# Achirality of Sol 3-Manifolds, Stevenhagen Conjecture
and Shimizu’s L-series

Ye Tian  Academy of Mathematics and System Science, Morningside Center of
Mathematics, Chinese Academy of Sciences, Beijing 100190<EMAIL_ADDRESS>,
Shicheng Wang  Department of Mathematical Sciences, Peking University, Beijing
100871, CHINA<EMAIL_ADDRESS>and  Zhongzi Wang  Department of
Mathematical Sciences, Peking University, Beijing 100871 CHINA
<EMAIL_ADDRESS>
###### Abstract.

A closed orientable manifold is achiral if it admits an orientation reversing
homeomorphism. A commensurable class of closed manifolds is achiral if it
contains an achiral element, or equivalently, each manifold in ${\mathcal{M}}$
has an achiral finite cover. Each commensurable class containing non-
orientable elements must be achiral. It is natural to wonder how many
commensurable classes are achiral and how many achiral classes have non-
orientable elements.

We study this problem for Sol 3-manifolds. Each commensurable class
${\mathcal{M}}$ of Sol 3-manifold has a complete topological invariant
$D_{{\mathcal{M}}}$, the discriminant of ${\mathcal{M}}$. Our main result is:

(1) Among all commensurable classes of Sol 3-manifolds, there are infinitely
many achiral classes; however ordered by discriminants, the density of achiral
commensurable classes is 0.

(2) Among all achiral commensurable classes of Sol 3-manifolds, ordered by
discriminants, the density of classes containing non-orientable elements is
$1-\rho$, where

$\rho:=\prod_{j=1}^{\infty}\left(1+2^{-j}\right)^{-1}=0.41942\cdots.$

###### Key words and phrases:

Class groups, Pell equations, Achirality of Sol 3-Manifolds

###### Contents

  1. 1 Introduction
  2. 2 Commensurable Classes of Sol 3-manifolds and Their Discriminants
  3. 3 Achirality, Gauss’ Theory of Class Groups and Genus
  4. 4 Density Results
  5. 5 Shimizu L-functions
  6. 6 Appendix: Proof of Lemma 3.3

##  1\. Introduction

In this paper, we assume that all manifolds $M$ are closed. A closed
orientable manifold is called achiral, if it admits an orientation reversing
homeomorphism, and is called virtually achiral if it has an achiral finite
cover. We call two manifolds $M_{1}$ and $M_{2}$ commensurable, if they have a
common finite cover. A commensurable class of closed manifolds is called
achiral if it contains an achiral element. A commensurable class
${\mathcal{M}}$ is achiral is equivalent to that each manifold in
${\mathcal{M}}$ is virtually achiral (Lemma 2.9). Since each non-orientable
manifold has an orientable double cover which is achiral, each commensurable
class containing non-orientable elements must be achiral.

The achirality is fundamental in topology for the classification of oriented
manifolds, specially in dimension 3, where the concept achirality is
originated. The study of various virtual properties, say virtually Haken,
virtually positive volume, virtual dominations, are important and active
topics in 3-manifolds, see a survey [LS].

It will be natural to wonder the following questions: In cartain range,

1\. What is the density of achiral manifolds among all?

2\. What is the density of manifolds have an achiral finite cover among all?
or what is the density of achiral classes among all commensurable classes?

3\. What is the density of classes containing non-orientable element among all
achiral classes?

Question 1 is for manifolds and Question 2 and Question 3 are for
commensurable classes. Note to study those questions, one need first to order
the manifolds or classes in the range considered, which is subtle, and often
difficult.

In dimension 1 and 2, the answers to those questions are direct: There is only
one closed 1-manifold, the circle, which is achiral. All orientable
2-manifolds are achiral. Denote ${\mathbb{H}}^{n}$, ${\mathbb{E}}^{n}$,
$S^{n}$ the $n$-dimensional hyperbolic, Euclidean, and spherical geometries,
respectively. In dimension two case, every closed surface supports one of the
three geometries: $S^{2},{\mathbb{E}}^{2},{\mathbb{H}}^{2}.$ Each geometry has
only one commensurable class which also contains non-orientable elements.

Once we come to dimension 3, those questions become highly non-trivial. Recall
that Thurston’s eight geometries among 3-manifolds are [Th] [Sc],

$S^{3},\quad{\mathbb{E}}^{3},\quad
S^{2}\times{\mathbb{E}}^{1},\quad{\mathbb{H}}^{2}\times{\mathbb{E}}^{1},\quad\mathrm{Nil},\quad\widetilde{{\mathrm{PSL}}(2,\mathbb{R})},\quad{\mathbb{H}}^{3},\quad\mathrm{Sol}.$

It is neither meaningful nor possible for us to order all those manifolds from
those eight geometries, and to get answers to those questions. Let’s see what
happen when we restrict to each individual geometry.

We first have a look of Question 1. Four cases are essentially known: Each
manifolds supports either $\mathrm{Nil}-$ or
$\widetilde{{\mathrm{PSL}}(2,\mathbb{R})}-$ geometry is non-achiral, so the
density is 0. There are only two orientable manifolds supporting $S^{2}\times
E^{1}$-geometry, both are achiral, so the density is 1. There are only seven
orientable manifolds supporting the $E^{3}$-geometry, four of them are achiral
and three are not ([SWW, Theorem 1.7] and Lemma 2.7), so the density is $4/7$.
It is not clear for us how to order the manifolds for the remaining geometries
to get the density.

Now we discuss Questions 2 and 3, which is the issue of the paper. The
situation is known for the first six classes: 3-manifolds supporting one of
the first six geometries form only one commensurable class ([Sc]), and
commensurable classes are achiral for the manifolds supporting first four
geometries, and non-achiral for manifolds supporting next two geometries.
Moreover there is no non-orientable 3-manifolds supporting $S^{3}$-geometric,
and there are non-orientable 3-manifolds supporting the next three geometries.
On the other hand 3-manifolds supporting one of the last two geometries, Sol
or ${\mathbb{H}}^{3}$, form infinitely many commensurable classes, and the
achirality of those commensurable classes are not addressed.

We study the achirality of commensurable classes of Sol 3-manifolds in this
paper.

Each commensurable class ${\mathcal{M}}$ of Sol 3-manifold contains an
orientable torus bundle $M=M_{\phi}$, eigenvalues of whose monodromy map
$\phi$ generates over ${\mathbb{Q}}$ a real quadratic field. The discriminant
of the real quadratic field is the complete topological invariant of
${\mathcal{M}}$, denoted by $D_{{\mathcal{M}}}$ and called the discriminant of
${\mathcal{M}}$

Our main result is:

######  Theorem 1.1.

(1) Among all commensurable classes of Sol 3-manifolds, there are infinitely
many achiral classes; however ordered by discriminants, the density of achiral
commensurable classes:

$\lim_{X\rightarrow\infty}\frac{\\#\\{\text{achiral}\ {\mathcal{M}}\ \
|D_{{\mathcal{M}}}<X\\}}{\\#\\{{\mathcal{M}}\ |\ D_{{\mathcal{M}}}<X\
{}\\}}=0,$

(2) Among all achiral commensurable classes of Sol 3-manifolds ordered by
discriminants, the density of classes containing non-orientable elements:

$\lim_{X\rightarrow\infty}\frac{\\#\\{{\mathcal{M}}\ {}\text{contains an non-
orientable element \,}|\,D_{\mathcal{M}}<X\ \\}}{\\#\\{\text{achiral}\
{\mathcal{M}}\ \ |D_{{\mathcal{M}}}<X\\}}=1-\rho,$

where

$\rho:=\prod_{j=1}^{\infty}\left(1+2^{-j}\right)^{-1}=0.41942\cdots.$

The project is started by wondering if each Sol 3-manifold has an achiral
finite cover. Theorem 1.1 (1) is unexpected to us which claims that to have
achiral finite covers are rare among Sol 3-manifolds. Theorem 1.1 (2) is a big
surprise that the Stevenhagen’s conjecture, which was solved by P. Koymans and
C. Pagano [KP] in 2022, has its avatar in topology.

For each orientable Sol torus bundle $M=M_{\phi}$, we introduce the
discriminant $D_{M}$ of $M$ to be the discriminant of the fixed points of its
monodromy $\phi$. The integer $D_{M}$ is a topological invariant of $M$ so
that ${\mathbb{Q}}(\sqrt{D_{M}})$ is the real quadratic field of the
commensurable class containing $M$. Another result, Theorem 3.7, in this paper
is that for any given $D\equiv 0,1\ \mathrm{mod}\ 4$ non-square positive
integer, the following conditions are equivalent

  * •

there exists an achiral Sol torus bundle $M$ with discriminant $D$;

  * •

no prime factor of $D$ is $\equiv 3\ \mathrm{mod}\ 4$ and $16\nmid D$.

Consequences of Theorem 3.7 including: among Sol 3-manifolds, a commensurable
class ${\mathcal{M}}$ is achiral if and only if $D_{{\mathcal{M}}}$ contains
no prime $\equiv 3\ \mathrm{mod}\ 4$ (Theorem 3.8); each achiral commensurable
class contains non-achiral manifolds (Corollary 3.9); and ${\mathcal{M}}$
contains an non-orientable element if and only if it contains an achiral torus
semi-bundle (Corollary 4.2).

In his study on cusp cross-sections of Hilbert modular varieties, Hirzebruch
made a conjecture relates the signature defects of an orientable Sol torus
bundle to the special value of its corresponding Shimizu’s L-function. In this
paper, we also show that, Theorem 5.1 (1), for an oriented Sol torus bundle
$M$, $M$ is achiral if and only if the Shimizu’s L-series of $M$ is
identically zero. If monodormies of two oriented Sol torus bundles are
negative to each other, they are not homeomorphic but their Shimuz L-series
are the same. We have that, Theorem 5.1 (2), non-achiral oriented Sol tours
bundles are determined by their Shimizu L-functions up to such a relation.

Theorem 1.1 follows from the positive answer of Stevenhagen Conjecture,
Theorem 3.8, and Corollary 4.2 which claims that a commensurable class
contains a non-orientable element if and only if the corresponding negative
Pell equation has a solution. Both the proofs of Theorem 3.7 and 5.1 are based
on Theorem 3.5 which claims that a Sol 3-manifold $M_{\phi}$ is achiral if and
only if $[-Q_{\phi}]=\pm[Q_{\phi}]$ in class group $C(D_{M})$ for its
corresponding quadratic form $Q_{\phi}$. The proof of Theorem 3.7 also highly
relies on Gauss’ genus theory.

##  2\. Commensurable Classes of Sol 3-manifolds and Their Discriminants

Classification of Sol 3-manifolds. Recall the Sol geometry is the Lie group
${\mathbb{R}}^{2}\rtimes{\mathbb{R}}^{1}$ with structure given by the
representation
${\mathbb{R}}^{1}\rightarrow{\mathrm{GL}}_{2}({\mathbb{R}}),z\mapsto\begin{pmatrix}e^{z}&\\\
&e^{-z}\end{pmatrix}$, together the invariant Riemann metric
$e^{2z}dx^{2}+e^{-2z}dy^{2}+dz^{2}$. A connected closed manifold is called Sol
3-manifold if it is modelled on $\mathrm{Sol}$, i.e. is of form
$\mathrm{Sol}/\Gamma$ for some discrete subgroup $\Gamma$ of
${\mathrm{Isom}}(\mathrm{Sol})$. There are two types Sol 3-manifolds:

  * •

let $T={\mathbb{R}}/2\pi{\mathbb{Z}}\times{\mathbb{R}}/2\pi{\mathbb{Z}}$ be a
torus, $I=[0,1]$, and $\phi\in{\mathrm{GL}}_{2}({\mathbb{Z}})$ Anosov matrix,
i.e. its eigenvalues are real but not $\\{\pm 1\\}$. Define

$M_{\phi}=T\times I/((x,y)\phi,0)\sim(x,y,1).$

Thus $M_{\phi}$ is a tours bundle over a circle.

  * •

let

$N=T\times I/(x,y,z)\sim(x+\pi,-y,1-z)$

a twisted $I$-bundle of the Klein bottle $K=T/(x,y)\sim(x+\pi,-y)$. For each
$\psi=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\in{\mathrm{SL}}_{2}({\mathbb{Z}})$ with $abcd\neq 0$, define
the semi-torus bundle by gluing two $N$’s along their boundary
$T\times\\{1\\}$ via $\psi$:

$N_{\psi}=N\cup N\big{/}(x,y,1)\sim((x,y)\psi,1),$

The following classification result of Sol 3-manifolds is basically known
except the non-orientable case needs some explanations.

######  Proposition 2.1.

Any Sol 3-manifold is homeomorphic to one of the following

  1. (1)

Torus bundles $M_{\phi}$: two $M_{\phi}$ and $M_{\phi^{\prime}}$ are
homeomorphic if and only if $\phi^{\prime}$ is
${\mathrm{GL}}_{2}({\mathbb{Z}})$-conjugate to $\phi^{\pm 1}$. Moreover,
$M_{\phi}$ is orientable if and only if
$\phi\in{\mathrm{SL}}_{2}({\mathbb{Z}})$.

  2. (2)

Torus semi-bundles $N_{\psi}$: two $N_{\psi}$ and $N_{\psi^{\prime}}$are
homeomorphic if and only if

$\psi^{\prime}=\begin{pmatrix}\pm 1&0\\\ 0&\pm 1\end{pmatrix}\psi^{\pm
1}\begin{pmatrix}\pm 1&0\\\ 0&\pm 1\end{pmatrix}.$

Moreover, $N_{\psi}$ is always orientable.

###### Proof.

Orientable Sol 3-manifolds are classified by ([Ha], Theorems 2.6 and 2.8, see
also [SWW] Theorems 2.3 and 2.4). Now let $M$ be a non-orientable Sol
3-manifold. Then the orientable double cover of $M$ must be either a torus
bundle or torus semi-bundle. In the first case, $M$ must also be a torus
bundle by the classification of free involutions on torus bundles (see [Sak]).
In fact, it is exactly the Case (II) in [Sak] p 168. In the second case, $M$
is given by a free involution on torus semi-bundle which must be orientable by
[Theorem 2] in [BGV]. ∎

Fix orientations on $T$ and $I$, each orientable torus bundle
$M_{\phi},\phi\in{\mathrm{SL}}_{2}({\mathbb{Z}})$, inherits an orientation,
and also denote by $M_{\phi}$ the oriented torus bundle.

######  Lemma 2.2.

For each oriented $M_{\phi}$, $M_{\phi^{-1}}=-M_{\phi}.$ where $-M$ is the $M$
with opposite orientation.

###### Proof.

By definition we have

$M_{\phi^{-1}}=\frac{T\times[0,1]}{(x,1)\sim(\phi^{-1}(x),0)}.$

Note identify the top to the bottom via $\phi^{-1}$ is the same as identify
the bottom to top via $\phi$, and later is homeomorphic to $M_{\phi}$ via an
orientation reversing homeomorphism which is obtained by changing the
orientation of $[0,1]$. We proved the lemma. ∎

######  Lemma 2.3.

Two oriented tours bundles $M_{\phi}$ and $M_{\phi^{\prime}}$ are homeomorphic
if and only if $\phi^{\prime}$ is ${\mathrm{SL}}_{2}({\mathbb{Z}})$-conjugate
to $\phi$ or $w\phi^{-1}w$, where $w=\begin{pmatrix}&1\\\ 1&\end{pmatrix}$.

###### Proof.

By Proposition 2.1 (1), $M_{\phi^{\prime}}$ is homeomorphic to $M_{\phi}$ if
and only if
$\phi^{\prime}=A\phi^{\pm}A^{-1},A\in{\mathrm{GL}}_{2}({\mathbb{Z}})$. If
$A\notin SL_{2}({\mathbb{Z}})$, we write $A=Bw$ with $B\in
SL_{2}({\mathbb{Z}})$. So $M_{\phi^{\prime}}$ is homeomorphic to $M_{\phi}$ if
and only if $\phi^{\prime}$ is ${\mathrm{SL}}_{2}({\mathbb{Z}})$ conjugate to
$\phi^{\pm 1}$ or $w\phi^{\pm 1}w$. The conclusion follows by Lemma 2.2 and
that $\det w=-1$, that is $w$ is orientation reversing on $T$. ∎

Commensurable classes of Sol 3-manifolds. We call two manifolds $M_{1}$ and
$M_{2}$ commensurable, if they have a common finite cover. It is easy to see
that commensurable relation is an equivalent relation. We will use
${\mathcal{M}}$ to denote the commensurable class of $M$.

Suppose $K={\mathbb{Q}}(\sqrt{d})$ is a real quadratic field, where $d>0$ is
square free. Define the fundamental discriminant $D_{K}$ of $K$ by $D_{K}=d$
if $d\equiv 1\,\text{mod}\,4$ and $D_{K}=4d$ otherwise [EW, p.92].

Each commensurable class ${\mathcal{M}}$ of Sol 3-manifold contains an
orientable torus bundle, eigenvalues of whose monodromy map generates over
${\mathbb{Q}}$ a real quadratic field $K$. The fundamental discriminant
$D_{K}$, denoted by $D_{\mathcal{M}}$ and called the discriminant of
${\mathcal{M}}$.

######  Proposition 2.4.

There is a one-to-one bijection between

  * •

commensurable classes ${\mathcal{M}}$ of Sol 3-manifolds;

  * •

real quadratic fields ${\mathbb{Q}}(\sqrt{D_{\mathcal{M}}})$ (or equivalently,
fundamental positive discriminants $D_{\mathcal{M}}$),

For each torus bundle $M_{\phi}:T\to M_{\phi}\to S^{1}$, we have a short fiber
exact sequence

$1\to\pi_{1}(T)\to\pi_{1}(M_{\phi})\to\pi_{1}(S^{1})\to 1.\qquad(2.1)$

For each covering map $p:T\to T$, we can homotopy $p$ to be non-degenerated
linear map, that is $p\in M_{2}({\mathbb{Z}})$ and $p$ is of rank $2$. For
each fiber preserving covering map $f:M_{\psi}\to M_{\phi}$ between torus
bundles, there is an induced commutative diagram between of these exact
sequences.

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
5.5pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\pi_{1}(T)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
68.05663pt\raise 6.32918pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.39862pt\hbox{$\scriptstyle{i^{\prime}_{*}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 83.01071pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
44.25536pt\raise-20.83554pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.75pt\hbox{$\scriptstyle{f|_{*}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
44.25536pt\raise-31.17108pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
83.01071pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\pi_{1}(M_{\psi})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
125.25745pt\raise 5.74306pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.38196pt\hbox{$\scriptstyle{q^{\prime}_{*}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 143.72824pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
101.36948pt\raise-20.83554pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.75pt\hbox{$\scriptstyle{f_{*}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
101.36948pt\raise-31.17108pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
143.72824pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces\pi_{1}(S^{1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
159.62143pt\raise-20.83554pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.37361pt\hbox{$\scriptstyle{\bar{f}_{*}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
159.62143pt\raise-30.03333pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
199.51462pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
199.51462pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{1}$}}}}}}}{\hbox{\kern-5.5pt\raise-41.67108pt\hbox{\hbox{\kern
0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
29.5pt\raise-41.67108pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
29.5pt\raise-41.67108pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\pi_{1}(T)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
67.60661pt\raise-35.89746pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.84306pt\hbox{$\scriptstyle{i_{*}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
83.16628pt\raise-41.67108pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
83.16628pt\raise-41.67108pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{\pi_{1}(M_{\phi})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
124.80743pt\raise-36.48358pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-0.8264pt\hbox{$\scriptstyle{q_{*}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
143.72824pt\raise-41.67108pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
143.72824pt\raise-41.67108pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{\pi_{1}(S^{1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
199.51462pt\raise-41.67108pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
199.51462pt\raise-41.67108pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{1}$}}}}}}}\ignorespaces}}}}\ignorespaces\qquad(2.2)$

Call the covering $f$ is vertical, if the left side vertical map $f|_{*}$ is
an isomorphism, that is to say the covering is from the circle direction; and
the covering is horizontal, if the right side vertical map $\bar{f}_{*}$ is an
isomorphism, that is to say the covering is from the torus direction.

Call a closed orientable 3-manifold $M$ Haken, if each embedded 2-sphere in
$M$ bounds a 3-ball in $M$, and there exists a closed orientable embedded
surface of genus $\geq 1$ in $M$ which induce an injection $\pi_{1}$. Each Sol
3-manifold is Haken.

######  Lemma 2.5.

Suppose $M_{\phi}$ and $M_{\psi}$ are orientable torus bundle supporting Sol
geometry. Then

(1) Each Sol $M_{\phi}$ has unique torus bundle structure up to isotopy.

(2) Any covering $f:M_{\psi}\to M_{\phi}$ is fiber preserving up to isotopy.

###### Proof.

(1) Suppose $T^{\prime}$ is a torus fiber of a fiberation of $M_{\phi}$.
According to (2.1), $T\to M_{\phi}$ induces an injection on $\pi_{1}$, hence
$T^{\prime}$ is incompressible. Since $|{\mathrm{tr}}(\phi)|>2$, $\phi$ is not
conjugate to $\begin{pmatrix}1&0\\\ n&1\end{pmatrix}$, by [Ha, Lemma 5.2],
$T^{\prime}$ is isotopy to $T$, the torus fiber of the fiberation determined
by $\phi$.

(2) If we fiber $M_{\psi}$ with lifted torus fiberation of $M_{\phi}$, then
the covering is fiber preserving. Since $M_{\psi}$ also supports Sol geometry,
by (1) its original torus fiberation is isotopic to the lifted torus
fiberation, and (2) follows. ∎

######  Lemma 2.6.

For each fiber preserving covering map $f:M_{\psi}\to M_{\phi}$ between torus
bundles, $f=f_{v}\circ f_{h}$, where $f_{h}$ is a horizontal covering and
$f_{v}$ is a vertical covering. Moreover

  * (1)

there is a horizontal covering $f:M_{\psi}\rightarrow M_{\phi}$ if and only if
$\psi$ is ${\mathrm{GL}}_{2}({\mathbb{Q}})$-conjugate to $\phi^{\pm 1}$, or
equivalently, ${\mathrm{tr}}\phi={\mathrm{tr}}\psi$

  * (2)

there is a vertical covering $f:M_{\psi}\rightarrow M_{\phi}$ of degree $n$ if
and only if $\psi$ is ${\mathrm{GL}}_{2}({\mathbb{Z}})$-conjugate to
$\phi^{\pm n}$.

It follows that any fiber preserving covering $f:M_{\psi}\rightarrow M_{\phi}$
has that $\psi$ is ${\mathrm{GL}}_{2}({\mathbb{Q}})$-conjugate to $\phi^{n}$
for some integer $n\neq 0$.

###### Proof.

Let $G=(q^{*})^{-1}\bar{f}_{*}(\pi_{1}(S^{1}))$, $G$. By (2.2), $G$ is
subgroup of $\pi_{1}(M_{\phi})$ and $f_{*}(\pi_{1}(M_{\psi}))$ is a subgroup
of $G$. Since $f:M_{\psi}\to M_{\phi}$ is a finite covering, we have that $G$
is a finite index subgroup of $\pi_{1}(M_{\phi})$. By covering space theory,
there is a 3-manifold $M$ with $\pi_{1}(M)=G$ and the finite covering $f$ is
docomposed into the finite covering $f_{h}:M_{\psi}\to M$ and $f_{v}:M\to
M_{\phi}$. If we lift the torus bundle structure of $M_{\phi}$ to $M$, then
both $f_{h}$ and $f_{v}$ become a fiber preserving covering. One can verify
that $f_{h}$ is horizontal and $f_{v}$ is vertical.

We now show (1). Let $T$ be a fiber torus of $M_{\phi}$. Since the fiber
preserving covering $f$ is horizontal, $\tilde{T}=f^{-1}(T)$ is a fiber torus
of $M_{\psi}$. Cutting $M_{\phi}$ along $T$ and $M_{\psi}$ along $\tilde{T}$,
we get an induced a fiber preserving covering map

$\bar{f}:\tilde{T}\times[0,1]\longrightarrow T\times[0,1],$

where $f(x,t)=(p(x),\epsilon t)$, where $p:\tilde{T}\to T$ is a covering map,
and $\epsilon=\pm 1$ which depends on the map preserving or reversing the
orientation of $[0,1]$. Then from the constructions of $M_{\psi}$ and
$M_{\phi}$, we have the following commutative diagram

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
17.1667pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{\tilde{T}\times\\{1\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
25.38057pt\raise 5.1875pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-0.8264pt\hbox{$\scriptstyle{p}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 41.1667pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
0.0pt\raise-20.0pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern 0.0pt\raise-1.75pt\hbox{$\scriptstyle{\psi}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 0.0pt\raise-29.5pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
41.1667pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{T\times\\{1\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
60.28304pt\raise-20.0pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.18056pt\hbox{$\scriptstyle{\phi^{\epsilon}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 60.28304pt\raise-29.5pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx<EMAIL_ADDRESS>0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{\tilde{T}\times\\{0\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
25.38057pt\raise-34.8125pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-0.8264pt\hbox{$\scriptstyle{p}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 41.1667pt\raise-40.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
41.1667pt\raise-40.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{T\times\\{0\\}}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

That is to say for any $(x,1)\in\tilde{T}\times\\{1\\}$, we have
$\phi^{\epsilon}\circ p(x,1)=p\circ\psi(x,1),$ where we define
$p(x,1)=(p(x),1)$. That is $(\phi^{\epsilon}\circ p(x),0)=(p\circ\psi(x),0).$
So we have $\phi^{\epsilon}\circ p=p\circ\psi$, or

$A\circ\psi=\phi^{\epsilon}\circ A,\qquad(2.3)$

where $A$ is the rank 2 element in $M_{2}({\mathbb{Z}})$ induced by $p$.
Another direction is a direct construction.

It is clear that $\psi=A^{-1}\circ\phi^{\pm 1}\circ A$ implies that
${\mathrm{tr}}(\phi)={\mathrm{tr}}(\psi)$. On the other hand, if
${\mathrm{tr}}(\phi)={\mathrm{tr}}(\psi)$, then $\phi$ and $\psi$ are
conjugate in ${\mathrm{GL}}_{2}({\mathbb{R}})$ and thus in
${\mathrm{GL}}_{2}({\mathbb{Q}})$ and in ${\mathrm{GL}}_{2}({\mathbb{Z}})$
(the equation $p\phi=\psi p$ in $p$ is linear). In fact, let
$\phi=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}$ and
$\psi=\begin{pmatrix}a+u&w\\\ v&d-u\end{pmatrix}$, then $cv\neq 0$ and one may
take $p=\begin{pmatrix}1&u/c\\\ 0&v/c\end{pmatrix}$.

We now show (2). First $M_{\phi^{n}}$ has a vertical covering structure of
$M_{\phi}$ of degree $n$. On the other hand,
$H_{1}(M_{\phi},{\mathbb{Z}})={\mathbb{Z}}\oplus\frac{H_{1}(T,{\mathbb{Z}})}{{\mathrm{Im}}(\phi_{*}-\mathrm{Id})}$
and a vertical covering $M_{\psi}\rightarrow M_{\phi}$ of degree $n$
corresponds to the index $n$ subgroup of $H_{1}(M_{\phi},{\mathbb{Z}})$ given
by the ${\mathbb{Z}}$-component. By Lemma 2.5 (1), the uniqueness of
fiberation, $\psi$ is conjugate to $\phi^{\pm n}$. ∎

Now we recall some facts about real quadratic field $K={\mathbb{Q}}(\sqrt{d})$
[EW, Chapters 3 and 4]which will be used in the proof of Proposition 2.4 and
thereafter. For each $x=a+b\sqrt{d}\in F$, define its norm
$N(x)=a^{2}-b^{2}d$. Call $x$ absolutely positive if both $x$ and $N(x)$ are
positive.

Let $O_{K}$ be the ring consists of all algebraic integers of $K$, or
equivalently, the $x\in K$ whose minimal polynomial hase form $x^{2}+bx+c$ for
$b,c\in{\mathbb{Z}}$.

Call a subring ${\mathcal{O}}\subset O_{K}$ of rank 2 an order of $K$. It is
known that ${\mathcal{O}}={\mathbb{Z}}\oplus c^{2}O_{K}$ for
$c\in{\mathbb{Z}}$. Define the discriminant of ${\mathcal{O}}$ to be
$c^{2}D_{F}$. Note $O_{K}$ itself is the maximum order of $K$. Let
${\mathcal{O}}^{\times}$ be group of all units of ${\mathcal{O}}$. Then
Dirichlet Unit Theorem implies
${\mathcal{O}}^{\times}={\mathbb{Z}}\oplus{\mathbb{Z}}_{2}$, and call a
positive generator of ${\mathbb{Z}}$ a fundamental unit of ${\mathcal{O}}$.

###### Proof of Proposition 2.4.

Suppose $M_{\phi_{1}},M_{\phi_{2}}$ are commensurable. Then there are two
coverings $M_{\phi}\rightarrow M_{\phi_{1}},M_{\phi_{2}}$ between Sol torus
bundles. By Lemma 2.5, we may assume that they are fiber preserving coverings.
By Lemma 2.6, we have

$\phi=A_{i}\phi_{i}^{n_{i}}A_{i}^{-1},\qquad i=1,2.$

Here $A_{i}\in{\mathrm{GL}}_{2}({\mathbb{Q}})$ and $n_{i}\neq 0$ integers. It
follows that

$\phi_{2}^{n_{2}}=(A_{2}^{-1}A_{1})\phi_{1}^{n_{1}}(A_{1}^{-1}A_{2}).$

Therefore, $\phi_{1},\phi_{2}$ give rise to the same real quadratic fields.

On the other hand, suppose $\phi_{1},\phi_{2}$ give rise to the same real
quadratic field $K$, and let $\lambda_{1},\lambda_{2}$ be their eigenvalues,
respectively. As the roots of characteristic polynomial of element in
$SL_{2}({\mathbb{Z}})$, both $\lambda_{1},\lambda_{2}$ are units in $O_{F}$.
Let $\epsilon$ be a fundamental unit of $O_{K}$. Since
$O_{K}^{\times}={\mathbb{Z}}\oplus{\mathbb{Z}}_{2}$, we have

$\lambda_{1}=\pm\epsilon^{n_{1}},\quad\lambda_{2}=\pm\epsilon^{n_{2}}$

for some integers $n_{1},n_{2}\neq 0$. This implies
$\lambda_{1}^{2n_{2}}=\lambda_{2}^{2n_{1}}$, and therefore
${\mathrm{tr}}(\phi_{1}^{2n_{2}})={\mathrm{tr}}(\phi_{2}^{2n_{1}})$. By Lemma
2.6 (1), there exists a covering $M_{\phi_{1}^{2n_{2}}}\rightarrow
M_{\phi_{2}^{2n_{1}}}$. Since $M_{\phi_{2}^{2n_{1}}}$ covers $M_{\phi_{2}}$,
$M_{\phi_{1}^{2n_{2}}}$ covers both $M_{\phi_{1}}$ and $M_{\phi_{2}}$. It
follows that $M_{\phi_{1}}$ and $M_{\phi_{2}}$ are commensurable. ∎

######  Lemma 2.7.

For a Haken manifold $M$, $M$ is achiral if and only if $-1$ is a mapping
degree of $M$.

###### Proof.

Suppose $f:M\to M$ is a map of degree $-1$. Then $f$ induces a surjection on
$f_{*}:\pi_{1}(M)\to\pi_{1}(M)$ [He1, page 175]. Since $M$ is Haken,
$\pi_{1}(M)$ is Hopfian, which first implies that $f_{*}$ is an isomorphism
[He1, page 177], [He2], then implies that $f$ is homotopic to a homeomorphism
[He1, Chap. 13], which must be orientation reversing. Another direction is
obvious. ∎

######  Lemma 2.8.

Each torus semi-bundle $N_{\psi}$ has a unique torus bundle double cover
$M_{\phi}$. The covering is characteristic and has a realization as
$M_{\phi}\rightarrow N_{\psi}$ with $\psi=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}$ and $\phi=\begin{pmatrix}2bc+1&2ab\\\
2cd&2bc+1\end{pmatrix}$.

###### Proof.

It is known that each $N_{\psi}$ is doubly covered by a torus bundle
$M_{\phi}$ ([Ha], Theorems 2.6 and 2.8). We just verify that $M_{\phi}$ is
unique and characteristic.

For each torus semi-bundle $N_{\psi}$, we have a short fiber exact sequence

$1\to\pi_{1}(T)\to\pi_{1}(N_{\psi})\to{\mathbb{Z}}_{2}*{\mathbb{Z}}_{2}\to
1.\qquad(2.4)$

The unique torus bundle double cover $f:M_{\phi}\to N_{\psi}$ is given by the
unique index 2 subgroup
${\mathbb{Z}}\subset{\mathbb{Z}}_{2}*{\mathbb{Z}}_{2}$, which is also
characteristic, indeed ${\mathbb{Z}}_{2}*{\mathbb{Z}}_{2}$ has only three
index 2 subgroups, two of them isomorphic to
${\mathbb{Z}}_{2}*{\mathbb{Z}}_{2}$, one is isomorphic to ${\mathbb{Z}}$.
There is an induced commutative diagram between of these exact sequences.

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
5.5pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\pi_{1}(T)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
67.97885pt\raise 6.32918pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.39862pt\hbox{$\scriptstyle{i^{\prime}_{*}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 83.01071pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
44.25536pt\raise-20.26666pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.75pt\hbox{$\scriptstyle{f|_{*}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
44.25536pt\raise-30.03333pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
83.01071pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\pi_{1}(M_{\phi})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
124.28307pt\raise 5.74306pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.38196pt\hbox{$\scriptstyle{q^{\prime}_{*}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 151.77267pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
101.21391pt\raise-20.26666pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.75pt\hbox{$\scriptstyle{f_{*}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
101.21391pt\raise-30.03333pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
151.77267pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces{\mathbb{Z}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
157.82823pt\raise-20.26666pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.37361pt\hbox{$\scriptstyle{\bar{f}_{*}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
157.82823pt\raise-30.70001pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
196.23935pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
196.23935pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{1}$}}}}}}}{\hbox{\kern-5.5pt\raise-40.53333pt\hbox{\hbox{\kern
0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
29.5pt\raise-40.53333pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
29.5pt\raise-40.53333pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\pi_{1}(T)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
67.52884pt\raise-34.7597pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.84306pt\hbox{$\scriptstyle{i_{*}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
83.68849pt\raise-40.53333pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
83.68849pt\raise-40.53333pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{\pi_{1}(N_{\psi})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
123.83305pt\raise-35.34583pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-0.8264pt\hbox{$\scriptstyle{q_{*}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
143.41711pt\raise-40.53333pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
143.41711pt\raise-40.53333pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{{\mathbb{Z}}_{2}*{\mathbb{Z}}_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
196.23935pt\raise-40.53333pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
196.23935pt\raise-40.53333pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{1}$}}}}}}}\ignorespaces}}}}\ignorespaces\qquad(2.5)$

where $f|_{*}$ is an isomorphism. Then we can identify
$\pi_{1}(T)\subset\pi_{1}(M_{\phi})$ and $\pi_{1}(T)\subset\pi_{1}(N_{\psi})$.
Note any homeomorphism on $\tau$ on $N_{\psi}$ keeps $T=\partial N$ invariant
up to isotopy and $\pi_{1}(T)$ is a normal subgroup of $\pi_{1}(N_{\psi})$. It
follows that $\pi_{1}(T)\subset\pi_{1}(N_{\psi})$ is characteristic. Then for
any automorphism $\eta$ on $\pi_{1}(N_{\psi})$, $\eta$ induces an automorphism
$\bar{\eta}$ on ${\mathbb{Z}}_{2}*{\mathbb{Z}}_{2}$. Since
$\bar{\eta}({\mathbb{Z}})={\mathbb{Z}}$, we have
$q^{-1}(\bar{\eta}({\mathbb{Z}}))=q^{-1}({\mathbb{Z}})$. Since
$q^{-1}(\bar{\eta}({\mathbb{Z}}))=\eta(q^{-1}({\mathbb{Z}}))$, we have
$\eta(q^{-1}({\mathbb{Z}}))=q^{-1}({\mathbb{Z}})$. That is
$\pi_{1}(M_{\phi})=q^{-1}({\mathbb{Z}})$ is characteristic. ∎

######  Lemma 2.9.

${\mathcal{M}}$ is achiral if and only if each element in ${\mathcal{M}}$ is
virtually achiral.

###### Proof.

Suppose ${\mathcal{M}}$ is achiral, we may assume that $M$ is achiral, that is
there is an orientation reversing homeomorphism $\tau$ on $M$. For any
$M^{\prime}\in{\mathcal{M}}$, let $\tilde{M}$ be a common finite cover of $M$
and $M^{\prime}$. Suppose the covering degree $\tilde{M}\to M$ is $d$, then
the index $[\pi_{1}(M):\pi_{1}(\tilde{M})]=d$. Since $G=\pi_{1}(M)$ is
finitely generated, by group theory $G$ has only finitely many subgroup of
indes $d$. Let

$H=\bigcap_{\\{G_{i},[G:G_{i}]=d\\}}G_{i}.$

Then $H$ is a finite index charecteristic subgroup of $G$. Let $M_{H}\to M$ be
the covering corresponding to $H$. Since $\tau_{*}(H)=H$, $\tau$ lifts to a
map $\tau_{H}$ on $M_{H}$, which must be an orientation reversing
homeomorphism, that is $M_{H}$ is achiral. Clearly $M_{H}$ covers $\tilde{M}$,
so covers $M_{2}$. That is $M_{2}$ is virtually achiral. We prove one
direction.

Another direction is from the definition. ∎

##  3\. Achirality, Gauss’ Theory of Class Groups and Genus

A closed oriented manifold is called achiral if it admits an orientation
reversing homeomorphism.

######  Lemma 3.1.

Let $M=M_{\phi}$ be an oriented Sol torus bundle. Then $M_{\phi}$ is achiral
if and only if there exists $A\in{\mathrm{GL}}_{2}({\mathbb{Z}})$ such that
one of the following holds.

  1. (i)

$\phi A=A\phi$, $\det A=-1$.

  2. (ii)

$\phi A=A\phi^{-1}$, $\det A=1$.

###### Proof.

Let $f$ be a self homeomorphism on $M_{\phi}$. By Lemma 2.5, we assume that
$f$ is fiber preserving, so $f$ is a fiber preserving horizontal covering, and
(2.3) in the proof of Lemma 2.6 become

$A\circ\phi=\phi^{\epsilon}\circ A,$

where $A\in{\mathrm{GL}}_{2}({\mathbb{Z}})$ since the restriction of $f$ on
the fiber $T$ is a homeomorphism.

It is clear that $\epsilon\cdot\det A=-1$ if and only if $f$ is orientation
reversing. ∎

We now recall Gauss’ of class group theory [Gauss], [Fl, Chapter 5], [EW,
Chapters 3, 4].

Recall ${\mathrm{SL}}_{2}({\mathbb{Z}})$ acts on the complex plane by

$\phi z=\frac{az+b}{cz+d},\qquad\forall\phi=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\in{\mathrm{SL}}_{2}({\mathbb{Z}}),\ z\in{\mathbb{C}},$

which induces an action of ${\mathrm{SL}}_{2}({\mathbb{Z}})$ on
${\mathbb{R}}$. There are two conjugate real quadratic numbers
$z_{\phi},\bar{z}_{\phi}$ fixed by $\phi$ when $|{\mathrm{tr}}\phi|>2$.

Let $M=M_{\phi},\phi=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\in{\mathrm{SL}}_{2}({\mathbb{Z}}),$ $|{\mathrm{tr}}\phi|>2$.
Let

$u={\mathrm{sign}}{\mathrm{tr}}(\phi)\gcd(b,c,(d-a)),\,\text{and}\,\,\alpha=\frac{c}{u},\,\beta=\frac{d-a}{u},\,\gamma=\frac{-b}{u}.$

We define the discriminant of $\phi$ and of $M$ to be

$D_{M}=D_{\phi}:=\frac{(d+a)^{2}-4}{u^{2}}=\beta^{2}-4\alpha\gamma>0,\qquad(3.1)$

It is easy to see that $D=D_{\phi}\equiv 0,1\ \mathrm{mod}\ 4$ is a non-square
positive integer.

######  Lemma 3.2.

Let $D\equiv 0,1\ \mathrm{mod}\ 4$ be a non-square positive integer. Then
there exist $M_{\phi}$ such that $D_{\phi}=D$.

###### Proof.

It is known that the Pell equation $x^{2}-dy^{2}=1$, for each integer $d>0$,
has (infinitely many) non-trivial integer solutions. Consider the following
Pell equations:

$x^{2}-\frac{D}{4}y^{2}=1\quad(4|D),\qquad x^{2}-Dy^{2}=1\quad(4\nmid D).$

Let

$\phi=\begin{pmatrix}w&\frac{D}{4}v\\\
v&w\end{pmatrix}\quad(4|D),\qquad\begin{pmatrix}w-v&\frac{D-1}{2}v\\\
2v&w+v\end{pmatrix}\quad(4\nmid D),$

where $(w,v)$ is a non-trivial integer solution of the corresponding Pell
equation. Clearly $\phi\in{\mathrm{SL}}_{2}({\mathbb{Z}})$. One can also chech
by the definition that $D_{\phi}=D$. ∎

Let $D\equiv 0,1\ \mathrm{mod}\ 4$ be a non-square integer. Define primitiive
binary quadratic form $Q_{\phi}$ with discrimiment $D$ given by

$Q_{\phi}(x,y):=\frac{cx^{2}+(d-a)xy-by^{2}}{u}=\alpha x^{2}+\beta xy+\gamma
y^{2}.\qquad(3.2)$

Note $\alpha x^{2}+\beta x+\gamma$ is a minimal polynomial of the fixed point
of $\phi$, which is unique up to a sign, and is determined by the fixed point
of $\phi$.

######  Lemma 3.3.

For each $\phi\in SL_{2}({\mathbb{Z}})$, (i) $Q_{-\phi}=Q_{\phi}$, (ii)
$Q_{\phi^{-1}}=-Q_{\phi}$, and (iii) $Q_{\phi^{n}}=Q_{\phi}$ for $n>0$, which
is equivalent to $Q_{\pm\phi^{n}}=\text{sign}(n)Q_{\phi}$ for $n>0$.

###### Proof.

It follows from a direct calculation. A quick way to prove
$Q_{\phi^{n}}=Q_{\phi}$ for $n>0$ is apply the sentence before the lemma and
the fact $\phi$ and $\phi^{n},n\neq 0$, has the same fixed points. ∎

For each $A\in{\mathrm{GL}}_{2}({\mathbb{Z}})$ one can verify that

$Q_{A\phi A^{-1}}(x,y)=Q_{\phi}((x,y)(A^{t})^{-1}).\qquad(3.3)$

Call two primitive binary quadratic forms $Q_{1}$ and $Q_{2}$ with
discriminant $D$ are $SL_{2}({\mathbb{Z}})$-equivalent, if there exists $A\in
SL_{2}({\mathbb{Z}})$, such that

$Q_{1}(x,y)=Q_{2}((x,y)A).$

Those $SL_{2}({\mathbb{Z}})$-equivalent classes of primitive binary quadratic
forms of discriminant $D$ is the class group $C(D)$. For a quadratic form $Q$,
denote by $[Q]$ its class in $C(D)$. Let $Q_{0,D}$ denote the principal form
of discriminant $D$, i.e.

$x^{2}-\frac{D}{4}y^{2},\ (\ \text{if}\ 4|D\ );\qquad
x^{2}+xy+\frac{1-D}{4}y^{2}\ (\ \text{if}\ 4\nmid D\ ).$

Then $[Q_{0,D}]$ is the zero element in $C(D)$. For any form
$Q=(\alpha,\beta,\gamma):=\alpha x^{2}+\beta xy+\gamma y^{2}$, let
$-Q=(-\alpha,-\beta,-\gamma)$, for $[Q]\in C(D)$, let $-[Q]$ denote its
negative in $C(D)$.

######  Lemma 3.4.

For forms of discirminant $D$,

(i) $[Q]=[Q_{0,D}]$ is equivalent to $Q$ represents $1$, $[Q]=[-Q_{0,D}]$ is
equivalent to $Q$ represents $-1$;

(ii) $[-Q_{0,D}]=[Q_{0,D}]$ is equivalent to $x^{2}-Dy^{2}=-4$ has an integral
solution;

(iii) $[-Q]=-[Q]+[-Q_{0,D}]$;

(iv) $[-Q]=\pm[Q]$ is equivalent to either $[Q_{0,D}]=[-Q_{0,D}]$ or
$2[Q]=[-Q_{0,D}]$;

(v) $[Q_{1}]=-[Q_{2}]$ if and only if $Q_{1}(x,y)=Q_{2}((x,y)B)$, where
$B\in{\mathrm{GL}}_{2}({\mathbb{Z}})$ and $\det B=-1$.

Lemma 3.4 will be used repeatedly below. Its proof is given in the appendix.

######  Theorem 3.5.

Let $M=M_{\phi}$ be an oriented Sol torus bundle with $D=D_{\phi}$ defined in
(3.1) and $Q_{\phi}$ defined in (3.2). The following are equivalent.

  1. (1)

the Sol 3-manifold $M$ is achiral.

  2. (2)

either $[Q_{0,D}]=[-Q_{0,D}]$ or $2[Q_{\phi}]=[-Q_{0,D}]$ (or quivalently
$[-Q_{\phi}]=\pm[Q_{\phi}]$).

We first to prove the following

######  Lemma 3.6.

If $[Q_{\phi_{1}}]=[Q_{\phi_{2}}]$ for two anosov maps $\phi_{1},\phi_{2}\in
SL_{2}({\mathbb{Z}})$, then there exists $\alpha\in SL_{2}({\mathbb{Z}})$ and
integers $m,n>0$ such that $\phi_{1}$ is $SL_{2}({\mathbb{Z}})$-conjugates
$\pm\alpha^{m}$ and $\phi_{2}$ is $SL_{2}({\mathbb{Z}})$-conjugates
$\pm\alpha^{n}$.

###### Proof.

We first claim: If $Q_{\phi_{1}}=Q_{\phi_{2}}$ for two anosov maps
$\phi_{1},\phi_{2}\in SL_{2}({\mathbb{Z}})$, then there exists $\alpha\in
SL_{2}({\mathbb{Z}})$ and integers $m,n>0$ such that
$\phi^{t}_{1}=\pm\alpha^{m},\,\phi^{t}_{2}=\pm\alpha^{n}$.

For each primitive binary quadratic form $Q$, define

$Aut^{+}(Q)=\\{\beta\in SL_{2}({\mathbb{Z}})|Q(x,y)=Q((x,y)\beta)\\}$

Suppose first $Q_{\phi_{1}}=Q_{\phi_{2}}=Q$, then it is a direct verification
that $\phi_{1}^{t},\phi_{2}^{t}\in Aut^{+}(Q)$. By [Fl, Corollary 5.9], which
claims that $Aut^{+}(Q)={\mathbb{Z}}\times\\{\pm 1\\}=\\{\pm\alpha^{n}\\},$ ,
we have

$\phi^{t}_{1}=\pm\alpha^{m},\,\phi^{t}_{2}=\pm\alpha^{n}$

for integers $m,n\neq 0$. Now $m$ and $n$ must have the same sign, otherwise
may assume that $m>0$ and $n<0$, we have

$Q=Q_{\phi_{1}}=Q_{\pm(\alpha^{t})^{m}}=Q_{(\alpha^{t})^{m}}=Q_{\alpha^{t}}=-Q_{(\alpha^{t})^{-1}}=-Q_{(\alpha^{t})^{(-1)(-n)}}=-Q_{(\alpha^{t})^{n}}=-Q_{\phi_{2}}=-Q,$

where the 3rd, 4th, 5th, and 6th equalities follow from Lemma 3.3 (i), (iii)
(ii) and (iii) respectively. and we reach a contradiction. We may assume that
$m,n>0$, otherwise replace $\phi$ by $\phi^{-1}$. So the claim is verified.

Now suppose $[Q_{\phi_{1}}]=[Q_{\phi_{2}}]$. From the definition, we have
$Q_{\phi_{1}}(x,y)=Q_{\phi_{2}}((x,y)\beta^{t})$ for some $\beta\in
SL_{2}({\mathbb{Z}})$. By (3.3) we have
$Q_{\phi_{1}}((x,y)\beta^{t})=Q_{\beta^{-1}\phi_{2}\beta}(x.y)$. So we have
$Q_{\phi_{1}}=Q_{\beta^{-1}\phi_{2}\beta}$. Now apply the claim to
$Q_{\phi_{1}}=Q_{\beta^{-1}\phi_{2}\beta}$, we proved Lemma 3.6. ∎

###### Proof of Theorem 3.5.

Assume that $M_{\phi}$ is achiral. With notations in Lemma 3.1, let

$A=\begin{pmatrix}x&z\\\ y&w\end{pmatrix},\phi=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}.$

The conditions (i) and (ii) is equivalent to that

$A=\begin{cases}\begin{pmatrix}x&\frac{by}{c}\\\
y&\frac{(d-a)y+cx}{c}\end{pmatrix},\quad\det A=-1,\qquad&\text{for (i)}\\\
\begin{pmatrix}x&\frac{(d-a)x-by}{c}\\\ y&-x\end{pmatrix},\quad\det
A=1\qquad&\text{for (ii)}.\end{cases}$

It is easy to see that the condition (i) is equivalent to that $Q_{0,D}$
represents $-1$, therefore equivalent to $[Q_{0,D}]=[-Q_{0,D}]$ by Lemma 3.4
(i).

Below we prove that Lemma 3.1 (ii) is equivalent to that
$[Q_{\phi}]=[-Q_{\phi}]$, that is, $[Q_{\phi}]=[-Q_{\phi}]$ if and only if
$\phi=\beta\phi^{-1}\beta^{-1}$ for some $\beta\in SL_{2}({\mathbb{Z}})$.

The ”if” part follows from the definition and formula (3.3).

Now suppose $[Q_{\phi}]=[-Q_{\phi}]$. Then $[Q_{\phi}]=[Q_{\phi^{-1}}]$ by
Lemma 3.3. Then by Lemma 3.6, there exist $\alpha\in SL_{2}({\mathbb{Z}})$ and
integers $m,n>0$ such that $\phi$ is $SL_{2}({\mathbb{Z}})$-conjugates
$\pm\alpha^{m}$ and $\phi^{-1}$ is $SL_{2}({\mathbb{Z}})$-conjugates
$\pm\alpha^{n}$. Since $\phi$ and $\phi^{-1}$ have the same eigenvaluse, it
concludes that $m=n$ and the sign before $\alpha$ are the same. So
$\phi=\beta\phi^{-1}\beta^{-1}$ for some $\beta\in SL_{2}({\mathbb{Z}})$. ∎

We now recall Gauss’ genus theory (see [Gauss], [Fl, Chapter 5] and [BS]). Let
$D\equiv 0,1\ \mathrm{mod}\ 4$ be a non-square positive integer, the Kronecker
symbol $\chi_{D}:({\mathbb{Z}}/D{\mathbb{Z}})^{\times}\rightarrow\\{\pm 1\\}$
is a homomorphism such that $\chi_{D}(p\ \mathrm{mod}\
D)=\left(\frac{D}{p}\right)$ for all primes $p\nmid 2D$, where
$\left(\frac{D}{p}\right)$, the Legendre symbol, is 1 if and only if $D$ is
quadratic residue modulo $p$. Let $H_{D}\subset{\mathrm{Ker}}\chi_{D}$ be the
subgroup consisting of values represented by the principal form of
discriminant $D$. Then there is an exact sequence

$0\rightarrow 2C(D)\longrightarrow
C(D)\stackrel{{\scriptstyle\omega}}{{\longrightarrow}}{\mathrm{Ker}}\chi_{D}/H_{D}\longrightarrow
0,\qquad(3.6)$

where $\omega$ sends a class to the coset of $H_{D}$ in
${\mathrm{Ker}}\chi_{D}$ it represents.

The group $H_{D}$ consists of elements
$[a]\in({\mathbb{Z}}/D{\mathbb{Z}})^{\times}$ such that

$\left(\frac{a}{p}\right)=1\,\,\text{for all odd prime divisors $p$ of
$D$}\qquad(3.7)$

and such that

$a\equiv\begin{cases}1,7\ \mathrm{mod}\ 8,\qquad&\text{if $D\equiv 8\
\mathrm{mod}\ 32$},\\\ 1\ \mathrm{mod}\ 4,\qquad&\text{if $D\equiv 12,16,28\
\mathrm{mod}\ 32$},\,\,\,\qquad(3.8)\\\ 1,3\ \mathrm{mod}\ 8,&\text{if
$D\equiv 24\ \mathrm{mod}\ 32$},\\\ 1\ \mathrm{mod}\ 8,&\text{if
$32|D$}.\end{cases}$

######  Theorem 3.7.

Let $D\equiv 0,1\ \mathrm{mod}\ 4$ be a positive non-square integer. Then the
following are equivalent:

  1. (1)

there exists an achiral oriented Sol torus bundle $M$ of discriminant $D$,

  2. (2)

$16\nmid D$ and no prime factor of $D$ is $\equiv 3\ \mathrm{mod}\ 4$.

###### Proof.

We first verity that (1) is equivalent to the following

(3) there exists a $[Q]\in C(D)$ such that $2[Q]=[-Q_{0,D}]$ (or equivalently
$2[Q]$ represents $-1$).

The equivalence of (1) and (3) follows from Theorem 3.5: If there exists a
$[Q]\in C(D)$ such that $2[Q]=[-Q_{0,D}]$, then the Sol 3-manifold $M_{\phi}$
with $Q_{\phi}=Q$ is achiral by Theorem 3.5. On the other hand if $M_{\phi}$
with $D_{\phi}=D$ is achiral, then either $[Q_{0,D}]=[-Q_{0,D}]$ or
$2[Q_{\phi}]=[-Q_{0,D}]$ by Theorem 3.5. Note if $[Q_{0,D}]=[-Q_{0,D}]$,
choose $\phi_{0}$ such that $Q_{\phi_{0}}=Q_{0,D}$. Then
$2[Q_{\phi_{0}}]=2[Q_{0,D}]=[Q_{0,D}]=[-Q_{0,D}]$.

Below we will prove that (2) and (3) are equivalent in two steps by using
Gauss’ genus theory.

Step 1: To verify that (3) $2[Q]=[-Q_{0,D}]$ ($2[Q]$ represents $-1$) is
equivalent to $[-1]\in H_{D}$ by using the short exact sequence (3.6):

Suppose $-1$ is represented by $2[Q]$. Since $2[Q]\in 2C(D)$, we have
$\omega(2[Q])=\bar{1}\in{\mathrm{Ker}}\chi_{D}/H_{D}$, which implies $[-1]\in
H_{D}$.

Suppose $[-1]\in H_{D}$. We only need to prove
$\omega([-Q_{0,D}])==\bar{1}\in{\mathrm{Ker}}\chi_{D}/H_{D}$, therefore
$[-Q_{0,D}]=2[Q]$ for some $[Q]\in C(D)$. If not, we have
$\omega([-Q_{0,D}])=gH_{D}\in{\mathrm{Ker}}\chi_{D}/H_{D}$, where $g\notin
H_{D}$. Since $[-1]\in H_{D}$, $[-1]\notin gH_{D}$, which implies $-Q_{0,D}$
does not represent any $a\equiv-1\ \mathrm{mod}\ D$, contradicting that
$-Q_{0,D}$ represents $-1$.

Step 2: To verify $[-1]\in H$ and (2) are equivalent by using (3.7) and (3.8).

For any odd prime $p|D$, by Euler’s Theorem, $(\frac{-1}{p})=1$ if and only if
$p$ is not $\equiv 3\ \mathrm{mod}\ 4$.

Suppose $[-1]\in H_{D}$. By Euler’s Theorem and (3.6) (a characterization of
$H_{D}$), we have that no prime factor of $D$ is $\equiv 3\ \mathrm{mod}\ 4$.
Since $-1$ is not $\equiv 1\ \mathrm{mod}\ 8$, by the fourth line of (3.7)
(another characterization of $H_{D}$), we have $D$ is not a multiple of 32.
Since $-1$ is not $\equiv 1\ \mathrm{mod}\ 4$, by the second line of (3.7), we
have $D$ is not congruent to 16 mod 32. Hence $D$ is not a multiple of 16.

Now suppose $16\nmid D$ and no prime factor of $D$ is $\equiv 3\ \mathrm{mod}\
4$. Then no prime factor of $D$ is $\equiv 3\ \mathrm{mod}\ 4$ implies that
for any odd prime $p|D$, $\left(\frac{-1}{p}\right)=1$ [Fl, Page 71]. Moreover
suppose

$D=2^{\alpha}\prod_{i=1}^{k}(4n_{i}+1)^{\beta_{i}}$

is the prime decomposition of $D$. Then $D=2^{\alpha}(4n+1)$ for some
$n\in{\mathbb{Z}}$. Since $16\nmid D$, we have $\alpha\leq 3$. We have $4\nmid
D$ if $\alpha\leq 1$, and $D\equiv 4\ \mathrm{mod}\ 16$ if $\alpha=2$, and
$D\equiv 8\ \mathrm{mod}\ 32$ if $\alpha=3$. Overall we have that $D$ is not
$0,16,12,24,28,\ \mathrm{mod}\ 32$. Then by $-1$ appears in (3.8) for other
$D$. And by the charaterization of $H_{D}$, $[-1]\in H_{D}$. ∎

######  Theorem 3.8.

A commensurable class ${\mathcal{M}}$ of Sol 3-manifolds is achiral if and
only if $D_{{\mathcal{M}}}$ contains no prime $\equiv 3\ \mathrm{mod}\ 4$.

###### Proof.

Suppose ${\mathcal{M}}$ is achiral, then there is achiral Sol 3-manifold
$M\in{\mathcal{M}}$. If $M$ is a semi-bundle. Let $\tilde{M}$ be the torus
bundle double cover of $M$. Then $\tilde{M}\in\mathcal{M}$. Since $M$ is
achiral and $\tilde{M}\rightarrow M$ is a characteristic cover (Lemma 2.8),
$\tilde{M}$ is also achiral. So way assume that $M$ is a torus bundle. Then
$D_{M}$ contains no prime $\equiv 3\ \mathrm{mod}\ 4$ by Theorem 3.7. Since
$D_{M}=c^{2}D_{{\mathcal{M}}}$ $(c\in{\mathbb{Z}})$, $D_{{\mathcal{M}}}$
contains no prime $\equiv 3\ \mathrm{mod}\ 4$.

From the definition of $D_{{\mathcal{M}}}$ as fundamental discriminant we know
that $16\nmid D_{{\mathcal{M}}}$. If $D_{{\mathcal{M}}}$ contains no prime
$\equiv 3\ \mathrm{mod}\ 4$, then there is an achiral Sol 3-manifold
$M\in{\mathcal{M}}$ with $D_{M}=D_{{\mathcal{M}}}$ by Theorem 3.7. So
${\mathcal{M}}$ is achiral. ∎

######  Corollary 3.9.

Each commensurable class of Sol 3-manifolds contains non-achiral manifolds.

###### Proof.

For each commensurable class ${\mathcal{M}}$ of Sol 3-manifolds and each
integer $c\neq 0$, $c^{2}D_{{\mathcal{M}}}$ is a non-square positive integer
$\equiv 0,1\ \mathrm{mod}\ 4$. Then there is a $M\in{\mathcal{M}}$ such that
$D_{M}=c^{2}D_{{\mathcal{M}}}$ by Lemma 3.2. If we choose $c$ to be either 3
or 4, then $M$ is non-achiral by Theorem 3.7. ∎

##  4\. Density Results

######  Proposition 4.1.

Let $D\equiv 0,1\ \mathrm{mod}\ 4$ be a positive non-sqaure integer. Then the
following are equivalent:

  1. (1)

The negative Pell equation $x^{2}-Dy^{2}=-4$ has solutions;

  2. (2)

There is a non-oriented Sol 3-manifold with characteristic double cover of
discriminant $D$;

  3. (3)

There is an achiral semi-tours bundle with characteristic double cover of
discriminant ${\mathrm{lcm}}(D,4)$.

###### Proof.

Assume (1), or equivalently, by Lemma 3.4 (i) and (ii), the solvability of

$u^{2}-\frac{D}{4}v^{2}=-1\quad(4|D),\qquad u^{2}-Dv^{2}=-1\quad(4\nmid
D).\qquad(4.1)$

Here for the later one, note that if $x^{2}-Dy^{2}=-4$ has solution then has
solution with $x$ even. Let

$\psi=\begin{pmatrix}u&\frac{D}{4}v\\\
v&u\end{pmatrix}\quad(4|D),\qquad\begin{pmatrix}u-v&\frac{D-1}{2}v\\\
2v&u+v\end{pmatrix}\quad(4\nmid D).\qquad(4.2)$

Then the determinant of $\psi$ is $-1$ by (4.1), so the torus bundle
$M_{\psi}$ is non-orientable. It is the direct calculation that $D_{\psi}=D$.
Then $M=M_{\psi^{2}}$ is the unique orientable double cover of $M_{\psi}$.
Since $\psi^{2}$ and $\psi$ have the same fixed points as actions on
$\mathbb{R}\cup\infty$, we have $D_{\psi^{2}}=D_{\psi}$ by definition. So
$D_{M}=D$.

On the other hand, assume (2), i.e. the double cover $M_{\psi^{2}}$ of
$M_{\psi}$, $\det\psi=-1$, has discriminant $D$, then by the equality

$\psi^{2}=(w\psi)^{-1}w\psi^{2}w(w\psi),\qquad(4.3)$

where $w=\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}$ with $\det w=-1$. Since $\det
w\psi=1$, by (4.3) we have

$[Q_{\psi^{2}}]=[Q_{w\psi^{2}w}].\qquad(4.4)$

It is a direct verification that

$Q_{w\phi w}((x,y)w)=-Q_{\phi}(x,y).\qquad(4.4+)$

So we have

$-Q_{\psi^{2}}(x,y)=Q_{w\psi^{2}w}(y,x)$

which implies that

$[-Q_{\psi^{2}}]=-[Q_{w\psi^{2}w}]\qquad(4.5)$

Combing (4.4), (4.5) and $[-Q]=-[Q]+[-Q_{0,D}]$ for any $[Q]\in C(D)$ (Lemma
3.4 (iii)), we have $[-Q_{0,D}]=0$, that is $[-Q_{0,D}]=[Q_{0,D}]$, which is
equivalent to the negative Pell equation $x^{2}-Dy^{2}=-4$ has integral
solution (Lemma 3.4 (ii)). This shows the equivalence between (1) and (2).

Note that the double cover $M_{\phi}$ of $N_{\psi}$ is given as follows:

$\phi=\begin{pmatrix}2bc+1&2bd\\\ 2ac&2bc+1\end{pmatrix},\qquad\text{if}\
\psi=\begin{pmatrix}a&b\\\ c&d\end{pmatrix},\ abcd\neq 0.$

It is clear that $M_{\phi}$ has discriminant
$D_{\phi}=4\frac{abcd}{\gcd(ac,bd)^{2}}$. Note that

  * •

$2[Q_{\phi}]\in C(D_{\phi})$ is the zero class by Theorem 3.7, and thus
$M_{\phi}$ is achiral if and only if the negative equation
$x^{2}-D_{\phi}y^{2}=-4$ has solution.

  * •

$N_{\psi}$ is achiral if and only $-1$ is a self-mapping degree of $N_{\psi}$
by Lemma 2.7, and if and only if $a+d=0$ by [SWW, Theorem 1.7].

  * •

if $a+d=0$, then $D_{\phi}=-\frac{4bc}{\gcd(b,c)^{2}}$, and therefore
$-4=4a^{2}+4bc=4a^{2}-D_{\phi}\gcd(b,c)^{2}$.

  * •

if if $N_{\psi}$ is achiral, then $M_{\phi}$ is achiral, since the cover
$M_{\phi}\to N_{\psi}$ is characteristic by the ”Moreover” part of Proposition
2.1

It follows that if $N_{\psi}$ is achiral with $D_{\phi}={\mathrm{lcm}}(D,4)$,
then $x^{2}-Dy^{2}=-4$ has solutions.

On the other hand, if the negative Pell equation $x^{2}-Dy^{2}=-4$ has
solution, then there is always solution with $x$ even, say $(2a,b)$. Take

$\psi=\begin{cases}\begin{pmatrix}a&b\\\ -Db/4&-a\end{pmatrix},\qquad&\text{if
$4|D$},\\\ \begin{pmatrix}a&b/2\\\ -Db/2&-a\end{pmatrix},&\text{if $D\equiv 1\
\mathrm{mod}\ 4$}.\end{cases}$

Then $M_{\phi}$ is achiral of discriminant $D$ or $4D$. This shows the
equivalence between (1) and (3). ∎

######  Corollary 4.2.

Let ${\mathcal{M}}$ be a commensurable class of Sol 3 manifolds of
discriminant $D$. Then the following conditions are equivalent.

  1. (1)

the negative Pell equation $x^{2}-Dy^{2}=-4$ has solutions;

  2. (2)

${\mathcal{M}}$ contains a non-orientable Sol 3-manifolds;

  3. (3)

${\mathcal{M}}$ contains an achiral tour semi-bundle.

The equation $x^{2}-Dy^{2}=-4$ is called the negative Pell equation. The
solvability of negative Pell equation has a long history and P. Stevenhagen
[Ste] made the following conjecture.

######  Conjecture 4.3 (Stevenhagen).

The density of fundamental positive discriminants $D$, for which
$x^{2}-Dy^{2}=-4$ have solutions, among all fundamental positive discriminants
$D$ without prime factors $\equiv 3\ \mathrm{mod}\ 4$ is $1-\rho$ with

$\rho:=\prod_{j=1}^{\infty}\left(1+2^{-j}\right)^{-1}=0.41942\cdots.$

This conjecture is recently proved by P. Koymans and C. Pagano [KP].

######  Lemma 4.4.

$\lim_{X\rightarrow\infty}\frac{\\#\\{D_{{\mathcal{M}}}<X|D_{{\mathcal{M}}}\,\text{contains
no prime $\equiv 3\ \mathrm{mod}\ 4$}\\}}{X}=0,\qquad(4.6)$
$\lim_{X\rightarrow\infty}\frac{\\#\\{D_{{\mathcal{M}}}<X\\}}{X}=C>0,\qquad(4.7)$

###### Proof.

Note that a integer $n>0$ is the sum of two integral squares if and only if
the square-free part of $n$ contains no prime $\equiv 3\ \mathrm{mod}\ 4$ [Fl,
Chap. 2]. So we have

$\\{D_{{\mathcal{M}}}|D_{{\mathcal{M}}}\,\text{contains no prime $\equiv 3\
\mathrm{mod}\
4$}\\}=\\{D_{{\mathcal{M}}}|D_{{\mathcal{M}}}=a^{2}+b^{2}\\}.\qquad(4.8)$

It is a classical fact that

$\lim_{X\rightarrow\infty}\frac{\\#\\{0<D<X|D=a^{2}+b^{2}\\}}{X}=0,\qquad(4.9)$

Then (4.6) follows from (4.8) and (4.9).

Let $S$ be the set of fundamental discriminants. Then for each $D\in
S$,$D>0,\,D\equiv 0,1\ \mathrm{mod}\ 4,$ which is equivalent to

$D\equiv 1\ \mathrm{mod}\ 4\,\,{\it or}\,\,D\equiv 0\ \mathrm{mod}\ 8\,\,{\it
or}\,\,D\equiv 4\ \mathrm{mod}\ 8.$

Therefore, the set $S$ equals the disjoint union of the following three
subsets:

$S_{1}=\\{D\equiv 1\ \mathrm{mod}\ 4\mid D\in S\\},\,S_{2}=\\{D\equiv 0\
\mathrm{mod}\ 8\mid D\in S\\},\,S_{3}=\\{D\equiv 4\ \mathrm{mod}\ 8\mid D\in
S\\},$

Now we can cite [FK, p12] to prove (4.7): all densities of $S_{1}$, $S_{2}$
and $S_{3}$ exist and equal to $2/\pi^{2}$, $1/2\pi^{2}$, $1/2\pi^{2}$
respectively. So the density of $S$ exists and equals
$C=\frac{2}{\pi^{2}}+\frac{1}{2\pi^{2}}+\frac{1}{2\pi^{2}}=\frac{3}{\pi^{2}}>0.$
∎

###### Proof of Theorem 1.1.

(1) Let $D$ be prime in the form of $4k+1$. The $D=D_{\mathcal{M}}$ for some
commensurable class ${\mathcal{M}}$ by Proposition 2.4. Then ${\mathcal{M}}$
is achiral by the ”only if” part of Corollary 3.8. It is a well known fact
that there are infinitely many primes in this form of $4k+1$ [EW, Theorem
10.5].

Now we prove the ”however” part. (4,7) and (4.8) implies that

$\lim_{X\rightarrow\infty}\frac{\\#\\{D_{{\mathcal{M}}}<X|D_{{\mathcal{M}}}\,\text{contains
no prime $\equiv 3\ \mathrm{mod}\
4$}\\}}{\\#\\{D_{{\mathcal{M}}}<X\\}}=0.\qquad(4.9)$

By Theorem 2.4, we have

${\\#\\{{\mathcal{M}}\,|\,D_{{\mathcal{M}}}<X\\}}={\\#\\{D_{{\mathcal{M}}}<X\\}}\qquad(4.10)$

By Theorem 2.4 and Theorem 3.8, we have

${\\#\\{\text{achiral}\ {\mathcal{M}}\
|D_{{\mathcal{M}}}<X\\}}={\\#\\{D_{{\mathcal{M}}}<X|{\mathcal{M}}\
\text{achiral}\ \\}}$
$=\\#\\{D_{{\mathcal{M}}}<X|D_{{\mathcal{M}}}\,\text{contains no prime $\equiv
3\ \mathrm{mod}\ 4$}\\}\qquad(4.11)$

Then (4.10), (4.11) and (4.9) imply

$\lim_{X\rightarrow\infty}\frac{\\#\\{\text{achiral}\ {\mathcal{M}}\ \
|D_{{\mathcal{M}}}<X\\}}{\\#\\{{\mathcal{M}}\ |\ D_{{\mathcal{M}}}<X\
{}\\}}=0,$

(2) By Theorem 2.4 and Corollary 4.2, we have

$\\#\\{{\mathcal{M}}\,\text{contains an non-orientable element
}\,|\,D_{\mathcal{M}}<X\ \\}$
$=\\#\\{D_{\mathcal{M}}<X\,|\,{\mathcal{M}}\,\text{contains an non-orientable
element}\\}$
$=\\#\\{D_{\mathcal{M}}<X\,|\,\text{$x^{2}-D_{\mathcal{M}}y^{2}=-4$ has
solutions}\\}.\qquad(4.12)$

(4.11) and (4.12) and the positive answer of Stevenhagen Conjecture [KP] imply

$\lim_{X\rightarrow\infty}\frac{\\#\\{{\mathcal{M}}\ {}\text{contains an non-
orientable element \,}|\,D_{\mathcal{M}}<X\ \\}}{\\#\\{\text{achiral}\
{\mathcal{M}}\ \ |D_{{\mathcal{M}}}<X\\}}=1-\rho.$

∎

######  Proposition 4.5.

There exists $M\in\mathrm{Sol}_{D}$ double cover a tours semi-bundle if and
only if $4|D$. Moreover, $M_{\phi}\in\mathrm{Sol}_{D}$ is the double cover of
a tours semi-bundle if and only if $2[Q_{\phi}]=0$ and $\phi\equiv I\
\mathrm{mod}\ 2{\mathbb{Z}}$, in this case, there is a unique tours semi-
bundle double covered by $M_{\phi}$.

###### Proof.

By [Ma], $M_{\phi}\in\mathrm{Sol}_{D}$ is the double of a tours semi-bundle if
and only if there exists $A\in{\mathrm{GL}}_{2}({\mathbb{Z}})$ such that

$A\phi=\phi^{-1}A,\qquad\det A=-1,\qquad A\equiv\phi\equiv I\ \mathrm{mod}\
2{\mathbb{Z}}.$

Let $A=\begin{pmatrix}x&y\\\ z&w\end{pmatrix}$ and assume that
$Q_{\phi}=\alpha x^{2}+\beta xy+\gamma y^{2}$. Then the condition
$A\phi=\phi^{-1}A$ and $\det A=-1$ is equivalent to

$A=\begin{pmatrix}x&y\\\ z&-x\end{pmatrix},\quad x^{2}+yz=1,\quad\alpha
z=\beta x+\gamma y.$

One can check that

$\begin{pmatrix}-x&y\\\ -z&-x\end{pmatrix}\begin{pmatrix}\alpha&-\beta/2\\\
-\beta/2&\gamma\end{pmatrix}\begin{pmatrix}-x&-z\\\
y&-x\end{pmatrix}=\begin{pmatrix}\alpha&\beta/2\\\
\beta/2&\gamma\end{pmatrix}.$

It follows that the condition that there exists
$A\in{\mathrm{GL}}_{2}({\mathbb{Z}})$ such that $A\phi=\phi^{-1}A$ and $\det
A=-1$ if and only if $2[Q_{\phi}]=0$.

Note that $\phi\equiv I\ \mathrm{mod}\ 2$ implies $4|D$. Assume that $4|D$
take $M_{\phi}\in\mathrm{Sol}_{D}$ such that $2[Q_{\phi}]=0$, we have that
$\phi$ is ${\mathrm{SL}}_{2}({\mathbb{Z}})$-conjugate to
$\phi_{0}=\begin{pmatrix}a&b\\\ c&a\end{pmatrix}$. It is easy to see that
$A_{0}:=\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}\equiv I\ \mathrm{mod}\ 2$
satisfies $A_{0}\phi_{0}=\phi_{0}^{-1}A_{0}$, $\det A_{0}=-1$, and that
$\phi_{0}^{2}\equiv I\ \mathrm{mod}\ 2$. Thus
$M_{\phi^{2}}\in\mathrm{Sol}_{D}$ is the double cover of a tours semi-bundle.

Let $A,A^{\prime}$ be two such matrices, then $A^{\prime}=AC$ with
$C\in{\mathrm{SL}}_{2}({\mathbb{Z}}),\equiv I\ \mathrm{mod}\ 2$ and
$C\phi=\phi C$. One can easily check that $A^{\prime}$ is equivalent to $A$ in
the sense of [Ma]. Thus if $M_{\phi}$ is the double cover of at most one tours
semi-bundle. ∎

##  5\. Shimizu L-functions

In the 1970s, Hirzebruch [Hr] formulated a conjecture on the signature of the
Hilbert modular varieties whose cusp cross-sections are solvmanifolds, which
relates the signature defects of these Sol manifolds to special values of
Shimizu’s L-functions of totally real fields. We now recall the definition of
Shimizu L-functions for Hilbert modular surfaces. In this case, these
solvmanifolds are oriented tours bundle $M_{\phi}$.

Given $\phi=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\in{\mathrm{SL}}_{2}({\mathbb{Z}})$ with discrimimant
$D_{\phi}=D$ and quadratic form $Q_{\phi}$:

the Shimizu L-function is defined as

$L(M,s):=\sum_{(0,0)\neq(x,y)\in{\mathbb{Z}}^{2}/\phi^{\mathbb{Z}}}\frac{{\mathrm{sign}}\
Q_{\phi}(x,y)}{|Q_{\phi}(x,y)|^{s}},\qquad{\mathrm{Re}}(s)>1$

only depends on its oriented equivalence class. In fact $L(M,s)$ has an entire
continuation to ${\mathbb{C}}$ (See Proposition 5.6).

######  Theorem 5.1.

Let $M=M_{\phi}$ be an oriented Sol torus bundle.

(1) $M$ is achiral if and only if the Shimizu L-function $L(M,s)\equiv 0$.

(2) If $L(M_{\phi_{1}},s)=L(M_{\phi_{2}},s)$ for two non-achiral oriented
bundles $M_{\phi_{1}},M_{\phi_{2}}$ with same discriminant, then
$M_{\phi_{1}}$ is homeomorphic to either $M_{\phi_{2}}$ or $M_{-\phi_{2}}$ as
oriented torus bundle.

We list some facts for the proof of Theorem 5.1.

######  Proposition 5.2.

[Web] Each $[Q]\in C(D)$ represents infinitely many primes.

######  Lemma 5.3.

It both $Q_{1},Q_{2}\in C(D)$ represent a prime $p$ with $(p,2D)=1$, then
$[Q_{1}]=\pm[Q_{2}]$.

###### Proof.

By [EW, Exercise 3.30], each $Q_{i}$ is $SL_{2}({\mathbb{Z}})$-equivalent to a
form $Q_{i}^{\prime}(x,y)=px^{2}+d_{i}xy+e_{i}y^{2}$, $d_{i}\leq p$. Note
$d_{i}-4pe_{i}=D$. Clearly $d_{1}$ and $d_{2}$ has the same parity. Therefore
$\frac{|d_{1}|-|d_{2}|}{2},\frac{|d_{1}|+|d_{2}|}{2}\in\mathbb{Z}$. By $p\nmid
D$, we have $p\nmid d_{i}$ and therefore $0<|d_{i}|<p$. Therefore
$0<\frac{|d_{1}|+|d_{2}|}{2}<p$. Since

$\frac{d_{1}^{2}-d_{2}^{2}}{4}=\frac{|d_{1}|-|d_{2}|}{2}\frac{|d_{1}|+|d_{2}|}{2}=p(e_{1}-e_{2}).$

Therefore $p|\frac{|d_{1}|-|d_{2}|}{2}.$ It follows
$|d_{1}|=|d_{2}|,e_{1}=e_{2}$. So $[Q_{1}^{\prime}]=\pm[Q_{2}^{\prime}]$ and
$[Q_{1}]=\pm[Q_{2}]$. ∎

######  Lemma 5.4.

Suppose $\phi,\psi\in{\mathrm{SL}}_{2}({\mathbb{Z}})$ with
$D_{\phi}=D_{\psi}=D$. If $[Q_{\phi}]=\pm[Q_{\psi}]$ and
${\mathrm{tr}}(\phi)={\mathrm{tr}}(\psi)$. Then $\phi$ is
${\mathrm{SL}}_{2}({\mathbb{Z}})$-conjugate to either $\psi$ or $w\psi^{-1}w$.

###### Proof.

(i) First note that $\phi$ is determined by $Q_{\phi}$ and
${\mathrm{tr}}(\phi)$: Suppose $Q_{\phi}(x,y)=\alpha x^{2}+\beta xy+\gamma
y^{2}$, one can solve that
$\phi={\mathrm{sign}}{\mathrm{tr}}(\phi)\begin{pmatrix}\frac{t+\beta
s}{2}&-\gamma s\\\ \alpha s&\frac{t-\beta s}{2}\end{pmatrix}$, where
$t=|{\mathrm{tr}}(\phi)|$ and $s=\sqrt{\frac{t^{2}-4}{D}}$.

(ii) If $[Q_{\phi}]=[Q_{\psi}]$, then $Q_{\phi}(x,y)=Q_{\psi}((x,y)A)$; If
$[Q_{\phi}]=-[Q_{\psi}]$, then $Q_{\phi}(x,y)=Q_{\psi}((x,y)A^{\prime}w)$;
both $A,A^{\prime}\in{\mathrm{SL}}_{2}({\mathbb{Z}})$. Then one can verifies
the Lemma by (i), (3.3) and Lemma 3.3. ∎

By Lemma 3.3 we have

######  Lemma 5.5.

$L(M_{-\phi},s)=L(M_{\phi},s)$, $L(M_{\phi^{-1}},s)=-L(M_{\phi},s)$ and
$L(M_{\phi},s)=-L(M_{w\phi w},s)$.

###### Proof of Theorem 5.1.

Suppose $M$ is achiral, then $M_{\phi}=-M_{\phi}$, and furthermore
$M_{\phi}=M_{\phi^{-1}}$ by Lemma 2.2. Then $\phi^{-1}$ is
$SL_{2}({\mathbb{Z}})$ conjugate to either $\phi$ or $w\phi^{-1}w$ by Lemma
2.3. That is either

$L(M_{\phi^{-1}},s)=L(M_{\phi},s)\ \ \text{or}\ \
L(M_{\phi^{-1}},s)=L(M_{w\phi^{-1}w},s)\qquad(5.1)$

Combining (5.1) and Lemma 5.5, we have $L(M,s)\equiv 0$.

On the other hand, let
$K_{n,\phi}=\\{(x,y)\in{\mathbb{Z}}\oplus{\mathbb{Z}}|Q(x,y)=n\\}/\phi$ for
each $n\in{\mathbb{Z}}$. $K_{n,\phi}$ is finite. In $L(M_{\phi},s)$, the term
$1/n^{s}$ occurs $|K_{n,\phi}|$ times since each orbit in $K_{n,\phi}$
contribute a single $1/n^{s}$. Similarly the term $-1/n^{s}$ occurs
$|K_{-n,\phi}|$ times. So we have

$L(M_{\phi},s)=\sum_{n=1}^{\infty}\frac{-|K_{-n,\phi}|+|K_{n,\phi}|}{|n|^{s}}.\qquad(5.2)$

Hence $L(M_{\phi},s)\equiv 0$ if and only if $|K_{-n,\phi}|=|K_{n,\phi}|$ for
each $n$, which is equivalent to $Q$ and $-Q$ represent the same set of
integers. Then both $Q$ and $-Q$ represent some prime $p$ with $(p,2D)=1$ by
Proposition 5.2, and then $[-Q_{\phi}]=\pm[Q_{\phi}]$ by Lemma 5.3. By Theorem
3.5, this is equivalent to that $M$ is achiral.

We have proved (1). Below we prove (2).

Suppose both ${\mathrm{tr}}(\phi_{1}),{\mathrm{tr}}(\phi_{2})>0$ (otherwise
replace $\phi$ by $-\phi$). Let $\psi_{1}=\phi_{1}^{n_{1}}$ and
$\psi_{2}=\phi_{2}^{n_{2}}$ such that
${\mathrm{tr}}{\psi_{1}}={\mathrm{tr}}{\psi_{2}}$, where $n_{i}$ a positive
integer, $i=1,2$. Then from the definition, it is easy to see

$L(M_{\psi_{i}},s)=n_{i}L(M_{\phi_{i}},s),i=1,2.\qquad(5.3)$

Following (5.2) and $L(M_{\phi_{1}},s)=L(M_{\phi_{2}},s)$ we have

$L(M_{\psi_{1}},s)=\frac{n_{2}}{n_{1}}L(M_{\psi_{2}},s)=\sum_{n=1}^{\infty}\frac{-|K_{-n}|+|K_{n}|}{|n|^{s}}\qquad(5.4)$

By Proposition 5.2, $Q_{\psi_{1}}$ represents a prime $p$ with $(p,2D)=1$.
Then $-p$ is not represented by $Q_{\psi_{1}}$ otherwise
$[Q_{\psi_{1}}]=\pm[-Q_{\psi_{1}}]$ by Lemma 5.3, and therefore $M_{\psi_{1}}$
is achiral by Theorem 3.5. So we have $K_{p}\neq 0$ and $K_{-p}=0$. By (5.4)
$Q_{\psi_{2}}$ also represents $p$. Then $[Q_{\psi_{2}}]=\pm[Q_{\psi_{1}}]$
again by Lemma 5.3. Since ${\mathrm{tr}}{\psi_{1}}={\mathrm{tr}}{\psi_{2}}$,
$\psi_{1}$ is $SL_{2}({\mathbb{Z}})$-conjugated to either $\psi_{2}$ or
$w\psi_{2}^{-1}w$ by Lemma 5.4. Therefore $L(\psi_{1},s)=L(\psi_{2},s)$ by
Lemma 5.5. That is to say $n_{1}=n_{2}$, and therefore $\phi_{1}$ is
$SL_{2}({\mathbb{Z}})$-conjugated to either $\phi_{2}$ or $w\phi_{2}^{-1}w$.
Since $M_{\phi_{2}}$ and $M_{w\phi_{2}^{-1}w}$ are homeomorphic as oriented
torus bundles, in either case, $M_{\phi_{1}}$ and $M_{\phi_{2}}$ are
homeomorphic as oriented torus bundles.

When requiring $tr(\phi_{i})>0$, we replace $\phi_{i}$ by $-\phi_{i}$, so in
general we have $M_{\phi_{1}}$ is homeomorphic to either $M_{\phi_{2}}$ or
$M_{-\phi_{2}}$ as oriented torus bundles. ∎

######  Proposition 5.6.

For an oriented tours bundle $M=M_{\phi}$ of discriminant $D$, $L(M,s)$ has an
entire continuation to ${\mathbb{C}}$ and satisfies a functional equation:

$\Lambda(M,s):=\Gamma_{\mathbb{R}}(s+1)^{2}D^{s/2}L(M,s)=\Lambda(M,1-s).$

###### Proof.

It follows (see [ADS]) that $L(M,s)$ ha s entire continuation and satisfies
the functional equation:

$\Lambda(M,s)=-\Lambda(M_{\phi^{t}},1-s).$

Since for any $\phi\in{\mathrm{SL}}_{2}({\mathbb{Z}})$, $\phi^{t}$ is
${\mathrm{SL}}_{2}({\mathbb{Z}})$ conjugate to $\phi^{-1}$:

$\phi^{t}=\begin{pmatrix}0&1\\\
-1&0\end{pmatrix}\phi^{-1}\begin{pmatrix}0&1\\\ -1&0\end{pmatrix}.$

By Lemma 5.5, we have $L(M_{\phi^{t}},s)=L(M_{\phi^{-1}},s)=-L(M,s)$. So
$\Lambda(M,s)=\Lambda(M,1-s)$. ∎

##  6\. Appendix: Proof of Lemma 3.3

Lemma 3.3 For forms of discirminant $D$,

(i) $[Q]=[Q_{0,D}]$ is equivalent to $Q$ represents $1$, $[Q]=[-Q_{0,D}]$ is
equivalent to $Q$ represents $-1$;

(ii) $[-Q_{0,D}]=[Q_{0,D}]$ is equivalent to $x^{2}-Dy^{2}=-4$ has an integral
solution;

(iii) $[-Q]=-[Q]+[-Q_{0,D}]$;

(iv) $[-Q]=\pm[Q]$ is equivalent to either $[Q_{0,D}]=[-Q_{0,D}]$ or
$2[Q]=[-Q_{0,D}]$;

(v) $[Q_{1}]=-[Q_{2}]$ if and only if $Q_{1}(x,y)=Q_{2}((x,y)B)$, where
$B\in{\mathrm{GL}}_{2}({\mathbb{Z}})$ and $\det B=-1$.

###### Proof.

Dirichlet gave an approach to the composition law of $C(D)$. Let
$[\alpha_{i},\beta_{i},\gamma_{i}]$, $i=1,2$, be two primitive forms of
discriminant $D$ satisfying
$\gcd(\alpha_{1},\alpha_{2},(\beta_{1}+\beta_{2})/2)=1$. Let $\beta$ be an
integer (unique mod $2\alpha_{1}\alpha_{2}$) such that

$\beta\equiv\beta_{1}\ \mathrm{mod}\ 2\alpha_{1},\quad\beta\equiv\beta_{2}\
\mathrm{mod}\ 2\alpha_{2},\quad\beta^{2}\equiv D\ \mathrm{mod}\
4\alpha_{1}\alpha_{2}.$

Then the composition

$[\alpha_{1},\beta_{1},\gamma_{1}]+[\alpha_{2},\beta_{2},\gamma_{2}]=[\alpha_{1}\alpha_{2},\beta,(\beta^{2}-D)/4\alpha_{1}\alpha_{2}].\qquad(2.1)$

In fact, any two classes have concordant forms, which means

$(i)\,\alpha_{1}\alpha_{2}\neq 0,\ \ (ii)\,\beta_{1}=\beta_{2},\ \
(iii)\,\alpha_{2}|\gamma_{1},\alpha_{1}|\gamma_{2}\qquad(2.2)$

.

In this case, $\beta=\beta_{i}$. Moreover, we may even require that
$\gcd(\alpha_{1},\alpha_{2})=1$ and $\gcd(\alpha_{1}\alpha_{2},u)=1$ for any
given $u\neq 0$ (See Chapter 5 Lemma 2.5 in [Fl]). It is easy to see that

$-[\alpha,\beta,\gamma]=[\gamma,\beta,\alpha]=[\alpha,-\beta,\gamma].\qquad(2.3)$

Recall $Q_{0,D}$ is the principal form $x^{2}-\frac{Dy^{2}}{4}$ if $4|D$ and
$x^{2}+xy+\frac{1-D}{4}y^{2}$ otherwise. In the proof of (i) and (ii), we just
verify the case $4|D$, and the verification for $D=4k+1$ is similar.

(i) It is clear that $[Q]=[Q_{0,D}]$ if and only if $Q$ representing $1$ is
equivalently that $[Q]=[-Q_{0,D}]$ if and only if $Q$ representing $-1$. We
just prove the former: Clearly $Q_{0,D}$ represents 1. On the other hand,
Suppose $Q(x_{0},y_{0})=1$ for some integers $x_{0},y_{0}$. Then
$(x_{0},y_{0})=1$, and $(x_{0},y_{0})=(1,0)A$ for some $A\in
SL_{2}{\mathbb{Z}}$ by Bezout’s Theorem. Then for $Q^{\prime}(x,y)=Q((x,y)A)$,
$Q^{\prime}(1,0)=1$, which implies that $Q^{\prime}(x,y)=x^{2}+bxy+cy^{2}$. If
$4|D$, $b$ is even, then
$Q^{\prime}(x,y)=(x-\frac{b}{2}y)^{2}-\frac{D}{4}y^{2}$.

(ii) Suppose $[-Q_{0,D}]=[Q_{0,D}]$. By (i), when $4|D$,
$Q_{0,D}(x_{0},y_{0})=x_{0}^{2}-\frac{D}{4}y_{0}^{2}=-1$ for some integers
$x_{0},y_{0}$, so $(2x_{0})^{2}-Dy_{0}^{2}=-4$; So $x^{2}-Dy^{2}=-4$ has an
integral solution.

Suppose $x^{2}-Dy^{2}=-4$ has an integral solution $(x_{0},y_{0})$. If $4|D$,
then $0\equiv-4=x_{0}^{2}-Dy_{0}^{2}\equiv x_{0}^{2}({\rm mod\,\,}4)$. So
$2|x_{0}$, and
$Q_{0,D}(\frac{x_{0}}{2},y_{0})=\frac{x_{0}^{2}}{4}-\frac{D}{4}y_{0}^{2}=\frac{x_{0}^{2}-Dy_{0}^{2}}{4}=-1$.
By (i), $[-Q_{0,D}]=[Q_{0,D}]$.

(iii) For any $[Q]=[\alpha,\beta,\gamma]$. Then

$[-\alpha,-\beta,-\gamma]+[\alpha,\beta,\gamma]=[-\alpha,-\beta,-\gamma]+[\gamma,-\beta,\alpha]=[-\alpha\gamma,-\beta,-1]=[-Q_{0,D}].\qquad(2.4)$

The secand, the third, and last equality are based on (2.3), (2.1) and (i).
repectively.

(iv) If $[-Q]=-[Q]$, then by (iii) we have $-[Q]=-[Q]+[-Q_{0,D}]$, that is
$[Q_{0,D}]=[-Q_{0,D}]$. If $[-Q]=[Q]$, then by (iii) we have
$[Q]=-[Q]+[-Q_{0,D}]$, that is $2[Q]=[-Q_{0,D}]$.

(v) Suppose $Q_{1}(x,y)=Q_{2}((x,y)B)$, where
$B\in{\mathrm{GL}}_{2}({\mathbb{Z}})$ and $\det B=-1$. Let
$w=\begin{pmatrix}&1\\\ 1&\end{pmatrix}$. Since $w^{2}B=B$ and
$wB\in{\mathrm{SL}}_{2}({\mathbb{Z}})$, we may assume that $B=w$. If
$Q_{2}=[\alpha,\beta,\gamma]$, then $Q_{1}=[\gamma,\beta,\alpha]$. Then by
(2.3) $[Q_{1}]=-[Q_{2}]$.

Suppose $[Q_{1}]=-[Q_{2}]$. Let $Q^{\prime}_{1}(x,y)=Q_{2}((x,y)B^{\prime})$,
where $B^{\prime}\in{\mathrm{GL}}_{2}({\mathbb{Z}})$ and $\det B=-1$. Then we
just proved that $[Q^{\prime}_{1}]=-[Q_{2}]$. So $[Q_{1}]=[Q^{\prime}_{1}]$.
The remaining follows from the definition. ∎

## References

  * [ADS] M.F.Atiyah, H.Donnelly, I.M.Singer, Eta invariants, signature defects of cusps, and values of L-functions, Annals of Math. 118 (1983), 131-177.
  * [BGV] A. P. Bareto, D. L. Goncalves and D. Vedruscolo, Free involutions on torus semi-bundles and the Borsuk-Ulam theorem for maps into ${\mathbb{R}}^{n}$, Hiroshima Math. J. 46 (2016), 255-270.
  * [BS] W. Bosma and P. Stevenhagen, On the computation of quadratic 2-class groups, Journal de Theorie des Nombres de Bordeaux, tome 8, $n^{o}2$ (1996), p.283-313.
  * [EW] G. Everest, T. Ward, An introduction to number theory. Graduate Texts in Mathematics, 232. Springer-Verlag London, Ltd., London, 2005.
  * [Fl] D. Flath, Introduction to Number Theory, Wiley, New York, 1988.
  * [Gauss] C.F. Gauss, Disquisitiones Arithmeticae, Gerhard Fleischer, Leipzig, 1801.
  * [Ha] A. Hatcher, Notes on basic $3$-manifold topology. http://pi.math.cornell.edu/~hatcher/.
  * [He1] J. Hempel, $3$-manifolds, Princeton University Press and University of Tokyo Press, 1976.
  * [He2] J. Hempel, Residual finiteness for 3-manifolds. Combinatorial group theory and topology (Alta, Utah, 1984), 379-396, Ann. of Math. Stud., 111, Princeton Univ. Press, Princeton, NJ, 1987.
  * [Hr] F. Hirzebruch, Hilbert modular surfaces. L’Enseign. Math. 19 (1973), 183-281.
  * [KP] P. Koymans and C. Pagano, On Stevenhagen’s conjecture, arXiv: 2201.13424v1 [math.NT] 31 Jan 2022.
  * [LS] Y. Liu and H. Sun, Toward and after virtual specialization in 3-manifold topology. Surveys in differential geometry 2020. Surveys in 3-manifold topology and geometry, 215-252, Surv. Differ. Geom., 25, Int. Press, Boston, MA.
  * [Ma] B. Martelli An Introduction to Geometric Topology arXiv:1610.02592 math.GT math.DG (2016)
  * [Sak] M. Sakuma, Involutions on torus bundles over $S^{1}$, Osaka J. Math. 22 (1985), 163-185.
  * [Sc] P. Scott, The geometries of 3-manifolds, Bull. London Math. Soc. 15 (1983), 401–487.
  * [Sh] H. Shimizu, On discontinuous groups operating on the product of the upper half planes. Ann. of Math. (2) 77 (1963), 33-71.
  * [Smi2] A. Smith. _$2^{\infty}$ -Selmer groups, $2^{\infty}$-class groups, and Goldfeld’s conjecture_. Preprint.
  * [Smi3] A. Smith, _$\ell^{\infty}$ -Selmer groups in degree $\ell$ twist families_. Ph. D. Thesis.
  * [Ste] P. Stevenhagen, The number of real quadratic fields having units of negative norm. Experiment. Math. 2:121-136, 1993.
  * [SWW] H. Sun, S. Wang, J. Wu, Self-mapping degrees of trus bundles and torus semi-bundles, Osaka J. Math. 47 (2010), 131–155.
  * [Th] W. Thurston, Three-dimensional manifolds, Kleinian groups and hyperbolic geometry, Bull. Am. Math. Soc. 6 (1982), 357–381.
  * [Th1] W. P. Thurston, Three-dimensional geometry and topology. Vol. 1. Edited by Silvio Levy. Princeton Mathematical Series, 35. Princeton University Press, Princeton, NJ, 1997.
  * [Web] H. Weber, Beweis des Satzes. Math. Ann. 20 (1882), 301-329.
  * [Zhang] W.P. Zhang, Lectures on Chern-Weil Theory and Witten Deformations, Nankai Tracts in Mathematics Vol. 4, World Scientific.