi \cap \Solmn$ is a lattice of $\Solmn$
with finite quotient group $\Phi \subset \bbz_2^3$. We have the short exact sequence
\[1 \ra \wt \Gamma \ra \varPi \ra \Phi \ra 1.\]

The arguments in this section are nearly identical to those used in our classification
of $\Sol$ crystallographic groups. For $u\in \bbr$, let $E^{u}$ denote the action of $\bbr$ on $\bbr^3$ in $\Solmn$.

\step[{\bf Lattices of $\Solmn$}]
\label{latsolmn}

First we will describe the lattices of $\Solmn$. Let $\cals \in \on{SL}(3,\bbz)$ have three distinct real eigenvalues $\lambda_1, \lambda_2, \frac{1}{\lambda_1 \lambda_2}$, satisfying $\lambda_1=\lambda_2^{\theta}$ and
\[
\lambda_1 > \lambda_2 > 1 > \frac{1}{\lambda_1 \lambda_2}.
\]


\begin{notation}
We always take
$$\Delta
=\left[
\begin{matrix} \lambda_1& 0 & 0\\
0 & \lambda_2 & 0 \\
0 & 0 &  \frac{1}{\lambda_1 \lambda_2} \end{matrix}
\right],
$$
with $\lambda_1 > \lambda_2 > 1 > \frac{1}{\lambda_1 \lambda_2}$.
\end{notation}

We form the semi-direct product $\bbz^3 \rx_{\cals} \bbz$ and embed 
this group in $\Solmn$ as a lattice. Let $P$ be a matrix with $\det(P)=1$ diagonalizing
$\cals$:
\[
P S P \inv = \Delta = E^{\ln(\lambda_2)},
\]
where the last equality follows from $\lambda_2^{\theta} = \lambda_1$.

With such $P$ and $\Delta$ for $\cals$, we define a map
\begin{align}
\label{def-phi2}
\phi: \bbz^3\rx_\cals\bbz &\lra \kern24pt\Solmn\\
\notag \left(\bfn,u\right)
&\longmapsto
\notag \left(P \bfn,u\ln(\lambda_2)\right)
\end{align}
The equality $P \cals P\inv =\Delta = E^{\ln(\lambda_2)}$ guarantees that $\phi$ is
a homomorphism. The image is a lattice of $\Solmn$, and
it maps the generators as follows.
\begin{align}
\label{sol-iso2}
\notag
\bfe_1 &\mapsto \bft_1=P \bfe_1\\
\bfe_2 &\mapsto \bft_2=P \bfe_2\\
\notag
\bfe_3 &\mapsto \bft_3=P \bfe_3\\
\notag
\bfe_4 &\mapsto \bft_4=(\bfzero,\ln(\lambda_2)).
\end{align}
We denote image of $\bbz^3\rx_\cals\bbz$ as $\gamsig$:
$$
\gamsig=\angles{\bft_1,\bft_2,\bft_3,\bft_4} \subset\Solmn.
$$



\begin{notation}
A lattice of $\Solmn$ generated by $\bft_1, \bft_2, \bft_3, 
\bft_4$ of the form
in assignment (\ref{sol-iso2}) is a \emph{standard lattice} of $\Solmn$.
\end{notation}

Conversely, we show any lattice of $\Solmn$ is isomorphic to such a $\gamsig$
as the following proposition shows.


\begin{proposition}[cf. Proposition \ref{lattice-a=0}]
\label{Solmnlattice}
There is a one-one correspondence between the isomorphism classes of
$\Solmn$-lattices and the conjugacy classes of
$\cals \in \on{SL}(3,\bbz)$ with three distinct real eigenvalues $\lambda_1 > \lambda_2 > 1 > \frac{1}{\lambda_1 \lambda_2}$ such that $\lambda_2^{\theta} = \lambda_1$.
Therefore, any lattice  of $\Solmn$ is conjugate to $\gamsig$,
for some $\cals$, by an inner automorphism of $\Solmn$.
\end{proposition}
\begin{proof}
We show that any lattice $\Gamma$ is conjugate to $\gamsig$ for some $\cals$.
The intersection of
$\Gamma$ with the nilradical $\bbr^3$, $\Gamma\cap\bbr^3$, is a lattice
of $\bbr^3$, \cite[Corollary 8.28]{rag}.
Let $\bft_1,\bft_2, \bft_3 \in\bbr^3$ generate $\Gamma\cap\bbr^3$,
and let $\{\bft_1,\bft_2, \bft_3, \bft\}$ generate $\Gamma$.
Then $\bft$ is of the form $(\bfa,\ln(\lambda_2))$, for some $\lambda_2$. We can assume
that $\lambda_2>1$ by taking the inverse of $\bft$, if needed. Conjugation by $((E^{\ln(\lambda_2)}-I)\inv \bfa,0)$ maps this lattice to
$$
\Gamma=\angles{(\bft_1,0),\
(\bft_2,0), (\bft_3,0)\
(\bfzero,\ln(\lambda_2))}.
$$
Define $P$ by $P \bfe_i = \bft_i$, for $i=1,2,3$, and set
$\cals=P\inv E^{\ln(\lambda_2)} P$.
As $E^{\ln(\lambda_2)}$ acts on $\angles{\bft_1,\bft_2,\bft_3}$ as an integral matrix,
$\cals \in \on{SL}(3,\bbz)$ necessarily. Notice that, in case $\det(P)<0$, we can make it positive by taking
$-\bft_1$ in place of $\bft_1$.
Furthermore, we can assume that $\det(P) = 1$. For conjugation by $(0,
{\frac {1} {\sqrt[3]{\det(P)}}} I) \in \aff(\Solmn)$ maps this lattice to
another which has $\det(P)=1$. Therefore $\Gamma$ is conjugate to $\gamsig$ in
$\aff(\Solmn)$.

We now show the isomorphism class statement.
If $\cals' = B {\cals} B \inv$ with $B\in \on{SL}(3,\bbz)$, then an isomorphism between $\bbz^3 \rx_{\cals}
\bbz$ and $\bbz^3 \rx_{\cals'} \bbz$ is given by $\bfe_i \ra B \bfe_i$ for
$i = 1,2, 3$ and $\bfe_4 \ra \bfe_4$.

Conversely,
suppose two lattices of $\Solmn$ are isomorphic. We may assume these lattices
are of the form $\Gamma_{\cals}$ and $\Gamma_{\cals'}$.
We have an isomorphism
\[
\phi: \bbz^3 \rx_{\cals} \bbz \lra \bbz^3 \rx_{\cals'}\bbz.
\]
Since $\bbz^3$ is the discrete nil-radical of both groups, $\phi$ restricts to an isomorphism on the
$\bbz^3$ factors, say $B\in \on{GL}(3,\bbz)$. Also, $\phi$ induces an isomorphism
$\epsilon$ on the quotients, so that $\epsilon=\pm 1$. Therefore, $\phi$ is of the form
\[
\phi(\bfx,m)=(B\bfx +\eta(\bfx,m),\epsilon m).
\]
Evidently, $\eta$ is independent
of $\bfx$. So $\eta(\bfx,m)=\eta(m)$. Then applying $\phi$ to $(0,1)(\bfx,0)(0,-1)\in \bbz^3 \rx_{\cals} \bbz$
in two different ways, we get
\begin{align*}
\phi((0,1)(\bfx,0)(0,-1))
&=\phi(\cals\bfx,0)\\
&=(B(\cals\bfx),0)\\
\phi(0,1) \phi(\bfx,0) \phi(0,-1)
&=(\eta(1),\epsilon)(B\bfx,0)(\eta(1),\epsilon)^{-1}\\
%&=(\cals' B\bfx,0)\quad\text{(since $\epsilon=+1$).}
&=(\cals'^{\epsilon} B\bfx,0)\quad
\end{align*}
Thus, $B\cals\bfx=\cals'^{\epsilon} B\bfx$ for every $\bfx$, and we have
$\cals'^{\epsilon}=B\cals B\inv$. Now, $\epsilon=+1$ is forced since, by
definition, both $\cals$ and $\cals'$ have two eigenvalues greater than $1$, and
one eigenvalue less than $1$. Furthermore, by taking $-B$ in place of $B$ if needed,
we can assume that $B \in \on{SL}(3,\bbz)$. Thus, $\cals$ and $\cals'$ are conjugate in 
$\on{SL}(3,\bbz)$.
\qedhere
\end{proof}

\step[{\bf Classification of $\Solmn$ Crystallographic Groups}]

\begin{proposition}
\label{puttingToLattice2}
Any crystallographic group $\varPi$ of $\Solmn$ can be conjugated in $\aff(\Solmn)$
to $\varPi \subset \Solmn \rx \bbz_2^3$ so that:
\end{proposition}

\noindent
{\rm{(1)}}
We have $\varPi \cap \Solmn = \gamsig$. That is, the translation subgroup of $\varPi$ is a standard lattice of $\Solmn$, generated by $\bft_1$, $\bft_2$, $\bft_3$, and $\bft_4$ as in assignment (\ref{sol-iso2}) in Subsection \ref{latsolmn} .

\noindent
{\rm{(2)}} The holonomy group $\Phi$ is generated by at most three elements of $\bbz_2^3$, and therefore $\varPi$ is generated by $\angles{\bft_1, \bft_2, \bft_3, \bft_4}$, and at most three isometries of the form $(\bft_1^{a_1} \bft_2^{a_2} \bft_3^{a_3} \bft_4^{a_4}, A)$, for $A \in \bbz_2^3$ and real numbers $a_i$.


\begin{proof}
Any maximal compact subgroup of $\aut(\Solmn)$ is conjugate to $\bbz_2^3 \subset \aut(\Solmn)$ (see Proposition \ref{autosolmn}). Observe that the affine conjugations
in Proposition \ref{Solmnlattice} leave this maximal compact subgroup $\bbz_2^3$ fixed,
while putting the lattice in standard form.
\qedhere
\end{proof}

We will assume any $\Solmn$ crystallographic group is embedded in $\Solmn \rx \bbz_2^3$ as in Proposition \ref{puttingToLattice2}. In fact, we will see that the group $\bbz_2^3$ can never occur as a holonomy group of a $\Solmn$ crystallographic group. In fact, the maximal holonomy of a $\Solmn$ crystallographic group is $\bbz_2^2$.

\begin{notation}
Let
\[\tau_1 = \left[ \begin{matrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right],
\tau_2 = \left[\begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{matrix} \right],
\tau_3 = \left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix}\right],\]
denote the automorphisms generating the maximal compact subgroup $\bbz_2^3$ of $\aut(\Solmn)$ (see Proposition \ref{autosolmn}).
\end{notation}

Note that $\varPi \cap \bbr^3 = \angles{\bft_1, \bft_2, \bft_3}$ is a lattice of $\bbr^3$.
Denote the quotient $\varPi/\bbz^3$ by $\bbzphi$ so that we have the diagram:
\begin{equation}
\label{coreClass2}
\CD
@. 1 @. 1 @.  @.\\
@. @VVV @VVV @. @.\\
@. \bbz^3 @=  \bbz^3 @. @.\\
@. @VVV @VVV @. @.\\
1 @>>> \gamsig @>>> \varPi @>>> \Phi @>>> 1\\
@. @VV{/ \bbz^3}V @VV{/ \bbz^3}V @| @.\\
1 @>>> \bbz @>>> \bbzphi @>>> \Phi @>>> 1\\
@. @VVV @VVV @. @.\\
@. 1 @. 1 @.  @.\\
\endCD
\end{equation}

Note that the quotient $\bbz$ in the above diagram is generated by $\bft_4$, not $\bft_3$, as was the case for $\Sol$. First we will classify all possible $\bbzphi$, as well as abstract kernels $\varphi: \bbzphi \ra \on{GL}(3,\bbz)$, which can arise as quotients of $\Solmn$ crystallographic groups. The holonomy $\Phi$ is a subgroup of $\bbz_2^3 = \angles{\tau_1, \tau_2, \tau_3}$. 

\begin{lemma}[cf. Lemma \ref{abarMinus}]
\label{abarMinus2}
For $\bar\alpha=(\bft_4^{a_4},A) \in \bbzphi$, $a_4=0$ or $\tfrac12$  {\rm (\mod $\bbz$)}.
\end{lemma}
\begin{proof}
Note that $\abar = +1$ for all $A \in \bbz_2^3$. Thus,
\[
(\bft_4^{a_4},A)^2 = (\bft_4^{2a_4},I).
\]
For this to be in $\bbz = \angles{\bft_4}$, $a_4=0$ or $\frac12$ is forced.
\qedhere
\end{proof}

\begin{lemma}[cf. Lemma \ref{PPM-A30}]
\label{simp1}
If $\bbzphi$ is an extension of $\bbz$ by $\Phi$ that is the quotient of a crystallographic
group $\varPi$ of $\Solmn$, then $\bbzphi$ cannot contain an element of the form
$\bar\alpha=(\bft_4^{0},\pm \tau_i)$.
\end{lemma}
\begin{proof}
Recall that the lattice of $\varPi$ is of the form $\gamsig$. The trace of $\cals$ must be integral:
\[
\lambda_1 + \lambda_2 + \frac{1}{\lambda_1 \lambda_2} \in \bbz. 
\]
Note that the matrix
\[
\varphi(\bft_4) \varphi(\bar\alpha) = \cals \varphi(\bar\alpha) = P \inv \Delta P P \inv (\pm \tau_i) P = P \inv \Delta (\pm \tau_i) P.
\]
must be integral, and so must have integral trace.  We will argue the claim for $\pm \tau_1$. Then,
\[
\mp \lambda_1 \pm \lambda_2 \pm \frac{1}{\lambda_1 \lambda_2} \in \bbz.
\]
Of course, the Rational Zero Theorem implies that all eigenvalues of $\cals$ must be irrational. From the above equations, we infer that $2\lambda_1 \in \bbz$, so that
$\lambda_1$ is rational, a contradiction. Therefore, $\bbzphi$ cannot contain $(\bft_4^{0},\pm \tau_1)$.
The proof of the claim for $\pm \tau_2$ and $\pm \tau_3$ is identical.
\qedhere
\end{proof}


\begin{remark}
\label{abstractKernelSolmn}
From Theorem \ref{H1-cokernel-theorem},
an element of $H^1(\Phi;\coker(I-\cals))$ fixes the exponents on the $\bft_i$ in Proposition \ref{puttingToLattice2}, and hence defines an abstract kernel $\Phi \ra \out(\gamsig)$. As $\gamsig$ has trivial center, $H^2(\Phi; \calz(\gamsig))$ is trivial, and we have
a unique (up to isomorphism) extension of $\gamsig$ by $\Phi$, which we can also
view as 
\[1 \ra \bbz^3 \ra \varPi \ra \bbzphi \ra 1.\]
\end{remark}

\begin{proposition}
\label{a-conj-zero3mn}
Let $\varPi = \angles{\gamsig,
  (\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3} \bft_4^{\frac12},A)}$ be
an extension in {\rm Theorem \ref{H1-cokernel-theorem}} defining a $\Solmn$ crystallographic group with lattice $\Gamma_\cals =
\angles{\bft_1,
\bft_2, \bft_3, \bft_4}$.
Then $H^1(\Phi;\coker(I-\cals))=0$.
Therefore, there exists $\bft_1^{v_1}\bft_2^{v_2}\bft_3^{v_3}$ which conjugates
$(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3} \bft_4^{\frac12},A)$ to $(\bft_4^{\frac12},A)$,
and leaves $\gamsig$ invariant.
\end{proposition}

\begin{proof}
Note that
$A$ has order 2 and $\abar=1$.
Further,
$\varphi(\alphabar) = P \inv \sqrt{\Delta} A P$. As $I-\sqrt{\Delta} A$ is non-singular,
\[P \inv  (I-\sqrt{\Delta} A ) P = I - P \inv \sqrt{\Delta} A P   =   I-\varphi(\alphabar)        \] is non-singular and we may take
$\bfv=(I-\varphi(\alphabar))\inv\bfa$. Then our $\bfv$ conjugates
$(\bft_3^{\frac 1 2}, A)$ to $(\bft_1^{a_1} \bft_2^{a_2} \bft_3^{a_3}\bft_4^{\frac 1 2},
A)$.
It remains to show $\bfv\in (I-\cals)\inv\bbz^3$.
First observe that $\varphi(\alphabar)$ is a square root of $\cals$.
\begin{align*}
\varphi(\alphabar)^2
&=P\inv\sqrt{\Delta} A \sqrt{\Delta} A P =P\inv\Delta  A^2 P\ \text{(since $A$ is diagonal)}\\
&=P\inv\Delta P=\cals.
\end{align*}
Now
\[
(I-\cals) \bfv = (I+\varphi(\alphabar))(I-\varphi(\alphabar)) \bfv =
(I+\varphi(\alphabar))\bfa \in \bbz^3.\qedhere
\]
\end{proof}

\begin{lemma}[cf. Lemma \ref{EasyTor}]
\label{EasyTor2}
When $\alpha =(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3} \bft_4^{a_4}, A) \in \varPi$ satisfies
$a_4 = \frac{1}{2}$, $\gamma \alpha$ is infinite order for all $\gamma \in
\gamsig$.
\end{lemma}
\begin{proof}
Let $\pr: \Solmn \ra \bbr$ denote the
quotient homomorphism of $\Solmn$ by its nil-radical $\bbr^3$. Write $\gamma \in \gamsig$
as $\bft_1^{n_1}\bft_2^{n_2}\bft_3^{n_3}\bft_4^{n_4}$. Application of $\pr$ to $(\gamma \alpha)^2$
yields
\[
\pr(\gamma \alpha)^2 = 2 n_4+1,
\]
from which we infer $\gamma \alpha$ is not torsion.
\qedhere
\end{proof}

\begin{theorem}[Classification of $\Solmn$-geometry]
\label{ClassificationSolmn-geometry}
The following is a complete list of
crystallographic groups of $\Solmn$. In each case,
we give $\Phi$, $\bbzphi$, the abstract kernel $\varphi:\bbzphi\ra\gl(3,\bbz)$ (which
gives the action of $\bbzphi$ on $\angles{\bft_1,\bft_2, \bft_3}$), equations defining $\bfa=\left[ \begin{matrix} a_1\\ a_2\\ a_3 \end{matrix}  \right]$, and $\bfb=\left[ \begin{matrix} b_1\\ b_2\\ b_3 \end{matrix}  \right]$
from $H^1(\Phi;\coker(I-\cals))$, and whether $\varPi$ is torsion-free.
Note that $\varphi(\bft_4) = \cals$ by definition, where $\cals\in \on{SL}(3, \bbz)$ has three distinct real eigenvalues \[\lambda_1 > \lambda_2 > 1 > \frac{1}{\lambda_1 \lambda_2}\] such that $\lambda_2^{\theta} = \lambda_1$.
\end{theorem}
\begin{enumerate}
\bigskip
\item[\gr{0}]
$\Phi=\text{trivial}$\parn
$\bullet$
$\varPi = \gamsig$
\par\noindent
$\bullet$ Torsion-free
%
\item[\gr{1a}]
$\Phi=\bbz_2$: $A=\tau_i$,
$\bbzphi=\bbz=\angles{\bft_4,\bar\alpha=(\bft_4^{\frac12},A)}$.\parn
$\bullet$
$\vara=K$ is a square root of $\cals$, with one negative eigenvalue.\parn
$\bullet$ $H^1(\Phi;\coker(I-\cals))$ is trivial so that $\bfa=\bzero$. \parn
%
$\bullet$
$\varPi=\angles{\gamsig,
\alpha =(\bft_4^{\frac12},A)}=\bbz^3 \rx_{K} \bbz$\parn
$\bullet$
Torsion-free
%
\item[\gr{1b}]
$\Phi=\bbz_2$: $A=-\tau_i$,
$\bbzphi=\bbz=\angles{\bft_4,\bar\alpha=(\bft_4^{\frac12},A)}$.\parn
$\bullet$
$\vara=K$ is a square root of $\cals$, with two negative eigenvalues.\parn
$\bullet$ $H^1(\Phi;\coker(I-\cals))$ is trivial so that $\bfa=\bzero$. \parn
%
$\bullet$
$\varPi=\angles{\gamsig,
\alpha =(\bft_4^{\frac12},A)}=\bbz^3 \rx_{K} \bbz$\parn
$\bullet$
Torsion-free

\item[\gr{2a}]
$\Phi=\bbz_2$: $A=-I$,
$\bbzphi=\bbz \x \bbz_2=\angles{\bft_4,\bar\alpha=(\bft_4^{0},A)}$.\parn
$\bullet$
$\vara=-I$\parn
$\bullet$ $H^1(\Phi;\coker(I-\cals)) = \frac {\coker(I-\cals)}{2\coker(I-\cals)} \subseteq \bbz_2^3$.
That is, all $\bfa$ with $(I-\cals)\bfa\equiv \bfzero$, taken modulo $2\bfa$.\parn
$\bullet$
$\varPi=\angles{\gamsig,
\alpha =(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3} \bft_4^{0},A)}$\parn
$\bullet$
Always has torsion

\item[\gr{2b}]
$\Phi=\bbz_2$: $A=-I$,
$\bbzphi=\bbz=\angles{\bft_4,\bar\alpha=(\bft_4^{\frac12},A)}$.\parn
$\bullet$
$\vara=K$ is a square root of $\cals$, with three negative eigenvalues.\parn
$\bullet$ $H^1(\Phi;\coker(I-\cals))$ is trivial so that $\bfa=\bzero$. \parn
$\bullet$
$\varPi=\angles{\gamsig,
\alpha =(\bft_4^{\frac12},A)}=\bbz^3 \rx_{K} \bbz$\parn
$\bullet$
Torsion-free

\item[\gr{3a}]
$\Phi=\bbz_2 \x \bbz_2$: $A=-I$, $B=\tau_i$,\parn
$\bbzphi=\bbz \x \bbz_2=\angles{\bft_4,\bar\alpha=(\bft_4^{0},A), \bar\beta=(\bft_4^{\frac12},B)}$.\par
$\bullet$
$\vara=-I$, $\varb=K$ is a square root of $\cals$, with one negative eigenvalue.\parn
$\bullet$ $H^1(\Phi;\coker(I-\cals)) \cong \frac {\coker(I-K)}{2\coker({I-K})} \subseteq \bbz_2^3$. That is,
all $\bfa$ with $(I-K)\bfa\equiv \bfzero$, taken modulo $2\bfa$.\parn
$\bullet$
$\varPi=\angles{\gamsig,
\alpha =(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3} \bft_4^{0},A), \beta =
( \bft_4^{\frac12},B)} $\parn
$\bullet$
Always has torsion

\item[\gr{3b}]
$\Phi=\bbz_2 \x \bbz_2$: $A=-I$, $B=-\tau_i$,\parn
$\bbzphi=\bbz \x \bbz_2=\angles{\bft_4,\bar\alpha=(\bft_4^{0},A), \bar\beta=(\bft_4^{\frac12},B)}$.\par
$\bullet$
$\vara=-I$, $\varb=K$ is a square root of $\cals$, with two negative eigenvalues.\parn
$\bullet$ $H^1(\Phi;\coker(I-\cals)) \cong \frac {\coker(I-K)}{2\coker({I-K})}  \subseteq \bbz_2^3$. That is,
all $\bfa$ with $(I-K)\bfa\equiv \bfzero$, taken modulo $2\bfa$.\parn
$\bullet$
$\varPi=\angles{\gamsig,
\alpha =(\bft_1^{a_1}\bft_2^{a_2}\bft_3^{a_3} \bft_4^{0},A), \beta =
( \bft_4^{\frac12},B) }$\parn
$\bullet$
Always has torsion
%
\end{enumerate}
\begin{proof}
Straightforward application of Lemmas \ref{abarMinus2} and \ref{simp1}
show the various possibilities for $\bbzphi$. In particular,
there is no $\bbzphi$ with $\Phi = \bbz_2^3$, as any such $\bbzphi$ cannot
occur as the quotient of a $\Solmn$ crystallographic group by Lemma \ref{simp1}.


In cases \gr{1a}, \gr{1b}, and \gr{2b}, recall from the proof of Proposition \ref{a-conj-zero3mn} that
\[\varphi(\alphabar) = P \inv \sqrt{\Delta} A P\]
is a square root of $\cals$. When $A=\tau_i$, $-\tau_i$, and $-I$, then $\varphi(\alphabar)$ has
 one, two, and three negative eigenvalues, respectively. We denote $\varphi(\alphabar)$ by $K$. Proposition \ref{a-conj-zero3mn} shows that we may take $\bfa=\bfzero$ in cases
\gr{1a}, \gr{1b}, and \gr{2b}. That is, $H^1(\Phi;\coker(I-\cals))$ is trivial in these cases.


In cases \gr{2a}, the cocycle condition on $\bfa$ vanishes since
\[(I+\varphi(\bar \alpha))=I-I=0.\]
The coboundaries are 
\[(I-\varphi(\bar \alpha))\bfa=2\bfa,\]
for $\bfa \in \coker(I-\cals)$.
Therefore,
\[H^1(\Phi;\coker(I-\cals)) = \frac {\coker(I-\cals)}{2\coker(I-\cals)} \subseteq \bbz_2^3,\]
where the last inclusion follows from the fact that $H^1(\Phi;\coker(I-\cals))$ is
generated by at most 3 elements of order 2.

In cases \gr{3a} and \gr{3b}, we can set $\bfb=\bfzero$ by Proposition \ref{a-conj-zero3mn}. The commutator cocycle condition then forces $\bfa \in \coker(I-K)$,
so that $(I-K)\bfa \equiv \bfzero$. Having fixed $\bfb=\bfzero$, the coboundaries are 
\[(I-\varphi(\bar \alpha))\bfa=2\bfa,\]
for $\bfa \in \coker(I-K)$. Hence,
\[H^1(\Phi;\coker(I-\cals)) = \frac {\coker(I-K)}{2\coker(I-K)} \subseteq \bbz_2^3.\]

Lemma \ref{EasyTor2} shows that \gr{1a}, \gr{1b}, and \gr{2b} all are torsion-free.
Indeed, $\varPi$ is isomorphic to $\bbz^3 \rx_{K} \bbz$ in these cases. In case \gr{2a},
$\alpha^2=e$, and cases \gr{3a} and \gr{3b} also have torsion, as they contain a type \gr{2a} subgroup.
\qedhere
\end{proof}

\begin{remark}
The quotient of $\Solmn$ by a crystallographic group $\varPi$
of type
 \gr{0}, \gr{1a}, \gr{1b}, or \gr{2b} is the mapping torus of a linear
 self-diffeomorphism of $T^3$. These were known to be all the infra-solvmanifolds of $\Solmn$ \cite{unimodular-groups}. On the other hand,
 our classification here includes the crystallographic groups with torsion, and shows that
 $\bbz_2^3$ cannot occur as the holonomy of any crystallographic group of $\Solmn$.
\end{remark}

\chapter{Curvature of 4-Dimensional Infra-Solvmanifolds}

Depending
on choice of left invariant metric, a 4-dimensional solvable geometry can have different
isometry groups and
Ricci signatures.
Here
are the possible Ricci signatures on the 4-dimensional
solvable geometries of the form $\bbr^3 \rx \bbr$ (all but $\Solof$.)
\cite{KremlevNikonorov}:

\begin{table}[H]
\caption{Ricci signatures}
\begin{center}
  \begin{tabular}{| l | l |}
    \hline
    $G$ & Possible Ricci signatures \\ \hline
    $\bbr^4$ & $(0,0,0,0)$ \\ \hline
    $\Nilr$ & $(0,+,-,-)$ \\ \hline
    $\Nilf$ & $(0,+,-,-), (+,+,-,-), (+,-,-,-) $ \\ \hline
    $\Solr$ & $(0,0,0,-), (0,+,-,-), (+,+,-,-), (+,-,-,-) $ \\ \hline
    $\Solmn$ & $(0,0,0,-), (0,+,-,-), (+,+,-,-), (+,-,-,-) $ \\ \hline
    $\Solo$ & $(0,0,0,-), (0,+,-,-)$ \\ \hline
  \end{tabular}
\end{center}
\end{table}

Suppose we only consider left invariant metrics on $G$ that induce a
particular Ricci signature. This may place restrictions on the size of the isometry group, and hence the compact isometric quotients of $G$ may not
account for all infra-solvmanifolds of $G$. Given an infra-solvmanifold $M$ of $G$, a Ricci signature
can be \emph{realized} on $M$ if there is a left invariant metric on $G$ with prescribed Ricci signature
such that $M=\varPi \bs G$, for some $\varPi \subset \isom(G)$.
 We will only consider 4-dimensional solvable geometries of the form $\bbr^3 \rx \bbr$.
 As there is only one possible Ricci signature for metrics on $\Nilr$, and $\Solo$ has no infra-solvmanifolds (though it has compact forms), we examine which Ricci signatures can be realized on infra-solvmanifolds with $\Nilf$, $\Solmn$, and $\Solr$ geometry. 

\section{Ricci Curvature}

Let $M$ be a Riemannian $n$-manifold. First, we recall the definition of
the Ricci quadratic form $r$ and Ricci transformation $\hat r$, which are both measures of curvature at a point $p \in M$. We will use Milnor's definitions \cite{milnor}.

\begin{definition}
 The \emph{Ricci quadratic form} at $p$
is a real valued quadratic function on $T_{p}(M)$. Fix an orthonormal basis $(e_1,
\ldots, e_n)$ of $T_{p}(M)$. For $v \in T_{p}(M)$,
\[r(v) = \sum_{i=1}^{n} \angles{R_{v e_i}(v), e_i}.\]
\end{definition}

If $v$ is a unit vector, then $r(v)$ is the \emph{Ricci curvature} in the
direction of $v$. It is $n-1$ times the average of the sectional curvatures of the 2-planes
containing $v$.

\begin{definition}
The \emph{Ricci transformation} at $p \in M$
is the self-adjoint map ${\hat r}: T_{p}(M) \ra T_{p}(M)$ defined by
\[{\hat r}(v) = \sum_{i=1}^{n} R_{e_i v}(e_i).\]
\end{definition}

The Ricci quadratic form $r$ and the Ricci transformation $\hat r$ are related
by the identity
\[\angles{ \hat r (v), v}=r(v). \]

\begin{definition}
The eigenvalues of the Ricci transformation $\hat r$ at $p$ are the \emph{principal Ricci curvatures}
of the metric at $p$.
\end{definition}

\begin{definition}
The collection of signs of the principal Ricci curvatures at $p$ is the \emph{Ricci signature} of the metric at $p$.
\end{definition}

\section{Computation of Ricci Signatures for Metric Lie Algebras}

When $G$ is a Lie group,
any left invariant metric can be specified by declaring
a basis $(v_1, v_2,\ldots,v_n)$ of the tangent space at the identity $T_{e}(G) \cong \fkg$
to be orthonormal.  We use left translation by $g$ to propagate the metric across the tangent bundle:
We declare $\ell_{g*}v_1,\ell_{g*}v_2,\ldots,\ell_{g*}v_n$ to be an orthonormal basis for $T_{g}(G)$.

\begin{proposition}[cf. \cite{shin}, Proposition 1.3]
There is a one-one correspondence between the set of all left
invariant metrics on a Lie group and the group of upper triangular matrices
with positive diagonal entries.
\end{proposition}
\begin{proof}
By choosing a basis for $\fkg$, we can identify $\fkg$ with $\bbr^n$, so
that the standard basis vectors $\bfe_1, \ldots, \bfe_n$ form an orthonormal basis
for $\fkg$. Any other metric on $G$ can be specified by
declaring $A \bfe_1, \ldots A \bfe_n$ to be an orthonormal basis, for $A \in \gl(n,\bbr)$.
We obtain the same metric if and only if $A$ is orthogonal.
By the Gram-Schmidt process, $A$ can be written uniquely as $K \mathcal{B}$,
where $K$ is orthogonal, and $\mathcal{B}$ is an upper triangular matrix with positive entries.
\end{proof}

Left translation by $g \in G$
is an isometry carrying $e \in G$ to $g$.
In particular, the Ricci signature at any point
$g \in G$ is the same as the Ricci signature at the identity. So we may speak of the
Ricci signature of the left invariant metric.
The Ricci signature of a left invariant metric can be computed from the structure of the Lie algebra $\fkg$ and the orthonormal
basis $v_1, v_2,\ldots,v_n$ of $\fkg$ \cite{milnor}.

\begin{notation}
Suppose a metric Lie algebra $\fkg$ has a codimension 1 ideal
$\fku$. Let $b$ be a unit vector orthogonal
to $\fku$, and let $L$ denote the restriction of $\ad(b)$ to $\fku$:
\[L(u)=\left[b,u\right].\]
Let $S$ be the self-adjoint transformation $\frac 12 (L + L^*)$.
The ideal $\fku$ may also be thought of as a metric Lie algebra;
let us denote the Riemannian connection for $\fku$ by $\overline \nabla$.
\end{notation}

\begin{lemma}[{\cite[Lemma 5.5]{milnor}}]
For any $u, v \in \fku$, we have the following expressions for $\nabla$:
\begin{align*}
\nabla_b b = 0 \hspace{1 cm} &\text{  and  } \hspace{1 cm} \nabla_b u = {\tfrac 1 2} (L-L^*) u\\
\nabla_u b = -S u \hspace{1 cm}&\text{ and } \hspace{1 cm} \nabla_u v
= {\overline \nabla_u} v + \angles{Su,v}b.
\end{align*}
\end{lemma}

\begin{proposition}[\cite{DM}]
\label{curU}
Suppose a Lie algebra $\fkg$ has an ideal
$\fku$ of codimension 1.
Assume that $\fku$ is abelian, so that the Lie bracket
and $\overline \nabla$ vanish on $\fku$, as well as $\tr(L)=0$,
so that $\tr(S)=0$. Then, for
$u \in \fku$, we have the following formula for the Ricci
transformation:
\[
{\hat r (u)} = \frac 12 \left[L, L^*\right](u) = {\frac 1 2} (LL^* - L^* L )(u).
\]
\end{proposition}
\begin{proof}
We compute, for an orthonormal basis $e_1, \ldots, e_n$ of $\fkg$,
with $e_n=b$,
\begin{align*}
{\hat r}(u)&=\left( \sum_{i=1}^{n-1} \nabla_{\left[e_i,u\right]} e_i - \nabla_{e_i}
\nabla_{u} e_i + \nabla_{u}
\nabla_{e_i} e_i \right) + \nabla_{\left[b,u\right]} b - \nabla_{b}
\nabla_{u} b + \nabla_{u}
\nabla_{b} b\\
&=\left(\sum_{i=1}^{n-1} - \angles{Su,e_i} \nabla_{e_i} b \right) + \left( \sum_{i=1}^{n-1}
 \angles{S e_i,e_i} \nabla_{u} b\right) - S L u + \tfrac 12 (L-L^*) S u\\
&=\left(\sum_{i=1}^{n-1}  \angles{S u,e_i} S e_i \right) - \tr(S) S u - S L u + \tfrac 12 (L-L^*)Su\\
&=SSu-SLu+ \tfrac 12 (L-L^*)Su\\
&=\tfrac 1 4 (L+L^*)(L+L^*)u - \tfrac 1 2 (L+L^*)Lu + \tfrac 1 4 (L-L^*)(L+L^*)u\\
&=\tfrac 1 2 \left(LL^*-L^*L\right) u =\tfrac 1 2 \left[L,L^*\right]u.\qedhere
\end{align*}
\end{proof}

\begin{proposition}[{\cite[Lemma 2.3]{milnor}}]
\label{curB}
Suppose a Lie algebra $\fkg$ has an ideal
$\fku$ of codimension 1. Let $b$ be a unit vector
orthogonal to $\fku$. Then
\[
{\hat r}(b)=-\tr(S^2)b,
\]
where $S^2$ is the self-adjoint transformation $\tfrac 1 2 (L+L^*)$.
\end{proposition}
\begin{proof}
Take an orthonormal basis $e_1, \ldots, e_n$ of $\fkg$,
with $e_n=b$.
\begin{align*}
{\hat r}(b)&=\left( \sum_{i=1}^{n-1} \nabla_{\left[e_i,b\right]} e_i - \nabla_{e_i}
\nabla_{b} e_i + \nabla_{b}
\nabla_{e_i} e_i \right) + \nabla_{\left[b,b\right]} b - \nabla_{b}
\nabla_{b} b + \nabla_{b}
\nabla_{b} b\\
&=\sum_{i=1}^{n-1} \nabla_{-L e_i} e_i -  \nabla_{e_i}
\tfrac 1 2 (L-L^*) e_i + \angles{S e_i, e_i}  \nabla_{b} b\\
 &=\sum_{i=1}^{n-1} \angles{-SL e_i, e_i}b - \angles{S e_i, \tfrac 1 2 (L-L^*) e_i} b\\
 &=\left(-\tr(SL)-\tfrac 1 2 \tr( SL-SL^*)\right) b.
 \end{align*}
 Now we have, using that $\tr(L)=\tr(L^*)$, $\tr(LL^*)=\tr(L^* L)$,
 \begin{align*}
 -\tr(SL)-\tfrac 1 2 \tr( SL-SL^*)
 &=\tr \left(-\tfrac 1 2 LL - \tfrac 1 2 L^* L - \tfrac 1 4 LL - \tfrac 1 4 L^* L +
 \tfrac 1 4 L L^* + \tfrac 1 4 L^* L^* \right)\\
 &=-\tfrac 1 2 \tr(LL) - \tfrac 1 2 \tr(L^* L).
 \end{align*}
 Now note that
  \begin{align*}
  -\tr(S^2) &= -\tfrac 1 4 \left(\tr(LL) + \tr(L^* L) + \tr(L L^*) + \tr(L^* L^*) \right) \\
  &=-\tfrac 1 4  \left( 2 \tr(LL) + 2 \tr(L^* L) \right)=-\tfrac 1 2 \tr(LL) - \tfrac 1 2 \tr(L^* L).\qedhere
  \end{align*}
\end{proof}



\section{Ricci Signatures of the 4-Dimensional Solvable Geometries}

From here on, we specialize to the case when $\fkg$ is
the Lie algebra of a 4-dimensional solvable  geometry of the form $\bbr^3 \rx_\psi \bbr$.
Note that $\bbr^3 \rx_\psi \bbr$ is completely determined by the matrix
$\psi(1)$:

\begin{align}
\label{listOfGeometriesLie}
\begin{array}{lllll}
&\bbr: &\psi(1)= \left[ \begin{matrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{matrix}\right]
\hspace {1 cm}  &\Nilr: &\psi(1)=
\left[ \begin{matrix} 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{matrix}\right]\\
&\Nilf: &\psi(1)= \left[ \begin{matrix} 0 & 1 & 0  \\ 0 & 0 & 1\\ 0 & 0 & 0 \end{matrix}\right]
\hspace {1 cm}  &\Solr: &\psi(1)=
\left[ \begin{matrix} -1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{matrix}\right]\\
&\Solo: &\psi(1)= \left[ \begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -2 \end{matrix}\right]
\hspace {1 cm}  &\Solmn: &\psi(1)=
\left[ \begin{matrix} \theta & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -(1+\theta) \end{matrix}\right]
\end{array}
\end{align}
Let $\bfe_1, \bfe_2, \bfe_3, \bfe_4$ denote
the standard basis vectors for $\fkg \cong \fku  \rx \bbr$:
\[ \bfe_1 = \left[ \begin{matrix} 1 \\ 0 \\ 0 \\ 0   \end{matrix} \right],
\bfe_2 = \left[ \begin{matrix} 0 \\ 1 \\ 0 \\ 0   \end{matrix} \right],
\bfe_3 = \left[ \begin{matrix} 0 \\ 0 \\ 1 \\ 0   \end{matrix} \right],
\bfe_4 = \left[ \begin{matrix} 0 \\ 0 \\ 0 \\ 1   \end{matrix} \right].
 \]
Any left-invariant metric on $G$ can be uniquely
specified by an upper triangular matrix:
\begin{align}
\label{newbasis}
\mathcal{B'}=
\left[\begin{matrix}
b_{11} & b_{12} & b_{13} & b_{14}\\
0& b_{22} & b_{23} & b_{24}\\
0 & 0 & b_{33} & b_{34}\\
0 & 0 & 0 & b_{44}
\end{matrix}\right],
\end{align}
with
$b_{11},b_{22},b_{33},b_{44}>0$, and declaring the vectors $\bfv_i = \mathcal{B'}\bfe_i$
to be an orthonormal basis for $\fkg$.
Note that $\bfv_1, \bfv_2, \bfv_3$
form an orthonormal basis of the ideal $\fku = \bbr^3$ of $\fkg$, while $\bfv_4$ is a
unit vector orthogonal to this ideal.

Let $L$ denote the transformation $\ad_{\bfv_4}$ restricted to the ideal
$\fku = \bbr^3 = \angles{\bfv_1, \bfv_2, \bfv_3}$, and set
\begin{align}
\mathcal{B}=\left[\begin{matrix}
b_{11} & b_{12} & b_{13}\\
0& b_{22} & b_{23}\\
0 & 0 & b_{33}\\
\end{matrix}\right].
\end{align}
With respect to the basis
$(\bfv_i)$, note that $L$ has matrix form
\begin{align*}
\label{formOfL}
\mathcal{B} \inv \psi(b_{44}) \mathcal{B}.
\end{align*}

By a theorem of Gordon and Wilson  \cite{GW}, with any left-invariant metric on a $4$-dimensional
solvable geometry, \[\isom(G) \subseteq G \rx K \subset \aff(G),\] where $K$ is a maximal compact subgroup of
$\aut(G)$. That is, the group of isometries fixing the identity, $\isom_{e}(G)$, 
must be a subgroup of $K \subset \aut(G)$. Given $\alpha \in \aut(G)$, let $\alpha_{*}$ denote the induced automorphism
of $\fkg$. Recall that we have identified $\fkg$ with $\bbr^4$ by taking the standard basis
for $\fkg \cong \bbr^3 \rx \bbr$, and so we may view $\alpha_{*}$ as an element of $\gl(4,\bbr)$. Of course, $\aut(G) \cong \aut(\fkg)$.

\begin{lemma}
\label{lemma1}
Suppose an orthonormal basis of $\fkg$ is specified by such a matrix $\mathcal{B'}$ 
in {\rm assignment (\ref{newbasis})}. Let $\alpha_{*} \in \aut(\fkg)$. Then $\alpha_{*}$ defines an isometry of $\fkg$ if and only if $\alpha_{*}$ is orthogonal when expressed in terms of the basis
\[\bfv_i = \mathcal{B'}\bfe_i.\]
That is,
\begin{align*}
\mathcal{B'}  \inv {\alpha_{*}} \mathcal{B'} \in \on{O}(4,\bbr).
\end{align*}
\end{lemma}
%\noindent
%\rm(1)
%Then $\alpha_{*} \in \aut(\fkg)$ defines an isometry of $\fkg$ precisely
%when 
%\begin{align*}
%\mathcal{B'}  \inv {\alpha_{*}} \mathcal{B'} \in \on{O}(4,\bbr)
%\end{align*}

%\noindent
%\rm(2)
%Suppose that $\alpha_{*} \in \isom(\fkg)$, and let $K$ denote a fixed maximal compact %subgroup of $\aut(\fkg)$.
%Then \[\alpha_{*} = \beta_{*} \inv \alpha'_{*} \beta_{*},\]
%for some $\alpha'_{*} \in K$ and some $\beta_{*} \in \aut(\fkg)$
%\end{lemma}
%\begin{proof}
 %For the second statement, note that $\alpha_{*}$ must belong to a maximal compact subgroup of $\aut(\fkg)$, and any two maximal compact subgroups are conjugate in $\aut(\fkg)$.
%\qedhere
%\end{proof}

We will restrict our attention to the ideal $\bbr^3$ of $\fkg \cong \bbr^3 \rx \bbr$.
This ideal is the nilradical of $\fkg$, and so we have the restriction homomorphism
\begin{align*}
p: \aut(\fkg) \ra \aut(\bbr^3) = \gl(3,\bbr).
\end{align*}
If $\alpha_{*} \in \aut(\fkg)$ is an isometry of $\fkg$, then $\alpha_{*}$ must
restrict to an isometry of the ideal $\bbr^3 = \angles{\bfv_1, \bfv_2, \bfv_3}$. Therefore, 
\begin{align*}
\mathcal{B}  \inv {p(\alpha_{*})} \mathcal{B} \in \on{O}(3,\bbr).
\end{align*}
%We will set $A := r(\alpha_{*})$, so that the equation becomes.
%\begin{align*}
%\mathcal{B}  \inv A \mathcal{B} \in \on{O}(3,\bbr).
%\end{align*}


The condition $\mathcal{B}  \inv {p(\alpha_{*})} \mathcal{B} \in \on{O}(3,\bbr)$ will put
rather strict restrictions on $\mathcal{B}$, for certain choices of ${p(\alpha_{*})}$ in the 
image of $p(\aut(\fkg))$. This in turn will place restrictions on the possible Ricci signatures.  Recall in our discussion of the crystallographic groups of $\Solmn$ that we have fixed the notation:
\[\tau_1 = \left[ \begin{matrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right],
\tau_2 = \left[\begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{matrix} \right],
\tau_3 = \left[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix}\right].\]
We also will set
\[
A=\left[\begin{matrix} 0 & -{a} & 0 \\ \frac{1}{a} & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right].\]

\begin{lemma}
\label{restrictions}

\hspace{1cm}

\noindent
\rm(1) The condition
$\mathcal{B}  \inv (\pm \tau_1) \mathcal{B} \in \on{O}(3,\bbr)$ forces $b_{12}=b_{13}=0$.

\noindent
\rm(2) The condition
$\mathcal{B}  \inv (\pm \tau_2)  \mathcal{B} \in \on{O}(3,\bbr)$ forces $b_{12}=b_{23}=0$.

\noindent
\rm(3) The condition
$\mathcal{B}  \inv (\pm \tau_3)  \mathcal{B} \in \on{O}(3,\bbr)$ forces $b_{13}=b_{23}=0$.

\noindent
\rm(4) The condition
$\mathcal{B}  \inv (\pm A)  \mathcal{B} \in \on{O}(3,\bbr)$
forces $b_{12}=b_{13}=b_{23}=0$.
\end{lemma}
\begin{proof}
For \rm(1),
\[
\mathcal{B}  \inv (\pm \tau_1) \mathcal{B}=
\pm \left[
\begin{array}{ccc}
 -1 & -\frac{2 b_{12}}{b_{11}} & -\frac {2 b_{13}}{b_{11}} \\
 0 & 1 & 0 \\
 0 & 0 & 1
\end{array}
\right] \in \on{O}(3,\bbr)\]
forces $b_{12}=b_{13}=0$.

For \rm(2),
\[
\mathcal{B}  \inv  (\pm \tau_2) \mathcal{B}=
\pm \left[
\begin{array}{ccc}
 1 & \frac{2 b_{12}}{b_{11}} & \frac{2 b_{12} b_{23}}{b_{11} b_{22}} \\
 0 & -1 & -\frac{2 b_{23}}{b_{22}} \\
 0 & 0 & 1
\end{array}
\right] \in \on{O}(3,\bbr)\]
forces $b_{12}=b_{23}=0$.

For \rm(3),
\[
\mathcal{B}  \inv (\pm \tau_3) \mathcal{B}=
\pm \left[
\begin{array}{ccc}
 1 & 0 &\frac{2 b_{13} b_{22}- 2 b_{12} b_{23}}{b_{11} b_{22}} \\
 0 & 1 & \frac{2 b_{23}}{b_{22}} \\
 0 & 0 & -1
\end{array}
\right] \in \on{O}(3,\bbr)\]
forces $b_{13}=b_{23}=0$.

For \rm(4), we compute that 
\[\mathcal{B}  \inv (\pm A) \mathcal{B}=
\pm \left[
\begin{array}{ccc}
 -\frac{b_{12}}{a b_{22}} & -\frac{b_{12}^2}{a b_{11} b_{22}}-\frac{a b_{22}}{b_{11}} & -\frac{b_{12} b_{13}}{a b_{11} b_{22}}-\frac{a b_{23}}{b_{11}}+\frac{-b_{13} b_{22}+b_{12} b_{23}}{b_{11} b_{22}} \\
 \frac{b_{11}}{a b_{22}} & \frac{b_{12}}{a b_{22}} & \frac{b_{13}}{a b_{22}}-\frac{b_{23}}{b_{22}} \\
 0 & 0 & 1
\end{array}
\right].
\]
As $A$ has positive determinant, this matrix must be in $\so(3,\bbr)$. As the upper
$2\x 2$ diagonal block must be in $\so(2,\bbr)$, we must have $b_{12}=0$. From the $(1,3)$ and $(2,3)$ entries of this matrix,
we obtain the equations
\begin{align*}
-b_{13}-a b_{23}&=0,\\
b_{13}-a b_{23}&=0,
\end{align*}
which force $b_{13}=b_{23}=0$, as well. 
%In all cases, the same computation goes through with $-A$ in place of $A$.
\qedhere
\end{proof}

\begin{definition}
Given an infra-solvmanifold $M$ of $G$, a Ricci signature
can be \emph{realized} on $M$ if there is a left invariant metric on $G$ with prescribed Ricci signature
such that $M=\varPi \bs G$, for some $\varPi \subset \isom(G)$.
\end{definition}

Suppose a basis of the Lie algebra $\fkg$ is specified by
$\mathcal{B'}$ as in assignment (\ref{newbasis}).
Recall that  $L$ denotes the transformation $\ad_{\bfv_4}$ restricted to the ideal
$\fku = \bbr^3 = \angles{\bfv_1, \bfv_2, \bfv_3}$.
With respect to this basis 
$(\bfv_i)$, note that $L$ has matrix form
\[
N=\mathcal{B} \inv \psi(b_{44}) \mathcal{B}.
\]
By Lemma \ref{curB}, \[r(\bfv_4) = -
\tr(S^2)  < 0.\]
We claim that $\tr(S^2)>0$ for all $G$ of the form $\bbr^3 \rx \bbr$, except for the trivial case $G=\bbr^4$. To see this, simply note that the matrix form of $S$ is $\frac 12 (N + N^ T)$, which is a symmetric matrix.
Thus, the $\bfv_4$ direction always corresponds to negative Ricci curvature, and we always put it as the last sign in the Ricci signature notation: \[(*,*,*,-).\]

By Proposition \ref{curU}, with respect to the basis $\bfv_1,\bfv_2, \bfv_3$ of $\bbr^3$, the matrix form 
of $\hat r$, restricted to $\bbr^3$, is given by
\[{\hat r} \big|_{\bbr^3} = \frac 12 \left[N, N^T\right]= {\frac 1 2} (NN^T- N^TN ).\] To obtain the Ricci signature of the metric with basis given by $\mathcal{B'}$, we need to compute the signs of the eigenvalues of ${\hat r} \big|_{\bbr^3}$.

\step[{\bf $\Nilf$}]

The crystallographic groups $\varPi$ of $\Nilf$, and their associated 
infra-nilmanifolds $\varPi \bs \Nilf$ are classified in \cite{Dekimpe}.
The maximal holonomy of an infra-nilmanifold of $\Nilf$
is $\bbz_2^2$. 
With standard basis for the Lie algebra of $\Nilf$, \[\isom(\Nilf)= \Nilf \rx \bbz^2_2, \]
where $\bbz_2^2$ is as defined in equation (\ref{maxnilf}) in section \ref{dimfix}. Note that $\bbz_2^2$ consists of diagonal matrices.

\begin{theorem}
\label{NilrSig}
If $M$ is an infra-nilmanifold of $\Nilf$, then any of the three Ricci signatures
$(0,+,-,-)$, $(+,+,-,-)$, $(+,-,-,-)$ can be realized on $M$.
\end{theorem} 
\begin{proof}
Declare the basis of the Lie algebra of $\Nilf$ given by
\[\mathcal{B'}=
\left[\begin{matrix}
b_{11} & 0 & 0 & 0\\
0& b_{22} & 0 & 0\\
0 & 0 & b_{33} & 0\\
0 & 0 & 0 & b_{44}
\end{matrix}\right],\]
to be orthonormal. That is, declare $\bfv_i=\mathcal{B'}\bfe_i$ to be orthonormal 
As $\mathcal{B'}$ is diagonal, \[\isom(\Nilf)= \Nilf \rx \bbz^2_2,\]
where $\bbz_2^2$ is the maximal compact subgroup in equation (\ref{maxnilf}) in section \ref{dimfix}, regardless of the diagonal entries of $\mathcal{B'}$.

Recall that with respect to the basis $\bfv_1,\bfv_2, \bfv_3$, the matrix form 
of $\hat r$ restricted to $\bbr^3$ is given by
\[{\hat r} \big|_{\bbr^3} = \frac 12 \left[N, N^T\right] = {\frac 1 2} (NN^T- N^TN ) \text{, where  } N= \mathcal{B} \inv \psi(b_{44}) \mathcal{B}.\]
We compute that the eigenvalues of this matrix are 
\[\left\{\frac{b_{22}^2 b_{44}^2}{2 b_{11}^2},-\frac{b_{33}^2 b_{44}^2}{2 b_{22}^2},\frac{1}{2} \left(-\frac{b_{22}^2 b_{44}^2}{b_{11}^2}+\frac{b_{33}^2 b_{44}^2}{b_{22}^2}\right) \right\}.\]
By choosing different values for $b_{11}, b_{22}, b_{33}$, and $b_{44}$, the third eigenvalue can be positive, negative, or vanishing. Therefore, 
any of the three Ricci signatures 
$(0,+,-,-)$, $(+,+,-,-)$, $(+,-,-,-)$ can be realized on $M$.
\qedhere
\end{proof}


\step[{\bf $\Solmn$}]

Recall that we have the family of Lie algebras $\bbr^3 \rx_{\psi(t)} \bbr$, where
\[\psi(1)=
\left[ \begin{matrix} \theta & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -(1+\theta) \end{matrix}\right].\]
The parameter $\theta > 1$ corresponds precisely to the isomorphism type of the Lie algebra. Let an orthonormal basis of $\fkg$ be specified by $\mathcal{B'}$ in assignment (\ref{newbasis}).
To realize metrics on certain infra-solvmanifolds of $\Solmn$, we will see that $\mathcal{B}$ must have a certain form.
For these various forms of $\mathcal{B}$, we record the eigenvalues of ${\hat r} \big|_{\bbr^3}$ in Table \ref{EigenvaluesSolmn}.

\begin{table}[H]
\caption{Eigenvalues of ${\hat r} \big|_{\bbr^3}$ for $\Solmn$}
\label{EigenvaluesSolmn}
\begin{center}
\begin{tabular}{|p{3cm}| p{10cm}|}
\hline
$\mathcal{B}$ & Eigenvalues of ${\hat r} \big|_{\bbr^3}=$ \newline $\frac 12 \left[N, N^T\right] = {\frac 1 2} (NN^T- N^TN )  \text{, where  } N= \mathcal{B} \inv \psi(b_{44}) \mathcal{B}$\\
\hline
$\left[\begin{matrix}
b_{11} & 0 & 0\\
0& b_{22} & b_{23}\\
0 & 0 & b_{33}\\
\end{matrix}\right]$  & $\left\{0,-\frac{b_{23} \sqrt{b_{22}^2+b_{23}^2} b_{44}^2 (2+\theta)^2}{2 b_{22}^2},\frac{b_{23} \sqrt{b_{22}^2+b_{23}^2} b_{44}^2 (2+\theta)^2}{2 b_{22}^2}\right\}$ \\
\hline
$\left[\begin{matrix}
b_{11} & 0 & b_{13}\\
0& b_{22} & 0\\
0 & 0 & b_{33}\\
\end{matrix}\right]$  &  $\left\{0,-\frac{b_{13} \sqrt{b_{11}^2+b_{13}^2} b_{44}^2 (1+2 \theta)^2}{2 b_{11}^2},\frac{b_{13} \sqrt{b_{11}^2+b_{13}^2} b_{44}^2 (1+2 \theta)^2}{2 b_{11}^2}\right\}$\\
\hline
$\left[\begin{matrix}
b_{11} & b_{12} & 0\\
0& b_{22} & 0\\
0 & 0 & b_{33}\\
\end{matrix}\right]$  & $ \left\{0,-\frac{b_{12} \sqrt{b_{11}^2+b_{12}^2} b_{44}^2 (-1+\theta)^2}{2 b_{11}^2},\frac{b_{12} \sqrt{b_{11}^2+b_{12}^2} b_{44}^2 (-1+\theta)^2}{2 b_{11}^2}\right\}$\\
\hline
\end{tabular}
\end{center}
\end{table}

\begin{lemma}
\label{quiteRigid}
Let \[M=\varPi \bs \Solmn\] for  $\varPi \subset \Solmn \rx \aut(\Solmn)$ be an infra-solvmanifold of $\Solmn$, diffeomorphic to 
an infra-solvmanifold $M'=\varPi' \bs \Solmn$ of type \gr{1a} (respectively, \gr{1b}) in {\rm Theorem \ref{ClassificationSolmn-geometry}}.

If the image of the holonomy of $\varPi'$ in $\aut(\bbr^3)$ is generated by $\tau_i$ (respectively, $-\tau_i$), then
the image of the holonomy of $\varPi$ in $\aut(\bbr^3)$ must also be generated by $\tau_i$ (respectively, $-\tau_i$).
\end{lemma}
\begin{proof}
Since $\varPi$ and $\varPi'$ are isomorphic, Theorem \ref{Rigidity} implies
that $\varPi$ and $\varPi'$ are conjugate in $\aff(\Solmn)$. Hence the holonomy groups
must be conjugate in $\aut(\Solmn)$. But the image of $\aut(\Solmn)$ (and hence also that of $\aut(\fkg)$) in $\aut(\bbr^3)$ under restriction is the group of diagonal matrices $(\bbr^*)^3$, and $\pm \tau_i$
is central in $(\bbr^*)^3$.
\qedhere
\end{proof}


\begin{theorem}
\label{SolmnSig}
For each type of torsion-free $\Solmn$ crystallographic group $\varPi$
in {\rm Theorem \ref{ClassificationSolmn-geometry}}, we list the Ricci signatures
which can be realized on the infra-solvmanifold $\varPi \bs \Solmn$.
\begin{table}[H]
\caption{Realizable Ricci signatures on infra-$\Solmn$ manifolds}
\begin{center}
\begin{tabular}{|p{2.5cm}| p{1.7cm} | p{6.8cm}| }
\hline
Infra-$\Solmn$ & Holonomy &  Realizable Ricci Signatures\\
\hline
\gr{0} & $\{e\}$ & (0,0,0,-), (0,+,-,-), (+,+,-,-), (+,-,-,-)\\
\hline
\gr{1a} & $\bbz_2$ & (0,0,0,-), (0,+,-,-)\\
\hline
\gr{1b} &$\bbz_2$ &  (0,0,0,-), (0,+,-,-)\\
\hline
\gr{2b} & $\bbz_2$ & (0,0,0,-), (0,+,-,-), (+,+,-,-), (+,-,-,-)\\
\hline
\end{tabular}
\end{center}
\end{table}
\end{theorem}
\begin{proof}
In cases \gr{0} and \gr{2b}, Lemma \ref{restrictions} does not provide any 
restriction on $\mathcal{B}$. Of course, Table 
\ref{EigenvaluesSolmn} shows that  $(0,0,0,-)$ and $(0,+,-,-)$  are realizable on $M$. In fact, we can explicitly realize the Ricci signature $(+,+,-,-)$ with
\[\mathcal{B'}=
\left[\begin{matrix}
b_{11} & 0 & b_{13} & 0\\
0& b_{22} & b_{23} & 0\\
0 & 0 & b_{33} & 0\\
0 & 0 & 0 & b_{44}
\end{matrix}\right].\]
With this choice of $\mathcal{B'}$, the determinant of ${\hat r} \big|_{\bbr^3}= {\frac 1 2} (NN^T- N^TN )$ is:
\[-\frac{b_{13}^2 b_{23}^2 b_{44}^6 (-1+\theta)^2 (2+\theta)^2 (1+2 \theta)^2}{8 b_{11}^2 {b_{22}}^2}<0.\]
Since $\theta>1$, when we take $b_{13}$ and $b_{23}$ to be nonzero,
the three eigenvalues of ${\hat r} \big|_{\bbr^3}$ must have signs 
$(+,+,-)$, as the trace of  $ {\frac 1 2} (NN^T- N^TN )$ vanishes.
On the other hand, we realize the Ricci signature $(+,-,-,-)$ with
\[\mathcal{B'}=
\left[\begin{matrix}
b_{11} & b_{12} & b_{13} & 0\\
0& b_{22} & 0 & 0\\
0 & 0 & b_{33} & 0\\
0 & 0 & 0 & b_{44}
\end{matrix}\right].\]
With this choice of $\mathcal{B'}$, the determinant of ${\hat r} \big|_{\bbr^3}= {\frac 1 2} (NN^T- N^TN )$ is:
\[\frac{b_{12}^2 b_{13}^2 b_{44}^6 (-1+\theta)^2 (2+\theta)^2 (1+2 \theta)^2}{8 b_{11}^4}>0.\]
When $b_{12}$ and $b_{13}$ are nonzero,the eigenvalues of
${\hat r} \big|_{\bbr^3}$   must have signs  $(+,-,-)$, as the trace of 
${\frac 1 2} (NN^T- N^TN )$ vanishes. 

By Lemma \ref{quiteRigid}, the image of the holonomy in $\aut(\bbr^3)$ of any infra-solvmanifold diffeomorphic to an infra-sovlmanifold
of type \gr{1a} (or \gr{1b}) is generated by $\tau_i$ (or $-\tau_i$).

In case \gr{1a}, when the holonomy of $\varPi$ is generated by $\tau_1$, $\tau_2$, or $\tau_3$, Lemma
\ref{restrictions} forces $b_{12}=b_{13}=0$, $b_{12}=b_{23}=0$, or $b_{13}=b_{23}=0$,
respectively. Table \ref{EigenvaluesSolmn} shows that the eigenvalues of ${\hat r} \big|_{\bbr^3}$
can only have signs $(0,0,0)$ or $(0,+,-)$.

 Likewise, in case \gr{1b}, when the holonomy of $\varPi$ is generated by $-\tau_1$, $-\tau_2$, or $-\tau_3$, Lemma
\ref{restrictions} forces $b_{12}=b_{13}=0$, $b_{12}=b_{23}=0$, or $b_{13}=b_{23}=0$,
respectively. Again, Table \ref{EigenvaluesSolmn} shows that the eigenvalues of ${\hat r} \big|_{\bbr^3}$
can only have signs $(0,0,0)$ or $(0,+,-)$. Consequently, in both cases \gr{1a} and \gr{1b}, only the Ricci signatures $(0,0,0,-)$ and $(0,+,-,-)$  can be realized on $\varPi \bs \Solmn$.
\qedhere
\end{proof}


\step[{\bf $\Solr$}]
When $G=\Solr$, the image of 
$\aut(\fkg)$
under the restriction homomorphism \[
\aut(\fkg) \ra \aut(\bbr^3),
\] is the subgroup of $\gl(3,\bbr)$ consisting of matrices of the form
\[
\left[\begin{matrix} A & \bfzero \\ \bfzero & k
\end{matrix}\right],
\]
where $A$ is of the form $\left[ \begin{matrix} a & 0 \\ 0 & d \end{matrix} \right]$
or $\left[ \begin{matrix} 0 & b \\ c & 0 \end{matrix} \right]$. See Proposition \ref{autosolr}. 

\begin{lemma}
\label{quiteRigid2}
Let \[M=\varPi \bs (\Solr)\] and \[M'=\varPi' \bs (\Solr)\] for $\varPi, \varPi' \subset \Solr \rx \aut(\Solr)$ be two diffeomorphic infra-solvmanifolds of $\Solr$.

\bigskip
\noindent
\rm(1)
If $\tau_i$ (respectively, $-\tau_i$), for $i=1,2$, is in the image of the holonomy of $\varPi'$ in $\aut(\bbr^3)$,  then $\tau_j$ (respectively, $-\tau_j$), for $j=1,2$, is in
the image of the holonomy of $\varPi$ in $\aut(\bbr^3)$.

\bigskip
\noindent
\rm(2)
If $\tau_3$ (respectively, $-\tau_3$) is in the image of the holonomy of $\varPi'$ in $\aut(\bbr^3)$,  then $\tau_3$ (respectively, $-\tau_3$) is in
the image of the holonomy of $\varPi$ in $\aut(\bbr^3)$ (In our classification of $\Solr$ crystallographic groups, we denoted $\tau_3$ by $\tau$).

\bigskip
\noindent
\rm(3)
If $\left[\begin{matrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right]$ is in the image of the holonomy of $\varPi'$ in $\aut(\bbr^3)$, then $\left[\begin{matrix} 0 & -{a} & 0 \\ \frac{1}{a} & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right]$, for some real number $a$, is in
the image of the holonomy of $\varPi$ in $\aut(\bbr^3)$.
\end{lemma}
\begin{proof}
Since $\varPi$ and $\varPi'$ are isomorphic, Theorem \ref{Rigidity} implies
that $\varPi$ and $\varPi'$ are conjugate in $\aff(\Solr)$. Hence their holonomy groups
must be conjugate in $\aut(\Solr)$. Moreover, the image of the holonomy groups
in $\aut(\bbr^3)$ must be conjugate in the image of $\aut(\Solr)$ in $\aut(\bbr^3)$.

\bigskip
\noindent
For \rm(1), note that the only conjugates of $\tau_1$ and $-\tau_1$   in the image of $\aut(\Solr)$ in $\aut(\bbr^3)$,
are $\tau_2$ and $-\tau_2$, respectively.

\bigskip
\noindent
For \rm(2), note that $\tau_3$ is central in the image of $\aut(\Solr)$ in $\aut(\bbr^3)$, and so has no conjugates.

\bigskip
\noindent
For \rm(3), note that in the image of $\aut(\Solr)$ in $\aut(\bbr^3)$, $\left[\begin{matrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right]$ is conjugate to any matrix of the form $\left[\begin{matrix} 0 & -{a} & 0 \\ \frac{1}{a} & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right]$, $a>0$.
\qedhere
\end{proof}



%\begin{table}[H]
%\begin{tabular}{|p{3cm}| p{10cm}| |}
%\hline
%$\mathcal{B}$ & Eigenvalues of \newline $\frac 12 \left[N, N^T\right](u) = {\frac 1 2} %(NN^T- L^TL ) u$\\
%\hline
%$\left[\begin{matrix}
%b_{11} & 0 & 0\\
%0& b_{22} & b_{23}\\
%0 & 0 & b_{33}\\
%\end{matrix}\right]$  & $\left\{0,-\frac{b_{23} \sqrt{b_{22}^2+b_{23}^2} b_{44}^2 }{2 %b_{22}^2},\frac{b_{23} \sqrt{b_{22}^2+b_{23}^2} b_{44}^2}{2 b_{22}^2}\right\}$ \\
%\hline
%\left[\begin{matrix}
%b_{11} & 0 & b_{13}\\
%0& b_{22} & 0\\
%0 & 0 & b_{33}\\
%\end{matrix}\right]$  &  $\left\{0,-\frac{b_{13} \sqrt{b_{11}^2+b_{13}^2} b_{44}^2}{2 %b_{11}^2},\frac{b_{13} \sqrt{b_{11}^2+b_{13}^2} b_{44}^2 }{2 b_{11}^2}\right\}$\\
%\hline
%$\left[\begin{matrix}
%b_{11} & b_{12} & 0\\
%0& b_{22} & 0\\
%0 & 0 & b_{33}\\
%\end{matrix}\right]$  & $ \left\{0,-\frac{b_{12} \sqrt{b_{11}^2+b_{12}^2} b_{44}^2 }{2 %b_{11}^2},\frac{b_{12} \sqrt{b_{11}^2+b_{12}^2} b_{44}^2}{2 b_{11}^2}\right\}$\\
%\hline
%\end{tabular}
%\end{table}


\begin{theorem}
\label{SolrSig}
For each type of torsion-free $\Solr$ crystallographic group $\varPi$
in {\rm Section \ref{hithere}}, we list the Ricci signatures
which can be realized on the infra-solvmanifold $\varPi \bs (\Solr)$.
\begin{table}[H]
\caption{Realizable Ricci signatures on infra-$(\Solr)$ manifolds}
\begin{center}
\begin{tabular}{|p{3cm}| p{1.7cm} | p{6.8cm}| }
\hline
Infra-$\Solr$ & Holonomy &  Realizable Ricci Signatures\\
\hline
\gr{0a} & $\{e\}$ & (0,0,0,-), (0,+,-,-), (+,+,-,-), (+,-,-,-) \\
\hline
\gr{0b}, \gr{0c} & $\bbz_2$ & (0,0,0,-), (0,+,-,-)\\
\hline
\gr{1a}, \gr{1b} & $\bbz_2$ & (0,0,0,-), (0,+,-,-)\\
\hline
\gr{1c} & $\bbz_2^2$ & (0,0,0,-)\\
\hline
\gr{2aa} & $\bbz_2$ & (0,0,0,-), (0,+,-,-)\\
\hline
\gr{2ad} & $\bbz_2^2$ & (0,0,0,-), (0,+,-,-)\\
\hline
\gr{2ba} & $\bbz_2$ & (0,0,0,-), (0,+,-,-)\\
\hline
\gr{2bb} & $\bbz_2$ &  (0,0,0,-), (0,+,-,-), (+,+,-,-), (+,-,-,-) \\
\hline
\gr{2bc} & $\bbz_2^2$ &  (0,0,0,-), (0,+,-,-) \\
\hline
\gr{3a} & $\bbz_2$ &  (0,0,0,-), (0,+,-,-), (+,+,-,-), (+,-,-,-) \\
\hline
\gr{3\ii a}, \gr{3\ii b} & $\bbz_2$ &  (0,0,0,-), (0,+,-,-), (+,+,-,-), (+,-,-,-) \\
\hline
\gr{3\ii c}, \gr{3\ii d} & $\bbz_2^2$ &  (0,0,0,-), (0,+,-,-)\\
\hline
\gr{4a}& $\bbz_4$ &  (0,0,0,-)\\
\hline
\gr{5a}, \gr{5b}& $\bbz_2^2$ &  (0,0,0,-)\\
\hline
\gr{6a\ii a}, \gr{6a\ii b} & $\bbz_2^2$ &  (0,0,0,-), (0,+,-,-)\\
\hline
\gr{6a\ii e} & $\bbz_2^3$ &  (0,0,0,-), (0,+,-,-)\\
\hline
\gr{6ba} & $\bbz_2^2$ &  (0,0,0,-), (0,+,-,-)\\
\hline
\gr{6b\ii a}, \gr{6b\ii b} & $\bbz_2^2$ &  (0,0,0,-), (0,+,-,-)\\
\hline
\gr{6b\ii c}& $\bbz_2^2$ &  (0,0,0,-), (0,+,-,-), (+,+,-,-), (+,-,-,-) \\
\hline
\gr{6b\ii d}& $\bbz_2^3$ &  (0,0,0,-), (0,+,-,-)\\
\hline
\gr{7\ii d} & $D_4$ &  (0,0,0,-)\\
\hline
\end{tabular}
\end{center}
\end{table}
\end{theorem}
\begin{proof}

We can view $\Solr$ as $\Solmn$ with $\theta=-1$. Therefore,
Table \ref{EigenvaluesSolmn} applies.

The image of the holonomy in $\aut(\bbr^3)$ of any infra-solvmanifold of $\Solr$ diffeomorphic to an infra-solvmanifold
of type \gr{1c} must contain $\tau_3$ and either $\tau_1$ or $\tau_2$, by Lemma \ref{quiteRigid2}. Then Lemma \ref{restrictions} forces the condition $b_{12}=b_{13}=b_{23}=0$ on $\mathcal{B}$. Now apply Table \ref{EigenvaluesSolmn}
with $\theta=-1$.  Therefore, only the Ricci signature $(0,0,0,-)$ can be realized.

By Lemma \ref{quiteRigid2}, the image of the holonomy in $\aut(\bbr^3)$ of any infra-solvmanifold diffeomorphic to an infra-solvmanifold
of type \gr{4a} or \gr{7\ii d} contains an element of the form 
\[\left[\begin{matrix} 0 & -{a} & 0 \\ \frac{1}{a} & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right].\]
Now Lemma \ref{restrictions} places the condition $b_{12}=b_{13}=b_{23}=0$ on
the matrix $\mathcal{B}$. Thus Table \ref{EigenvaluesSolmn} shows that the eigenvalues
of ${\hat r} \big|_{\bbr^3}$ all vanish in these cases. Therefore, only the Ricci signature $(0,0,0,-)$ can be realized in cases \gr{4a} and \gr{7\ii d}.

The image of the holonomy in $\aut(\bbr^3)$  of any infra-solvmanifold of $\Solr$ diffeomorphic to an infra-solvmanifold of type \gr{5a} must contain
 $-\tau_3$, $\tau_1$, and $\tau_2$. For type  \gr{5b}, the image of the holonomy in $\aut(\bbr^3)$ must contain  $-\tau_3$, $-\tau_1$, and $-\tau_2$.
In both cases, $b_{12}=b_{13}=b_{23}=0$ by Lemma \ref{restrictions}.
Now apply Table \ref{EigenvaluesSolmn}
with $\theta=-1$ to see that only the Ricci signature $(0,0,0,-)$ can be realized.

The image of the holonomy in $\aut(\bbr^3)$  of any infra-solvmanifold of $\Solr$ diffeomorphic to an infra-solvmanifold of type \gr{2ad} or \gr{2bc}, must contain both $\tau_3$, and $-\tau_3$. Then Lemma \ref{restrictions} forces the condition $b_{13}=b_{23}=0$ on $\mathcal{B}$. We can make the signs of the eigenvalues of  ${\hat r} \big|_{\bbr^3}$ either $(0,+,-)$ or $(0,0,0)$, by choice of $b_{12}$ (Table \ref{EigenvaluesSolmn}
with $\theta=-1$).


The argument for cases \gr{0b}, \gr{0c}, \gr{1a}, \gr{1b}, \gr{2aa}, and \gr{2ba} is similar.
In these cases, any infra-solvmanifold of $\Solr$ diffeomorphic to an infra-solvmanifold
of the above types must have a holonomy element projecting to $\tau_i$ or $-\tau_i$
in $\aut(\bbr^3)$. Lemma \ref{restrictions} places conditions on $\mathcal{B}$, and
Table \ref{EigenvaluesSolmn} with $\theta=-1$ yields the desired result. In case \gr{2bb}, no condition on $\mathcal{B}$ is forced, as the image of the holonomy in $\aut(\bbr^3)$ is $-I$.


 
It remains to argue for cases \gr{3a}, \gr{3\ii a}, and \gr{3\ii b}. We claim that
any of the four Ricci signatures of $\Solr$ are realizable. In fact,
with basis
\begin{align}
\label{yay} \mathcal{B'}=
\left[\begin{matrix}
4 & -3 & b_{13} & 0\\
0& 5 & b_{13} & 0\\
0 & 0 & b_{33} & 0\\
0 & 0 & 0 & b_{44}
\end{matrix}\right],\end{align}
we compute that
\[
\mathcal{B'} \inv \left[\begin{matrix}
0 & 1 & 0 & 0\\
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & -1
\end{matrix}\right] \mathcal{B'} = \left[\begin{matrix}
\frac{3}{5} & \frac{4}{5} & 0 & 0\\
\frac{4}{5} & -\frac{3}{5} & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & -1
\end{matrix}\right] \in \on{O}(4,\bbr).
\]
Thus, with any orthonormal basis of this form, the isometry group
contains $\matNpp$, in the notation of Theorems \ref{3family} and \ref{3ifamily}.
With $\mathcal{B'}$ of this form, the determinant of ${\hat r} \big|_{\bbr^3}$ is
\begin{align*}
\label{deter}
\frac{1}{800} b_{13}^2 \left(75-b_{13}^2\right) b_{44}^6.
\end{align*}
As ${\hat r} \big|_{\bbr^3}$ has vanishing trace, the eigenvalues of
${\hat r} \big|_{\bbr^3}$ have signs $(+,-,-)$ and $(+,+,-)$,
when $75-b_{13}^2>0$ ($b_{13}\neq 0$) and $75-b_{13}^2<0$, respectively.
Of course, when $b_{13}^2=75$, the eigenvalues of ${\hat r} \big|_{\bbr^3}$ have signs $(0,+,-)$. This proves that the four Ricci signatures can be realized in cases \gr{3a} and \gr{3\ii a}. For case \gr{3\ii b},
with basis
\begin{align}
 \mathcal{B'}=
\left[\begin{matrix}
4 & 3 & -b_{13} & 0\\
0& 5 & b_{13} & 0\\
0 & 0 & b_{33} & 0\\
0 & 0 & 0 & b_{44}
\end{matrix}\right],\end{align}
we compute that
\[
\mathcal{B'} \inv \left[\begin{matrix}
0 & 1 & 0 & 0\\
1 & 0 & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1
\end{matrix}\right] \mathcal{B'} = \left[\begin{matrix}
-\frac{3}{5} & \frac{4}{5} & 0 & 0\\
\frac{4}{5} & \frac{3}{5} & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1
\end{matrix}\right] \in \on{O}(4,\bbr).
\]


Thus, with orthonormal basis of this form, the isometry group
contains $\matNpp \tau$, in the notation of case \gr{3\ii b} of Theorem \ref{3ifamily}.
With $\mathcal{B'}$ of this form, the determinant of ${\hat r} \big|_{\bbr^3}$ is again
\begin{align*}
\label{deterz}
\frac{1}{800} b_{13}^2 \left(75-b_{13}^2\right) b_{44}^6.
\end{align*}
So, all four Ricci signatures realized in cases \gr{3a} and \gr{3\ii a} are also realized in case \gr{3\ii b}.


In cases \gr{3\ii c}, \gr{3\ii d}, \gr{6a\ii a}, \gr{6a\ii b}, \gr{6a\ii e}, \gr{6ba}, \gr{6b\ii a}, \gr{6b\ii b}, and  \gr{6b\ii d}, $b_{13}=b_{23}=0$, as the holonomy must contain either $\tau_3$ or $-\tau_3$  . Only two Ricci signatures are realizable depending on choice of $b_{12}$.
In fact, in these cases, we can use the basis (\ref{yay}) with $b_{13}=0$, to realize the Ricci signature $(0,+,-,-)$.

For case \gr{6b\ii c}, the holonomy generator $B\tau$ in Theorem \ref{6bifamily} restricts to $-I$ in $\gl(3,\bbr)$, and does impose any restriction on $\mathcal{B}$, and we argue just like in cases \gr{3a} and \gr{3\ii a}.\qedhere
\end{proof}

\chapter{Topics for Further Research}

There are remaining questions about the topology and geometry of 4-dimensional infra-solvmanifolds (Problems 
\ref{geodesc} and \ref{spin} below).
Another direction for further research is to extend known partial results 
for Conjecture \ref{almostFlatBound} to 
certain infra-solvmanifolds (Problems \ref{counter} and \ref{boundForSolv} below).
Another interesting project is to characterize which solvable Lie groups satisfy the First Bieberbach Theorem 
(Problem  \ref{ProbBieberbach} below).

\section{Geometric Description of Infra-Solvmanifolds}

%\addcontentsline{toc}{subsection}{Geometric description of 4-dimensional infra-solvmanifolds}

Hillman provides a geometric classification of infra-solvmanifolds of $\Solr$ in \cite{hillman3} as Seifert fiberings
over $2$-dimensional flat orbifolds.
We have provided a complete classification of all crystallographic groups of $\Solr$. The torsion-free crystallographic groups $\varPi$
correspond to infra-solvmanifolds $\varPi \bs G$. However, we do not give a correspondence between our classification
and Hillman's classification.

\begin{problem}
\label{geodesc}
Find the correspondence between the torsion-free crystallographic groups of $\Solr$ described here,
and the classification Hillman provides in {\rm \cite{hillman3}}. Also describe the infra-solvmanifolds
of $\Solof$, as Hillman does for $\Solr$ in {\rm \cite{hillman3}}.
\end{problem}


\section{$Pin^{\pm}$ Structures on Infra-Solvmanifolds}
%\addcontentsline{toc}{subsection}{$Pin^{\pm}$  structures on 4-dimensional infra-solvmanifolds}

The group $Spin(n)$ is the double cover of $SO(n)$. 
The groups $Pin^{+}(n)$ and $Pin^{-}(n)$ are two double covers of $O(n)$.
They are the same topologically, but have different group structures.
The groups $Pin^{\pm}(n)$ allow a generalization of $Spin$ structures to non-orientable manifolds;
an orientation together with a  $Pin^{+}$ (or $Pin^{-}$) structure is equivalent to a $Spin$ structure.
An orientable Riemannian manifold $M$ has a $Spin$ structure precisely when 
the Stiefel Whitney class $\omega_{2}(M)$ vanishes. 
A Riemannian manifold $M$ has a $Pin^{+}$ structure if and only if $\omega_{2}(M)$  vanishes and has
a $Pin^{-}$ structure if and only if $\omega_{2}(M) +\omega_{1}^2(M)$  vanishes.

Hillman begins to explore which 4-dimensional infra-solvmanifolds admit $Pin^{+}$ and $Pin^{-}$  structures in \cite{hillman4}. 
This is related to a question of Ron Stern: is there an orientable aspherical surface bundle over the torus $T^2$
which does not admit a $Spin$ structure? A 4-dimensional infra-solvmanifold
may provide such an example.  Putrycz and Szczepanski determined that
only three orientable flat 4-manifolds are not $Spin$ \cite{PutryczSzczepanski}. 
However, these are not surface bundles over 
the torus.

\begin{problem}
\label{spin}
Determine which 4-dimensional infra-solvmanifolds
admit $Pin^{+}$ and $Pin^{-}$ structures.
\end{problem}

\section{Bounding Problem for Infra-Solvmanifolds}

%\addcontentsline{toc}{subsection}{Bounding problem for infra-solvmanifolds}

Since all 4-dimensional infra-solvmanifolds bound, some $n$-dimensional infra-nilmanifolds bound, and all 
closed flat $n$-manifolds bound, 
it is reasonable to ask if $n$-dimensional infra-solvmanifolds bound.

\begin{conjecture}
\label{conjSolv}
Let $M = \varPi \bs G$ be an $n$-dimensional infra-solvmanifold. Then there exists a compact $(n+1)$-dimensional manifold 
$W$ with
$\partial W = M$.
\end{conjecture}

Hamrick and Royster used translational involutions to show that closed
flat $n$-manifolds bound \cite{HR}. Marc Gordon introduced translational involutions in his proof that certain
closed flat manifolds bound \cite{MGor}. 
%Recent work by Davis and Fang
%on infra-nilmanifolds also uses this technique \cite{DavisFang}.
When $G$ is solvable and has non-trivial center,  translational involutions can also be used,  as was done in here
in the proof of Theorem \ref{mybound}.

\begin{problem}
\label{boundForSolv}
Let $M = \varPi \bs G$ be an $n$-dimensional infra-solvmanifold. Assume that $G$ has non-trivial center.
Is there a compact $(n+1)$-dimensional manifold $W$ with
$\partial W = M$?
\end{problem}

\noindent
On the other hand, Conjecture \ref{conjSolv} may be false. To give a counterexample
to Conjecture \ref{conjSolv} or (Conjecture \ref{almostFlatBound}), we would need an infra-solvmanifold (or 
infra-nilmanifold) with
a non vanishing Stiefel-Whitney number. In \cite{Vasquez, ImKim} there are examples of flat manifolds with 
non-trivial Stiefel Whitney classes
and this should be useful to approach:

\begin{problem}
\label{counter}
Give an example of an infra-solvmanifold or infra-nilmanifold which does not bound.
\end{problem}

\section{Bieberbach's First Theorem on Solvable Lie Groups}
%\addcontentsline{toc}{subsection}{Bieberbach's First Theorem on solvable Lie groups}

It is not well understood when a solvable Lie group satisfies Bieberbach's First Theorem (Theorem 
\ref{Bieberbach1}). Even a 4-dimensional solvable geometry, $\Solo$, does not satisfy it.
See \cite{DLR} for more examples of solvable Lie groups which do not satisfy Bieberbach's First Theorem.

\begin{problem}
\label{ProbBieberbach}
Characterize the solvable Lie groups for which
Bieberbach's first theorem holds.
\end{problem}

Good progress has been made by Dekimpe, Lee, and Raymond, who give a 
sufficient, but not necessary, condition for a solvable Lie group
to satisfy Bieberbach's first theorem \cite{DLR}.
In \cite{Buser}, Buser gives a geometric proof of Bieberbach's First Theorem for $\bbr^n$, inspired
by Gromov's work on almost flat manifolds. Essentially, the idea is to show that the isometries in a 
crystallographic group of $\bbr^n$ with very small
rotation part are in fact pure translations.
Examining Buser's argument when $\bbr^n$ is replaced with a solvable Lie group should provide
a simple characterization of solvable Lie groups satisfying Bieberbach's First Theorem.


%\subsection{$\Solo$}
%When $G=\Solo$,
%recall from Proposition \ref{autosolo}, that the image of $\aut(G)$
%under the restriction homomorphism \[
%\aut(G) \ra \aut(\bbr^3)
%\]
%is $\gl(2,\bbr) \x \bbr^*$, identified
%with the group of block diagonal matrices in $\gl(3,\bbr)$ given by
%\[\left[
%\begin{matrix}
%A & \bfzero \\
%\bfzero & k
%\end{matrix}
%\right].\]
%Of course, the image of \[r:\aut(\fkg) \ra \aut(\bbr^3)\]
%is likewise $\gl(2,\bbr) \x \bbr^*$.




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% BIBLIOGRAPHY %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

{
\singlespacing

%\bibliography{Xbib}
\bibliographystyle{bibtex}
\begin{thebibliography}{20}

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