On The Characterization Problem of Alexander Polynomials of Closed 3-Manifolds
K. Alcaraz
\noindent We give a characterization for the Alexander polynomials $\Delta_{M$ of closed orientable 3-manifolds $M$ with first Betti number $b_{1M=1$ and some partial results for the characterization problem in the cases of $b_{1M>1$. We first prove an analogue of a theorem of Levine: that the product of an Alexander polynomial $\Delta_{M$ with a non-zero trace symmetric polynomial in $b_{1M$ variables is again an Alexander polynomial of a closed orientable 3-manifold. Using the fact that there exist $M$ with $\Delta_{M=1$ for $b_{1M=1,2,3$, we conclude that the symmetric polynomials of non-zero trace in 1, 2 or 3 variables are Alexander polynomials of closed orientable 3-manifolds. When $b_{1M=1$ we prove that non-zero trace symmetric polynomials are the only ones that can arise. Finally, for $b_{1M\geq 4$, we prove that $\Delta_{M\not=1$ implying that for such manifolds not all non-zero trace symmetric polynomials occur.
KarinPaper_final.tex
\documentclass[14pt]{amsart}

\newcommand\hmmax{0} 
\newcommand\bmmax{0} 
\usepackage{bm}
 \newcommand{\tv}[1]{\bm {\mathcal{#1}}}

 \usepackage{upgreek}

%\usepackage[mathlf,textlf,minionint]{MinionPro}
\usepackage{stmaryrd} 
\usepackage[T1]{fontenc}
\usepackage{textcomp}


\usepackage{amsopn, amsthm, amsgen, amscd,amsmath}
\usepackage[bbgreekl]{mathbbol}
\usepackage[mathcal]{euscript}
\DeclareMathAlphabet{\mathbbm}{U}{bbm}{m}{n}% from bbm.sty
\usepackage{color}
\usepackage{tikz}
  \usepackage{graphicx}
\usepackage{epstopdf}
\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `basename #1 .tif`.png}
%\usepackage[small,nohug,heads = LaTeX]{diagrams} \diagramstyle[labelstyle=\scriptstyle]
\usepackage[small,nohug,heads = LaTeX]{diagrams} \diagramstyle[labelstyle=\scriptstyle]

\definecolor{CadetBlue}{cmyk}{0.62, 0.57, 0.23, 0 }
\definecolor{black}{cmyk}{1, 0.5, 0, 0 }
\definecolor{RedViolet}{cmyk}{0.07, 0.9, 0, 0.34 }
\definecolor{SeaGreen}{cmyk}{0.69, 0, 0.5, 0}


\definecolor{RoyalBlue}{cmyk}{1, 0.5, 0, 0 }

\definecolor{RedViolet}{cmyk}{0.07, 0.9, 0, 0.34 }

\definecolor{BrickRed}{cmyk}{0, 0.89, 0.94, 0.28}

\DeclareMathAlphabet{\mathpzc}{OT1}{pzc}{m}{it}



\newcommand{\A}{\mathbb A}
\newcommand{\R}{\mathbb R}
\newcommand{\C}{\mathbb C}
\newcommand{\D}{\mathbb D}
\newcommand{\F}{\mathbb F}
\newcommand{\HP}{\mathbb H}
\newcommand{\LL}{\mathbb L}
\newcommand{\M}{\mathbb M}
\newcommand{\K}{\mathbb K}
\newcommand{\N}{\mathbb N}
\newcommand{\PR}{\mathbb P}
\newcommand{\Q}{\mathbb Q}
\newcommand{\Z}{\mathbb Z}
\newcommand{\SO}{\widehat{\mathbb S}}
\newcommand{\T}{\mathbb T}
\newcommand{\I}{\mathbb I}
\newcommand{\X}{\mathbb X}
\newcommand{\SI}{\mathbb S}

\newcommand*\circled[1]{\tikz[baseline=(char.base)]{
            \node[shape=circle,draw,inner sep=.5pt] (char) {#1};}}
            
%\newcommand\mathfrak[1]{\mbox{\usefont{U}{euf}{m}{n}#1}}

\newtheorem{theo}{Theorem}
\newtheorem{lemm}{Lemma}
\newtheorem{prop}{Proposition}
\newtheorem{coro}{Corollary}
\newtheorem*{nntheo}{Main Theorem}
\newtheorem*{conj}{Conjecture}
\newtheorem*{ques}{Question}
\newtheorem*{theon}{Theorem}
\newtheorem*{coronn}{Corollary}
\newtheorem*{coron}{Characterization Theorem}


\theoremstyle{definition}
\newtheorem{defi}{Definition}
\newtheorem{axio}{Axiom}
\newtheorem{assu}{Assumption}

\theoremstyle{remark}
\newtheorem{nota}{Notation}
\newtheorem{exam}{Example}
\newtheorem{case}{Case}
\newtheorem{note}{Note}
\newtheorem*{aside}{Aside}
\newtheorem{rema}{Remark}
\newtheorem*{clai}{Claim}
\newtheorem{obse}{Observation }



\title[Alexander Polynomial]{On The Characterization Problem of Alexander Polynomials of Closed 3-Manifolds}
\author{K. Alcaraz}
\address{The University of Texas at Austin, Department of Mathematics, 1 University Station C1200 Austin, TX 78712-0257}
\email{karin@math.utexas.edu}
\date{April 15, 2014}
\subjclass[2010]{Primary 57M27}
\keywords{Alexander polynomial, closed 3-manifolds}
\begin{document}
\vspace{2cm}
\begin{abstract} 

\noindent We give a characterization for the Alexander polynomials $\Delta_{M}$ of closed orientable 3-manifolds $M$ with first Betti number $b_{1}M=1$ and some partial results for the characterization problem in the cases of $b_{1}M>1$.  We first prove an analogue of a theorem of Levine: that the product of an Alexander
polynomial $\Delta_{M}$ with a non-zero trace symmetric polynomial in $b_{1}M$ variables is again an Alexander polynomial
of a closed orientable 3-manifold.  Using the fact that there exist $M$ with $\Delta_{M}=1$ for $b_{1}M=1,2,3$, we conclude
that the symmetric polynomials of non-zero trace in 1, 2 or 3 variables are Alexander polynomials of closed orientable 3-manifolds.
When  $b_{1}M=1$ we prove  that non-zero trace symmetric polynomials are the only ones that can arise.  Finally,
for $b_{1}M\geq 4$, we prove that $\Delta_{M}\not=1$ implying that for such manifolds not all non-zero trace symmetric polynomials occur.

 \end{abstract}
 
 \maketitle
\tableofcontents
\section{Introduction}

Let $M$ be a manifold, $b_{1}M$ its first Betti number.  For $b_{1}M\geq 1$, the Alexander polynomial of $M$, $\Delta_{M}$, is a polynomial  invariant in $b_{1}M$ variables
first defined by Alexander \cite{Alex} for $M= \SI^{3}-K$ a knot complement and more generally for $M$ having  finitely presented $\pi_{1}$ by Fox \cite{FoxI}. 

 In the case of knot complements, a characterization of Alexander polynomials was given by Seifert \cite{Ro}, and later,
in the case of link complements, necessary conditions on Alexander polynomials were described by Torres \cite{To}  (when $b_{1}=2$, these conditions are insufficient \cite{Hill1}, \cite{Pl}).  

In this paper, we consider the characterization problem for 
Alexander polynomials of closed orientable 3-manifolds, giving a characterization in the case $b_{1}M=1$ and some partial results concerning characterization in the cases of $b_{1}M> 1$.  

Our first result is the following analog for closed orientable 3-manifolds of a Theorem of Levine (see \S \ref{LevineSection} %\S \ref{LevLinks}
or \cite{Le2}), which allows us to produce by multiplication new Alexander polynomials from a given one.
By the {\it trace} of a polynomial we mean the sum of its coefficients.

\begin{theon} Let $M$ be a closed orientable $3$-manifold with $b_{1}M=n$ and let $\lambda$ be a symmetric Laurent polynomial in $n$ variables with non zero trace.
Then there exists a $3$-manifold $M'$ with $b_{1}M'=n$ for which
\[ \Delta_{M'}=\lambda\cdot\Delta_{M}.\]
\end{theon}

This Theorem is proved in \S \ref{LevineSection}.


For low Betti numbers $b_{1}=1, 2$ or $3$, the following closed orientable 
manifolds have 
$\Delta_{M}=1$:

\begin{itemize}
\item[-] $\SI^{1}\times\SI^{2}$, $b_{1}=1$.
\item[-] $\text{\sf H}_{3}(\R)/\text{\sf H}_{3}(\Z )$ = Heisenberg manifold \cite{Mc}, $b_{1}=2$. 
\item[-]  $\T^{3}$ = the 3-torus, $b_{1}=3$.
\end{itemize} 


Combining these examples with the above analog of Levine's Theorem we have

\begin{coronn}  Every symmetric Laurent polynomial in $1$, $2$ or $3$ variables having non-zero trace is the Alexander polynomial of a $3$-manifold with first Betti number $1$, $2$ or $3$.
\end{coronn}

It is natural to ask if the statement in the Corollary gives necessary conditions for a characterization for Betti numbers $b_{1}=1, 2$ or $3$.  In the case $b_{1}=1$, it does.
We say that a Laurent polynomial is {\it unit symmetric} if it is symmetric after multiplication with a unit of the ring
$\Z [t_{1}^{\pm 1},\dots ,t_{n}^{\pm 1}]$.

\begin{coron}  Let $\lambda$ be a Laurent polynomial in $1$ variable.  Then $\lambda =\Delta_{M}$
for some closed $3$-manifold $M$ with $b_{1}M=1$ $\Leftrightarrow$ $\lambda$ is unit symmetric and has non-zero trace.
\end{coron}


Finally, we consider the case of manifolds with $b_{1}\geq 4$.  Our main result is the following:

 \begin{theon} $\Delta_{M}\not =1$ for any closed $3$-manifold with $b_{1}M\geq 4$.  
 \end{theon}
 
  A consequence of this is the following negative result: not every symmetric polynomial occurs as the Alexander polynomial
 for a 3-manifold having $b_{1}\geq 4$. 


\vspace{3mm}

\noindent {\bf Acknowledgements:} This paper is based on results contained in my PhD thesis, which was supervised by Marc Lackenby. I would also like to thank Cameron Gordon for helpful conversations.





\section{Alexander Polynomial}

From this point on, all 3-manifolds will be assumed to be orientable.

 We begin by reviewing a definition of the Alexander polynomial which employs absolute homology \cite{Ma}, \cite{Ro}.  This definition differs from those that appear in \cite{Blan,FoxI,Mc,Turaev0}  which use either relative homology or Fox calculus.   A proof of the equivalence of all three definitions can be found in \cite{me} (the equivalence is in fact implicit in Theorem 2.7 of \cite{Ma} as well as Theorem 16.5 of \cite{Turaevbook}).
 %Bellow we give the absolute homology definition; our exposition follows roughly that given by \cite{Mc}
%for relative homology, {\it mutatis mutandis}.
 
 

Let $\psi :G\rightarrow H$ be an epimorphism of a finitely presented group $G$ onto a finitely generated free abelian group $H$ of rank $n$.  Denote by $\Z [H]$ the group ring associated to $H$ i.e.\ the ring of formal
finite linear combinations
\[ n_{1}h_{1}+\dots +n_{k}h_{k},\quad n_{i}\in\Z,\; h_{i}\in H,\]
and the product in $\Z [H]$ is defined by linear extension of the product of $H$.  Note that if we choose a basis
$t_{1},\dots ,t_{n}$ of $H$ then
we may write elements of $H$ as
\[  t^{I} = t_{1}^{i_{1}}\cdots t_{n}^{i_{n}} \] for $i_{1},\dots ,i_{n}\in\Z$, so that the elements of  $\Z [H]$ may be viewed as a multivariable Laurent polynomials in the 
multivariable $t$.

Let $(X,x)$ be a pointed $CW$-complex whose $0$-skeleton consists of the base point $x$, such that $G=\pi_{1}(X,x)$.  Let 
\[ p_{\psi} \colon \widehat{X}_{\psi}\longrightarrow X\] be the normal covering space corresponding to $\psi$: that is, the covering indexed by ${\rm Ker}(\psi)\subset G$ with deck group $G/{\rm Ker}(\psi)\cong H$. 

 The {\bf {\em Alexander module}} is defined to be 
\begin{equation}\label{firstdefi}
A_{\psi}=H_{1}(\widehat{X}_{\psi})=H_{1}(\widehat{X}_{\psi};\;\mathbb{Z}),\end{equation}
where its structure of $\Z [H]$-module comes from the action of $H$ on $\widehat{X}_{\psi}$ by deck transformations.

For any finitely presented module $A$ over $\mathbb{Z}[H]$ consider a free resolution 
 \[ \mathbb{Z}[H]^{m} \stackrel{P}{\longrightarrow}  \mathbb{Z}[H]^{n}\longrightarrow  A
\longrightarrow 0 .\]
Such a resolution may be defined using a presentation $A=\langle x_{1},\dots x_{n}|r_{1},\dots ,r_{m}\rangle$:
we take $ \mathbb{Z}[H]^{n}=\langle x_{1},\dots , x_{n}| \;\;\rangle$, $\mathbb{Z}[H]^{m} = \langle R_{1},\dots ,R_{m}| \;\;\rangle$
with $P(R_{i}) =r_{i}$. 
Note that we may assume, without loss of generality, that $m\geq n$.
One can represent $P$ by an $m\times n$ matrix
also denoted $P$.  
For each $d=0,\dots , n$ we define the {\bf {\em dth elementary ideal}} $I_{d}(A)\subset \mathbb{Z}[H]$ to be the ideal generated by the $(n-d)\times (n-d)$ minors of the matrix $P$.  These ideals are independent of the resolution of $A$
and form a chain
\[   I_{0}(A)\subset I_{1}(A)\subset \cdots \subset I_{n}(A)=\mathbb{Z}[H] .\]
The {\bf {\em dth order ideal}} is the smallest principal ideal containing $I_{d}(A)$, where any generator $\Delta_{d}(A)$
of it is called a {\bf {\em dth order}} of $A$; it is well-defined up to multiplication by units.  The $d$th order can also be defined as the greatest
common divisor of the $(n-d)\times (n-d)$ minors of the matrix $P$ (which is well-defined since $\mathbb{Z}[H]$ is a unique factorization domain).




When $A=A_{\psi}$ we denote by 
\begin{itemize}
\item $P_{\psi}$ any presentation matrix and call it
an {\bf {\em Alexander matrix}}.  
\item $I_{\psi}$ the $0$th elementary ideal of a presentation matrix $P_{\psi}$ and call it the {\bf {\em Alexander ideal}}.
\item $\Delta_{\psi}$ a $0$th order of $A_{\psi}$ and call it an {\bf {\em Alexander polynomial}}.
\end{itemize}




Let $M$ be a compact manifold with $\pi_{1}M$ finitely presented and with first Betti number $b_{1}M=n$.  
Let $H(M)=H_{1}M/{\rm Tor}(H_{1}M)\cong\Z^{n}$, where ${\rm Tor}(H_{1}M)$ is the torsion subgroup of $H_{1}M$, and consider the epimorphism
\[ \psi^{\text{fr-ab}}: \pi_{1}M\longrightarrow H_{1}M\longrightarrow H(M),\]
where the first map is the abelianization and the second map is the projection.
The normal covering associated to $K^{\text{fr-ab}}={\rm Ker}(\psi^{\text{fr-ab}})$
\[  p:\widehat{M}\longrightarrow M \]
is called the {\bf {\em universal free abelian cover}} of $M$.  



%\begin{lemm}\label{abelemma}
%Let $G$ be a group and suppose that $N$ is a normal subgroup of $G$.   If $G/N$ is abelian then 
%$[G,G]<N$.
%\end {lemm}

%\begin{proof}  Since $G/N$ is abelian, $G^{\rm ab}$ projects onto $G/N$. But this means $ [G,G]<N$.
%\end{proof}

\begin{prop}  Let $H$ be a finitely generated free abelian group and let $\psi :\pi_{1}M\rightarrow H$ be an epimorphism
with associated covering $p_{\psi}: \widehat{M}_{\psi}\rightarrow M$.  Then there exists a covering
\[  q_{\psi}:\widehat{M}\longrightarrow   \widehat{M}_{\psi}  \]
such that $p_{\psi}\circ  q_{\psi}=p$.
\end{prop}

\begin{proof}  Let $K_{\psi}= {\rm Ker}(\psi )$ which is the image of $\pi_{1} \widehat{M}_{\psi}$ by 
$(p_{\psi})_{\ast}$.  Since $H_{1}M$ is the abelianization
of $\pi_{1}M$, there is an epimorphism $r:H_{1}M\rightarrow H$ with $\psi= r\circ \psi^{\text{fr-ab}}$.
But $H$ is torsion free, therefore
all torsion elements of $H_{1}M$ are mapped to $0$.  This means that there exists
an epimorphism $r' :H(M)\rightarrow H$ so that $\psi= r'\circ \psi^{\text{fr-ab}}$.  We conclude
that $K^{\text{fr-ab}}\subset K_{\psi}$, and the Proposition follows from this.
 %herefore there
%is an inclusion $K^{\text{fr-ab}}\subset$.
%By Lemma \ref{abelemma}, $K_{\psi}\supset {\rm Ker}(\psi^{\rm ab})$.  
%The existence of $ \hat{q}_{\psi}$ follows from covering space theory. 
\end{proof}

\begin{coro}\label{Alexunique} Any cover of $M$ with deck group $H(M)$ is isomorphic to the universal free abelian cover $p:\widehat{M}\longrightarrow M$.
\end{coro}

When $\psi=\psi^{\rm fr-ab}$ we denote $A_{M}$, $I_{M}$ and $\Delta_{M}$,  and refer to them as the {\bf {\em Alexander module, ideal}} and {\bf {\em polynomial}}
of $M$.  By Corollary \ref{Alexunique}, these invariants depend only on $M$.









\section{An Analog of a Theorem of Levine for Closed 3-manifolds}\label{LevineSection}

First we define three notions of symmetry that will be relevant for us.  

Let $H\cong\Z^{n}$, denote $\Lambda=\Z [H]$
the group ring and by $\Lambda^{\times}$ the group of units.
The inversion map in $H$, $h\mapsto h^{-1}$, extends to an automorphism $\iota:\Lambda\rightarrow\Lambda$.  
%In what follows we will
%write \[ f%\circeq
 %g\] for $f,g\in\Lambda$ if there exists $u\in \Lambda^{\times}$ with $f=ug$.


We say that a polynomial
$f\in\Lambda$ is 
\begin{itemize}
\item {\bf {\em symmetric}} if $\iota (f)=f$.   
\item {\bf {\em unit symmetric}} if there exists a unit $u\in \Lambda^{\times}$ such
that $uf$ is symmetric i.e.\ $\iota (uf) =uf$.
\item {\bf {\em mod unit symmetric}} if 
%$ \iota (f) %\circeq f $i.e.\ 
 $\iota (f)=uf$ for some unit $u\in\Lambda^{\times}$.  
\end{itemize}
The definitions which we have just described are given in descending order of strength. For example, $t^{2}+t+1$ is unit symmetric
but not symmetric.  And $t-1$ is mod unit symmetric but not unit symmetric.

%Actually in most texts on this subject the word symmetric is used when unit symmetric is what is meant.  
%We will sometimes use the word symmetric for unit symmetric if the context makes things clear.

\vspace{5mm}

Now we recall Levine's method for generating link polynomials.



\begin{theo}[Levine, \cite{Le2}]\label{LevinesTheorem} Let $\Delta_{L}$ be the Alexander polynomial of an $n$-component link $L$, and let $\lambda = \Sigma \; c_{I}t^{I}\in \Z [H]$ satisfy the following conditions:
\begin{enumerate}
\item[{\rm 1.}] $\lambda$ is symmetric.
\item[{\rm 2.}] $\lambda (1,\cdots ,1)=1$.
\end{enumerate}
Then there exists an oriented link $L'$ with the same number of components as  $L$ such that 
\[ \Delta_{L'}=\Delta_{L}\cdot \lambda.\] 
\end{theo}
%EXPLAIN THE RELATION OF THE SECOND CONDITION $\lambda(1,1)=1$ AND THE USE OF A 1-SURGERY. How does this condition appears in Levine�s paper?



Now we prove an analogue of Levine's Theorem for closed 3-manifolds.  By this we mean the following:
let $M$ be a closed 3-manifold with $b_{1}M=n$ and let $\lambda$ be a  symmetric Laurent polynomial in $n$-variables with non-zero trace.  Then we will prove that
there exists a closed 3-manifold $M'$ with  $b_{1}M'=n$ and whose Alexander polynomial is $\lambda\cdot\Delta_{M}$.


\begin{prop}\label{1generator} 
 
Let $M$ be a 3-manifold with $b_{1}M=n$,
%maybe non-compact, for example the universal abelian cover, then we want a formula for $H_{1}(M-K)$
$S$ a simple closed curve in M which is homotopically trivial and $N$ a tubular neighborhood of $S$.
Let $p:\widehat{M}\rightarrow M$ be the universal free abelian cover and let $\widehat{X}=\widehat{M}-p^{-1}(N)$.
Then we have the following isomorphism of $\Lambda$-modules
\[H_{1}(\widehat{X})\cong H_{1}(\widehat{M}) \oplus \Lambda.\]
%\[ b_{1}(M-\text{\r{ N}})=  \left\{
                                                           %     \begin{array}{ll}
                                                            %     b_{1}M & \text{if $\mu=0$} \\
                                                             %    b_{1}M+1 & \text{otherwise}
                                                                 %\end{array}
                                              %    \right.
% \]                                                  


%CHECK HOW TO WRITE THIS CORRECTLY

%we want in the proof that the lift $\tilde{X}-\tilde{S}$ has one generator

\end{prop}

\begin{proof}  

Consider the pair $(\widehat{M},\widehat{X})$ and denote $\widehat{N} = p^{-1}(N)$.   By excision on the interior of $\widehat{X}$, we have that
\[ H_{*}(\widehat{M},\widehat{X}) \cong H_{*}(\overline{\widehat{N}}, \partial \overline{\widehat{N}}).\] 
Since the deck group $H$ acts by homeomorphisms on the pairs $(\widehat{M},\widehat{X})$, $(\overline{\widehat{N}}, \partial \overline{\widehat{N}})$,
the excision isomorphism is an isomorphism of $\Lambda$-modules.
If we let $\overline{\widehat{N}}_{0}\supset \partial \overline{\widehat{N}}_{0}$ be a pair of fixed components of $\overline{\widehat{N}}\supset \partial \overline{\widehat{N}}$
(i.e.\ a fixed lift of $\overline{N}\supset \partial \overline{N}$) then we may write
\[  \overline{\widehat{N}} =  \bigsqcup_{h\in H} h( \overline{\widehat{N}}_{0})   \supset \partial \overline{\widehat{N}} = 
\bigsqcup_{h\in H} h( \partial\overline{\widehat{N}}_{0}).
 \]  
On the level of homology groups we have then
\[ H_{k} (\overline{\widehat{N}}, \partial \overline{\widehat{N}})\cong\bigoplus_{h\in H}H_{k}(\overline{N}, \partial \overline{N}) , \]
a direct sum of copies of $H_{k}(\overline{N}, \partial \overline{N})$ indexed by $H$.
%Let $X=M-N$ and consider the pair $(M,X)$.  By excision, we have that
%\[ H_{*}(M,X) \cong H_{*}(\bar{N}, \partial \bar{N}).\]  
By Lefschetz duality, since $\overline{N}- \partial \overline{N}=N$, we have that
\begin{equation}\label{lefschetz} H_{k}(\overline{N}, \partial \overline{N}) \cong H^{3-k}(N)\cong H^{3-k}(\SI^{1}) =   \left\{
                                                                \begin{array}{ll}
                                                                \Z & \text{if $k = 2,3.$} \\
                                                                 0 & \text{otherwise.}
                                                                 \end{array}
                                                    \right.
\end{equation}
We note that a generator of $H_{2}(\overline{N}, \partial \overline{N})$ is given by a disk $D$ whose boundary is a meridian of $N$.
Thus as a $\Lambda$-module,  we have
\[   H_{2} (\overline{\widehat{N}}, \partial \overline{\widehat{N}})\cong \Lambda ,  \]
with generator a disk in $ \overline{\widehat{N}}_{0}$ whose boundary is a meridian.
The generator of $H_{2}(\widehat{M},\widehat{X})\cong H_{2} (\overline{\widehat{N}}, \partial \overline{\widehat{N}}) $ corresponding to the disk generator $\widehat{D}_{0}$ of $H_{2} (\overline{\widehat{N}}, \partial \overline{N})$
is denoted $\nu$.  Note that the boundary of $\nu$ is equal to $\partial \widehat{D}_{0}$.

The long exact sequence of the pair $(\widehat{M},\widehat{X})$ (an exact sequence of $\Lambda$-modules) can now be written
\[ \cdots \longrightarrow H_{2}(\widehat{M},\widehat{X})\cong\Lambda \stackrel{\alpha}{\longrightarrow}
H_{1}(\widehat{X}) \stackrel{\beta}{\longrightarrow} H_{1}(\widehat{M})  \stackrel{\gamma}{\longrightarrow}  H_{1}(\widehat{M}, \widehat{X}) =0\longrightarrow\cdots
\]

Then $\gamma$ is the zero map
%since $H_{1}(\bar{N}, \partial \bar{N})=0$
and by exactness $\beta$ is surjective, so that:
%By the third isomorphism theorem which requires, gamma surjective.
\[ H_{1}(\widehat{M})\cong H_{1}(\widehat{X})/ {\rm Ker}(\beta). \]
Again by exactness ${\rm Ker }(\beta)= {\rm Im }(\alpha)$ and we obtain a short exact sequence
\begin{equation}\label{shorty}
0 \longrightarrow {\rm Im}(\alpha) \longrightarrow H_{1}(\widehat{X}) \longrightarrow H_{1}(\widehat{M}) \longrightarrow 0.
\end{equation}
Since, as indicated above, $H_{2}(\widehat{M},\widehat{X})\cong\Lambda$, the map $\alpha$ is of the form \[ \alpha : \Lambda \longrightarrow  H_{1}(\widehat{X}).\]
Thus if we can show that $\alpha$ is injective and that the sequence (\ref{shorty}) is split by a $\Lambda$-module map, this
will imply that $H_{1}(\widehat{X})\cong H_{1}(\widehat{M})\oplus \Lambda$. 



%Thus the long exact sequence of the pair $(M,X)$ can be written
%\[ H_{2}(X) \stackrel{\alpha}{\longrightarrow} H_{2}(M) \stackrel{\beta}{\longrightarrow} H_{2}(M, X)\cong\Z \stackrel{\gamma}{\longrightarrow}
%H_{1}(X) \stackrel{\delta}{\longrightarrow} H_{1}(M)  \stackrel{\epsilon}{\longrightarrow}  H_{1}(M, X) =0\longrightarrow\cdots
%\]
%Then $\epsilon$ is the zero map
%since $H_{1}(\bar{N}, \partial \bar{N})=0$
%and by exactness $\delta$ is surjective, so that:
%By the third isomorphism theorem which requires, gamma surjective.
%\[ H_{1}(M)\cong H_{1}(X)/ Ker(\delta). \]
%Now, by exactness $Ker (\delta)= Im (\gamma)$ and we obtain a short exact sequence
%\begin{equation}\label{shorty}
%0 \longrightarrow Im(\gamma) \longrightarrow H_{1}(X) \longrightarrow H_{1}(M) \longrightarrow 0
%\end{equation}
%Since, as indicated above, $H_{2}(M, X)\cong\Z $, the map $\gamma$ is of the form \[ \gamma :  \Z \longrightarrow  H_{1}(X).\]

%Thus if we can show that $\gamma$ is injective and that the sequence (\ref{shorty}) is split, this
%will imply that $H_{1}(X)\cong H_{1}(M)\oplus \Z$.   \begin{color}{red}lifting the isomorphism doesn't makes sense \end{color} %along the $\Z^{n}$ cover
%gives the corresponding isomorphism described in the statement of the Proposition.


%aqui tenemos un dubujito, el generador de H_{2}(M-\text{\r{ N} es un disco en el toro solido.
%Note now that by exactness, $\gamma$ is injective $\Leftrightarrow$
%$\beta$ is the zero map$\Leftrightarrow$ $\alpha$ is surjective.
%is alpha surjective? why?
 

We begin by showing that $\alpha$ is injective i.e.\ we show that ${\rm Im}(\alpha)\cong \Lambda$.
%given that $K$ is homologically trivial in $M$.
Note first that ${\rm Im}(\alpha)$ is generated as a $\Lambda$-module by $\mu=\beta (\nu)$, and $\mu$ is in fact the meridian boundary of $\widehat{D}_{0}$.  Suppose that $\mu$ is a torsion element of $H_{1}(\widehat{X})$, i.e $f\mu=0$ for some 
$f=\sum c_{h}h\in\Lambda$.  We will show that $f=0$ i.e.\ that $c_{h}=0$ for all $h\in H$.
Now $f\mu$ bounds a compact surface in $\widehat{X}$ which when included in $\widehat{M}$ can be filled by meridian disks to form a closed orientable surface $\Sigma$ in $\widehat{M}$.
Write $\widehat{S}=\sqcup\widehat{S}_{h}$ where $\widehat{S}_{h}=h\widehat{S}_{0}$ and $\widehat{S}_{0}$ is a fixed lift of $S$ to $\widehat{M}$.  Since $S$ is homotopically trivial, each lift $\widehat{S}_{h}$ is also homotopically trivial.  
Then $\widehat{S}_{h}$ intersects $\Sigma$ $c_{h}$-times with the same orientation. But $\widehat{S}_{h}$ is homologically trivial, so its intersection number with an oriented surface is $0$.  Thus $c_{h}=0$ for all $h$.


%Let ${\rm Im}(\gamma)$ be generated as a module by $\mu$ and suppose that $\mu$ is a torsion element of $H_{1}(X)$, i.e $\mu^{n}=0$ for some $n>0$. %Then $\mu^{n}$ bounds a compact surface in $X$ which closes up to form a closed orientable surface $\Sigma$ in $M$. Then $K$ intersects $\Sigma$ $n$-% times with the same orientation. But $S$ is homologically trivial so the intersection number with an oriented surface is 0.  Thus $n=0$.

Now we want to show that the short exact sequence is split. Since $\widehat{S}_{h}$ is homologically trivial for each $h$, it bounds a compact oriented surface $\Sigma_{h}$ in $\widehat{M}$. Let 
\[ \phi : H_{1}(\widehat{X}) \longrightarrow {\rm Im}(\alpha ) =\Lambda\mu \] 
\[l \longmapsto \sum_{h\in H} (l\cdot \Sigma_{h})h\mu\] where  $l\cdot\Sigma_{h}$ is the intersection number with $\Sigma_{h}$.   Then since $h\mu$ has intersection number one with $\Sigma_{h}$ and $h\mu$ has intersection number zero with any $\Sigma_{h'}$ for $h' \ne h$,  
we have $\phi\circ \gamma (h\mu)=h\mu$.
Thus the exact sequence splits,  proving the Proposition.


 %Let $\gamma(1)=\mu $, $\mu \in H_{1}(M-\text{\r{ N}})$.  Then $Im (\gamma)= <\mu>=Ker (\delta)$ and
%  \[ H_{1}(M)\cong H_{1}(M-\text{\r{ N}})/ <\mu> .\]
% Thus \[ b_{1}(M-\text{\r{ N}})=  \left\{
                                                            %    \begin{array}{ll}
                                                               %  b_{1}M & \text{if $\mu=0$} \\
                                                               %  b_{1}M+1 & \text{otherwise}
                                                                 %\end{array}
                                               %   \right.
%\]                    



\end{proof}



%this is the same than saying that the lift of S has only one generator (this is the main point)


\begin{theo}\label{generalizedLevinestate}
Let $M$ be a closed 3-manifold with $b_{1}M=n$, $H\cong\Z^{n}$ the deck group
of its universal free abelian cover and let $\lambda = \Sigma \; c_{I}t^{I}\in \Z [H]$ be a symmetric Laurent polynomial with trace not equal to $0$.  Then there exists a closed 3-manifold $M'$ with $ b_{1}M'= b_{1}M$ and having Alexander polynomial \[ \Delta_{M'}=\Delta_{M}\cdot  \lambda .\]

%\[ b_{1}M' = \left\{ \begin{array}{ll}
   %                        b_{1}M & \text{if }{\rm tr}(\lambda )\not=0 \\
      %                     b_{1}M+1 &  \text{if }{\rm tr}(\lambda )=0 .
         %                   \end{array}\right.
 %\]
\end{theo}


\begin{proof}  Let    $\lambda = \Sigma \; c_{I}t^{I}$ be a symmetric Laurent polynomial.
We start by constructing a simple closed curve $S\subset M$ associated to the polynomial $\lambda$ with which we will modify $M$.  %which is homotopically trivial in $M$ ????
%ie it wont go aroun the "windows" WE CAN ALSO SEE THIS AS THE DEFINITION OF THE GENERAL S


%give it an orientation

%we pick a loop $\alpha_{i_{1},...,i_{n}}$ based at a point on $\partial D$ such that \[ [\alpha_{i_{1},...,i_{n}}]=t_{1}^{i_{1}}...t_{n}^{i_{n}} \in H_{1}(M)/Tor(H_{1}(M)). \]
%See Figure?.  
% PICTURE DISC WITH LOOPS

%We will now modify $\partial D$ in the following way. For each loop $\alpha_{i_{1},...,i_{n}}$ based at $\partial D$ we remove a 3-ball neighborhood of the base point and insert a new one as shown in Figure?. This modified $\partial D$ will be our curve $S$.

%PICTURE REMOVING AND PUTTING BACK BALL

% For some $\epsilon>0$ attach a band $[0,1-\epsilon] \times I$ THE ORIENTATION to $D$ to form a new disc $D_{1}$ isotopic to $D$. In a neighborhood of $(1-\epsilon) \times I$ enlarge the disc, as shown in Figure ?. The number of crossings and their signs depends on $a_{i,j}$ as shown in Figure ?. The result is an immersed disc. Let $C$ be its boundary.

%it is a manifold, locally it is like S3!!!!


Let $B\subset M$ be a 3-ball and choose a simple closed curve $S_{0}\subset B$ bounding an embedded disk $D$ in $M$. We will modify $S_{0}$ to obtain a new simple closed curve $S$ that will bound an immersed disk in $M$.
For each term pair $\{ t^{I}, t^{-I}\}$, $I\not=0$, of the polynomial $\lambda$ having non-zero coefficient $c_{I}$, pick
\begin{enumerate}
\item A point $q_{I}$ on $S_{0}$ such that $q_{I}\not=q_{I'}$ if $\{ t^{I}, t^{-I}\}\not= \{ t^{I'}, t^{-I'}\}$.
\item One of the two homology classes associated to $\pm I\in \Z^{n}\cong
H_{1}(M)/{\rm Tor}(H_{1}(M))$, which we denote  $\gamma_{\pm I}$.
\end{enumerate}
Then pick an embedded loop $u_{I}$ in $M$ based at $q_{I}$ such that \[ [u_{I}]=\gamma_{\pm I}. \] We may assume, after an isotopy, that the loops $u_{I}$ are disjoint.  Consider
a segment $\tilde{u}_{I}$ obtained from $u_{I}$ by cutting off a small piece of one of its ends, then thicken this segment to a band $\tilde{u}_{I}\times [0,1]\approx [0,1]\times[0,1]$.
See Figure \ref{Band}.  We assume that we have done this in such a way that
\begin{itemize}
\item

$(\tilde{u}_{I}\times [0,1])\cap D= (\tilde{u}_{I}\times [0,1])\cap S_{0} =\{0\} \times [0,1]$ and that the two end segments $\{0,1\} \times [0,1]$
lie in a small neighborhood of $q_{I}$.
\item Each component $[0,1] \times \{0,1\}$ is a copy of $u_{I}$ in the sense that the union of it with a small arc in $S_{0}$ is isotopic to $u_{I}$. We may again assume that all such bands are disjoint.
%\item Need to be careful about orientation of curves and endpoints of bands.
\end{itemize}
Finally, modify the end of the band as in Figure \ref{Band} so that its boundary links according to the coefficient $c_{I}$.  
We call the new curve $S$.  Note that $S$ bounds an immersed disc in $M$ and is therefore homotopically 
trivial.  

\begin{figure}[htbp]
\centering
\includegraphics[width=5in]{GenlLevineS1}
\caption{Modifying $S_{0}$ to obtain $S$.}
\label{Band}
\end{figure}

 %This will means that $S$ is homotopically trivial in $M$. Falta esta demostracion.

Remove a tubular neighborhood $N$  of $S$ and let $X=M-N$. We construct $M'$ by performing an $m$-surgery on $M$ along $S$,
where $m={\rm tr}(\lambda )$, and with respect to
a {\it preferred framing} \[ f=(\ell,\mu)\subset\partial N.\]  Here, $\ell$ is the {\it preferred longitude} -- the one
characterized by defining a homologically trivial element in $X$ -- and $\mu$ is a meridian.
We will show that $\Delta_{M'}=\Delta_{M}\cdot \lambda$.

It will be important for us to have an explicit understanding of the preferred longitude $\ell$.  First, consider a tubular neighborhood of the curve as it appears before the linking step; call that curve $S'$.
See the first image in Figure \ref{Band}.  Choose a preferred longitude  for the original curve $S_{0}$, cut at the beginning
of the band.  Continue this longitude along a pair of homotopic segments lying along tubular neighborhoods
of the band boundaries. The result will be a curve $\ell'$ which is homologically trivial in the complement of a tubular neighborhood of $S'$.

A tubular neighborhood for $S$ is obtained by cutting the tubular neighborhood of $S'$ just constructed and adding a cylinder that links about $S_{0}$.
If we were to continue $\ell'$ without twisting about this cylindrical piece, we obtain a longitude called the {\bf {\em obvious longitude}},
denoted $l$,
which is not homologically trivial.  In fact, $l$ bounds an immersed punctured disk, punctured twice for each linking that has been introduced.  More specifically,
$l$ is homologous in $X$ to
\[    (\sum_{I\not=0} c_{I})\mu   \]
where $\mu$ is a meridian of the tubular neighborhood of $S$.
The preferred longitude of $S$, denoted $\ell$, is therefore obtained from $l$ by introducing a
 {\it pair} of twists in $l$ (opposite in orientation to the direction of the linkings) for each of the  $c_{I}$ linkings coming from $u_{I}$.
 That is,
 \[\ell = l -(\sum_{I\not=0} c_{I})\mu. \]
%We assume moreover that the modification of $S_{0}$ obtained by adding the curve $u_{I}$ was performed in such a way
%that no additional twists are introduced into $f$.  
See Figure \ref{preferredframeS}.

\begin{figure}[htbp]
\centering
\includegraphics[width=5in]{PrefFramingSCurve}
\caption{The preferred longitude.}
\label{preferredframeS}
\end{figure}


%we won't care for now what it means to do an integer surgery on M, the point is that by doing surgery I'm indeed changing the manifold.
We will now calculate the Alexander module $H_{1}(\widehat{M}')$ as a $\Lambda$-module. Let $\widehat{M}$ be the $\Z^{n}$ cover of $M$, $p$ the covering map so that $\widehat{S}=p^{-1}(S)$ and $\widehat{N}=p^{-1}(N)$.
%Each component of $\tilde {S}$ bounds a copy of $D$.
%but in the cover the boundary will be homotopically equivalent to the sum of the punctures.
%each of this bounds a copy of D but it is punctured by the others.
Let $\widehat{X}=\widehat{M}-\widehat{N}$. By Proposition \ref{1generator} we have that
\[ H_{1}(\widehat{X})\cong H_{1}(\widehat{M}) \oplus \Lambda\]
as $\Lambda$-modules.
Then the only new generator in homology which one obtains by removing $\widehat{N}$ corresponds algebraically to the factor $ \Lambda$.  In particular, this factor as a $ \Lambda$-module is cyclic, generated by an element $\alpha$.

In order to obtain $\widehat{M}'$ from $\widehat{M}$, we will glue in solid tori to $\widehat{X}$, that is, to perform an $m$-surgery on each torus component.  We will
choose a longitude in each torus boundary $\widehat{N}$ of $\widehat{X}$ coming from the lift of the preferred longitude $\ell$
of $N$.  For a fixed boundary component of $\widehat{N}$, we denote this longitude by $\hat{\ell}$ and denote by $\mu$
a meridian.  We let $\hat{l}$ be a lift to  $\widehat{N}$ of the obvious longitude $l$ to the same boundary component containing $\hat{\ell}$.
 Observe that the homology class of $\mu$ also generates the factor $  \Lambda$ occurring
in $H_{1}(\widehat{X})$, so we may assume that $\mu=\alpha$.

The longitude $\hat{l}$ is not homologically trivial (as $l$ was not) owing to the fact that other lifts of $N$ link with $\widehat{N}$.  Instead, it bounds
an immersed punctured disk, in which we have a puncture for each such linking.  See Figure \ref{punctures}.

\begin{figure}[htbp]
\centering
\includegraphics[width=5in]{punctures}
\caption{The immersed punctured disk.}
\label{punctures}
\end{figure}



Let $\lambda_{0}=\lambda - c_{0}$ so that ${\rm tr}(\lambda_{0})= \sum_{I\not=0} c_{I}$.  Note then
that $\hat{l}=\lambda_{0}\cdot\mu$ i.e.\ is the sum of the punctures.  Then it follows by our construction of $\ell$ from $l$ 
that
\[   \hat{\ell}=(\lambda_{0}-{\rm tr}(\lambda_{0}))\mu = \hat{l}-{\rm tr}(\lambda_{0})\mu .  \]
For the preferred framing $f=(\ell,\mu)\subset \partial X$ an $m$-surgery is given by the formula $\ell+m\mu$.  Thus, in the manifold $M'$, $\ell+m\mu$ is trivial.  This relation produces the relation in $\widehat{M}'$
\[  \hat{\ell}+m\mu = (\lambda_{0}-{\rm tr}(\lambda_{0}) +m)\mu =0.  \]
Thus since $m={\rm tr}(\lambda )$ we have $m-{\rm tr}(\lambda_{0})=c_{0}$ then this relation becomes
\[   \lambda\mu =0.\]

Since this new relation only involves the new generator $\alpha =\mu$, then the presentation of the Alexander
module of $M'$ is of the form
\[ \langle x_{1},\dots ,x_{\alpha}, \mu |r_{1},\dots ,r_{b}, \lambda\mu \rangle   \]
where
$\langle x_{1},\dots ,x_{\alpha}|r_{1},\dots ,r_{b}\rangle$ is the presentation for the Alexander module of $M$.  It then follows
immediately that the presentation matrix for the Alexander module of $M'$ is of the form
\[P' = \left(
                        \begin{array}{cc}
                        P & {\bf 0} \\
                        {\bf 0} & \lambda
                        \end{array}
\right)\]
where ${\bf 0}$ are zero vectors.  It follows that 
$\Delta_{M'} = \Delta_{M}\cdot \lambda$.


\end{proof}

\vspace{5mm}

For low Betti numbers $b_{1}=1, 2$ or $3$, the following closed orientable 
manifolds have 
$\Delta_{M}=1$ (see \cite{Mc}):

\begin{itemize}
\item[-] $\SI^{1}\times\SI^{2}$, $b_{1}=1$.
\item[-] $\text{\sf H}_{3}(\R)/\text{\sf H}_{3}(\Z )$ = Heisenberg manifold, $b_{1}=2$.
\item[-]  $\T^{3}$ = the 3-torus, $b_{1}=3$.
\end{itemize} 

%The first and third examples are well-known, for a proof %\S \ref{0DehnWhitehead}  
%of $\Delta_{M_{W}}=1$ see?.
Combining these examples with Theorem \ref{generalizedLevinestate} we have

\begin{coro}\label{sufficiency}  Every symmetric Laurent polynomial in $1$,$2$ or $3$ variables having non-zero trace is the Alexander polynomial of a $3$-manifold with first Betti number $1$,$2$ or $3$.
\end{coro}




\section{Characterization for $b_{1}M=1$}



In this section we prove the  

\begin{coron}  Let $\lambda$ be a Laurent polynomial in $1$ variable.  Then $\lambda =\Delta_{M}$
for some closed $3$-manifold with $b_{1}M=1$ $\Leftrightarrow$ $\lambda$ is unit symmetric and has non-zero trace.
\end{coron}


The sufficiency of the condition ``$\lambda$ is unit symmetric and has non-zero trace'' follows from Corollary \ref{sufficiency}. Therefore we will dedicate this section to proving necessity, which will
be accomplished as follows:
  
\begin{enumerate}
\item[\S \ref{blanchfield}] We recall Blanchfield's symmetry result, which says that the Alexander polynomial of any closed 3-manifold is mod unit symmetric. 
\item [\S\ref{SeifSurf}] We show the existence of a ``Seifert surface'' $\Sigma$  for $M$ and use it to construct the universal infinite cyclic cover $\widehat{M}$.
\item[\S \ref{seifertmat}]  Using $\Sigma$ we define a ``Seifert matrix''  for $M$, and use it to show that  $\Delta_{M}$ is unit symmetric. 
\item[\S \ref{cha}] We prove in Lemma \ref{nonzero} that the trace of $\Delta_{M}$ is not $0$.
\end{enumerate}

 
  

\subsection{Blanchfield's Mod Unit Symmetry Theorem}\label{blanchfield}


 
 Let $M$ be a compact and orientable $3$-manifold with or without boundary. 
 %Check with Marc to see if we need compactness, finitely presented fundamental group, etc.
In \cite{Blan}, a general symmetry result about Alexander ideals was proved.
 As before, we denote by $\widehat{M}$ the universal free abelian cover and denote by $\Lambda =\Z [H]$ 
the group ring generated by the deck group $H$ of $\widehat{M}\rightarrow M$.  
Recall the automorphism $\iota: \Lambda\rightarrow \Lambda$ defined in \S \ref{generalizedLevinestate}.
We will say that an ideal $\mathfrak{a}\subset\Lambda$ is {\bf {\em symmetric}} if $\iota (\mathfrak{a})=\mathfrak{a}$.





%Ther is an exact sequence for the homology relative to the boundary
%we may use it to conclude the result for the case without the boundary.

%He defines the torsion module $T_{1}(M')$

%\begin{theo} The i- ideals (I guess Alexander Ideals) of $T_{1}(M')$ are symmetric for i=0,1,2,...
%\end{the

The following theorem is a special case of Corollary 5.6 of \cite{Blan}.

\begin{theo} The principal ideal $(\Delta_{M})$ is symmetric. 
\end{theo}

We have as an immediate corollary: 


\begin{coro}\label{modunitsymmetric} Any Alexander polynomial $\Delta_{M}$ of $M$ is mod unit symmetric.
\end{coro}

\begin{proof}  By symmetry $(\iota (\Delta_{M})) = (\Delta_{M})$ so there exists $u\in\Lambda^{\times}$ such that $\iota (\Delta)= u\Delta$ i.e.\ $\Delta$ is mod unit symmetric  \end{proof}




%For a proof he uses the previous theorem and Lemma 4.10

%I have to give another Corollary saying that if the ideals are symmetric then the Alexander polynomial will be as well.
%The ALexander Polynomial is the greatest common divisor of the generators of the ideal.






\subsection{An Analogue of the Seifert Surface Construction}\label{SeifSurf}

%\begin{color}{blue} Let $M$ be a closed oriented 3-manifold with $b_{1}M=1$.  For $G$ an abelian group which is finitely generated,  ${\rm Tor} (G)$
%is the torsion subgroup and ${\rm Free}(G) =G/{\rm Tor}(G)$.  We will identify ${\rm Free}(G)$ as a subgroup of $G$ so that $G={\rm Free}(G)\oplus {\rm Tor} (G)$. \end{color}



Let $M$ be closed and orientable with $b_{1}M=1$.  Recall  the map $\psi :\pi_{1}M\rightarrow \Z$ defined as the composition
\begin{diagram}\label{defofpsi}  \pi_{1}M & \rTo^{\text{abelianization}}  &H_{1}(M,\Z ) &\rTo^{\text{projection}} & {\rm Free}(H_{1}(M,\Z ))\cong \Z ,\end{diagram}
where for $A$ an abelian group, ${\rm Free}(A)=A/{\rm Tor}(A)$.  We will usually identify $ {\rm Free}(A)$ as a subgroup of $A$
by choosing a section of the projection $A\rightarrow {\rm Free}(A)$.
For an embedded closed oriented surface $\Sigma\subset M$ and a simple closed oriented curve $\gamma\subset M$ we
denote by $\gamma\cdot \Sigma\in \Z$ the signed intersection number.





\begin{theo}  There exists an oriented, embedded and non separating closed surface $\Sigma\subset M$
such that for all $\gamma\in\pi_{1}M$,
\[  \psi (\gamma ) =   \gamma \cdot \Sigma  .    \]
\end{theo}

\begin{proof}  
Note that $\psi$ is by definition trivial on $[\pi_{1}M,\pi_{1}M]$ so that it induces an element of ${\rm Hom}(H_{1}(M,\Z ),\Z )$: the projection appearing in (\ref{defofpsi}), which is a non trivial homomorphism.
%which is not a torsion element and therefore gives
 %a nontrivial element of ${\rm Free}({\rm Hom}(H_{1}M,\Z ))$.   
 The intersection pairing version of Poincar\'{e} duality 
provides in particular a non-degenerate pairing 
\[ {\rm Free}( H_{1}(M,\Z ) ) \times  {\rm Free}( H_{2}(M,\Z ) )\longrightarrow \Z \]  
given by the signed intersection number.  
%By non-degenerate we mean that if $\gamma\in{\rm Free}( H_{1}(M,\Z ) )\cong\Z$
%and $\Sigma \in  {\rm Free}( H_{2}(M,\Z ) )\cong\Z$ then $\gamma\cdot \Sigma\not=0$.

See \cite{GrHa}. N.B. $ H_{2}(M,\Z)\cong H_{1}(M,\Z )$ which is free by the universal coefficient theorem. This means that every homomorphism 
\[ {\rm Free}(H_{1}(M,\Z ))\longrightarrow\Z\] 
% which is not torsion
 is given by the intersection pairing
with some element of $H_{2}(M,\Z )$.  Thus
we may associate
to $\psi$ an element $[\Sigma ]\in H_{2}(M,\Z )$.   We note that there exists a representative
$\Sigma\in [\Sigma ]$ which is a closed connected embedded surface since $[\Sigma ]$ is of co-dimension 1.
 This surface is orientable because
it represents a non-trivial element of $H_{2}(M,\Z )$;  any closed non-orientable surface has trivial
$H_{2}$ so could not represent a non-trivial element of $H_{2}(M,\Z )$.    Thus we have $\psi (\gamma ) =\gamma\cdot \Sigma$
for all $\gamma\in \pi_{1}M$.
In particular for any curve $\gamma$ with $\psi (\gamma ) =1$ we have  $\gamma\cdot\Sigma =1$.  This, along with the fact that $\Sigma$ is orientable, implies that
$\Sigma$ is non-separating.  For if $M-\Sigma= Y_{1}\sqcup Y_{2}$ is a disjoint union, then since $\Sigma$ is 2-sided, then after
moving $\gamma$ by an isotopy, $\gamma-\Sigma = \gamma-\{ {\rm point}\}\approx (0,1)$ would connect points of $Y_{1}$ to points of $Y_{2}$, which is impossible.

\end{proof}

Since $\Sigma$ is orientable and non-separating, it has a collar which we denote $C(\Sigma )$.  Let
\[  X = \overline{M - C(\Sigma )} = M - {\rm int}(C(\Sigma )).\]

Note that $X$ has two boundary components $\Sigma^{-}$ and $\Sigma^{+}$.

%\begin{defi} Let $S^{+}$ and $S^{-}$ be closed, oriented connected surfaces.   We say that they are {\bf homology cobordant} %if there is a compact oriented 3-manifold $W$ such that $\partial W=S^{-}\sqcup S^{+}$ and if the inclusions of $ S^{\pm}$ in 
%$W$  induce isomorphisms $H_{*}(S^{\pm};\Z) \longrightarrow H_{*}(W;\Z)$. We refer to $W$ as a {\bf homology cobordism}.   %More generally, we say that $W$ defines a {\bf rational (mod $p$)} {\bf homology cobordism}
%if the maps on homology induced by the inclusions are isomorphisms with $\Q$ (with $\Z/p\Z$) coefficients.
%\end{defi}



\begin{prop}\label{freepart} If $\Sigma$ is of genus $g$ then
\[  H_{1}X \cong H_{1}\Sigma \oplus {\rm Tor}(H_{1}X)\cong \Z^{2g}\oplus  {\rm Tor}(H_{1}X). \]
In other words, ${\rm Free}(H_{1}X)\cong\Z^{2g}$.
\end{prop} 

\begin{proof}  Observe that the statement of the Proposition is equivalent to showing that
\[  {\rm dim} H_{1}(X;\Q ) =2g. \]  For the remainder of the proof, all homology and cohomology will be with $\Q$ coefficients.
Next, we have
\[   H_{1}(X) \cong H^{2}(M, {\rm int} (C(\Sigma ))   ) \cong H^{2}(M, \Sigma )\]
where the first isomorphism is by Lefschetz duality (see \cite{Sp}, page 297) and the second isomorphism follows because $\Sigma$ is a
deformation retract of ${\rm int} (C(\Sigma ))$.  Since we are working with $\Q$-coefficients,
the universal coefficient theorem implies that
\[  H^{2}(M, \Sigma )\cong H_{2}(M, \Sigma ).\]
(Since $H_{1}M\cong\Q$ is free, ${\rm Ext}(H_{1}M,\Q )=0$ which implies $H^{2}(M, \Sigma )\cong {\rm Hom}(H_{2}(M,\Sigma),\Q )$;
but $H_{2}(M,\Sigma)$ is a $\Q$-vector space, so ${\rm Hom}(H_{2}(M,\Sigma),\Q )\cong H_{2}(M,\Sigma)$.)
So it will be enough to show that $ H_{2}(M, \Sigma )$ has dimension $2g$.

Let us consider the long exact sequence in homology of the pair $(M, \Sigma )$ :
\[  \cdots\longrightarrow H_{2}\Sigma  \stackrel{i_{2}}{\longrightarrow} H_{2}M \stackrel{j_{2}}{\longrightarrow} H_{2}(M, \Sigma )  \stackrel{\partial}{\longrightarrow} H_{1}\Sigma \stackrel{i_{1}}{\longrightarrow} H_{1}M \stackrel{j_{1}}{\longrightarrow} H_{1}(M,\Sigma) \longrightarrow \cdots\]
Notice that 
\[ H_{2}M\cong H^{1}M\cong H_{1}M\cong\Q,\]
where the first isomorphism is by Poincar\'{e} duality, the second by the universal coefficient theorem and the last one because $b_{1}M=1$.
Note that $j_{1}$ is injective, since $\Sigma$ intersects once a generator of ${\rm Free}(H_{1}(M,\Z ))$  and therefore when this generator
is mapped to $H_{1}(M,\Sigma)$, it persists.  
%\marginpar{Give more detail.}

Thus ${\rm Ker}(j_{1})=0 = {\rm Im}(i_{1})$ by exactness, so that $i_{1}$ is the zero map.
Thus ${\rm Ker}(i_{1}) = H_{1}\Sigma = {\rm Im}(\partial )$ again by exactness.  Thus $\partial$ is onto and $H_{2}(M, \Sigma )$ has
dimension $\geq 2g$.  On the other hand, $[\Sigma]$ generates an infinite cyclic subgroup of $H_{2}M$ since it corresponds by duality to $\psi$ which
is a free cohomology class.  So $i_{2}$ is injective and $\Q \cong{\rm Im}(i_{2} ) = {\rm Ker}(j_{2} )$.  Since ${\rm Ker}(j_{2} )\not=0$, it follows since
$H_{2}M\cong\Q$ that we must have that ${\rm Ker}(j_{2} ) = H_{2}M$.  Hence $j_{2}$ is the 0 map, which implies that ${\rm Ker}(\partial ) =0$
i.e. $\partial$ is injective and therefore an isomorphism.
\end{proof}

The argument above can be modified to show that $\partial$ is an isomorphism modulo torsion in $\Z$-coefficients.  More precisely,

\begin{lemm}\label{almostandiso}   The homomorphism 
\[ \partial : H_{2}(M, \Sigma ;\Z)\longrightarrow  H_{1}(\Sigma ;\Z )\]
satisfies 
\begin{enumerate}
\item ${\rm Ker}(\partial)\subset {\rm Tor}(H_{2}(M, \Sigma ;\Z))$. 
\item ${\rm Coker}(\partial )$
is a finite group.
\end{enumerate}
\end{lemm}


\begin{proof}  Consider the long exact sequence of the pair appearing in the proof of Proposition \ref{freepart}, but now with $\Z$-coefficients.  The map
$i_{2}$ is still injective and its image is still free (by how we defined $\Sigma$), but here we can no longer assert that $i_{2}$ is onto.  Nevertheless, this implies
that ${\rm Ker}(j_{2})={\rm Im}(i_{2})$ is free.  We claim that this implies that ${\rm Im}(j_{2})$ is a finite group.  For $H_{2}( M; \Z)\cong \Z\oplus {\rm Tor}(H_{2}(M;\Z))$, since we saw in
the proof of Proposition \ref{freepart} that $H_{2}(M;\Q)\cong H_{2}(M;\Z)\otimes\Q\cong\Q$.  Therefore ${\rm Ker}(j_{2})\subset{\rm Free}(H_{2}(M;\Z))\cong\Z$ and $j_{2}$ induces a map having domain the finite group
$H_{2}(M;\Z)/{\rm Ker}(j_{2})$, so  ${\rm Im}(j_{2})$ is a finite group as claimed.  By exactness, ${\rm Ker}(\partial )$ is finite, which proves (1).  Now since $ H_{2}(M, \Sigma ;\Q)\cong \Q^{2g}$,
$H_{2}(M, \Sigma ;\Z)\cong\Z^{2g}\oplus {\rm Tor}(H_{2}(M, \Sigma ;\Z))$ which by (1) means that  ${\rm Ker}(\partial)\subset  {\rm Tor}(H_{2}(M, \Sigma ;\Z))$.  Therefore $\partial$ restricted to the free part
of $H_{2}(M, \Sigma ;\Z)\cong\Z^{2g}$ maps onto a subgroup of rank $2g$ of $H_{1}(\Sigma ;\Z)\cong\Z^{2g}$.  This implies (2).

\end{proof}

 \marginpar{} 
 \begin{exam}\label{rationalcobordism}
For an example where $\partial$ is not surjective, consider the case
where $M$ is the mapping torus 
\[ \T_{A} = \T^{2}\times [0,1]/\sim_{A},\quad (x,0)\sim_{A} (Ax,1)\] associated
to the hyperbolic matrix
 \[ A = \left(\begin{array}{cc}
                                               3 & 2 \\
                                               1 & 1
                                     \end{array}\right).\]
 Since there is no simple closed curve $c\subset\T^{2}$ such that $A^{n}(c)$
 is isotopic to $c$ for some $n$, we have $b_{1}  \T_{A}=1$.  The ``Seifert
 surface" $\Sigma$ in this case is the image of $\T^{2}\times \{ 0 \}$
 in $\T_{A}$.   On the other hand, we know
(see \cite{Mc}) that 
\[ \Delta_{\T_{A}}(t) = {\rm det}(A-It) =  t^{2}-4t+1.  \]
By Lemma \ref{nonzero} of \S \ref{cha} below, the order of ${\rm Tor}(H_{1}\T_{A} )$
is $|\Delta_{\T_{A}}(1)|=2$.  The map $i_{1}:H_{1}\T^{2}\rightarrow H_{1}\T_{A}$, which is induced by the identification $\T^{2}\approx\Sigma\subset \T_{A}$
 has image ${\rm Tor}(H_{1}\T_{A}) \cong\Z/2\Z$: indeed, due to the defining identifications
 if we denote by $x=(1,0),y=(0,1)$ the basis of $\Z^{2}=H_{1}\T^{2}$, then $i_{1}(x) =x=A(x)= 3x+2y$
 and $i_{1}(y)= y=A(y)=x+y$ which implies that $2(x+y)=0$ and $x=0$ and therefore $2y=0$.
 In particular, ${\rm Ker}(i_{1})={\rm Im}(\partial )$
is a proper subgroup of $ H_{1}\T^{2}$.  Therefore, $\partial $ is not onto.      
 \end{exam}
     
                                    

We now use the surface $\Sigma$ to construct the universal infinite cyclic cover of $M$.
Take a countable collection $\{ X_{i}\}$, $i\in\Z$, of copies of $X$, and glue them such that $\Sigma^{+}_{i-1}$ is identified with $\Sigma^{-}_{i}$
by the ``re-gluing" homeomorphisms.  Denote the result $\widehat{M}$.

\begin{theo}  $\widehat{M}$ is a universal infinite cyclic cover of $M$.
\end{theo}

\begin{proof}  We show that there exists an infinite cyclic covering $p:\widehat{M}\rightarrow M$.  Each point in $X_{i}$ is mapped
to its counterpart in $X\subset M$.  Now we map $X$ onto $\overline{X} = X/\sim$ where $x\sim y$ if $x,y\in\partial X$ correspond.  The gluing
used to define $\widehat{M}$ is compatible with the relation $\sim$ so we get a covering map
$\widehat{M}\rightarrow \overline{X}\approx M$ which is infinite cyclic.
\end{proof}




\subsection{An Analogue of the Seifert Matrix} \label{seifertmat}

A key fact which we will need in order to prove that $\Delta_{M}$ is unit symmetric is the existence of a basis
of the homology of $X$ dual to a given basis of the homology of $\Sigma$.  However,
in view of Lemma \ref{almostandiso}, 
we will not be able to do this for integral homology, as one
does for knot complements \cite{Ro}.   In order to address this complication, we will work instead with homology with $\Q$-coefficients.





We begin with a notion of linking number valid in $M$.  Consider disjoint oriented simple closed curves $l_{1},l_{2}\subset M$ whose integral homology classes belong to
${\rm Tor}(H_{1}(M;\Z))$.  Then there exists integers $n_{1},n_{2}$ such that $[n_{1}l_{1}] = 0 = [n_{2}l_{2}]$, and therefore there exist compact and oriented surfaces
$S_{1}, S_{2}\subset M$ with $\partial S_{1} = n_{1}l_{1}$, $\partial S_{2} = n_{2}l_{2}$.

\begin{defi}\label{linked}  The {\bf linking number} of $l_{1}$ with $l_{2}$ is 
\[   {\rm lk}(l_{1}, l_{2}) =  \frac{ ( n_{2}l_{2})\cdot S_{1}  }{n_{1}n_{2}}=  \frac{ l_{2}\cdot S_{1}  }{n_{1}} \in \Q .  \]
\end{defi}
Since we have divided by $n_{1}n_{2}$, the linking number does not depend on the $n_{i}$ chosen so that $[n_{i}l_{i}]=0$.  

\begin{note}   We have ${\rm lk}(l_{1}, l_{2}) =- {\rm lk}(l_{2}, l_{1})$, just as
in the case of the classical linking number.
%To see this, it is enough to suppose that $n_{1}=n_{2}=1$.  Then
%every essential intersection of $l_{1}$ with $S_{2}$ produces an essential intersection of $l_{2}$
%with $S_{1}$, but with $l_{2}$ intersecting in the opposite sense with respect
%to the orientation of $S_{1}$. 
\end{note}
The linking number defined here is for pairs of simple closed curves $l_{1}$ and $l_{2}$, and only depends
on the isotopy type of $l_{1}\cup l_{2}$.   We can in fact extend it bi-linearly to rational multiples $q_{1}l_{1}$, $q_{2}l_{2}$ by the formula
\[  {\rm lk}(q_{1}l_{1}, q_{2}l_{2}) := q_{1}q_{2}\cdot {\rm lk}(l_{1}, l_{2}) .\]



%So this linking number may be viewed as defining a function 
%\[ {\rm lk}:{\rm Tor}(H_{1}M)\times {\rm Tor}(H_{1}M)\longrightarrow\Q .\]


%This notion of linking number can be extended to $\Q$ homology.  That is, if $l_{i}$, $i=1,2$ are two
%closed curves and $\alpha_{i}=q_{i}l_{i}$ are trivial in $H_{1}(M;\Q )$, then there exist integers $n_{i}$ so that $n_{i}\alpha_{i}$ are %trivial integral homology classes, and we define 
%\[ {\rm lk}(\alpha_{1}, \alpha_{2}) : = \frac{ {\rm lk}(n_{1}\alpha_{1}, n_{2}\alpha_{2})}{n_{1}n_{2}} \]

%\begin{defi}  The {\bf linking number mod p} of $l_{1}$ with $l_{2}$ is 
 %\[   {\rm lk}(l_{1}, l_{2}) =  \frac{ ( n_{2}l_{2})\cdot S_{1}  }{n_{1}n_{2}}=  \frac{ l_{2}\cdot S_{1}  }{n_{1}} \in \Q .  \]
%\end{defi}



We now choose special generating sets for the homology of the boundary components of $X$.


Fix bases $\{ a_{i}\}$, $\{ a_{i}^\pm\}$ of $H_{1}(\Sigma;\Z )$, $H_{1}(\Sigma^\pm;\Z )$; where $\{ a_{i}^\pm\}$ is a push-off in the $\pm$ normal direction of $\{ a_{i}\}$. When viewed in $M$, they give elements of Tor$(H_{1}M)$. 
We can choose $n\in\Z$ so that 
$\{na_{i}\}$, $\{na_{i}^{\pm}\}$ are homologous to $0$ in $H_{1}(M;\Z )$ and not just torsion (for example we could take
$n=\prod n_{i}$).


Now let $\{\bar{a}_{i}\}$, $\{ \bar{a}_{i}^{\pm}\}$ be $\{ na_{i}\}$, $\{ n{a}_{i}^{\pm}\}$, these elements give bases of $H_{1}(\Sigma;\Q )$, $H_{1}(\Sigma^{\pm};\Q)$. 

We define a square matrix $V_{\Q}=(\bar{v}_{ij})$ by
\[    \bar{v}_{ij} = {\rm lk} ( \bar{a}_{i}^{+}, \bar{a}_{j} ) .          \]
Notice that this is well-defined since these curves are torsion as elements of $H_{1}(M;\Z)$, and therefore
their linking numbers are defined.  

  Also note that
\[ \bar{v}_{ij} =  {\rm lk}( \bar{a}_{i}, \bar{a}^{-}_{j} )  \] 
so that \[ V_{\Q}^{T}=(\bar{v}_{ij}^{T}), \quad \text{where} \quad \bar{v}^{T}_{ij} =  {\rm lk} ( \bar{a}_{j}, \bar{a}^{-}_{i} ).\]  
%\marginpar{Prove this}




We now specify a basis of $H_{1}(X;\Q)$ dual to the basis $\{ \bar{a}_{i}\}$ of $H_{1}( \Sigma;\Q )$ with respect to the bilinear pairing ${\rm lk} (\cdot ,\cdot )$.  

\begin{lemm}\label{Xduality}   There exists a  basis $\bar{\beta}_{1},\dots ,\bar{\beta}_{2g}$ of  $H_{1}(X;\Q)$ such that viewed in $M$
\[   {\rm lk}(\bar{a}_{i},\bar{\beta}_{j} ) = \delta_{ij}   \] 
\end{lemm}

\begin{proof}  The map $\partial : H_{2}(M,\Sigma ;\Q )\rightarrow H_{1}(\Sigma;\Q )$ is an isomorphism.  If $\bar{a}_{i}$ is one of the generators of $H_{1}(\Sigma;\Q )$ specified above, then $\partial^{-1}(\bar{a}_{i})$ is a multiple ${n}_{i}S_{i}$ of a surface $S_{i}$. We may assume ${n}_{i}$ are integers by choosing $n$ large enough, we write $\bar{S}_{i}={n}_{i}S_{i}$. Thus we obtain a generating set $\bar{S}_{1},\dots ,\bar{S}_{2g}$ of $H_{2}(M,\Sigma ;\Q )$ in
 which $\bar{S}_{i}$ is the image in $H_{2}(M,\Sigma ;\Q )$ of the class $n_{i}S_{i}\in H_{2}(M,\Sigma ;\Z)$.

But 
\[ H^{2}(M,\Sigma ;\Q )
%\cong  {\rm Hom}(H_{2}(M,\Sigma ; \Z/p\Z),  \Z/p\Z) 
\cong H_{2}(M,\Sigma ; \Q)\cong \Q^{2g}\]
where the first isomorphism is a consequence of the universal coefficient theorem 
(see Corollary 4, page 244 of \cite{Sp}) and the second isomorphism comes from composition of isomorphisms
\begin{diagram} H_{2}(M,\Sigma ;\Q )& \rTo^{\partial}_{\cong} H_{1}(\Sigma;\Q )\cong \Q^{2g}.\end{diagram}
Therefore, $H^{2}(M,\Sigma ;\Q )$
has a basis dual to $\{\bar{S}_{i}\}$: given by cohomology classes $f_{1},\dots , f_{2g}$ with $f_{i}({n}_{j}S_{j} ) ={\delta}_{ij}$.  Here we are identifying cohomology classes with functionals of homology.

By Lefschetz duality we have $H_{1}(X;\Q )\cong H^{2}(M,\Sigma ;\Q)$. The duality isomorphism is given by the intersection pairing.  Therefore, if $\bar{\beta}\in H_{1}(X;\Q) $ and $\bar{S}\in H_{2}(M,\Sigma;\Q )$ then the $\Q $-Lefschetz duality isomorphism is induced by 
\[ \bar{S} \mapsto \bar{\beta}\cdot \bar{S} .\]
In particular, if we let $\bar{\beta}_{i}$ be such that the above function coincides with $f_{i}$, then we have 
\[ \bar{\beta}_{i}\cdot \bar{S}_{j}=f_{i}(\bar{S}_{j} )={\delta}_{ij}.\]  But $\partial (\bar{S}_{j})=\bar{a}_{j}$.
This implies that $ {\rm lk}(\bar{\beta}_{i}, \bar{a}_{j})={\delta}_{ij}$.
\end{proof}


Recall the construction of $\widehat{M}$ given at the end of \S \ref{SeifSurf}.  The Mayer-Vietoris Theorem applied to $\widehat{M}$ shows us that as a $\Lambda$-module
\[ H_{1}(\widehat{M};\Z ) \cong (H_{1}(X;\Z )\otimes\Lambda )/{\rm relations} \] where the relations are given by the gluing identifications 
\begin{align}\label{system}  a_{i}^{-} - ta_{i}^{+}=0,\quad i=1,\dots ,2g  
\end{align}
plus the torsion relations (which do not involve $t$):
\[  m_{1}\mu_{1} =0,\dots , m_{k}\mu_{k} =0   \]
where the $\mu_{i}$ are generators of the torsion of $H_{1}(X;\Z )$.  Hence the Alexander matrix has the form
\[   \left(  \begin{array}{cccc}
A(t) & 0 &\cdots & 0 \\
0 & m_{1} & \cdots & 0 \\
\vdots &  \vdots & \ddots & \vdots \\
0 & 0 & \cdots &m_{k}
\end{array}
    \right)   \]
    where $A(t)$ is a $2g\times 2g$ matrix corresponding to the system (\ref{system}).
    
    Let $\Lambda_{\Q}$ be the group ring $\Q [t^{\pm 1}]$ with coefficients in $\Q$. Then $H_{1}(\widehat{M};\Q )$
    is a $\Lambda_{\Q}$-module, and its presentation is given by
    \[ H_{1}(\widehat{M} ;\Q ) \cong (H_{1}(X;\Z )\otimes\Lambda_{\Q}) /{\rm relations} \] 
   where the relations are 
   \[  na_{i}^{-} - tna_{i}^{+} ,\quad i=1,\dots ,2g,\] 
   or in other words
    \[  \bar{a}_{i}^{-} - t\bar{a}_{i}^{+} ,\quad i=1,\dots ,2g.\] 
    Thus these are the only relations we have and the $\Q$ Alexander matrix is 
  \[A_{\Q}(t)=nA(t).\]  
  
  Recall that if $f(t) = \sum_{i=m}^{n}b_{i}t^{i}$ is a Laurent polynomial, where $b_{m},b_{n}\not=0$, then the degree is defined ${\rm deg}(f) = n-m$.  This notion
of degree is invariant with respect to multiplication by units.
  If we let $\Delta_{\Q}(t)$ be the ``Alexander polynomial'' associated to $A_{\Q}(t)$, then 
  \[ \Delta_{\Q}(t) := det (A_{\Q}(t))=
\frac{n^{2g}}{m_{1}...m_{k}}\Delta (t).\]
   We can see that $\Delta (t)$ and $\Delta_{\Q}(t)$
  have the same degree.  
 Our strategy will be to show that $\Delta_{\Q}(t)$ has even degree.

    
    



\begin{theo}\label{seifertmatrix}   $V_{\Q} - tV_{\Q}^{\sc T}=A_{\Q}(t)$.
\end{theo}

\begin{proof}  
 $H_{1}(X;\Q )$ has generators the $\bar{\beta}_{i}$ and we may write therefore
\[ \bar{a}_{i}^{\pm} = \sum c_{ij}^{\pm}\bar{\beta}_{j } .  \]
If we take the linking number both sides of this equation with $\bar{a}_{j}$, we get by Lemma \ref{Xduality} that
\[ c_{ij}^{\pm} = -{\rm lk} ( \bar{a}_{i}^{\pm},  \bar{a}_{j} ).\] 
It follows that the relations (\ref{system}) may be re-written
\[   \sum {\rm lk} ( \bar{a}_{i}^{-}, \bar{a}_{j} ) \bar{\beta}_{j} - t   \sum {\rm lk} ( \bar{a}_{i}^{+},  \bar{a}_{j} ) \bar{\beta}_{j}=0. \]
By our definition of $V_{\Q}$ and our identification of $V_{\Q}^{T}$ the relations (\ref{system}) may be re-written
\[\sum (\bar{v}_{ij}-t\bar{v}_{ji})\bar{\beta}_{j} =0, \quad i=1,\dots ,2g.\]
So
\[ A_{\Q}(t) = V_{\Q}-tV_{\Q}^{T},\] as claimed.
%Interchanging the roles of the ends $\Sigma^{-}$, $\Sigma^{+}$ we get that $V_{p}^{T}-tV_{p}$ is also a relation matrix.
\end{proof}



\subsection{Characterization}\label{cha}

We begin with the following corollary of Theorem \ref{seifertmatrix}:

\begin{coro}\label{modpsymm}  $\Delta_{\Q}(t)$ is unit symmetric and in particular is of even degree.
\end{coro}

\begin{proof}  By Theorem \ref{seifertmatrix} we have
\[\Delta_{\Q}(t) = \det (V_{\Q}-tV^{T}_{\Q}) =\det (V_{\Q}^{T}-tV_{\Q}) \]
and therefore
\[\Delta_{\Q}(t^{-1}) = \det (V_{\Q}-t^{-1}V^{T}_{\Q}) =\det ((-t^{-1})\cdot( V^{T}_{\Q}-tV_{\Q})) =t^{-2g}\Delta_{\Q}(t) .\]
Then $f(t)=t^{-g}\Delta_{\Q}(t)$ is symmetric so $\Delta_{\Q}(t)$ is unit symmetric.  The last statement follows since odd degree polynomials cannot
be unit symmetric (they can be at most mod unit symmetric).
\end{proof}





\begin{theo}\label{unitsymmthm}  The Alexander polynomial $\Delta(t)$ of $M$ is unit symmetric.
\end{theo}

\begin{proof}  By Corollary \ref{modpsymm} $\Delta (t)$ is of even degree.
By Blanchfield we know that any $\Delta_{M}(t)$ is mod unit symmetric.  Since the degree is even, this implies
that it is unit symmetric.
\end{proof}



\begin{lemm}\label{nonzero} For $b_{1}M=1$ the trace of the Alexander polynomial $\Delta_{M}$ is non-zero and its absolute value is equal to the order
of ${\rm Tor}H_{1}(M;\Z)$.
\end{lemm}


\begin{proof}
Let $\widehat{P}(t)$ be a presentation matrix for the $\Lambda$-module $H_{1}(\widehat{M})$, then $P:= \widehat{P}(1)$ is a presentation matrix
for $G$ = $p_{\ast}(H_{1}\widehat{M}) $ = the image of $H_{1}\widehat{M}$ in $H_{1}M$ by the covering map $p:\widehat{M}\rightarrow M$.   
Since $\widehat{M}$ is an infinite cyclic covering of $M$, we have the exact sequence 


\[ 1\longrightarrow p_{\ast}(\pi_{1}(\widehat{M}))\subset \pi_{1}(M) \longrightarrow \Z \longrightarrow 0 . \]
By abelianizing $\pi_{1}M$, we obtain the sequence
\[   0\longrightarrow G\subset H_{1}(M) \longrightarrow \Z \longrightarrow 0   \]
which is exact: here we are using the fact that 
\begin{enumerate}
\item The image of the subgroup $p_{\ast}(\pi_{1}\widehat{M}))<\pi_{1}(M)$
by the abelianization map $\pi_{1}(M)\rightarrow H_{1}(M)$ is $G$.
\item The epimorphism $\psi :\pi_{1}(M)\rightarrow\Z$
satisfies ${\rm Ker}(\psi )\supset [\pi_{1}(X), \pi_{1}(X)]$.
\end{enumerate}

  Thus $H_{1}(M)\cong G\oplus\Z$ as abelian groups
 so that ${\rm Tor}(G)={\rm Tor}(H_{1}(M))$.  But $G$ is a finitely generated abelian group and so is isomorphic to $\Z^{r}\oplus \Z/n_{1}\Z\oplus\cdots \oplus \Z/n_{k}\Z$ for integers $r, n_{1},\dots n_{k}$, and $P$ must be equivalent to the diagonal presentation matrix ${\rm diag}(n_{1} ,\dots , n_{k})$.  Then
\[  \Delta_{M} (1) = {\rm det}(P)  = n_{1}\cdots n_{k} = | {\rm Tor}(H_{1}(M))  |. \]

\end{proof}

We can now conclude with the

\begin{proof}[Proof (Characterization Theorem)]  Immediate from Theorem \ref{SeifSurf} and Lemma \ref{nonzero}.
\end{proof}



\section{Manifolds with $b_{1}>1$}\label{>1}


In this chapter we consider Alexander polynomials of closed 3-manifolds with $b_{1}>1$.  

 
 

\subsection{Manifolds with $b_{1}=2,3$}\label{b1=2,3}

As mentioned in the Introduction, the following closed 3-manifolds $M$ have
$\Delta_{M}=1$:
\begin{itemize}
\item[-] $\text{\sf H}_{3}(\R)/\text{\sf H}_{3}(\Z )$ = Heisenberg manifold \cite{Mc}, $b_{1}=2$.
\item[-]  $\T^{3}$ = the 3-torus, $b_{1}=3$.
\end{itemize}

Applying the generalized Levine's  theorem to the above examples, we have the following corollary:


\begin{coro} Let $M$ be a closed $3$-manifold with first Betti number 2 or 3. Then the set of symmetric Laurent polynomials in $2$ or $3$ variables with ${\rm tr} ( \lambda ) \ne 0$, is contained in the set of Alexander polynomials $\Delta_{M}$ with first Betti number equal to $2$ or $3$. \end{coro}


  




\subsection{Manifolds with $b_{1}\geq 4$}\label{b1>1}

 In this section we will prove 
 
 \begin{theo}\label{b1grt3impDeltnot1} $\Delta_{M}\not =1$ for any closed $3$-manifold with $b_{1}M\geq 4$.
 \end{theo}  

The proof of this theorem requires several facts which we summarize now.
Let $p$ be a prime and $\F_{p} =\Z/p\Z$. 
Recall that for a manifold $N$, the mod $p$ Betti numbers are defined
 \[  b_{k}(N;\F_{p} ) := {\rm rank}( H_{k}(N;\F_{p})).  \]
We will use the abbreviated notation  
\[  d_{p}(N) :=b_{1}(N;\F_{p} )\]
for the first mod $p$ Betti number.

\vspace{3mm}

  \begin{itemize} 
\item[\bf Fact 1.]  Let $\tilde{M}_{p}$ be the finite abelian cover of $M$ associated to the epimorphism
 \[ \psi_{p}:\pi_{1}M\longrightarrow H_{1}(M;\Z) \longrightarrow H_{1}(M;\F_{p} ) .\]
Let $r= d_{p}(M) $.  Using an inequality of Shalen and Wagreich \cite{SW},
we will deduce in \S \ref{SWsec} that
 \begin{align*} d_{p}(\tilde{M}_{p} ) \geq  \dbinom{r}{2} .\end{align*}
 
 
 
 \item[\bf Fact 2.]  Suppose that $\Delta_{M}=1$ and let $M'\rightarrow M$ be a finite abelian
 cover with deck group $\F_{p_{1}}\oplus\cdots \oplus \F_{p_{k}}$, $p_{1},\dots , p_{k}$ primes.  Then
 \begin{itemize}  
 \item[a.] The torsion subgroup of $H_{1}(M';\Z )$ is trivial. 
 \item[b.] $b_{1}M'=b_{1}M$.  This is a consequence of an equality of E. Hironaka \cite{Hiro}.
 \end{itemize}
These statements will be proved in \S \ref{b1grt4sec}.
\end{itemize}

\vspace{3mm}
  
  Assuming for the moment the above facts, we can now give the 
  
  \begin{proof}[Proof of Theorem \ref{b1grt3impDeltnot1}] We consider a cover $\tilde{M}_{p}$ as in Fact 1 with $p$ prime to the order of the torsion subgroup of $H_{1}(M)$, so that $ d_{p}(M) =b_{1}M$ and
  the deck group of $\psi_{p}$ is $\F_{p}^{b_{1}M}$.  Taking $M' = \tilde{M}_{p}$, it follows from Fact 2. (parts a. and b.)  that \[d_{p}(\tilde{M}_{p} )=d_{p}(M)= r.\]  But 
the inequality in Fact 1.\  is satisfied only for $r\leq 3$, since $r<\dbinom{r}{2}$ for all $r\geq 4$.
  \end{proof}
 
 \subsection{$\F_{p}$-Homology and Finite Cyclic Covers}\label{SWsec}
 
 In this section we will derive the inequality of Fact 1. above. 
 
 Fix a prime $p$ and let $G$ be a group.  If $A<G$ is a subgroup then we define 
   \[G\# A = [G,A]A^{p} := \langle [g,a]b^{p} \mid g \in G,a,b\in A\rangle ,\]
   where $[g,a]=gag^{-1}a^{-1}$ and $\langle X\rangle$  means the group generated by $X$.  
      
  We note that if $A\vartriangleleft G$ then $G\# A\vartriangleleft A$ and $A/(G\# A)$ is 
   an elementary abelian $p$-group i.e\ a direct sum
   of copies of $\F_{p}$.  
   The {\bf mod $p$ lower central series} $\{ G_{i}\}$ of $G$ is defined by 
   \[ G_{i+1}=G\# G_{i}\] where $G_{0}=G$.  By the above comments
   we have $G_{i+1}\vartriangleleft G_{i}$ and $G_{i}/G_{i+1}$ is an elementary abelian $p$-group for all $i=0,1,2,\dots$.
 See \cite{St}. 
  
   

 Let $\Gamma=\pi_{1}M$ where $M$ is an orientable closed 3-manifold, and let $\{\Gamma_{i}\}$
 be its mod $p$ lower central series.  Let $r={\rm rank} (\Gamma/\Gamma_{1})$.   The following result
 appears as Lemma 1.3 in \cite{SW}:
 
 
  \begin{theo}[Shalen and Wagreich] $ {\rm rank}(\Gamma_{1}/ \Gamma_{2}) \geq  \dbinom{r}{2} $.
\end{theo}

\begin{proof}  See \cite{SW} or \cite{La}.
%I have to learn any of the proofs for example in Sahlen and Wagreich they use the Stallings exact sequence
\end{proof}

We now give a geometric interpretation of this theorem in terms of first Betti numbers.  We start by noting that 
\begin{prop}
  $\Gamma/\Gamma_{1} \cong H_{1}(M;\F_{p}).$
  \end{prop}

\begin{proof}  This will follow from showing that $\Gamma_{1}$ is the kernel of the projection $\psi_{p}:\Gamma \rightarrow H_{1}(M;\F_{p})$.  First, note that $\psi_{p}$ is the composition
 \[ \Gamma \longrightarrow H_{1}(M;\Z) \longrightarrow H_{1}(M;\F_{p}),\] and the image of $\Gamma_{1}=\Gamma\#\Gamma$ by the first
map in the composition is $p H_{1}(M;\Z)$ which is the kernel of the second map.  Thus $\Gamma_{1}\subset {\rm Ker}(\psi_{p})$. On the other hand, any element $\gamma\in  {\rm Ker}(\psi_{p})$ must belong to a coset of the form $b^{p}[\Gamma ,\Gamma ]$, and thus $\gamma\in\Gamma_{1}$.
\end{proof}

Let $\tilde{M}_{p}\rightarrow M$ be the finite abelian cover associated to the projection
$\psi_{p}: \Gamma\rightarrow H_{1}(M;\F_{p})$.
We have $\pi_{1}\tilde{M}_{p}\cong \Gamma_{1}$ and
\[ {\rm Deck}(\tilde{M}_{p}/M)\cong \Gamma/\Gamma_{1} \cong H_{1}(M;\F_{p}) \cong (\F_{p})^{r}. \]  

Recall the following notation that was used in the introduction to this chapter
\[ d_{p}(N):= {\rm rank} \; (H_{1}(N;\F_{p} )).\]
When $N=M$ we have $d_{p}(M)=r$.  We have the following Corollary to the Theorem of Shalen and Wagreich: 

\begin{coro} $d_{p}(\tilde{M}_{p})  \geq \dbinom{r}{2}  $.
\end{coro}

\begin{proof} We will show that $d_{p}(\tilde{M}_{p})  \geq {\rm rank} (\Gamma_{1}/\Gamma_{2} )$.  Applying the analysis
of the previous paragraphs 
to $\tilde{M}_{p} $ in place of $M$, we know that 
\[d_{p}(\tilde{M}_{p}) ={\rm rank} (H_{1}(\tilde{M}_{p};\F_{p})) ={\rm rank} (  \Gamma_{1} /(\Gamma_{1}\#\Gamma_{1})).\]
Now 
\[ \Gamma_{1} /(\Gamma_{1}\#\Gamma_{1}) = \Gamma_{1}/[\Gamma_{1},\Gamma_{1}](\Gamma_{1})^{p} .\]
But $\Gamma_{2} = \Gamma \# \Gamma_{1} = [\Gamma_{1},\Gamma](\Gamma_{1})^{p}$ so that
\[\Gamma_{1}/\Gamma_{2}=\Gamma_{1}/[\Gamma,\Gamma_{1}](\Gamma_{1})^{p}.\]
Thus $\Gamma_{1}/\Gamma_{2}$ is a quotient of $ \Gamma_{1} /(\Gamma_{1}\#\Gamma_{1})$ and 
the Corollary follows.
\end{proof}

 
 
% We have  $\Gamma_{2} = \Gamma \# \Gamma_{1} = [\Gamma_{1},\Gamma](\Gamma_{1})^{p}$  \begin{color}{blue} leave some spaces \end{color}
% $\Gamma_{1}/\Gamma_{2}=\Gamma_{1}/[\Gamma,\Gamma_{1}](\Gamma_{1})^{p}$. 
 
% Let $r$ be the rank of $(\Gamma/ \Gamma_{1})$ as a ?.


 
 %Let $M$ be a closed orientable 3-manifold with $b_{1}M=n$, $p$ a 
 %prime number and consider the epimorphism \[ \psi_{p} :\pi_{1}M\rightarrow H_{1}(M;\Z/p) \] defined as the composition 
%\[  \pi_{1}M\stackrel{ \text{\tiny abelianization}}{\longrightarrow} H_{1}M\stackrel{\text{\tiny tensor p}}{\longrightarrow} H_{1}(M,\Z/p). \]
% \begin{color}{blue}$H_{1}(M,\Z/ \Z p)$ ?  \end{color}

%Let $\tilde{M_{p}}$ be the finite cyclic cover indexed by Ker$( \psi_{p})$ with deck group an elementary abelian p-group.
 

%Let $d_{p}\tilde{M}:=$rk$(H_{1}(\tilde{M}; \Z/p))$ then with this notation 
%\[ r=d_{p}M={\rm rk}(\Gamma/ \Gamma_{1})={\rm rk}(H_{1}(M; \Z/p))\]

%\begin{coro} $d_{p}\tilde{M}  \geq \dbinom{r}{2}  $.
%\end{coro}

%\begin{proof}

%$\Gamma_{1}/\Gamma_{2}=\Gamma_{1}/[\Gamma,\Gamma_{1}](\Gamma_{1})^{p}$ while, 
 
% $H_{1}(\tilde{M_{p}}; \Z/p)= \Gamma_{1}/[\Gamma_{1},\Gamma_{1}](\Gamma_{1})^{p}$
 %This is the reason why in the definition of the Alexander invariant we consider the second commutator subgroup

%N.B. $\Gamma_{2}\ge [\Gamma_{1},\Gamma_{1}](\Gamma_{1})^{p}$ then 
%$H_{1}(\tilde{M}; \Z/p)$ has $\Gamma_{1}/ \Gamma_{2}$ as a quotient.  \begin{color}{blue} I have to prove this algebra note \end{color}
 
% So if we can find a lower bound of $rank (\Gamma_{1}/ \Gamma_{2})$ then we get a lower bound on rank$(H_{1}(\tilde{M}; \Z/p))$.

%\end{proof}


\subsection{Closed 3-manifolds with $\Delta= 1$}\label{b1grt4sec}

We first recall the following result which relates the order of torsion in the homology of finite abelian covers to values of the Alexander polynomial:

 

\begin{theo}\label{TorFinAbCov}  Let $M'\rightarrow M$ be a finite abelian cover lying below the universal
free abelian cover $\widehat{M}\rightarrow M$, with deck group
$\F_{p_{1}}\oplus \cdots \oplus \F_{p_{n}}$, where the $p_{i}$ are primes.  
Assume that $\Delta_{M}(t_{1},\dots ,t_{n})$ has no zero of the form $(\rho_{1}^{e_{1}},\dots ,\rho_{n}^{e_{n}}) $ 
where $\rho_{i}$ is a $p_{i}$th root of unity.  Then
\[ | {\rm Tor}(H_{1}M' ) | = \Big| \prod   \Delta_{M}(\rho_{1}^{e_{1}},\dots ,\rho_{n}^{e_{n}}) \Big|  \]
where the product is over all $(e_{1},\dots ,e_{n})$ with $0\leq e_{i}<p_{i}$.
\end{theo}

Theorem \ref{TorFinAbCov} was first proved in the case of knot complements by Fox (see \cite{Gordon}) and stated in the above generality by Turaev (\cite{Turaev}, page 136) but he only provides a proof for cyclic covers.  
A complete proof can be found in \cite{me}.  
%Although the result for finite abelian covers was known to Turaev (see page 136 of \cite{Turaev}), his paper contains only a proof of  the cyclic case, and we do not know of a reference where the proof of
%the result for finite abelian covers is given.  Since we require this general version, we give a proof of it in Chapter \ref{OrdTor}.
We have immediately part a. of Fact 2:

\begin{coro}\label{notor}  Let $M$ be a closed 3-manifold with $\Delta_{M} =1$ and let $M'\longrightarrow M$
be a finite abelian cover.  Then
\[ {\rm Tor}(H_{1}M') =1.\]
In particular, taking $M=M'$,  ${\rm Tor}(H_{1}M) =1$.
\end{coro}

To prove part b. of Fact 2, we will need a formula of E. Hironaka \cite{Hiro}, which we describe in our setting.  Let $M'\rightarrow M$
be a finite cover and assume that $\Delta_{M}=1$.  Let us denote
\begin{itemize}
\item $D=$ the deck group of $M'\rightarrow M$.
\item $\Gamma=  \pi_{1}M$.
\item $\alpha: \Gamma \rightarrow D$ the projection.
\end{itemize}

For any group $G$, the character group is denoted 
\[ \hat{G}={\rm Hom}_{\rm cont}(G,\C^{\ast} )\] 
where ${\rm Hom}_{\rm cont}$ means the group of continuous homomorphisms.  Write $\hat{1}$ for the trivial character. 
We recall that $\hat{G}$ is a topological group. 
In our case, $G=\Gamma$ or $D$, which are discrete groups, so
continuous homomorphisms are just homomorphisms.   
Since $\alpha: \Gamma \rightarrow D$ is an epimorphism, there is an induced inclusion
\[ \hat{\alpha} : \hat{D} \hookrightarrow \hat{\Gamma}. \]

Let $\chi \in \hat{\Gamma}$ and $\Lambda = \Z [t_{1}^{\pm 1},\dots,t_{r}^{\pm 1}]$.  Note that $\chi$ induces a homomorphism of $\Gamma^{\rm ab}\cong\Z^{r}$ (the last isomorphism is by Corollary \ref{notor}).  We may then extend $\chi$ linearly to a ring homomorphism $\chi :\Lambda \rightarrow\C$.

We now describe the formula of E. Hironaka, following \cite{Hiro}.  Before doing so, we remark that the definition of the Alexander polynomial used in \cite{Hiro} is the one formulated using the 
 {\it relative Alexander module} \[ A_{M}^{\rm rel}=H_{1}(\widehat{M},\hat{x}), \quad \hat{x}=p^{-1}(x), x\in M\]
(see \cite{Mc}).   If $P(t_{1},\dots ,t_{r})  $  is a presentation matrix of  $A_{M}^{\rm rel}$, the Alexander polynomial is defined in this setting to be a generator
of the smallest principal ideal containing the ideal generated by the $(r-1)$ minors of $P(t_{1},\dots ,t_{r})$.  A proof of the equivalence of the relative homology definition with the absolute
homology definition can be found in \cite{me} (the equivalence is in fact implicit in Theorem 2.7 of \cite{Ma} as well as Theorem 16.5 of \cite{Turaevbook}).

Now given  $\chi \in \hat{\Gamma}$ let  $P(\chi )$ denote the matrix with complex entries
obtained by evaluating each entry of $P(t_{1},\dots ,t_{r})$ at $\chi$.  
For each $i$, define
\[  V_{i} = \left\{ \chi\in \hat{\Gamma}|\;  {\rm rank}(P(\chi ))<r-i  \right\}.\]
Then Hironaka's formula (see \cite{Hiro},  page 16,  Proposition 2.5.6.) says that 
\[b_{1}M' = b_{1}M + \sum_{i=1}^{r-1}  |  \hat{\alpha} (\hat{D} \setminus\hat{1}) \cap V_{i}  |. \]
 










  %Before doing this, let us take a look at the three examples known so far of such $M$.
%When $b_{1}M=1$ or $3$, this property is trivial since the fundamental groups are abelian.  For example, 
%\begin{itemize}
%\item When $M=\SI^{1}\times\SI^{2}$, then every finite cyclic cover has $\pi_1\cong\Z$, and therefore first Betti number $1$.
%\item Similarly when $M=\T^{3}$, every finite cyclic cover has $\pi_{1}\cong\Z^{3}$ and so has first Betti number $3$.
%\end{itemize}
%In the above examples, note that the universal abelian cover is the universal cover, which may explain why this fact is trivial in these cases.

%The case of the $M_{W}$ = 0-surgery on the Whitehead link complement is more interesting, and will motivate the proof of the Lemma
%below.  We recall that the universal abelian cover $\tilde{M}_{W}$ was constructed by gluing together $X$-pieces in a grid pattern
%to obtain the 3-manifold with boundary $\tilde{H}_{W}$, and to that we glued a ``trivial" grid along the boundary components.  We saw
%that a presentation of the Alexander polynomial was given by 
%\[ (s-1)\alpha = 0 = (t-1)\alpha \]
%where $\alpha\in H_{1}\tilde{M}$ is the curve circling the tubular neighborhood of the surgery core and $s,t$ are the generators
%of the deck group of $\tilde{M}_{W}\rightarrow M_{W}$.
%of the group ring $\Lambda =\Z[s^{\pm 1}, t^{\pm 1}]$.  

%We construct a cyclic cover $\tilde{M}_{m}\rightarrow M_{W}$ whose deck group is the image of $t$ by the covering $\tilde{M}_{W}\rightarrow
%\tilde{M}_{m} $.  To do this, we take the $X$-piece used to construct $\tilde{H}_{W}$ and glue back $x_{-}$ and $x_{+}$, so that now there
%are only two boundaries $y_{\pm}$ along which we may glue.  We take this modified (surgered) building block and glue end-to-end $m$ copies to obtain the 3-manifold with boundary $\tilde{H}_{m}$.  To this we glue a trivial version $\tilde{N}_{m}$ in which no surgery has been performed
%(analogous to the trivial infinite grid) to obtain $\tilde{M}_{m}$.   The homology $H_{1}\tilde{M}_{m}$ has $m+1$ generators $\alpha, t\alpha\dots ,t^{m-1}\alpha$ plus the lift of $x$.  However the relation $(t-1)\alpha=0$ descends to $\tilde{M}_{m}$ and shows that in homology
%the classes $\alpha, t\alpha\dots ,t^{m-1}\alpha$ are equal.  Thus the first Betti number of  $\tilde{M}_{m}$ is $2$.


\begin{theo}\label{samebetti} Suppose that $\Delta_{M}=1$ and let $M'\rightarrow M$ be any finite abelian cover of $M$ with deck group $D=\F_{p_{1}}\oplus \cdots \oplus \F_{p_{k}}$,
$p_{1},\dots ,p_{k}$ primes.  Then $b_{1}M'=b_{1}M$.
\end{theo}

\begin{proof} We claim that $ |  \hat{\alpha} (\hat{D} \setminus\hat{1}) \cap V_i  |=0$ for all $i=1,\dots ,r-1$.  To do this, it is enough to show 
that 
\[  |  \hat{\alpha} (\hat{D} \setminus\hat{1}) \cap V_1  |=0,\] since $V_{1}\supset \cdots \supset V_{r-1}$.  So we must show that for every character $\chi \in  \hat{\alpha} (\hat{D} \setminus\hat{1})$, 
\[ {\rm rank}\; P(\chi )\geq r-1.\]   Let us suppose not, that there exists a $\chi$ with ${\rm rank}\; P(\chi)< r-1$.  Then every
$r-1$ minor of $P(\chi )$ is $0$.   But
$P(\chi )$ is obtained by evaluating each polynomial appearing in $P$ at 
\[ t_{1} = \rho_{1}^{e_{1}}, \dots,
t_{k} = \rho_{k}^{e_{k}},\] where $\rho_{j}=\exp (2\pi i/p_{j})$ and the exponents $e_{1}, ..., e_{k}$ depend on $\chi$.  This implies that 
\[\Delta( \rho_{1}^{e_{1}},\dots ,  \rho_{k}^{e_{k}}) =0,\] which contradicts
the fact that $\Delta=1$.  Indeed, the greatest common factor of the $(r-1)\times (r-1)$ minors of $P(\chi )$ is 
$1=\Delta( \rho_{1}^{e_{1}},\dots ,  \rho_{k}^{e_{k}})$, so the minors cannot all be $0$.  This contradicts our hypothesis, and
therefore $ {\rm rank}\; P(\chi )\geq r-1$.
\end{proof}



%Note that the results of the previous section imply that for any cyclic cover of prime $p$ degree
%$M'\rightarrow M$, we have $T' := {\rm Tor} (H_{1}(M';\Z ))$ is trivial.  We will need to extend this result
 %to $\tilde{M}_{p}$.   Let $\tilde{T}_{p} := {\rm Tor} (H_{1}(\tilde{M}_{p};\Z ))$.

%\begin{lemm} $\tilde{T}_{p} =1$.
%\end{lemm}

%\begin{proof}  We may write $\tilde{M}_{p}\longrightarrow M$ as a composition of cyclic covers
%\[  \tilde{M}_{p}=M'_{r}\longrightarrow M'_{r-1}\longrightarrow M'_{r-2}  \longrightarrow \cdots   \longrightarrow M'_{0}=M .\]
%It will be enough then to show that $\Delta_{M'_{r-i}}=1$ for each $i=1,\dots ,r$; and to show this, it is enough to show
%that $\Delta_{M'}=1$ for any cyclic cover $M'\rightarrow M$ of prime $p$ degree.  Note first that $\tilde{M'}$=
%the universal abelian cover of $M'$ is equal to $\tilde{M}$: this follows from Lemma \ref{samebetti}.  
%Denote by $D'={\rm Deck}(\tilde{M}/M')$ and by $D={\rm Deck}(\tilde{M}/M)$ so that $D'\subset D$.
%We may identify $D\cong\Z^{n}$ so that $D'=(p\Z)\times \Z^{n}$.  Then we have an isomorphism
%$D\rightarrow D'$, $g\mapsto g'$, which corresponds to the isomorphism of $\Z^{n}$ with
%$(p\Z)\times \Z^{n}$ given by multiplying the first coordinate by $p$.
%Let $x'$ be a base point for $M'$ lying over the base point $x\in M$.
%Then the Alexander module for $M'$ is $A'=H_{1}(\tilde{M}, \tilde{x}'; \Z )$, where $\tilde{x}'$
%is the pre-image of $x'$ in $\tilde{M}$, 
%Then the Alexander module $A'$ for $M'$ is $H_{1}\tilde{M}$ which is
%a module over $\Lambda' = \Z [D']$.  As a set, this coincides with $A$ the Alexander module
%for $M$, except that $A$ is a module over $\Lambda = \Z [D]$. 
%On the other hand, the Alexander module for $M$ is $A=H_{1}(\tilde{M}, \tilde{x}; \Z )$, a module over $\Lambda = \Z [D]$.
%The isomorphism $D\cong D'$ allows us to consider
%$A'$ as a $\Lambda$ module as well.  Then we want to show that $A\cong A'$ as $\Lambda$ modules.

%We want to show that as $\Lambda$ modules
%$A'$ is isomorphic to $A$.  We do this as follows.  First, we choose an identification of $\tilde{x}$
%with $\Z^{n}\cong D$, so that $\tilde{x}'$ is identified with $(p\Z )\times\Z^{n-1}$.  
%As a $\Lambda$ module, $A$ is generated by connected segments $\gamma$ based at the origin $0$.  Let $g\in D$
%such that $g\cdot 0 = \gamma (1)$. 
%Consider
%the map
%\[ \gamma\mapsto \gamma' = \gamma + g\cdot \gamma + \cdots +g^{p}\gamma . \]
%Then $\gamma'$ has endpoint belonging to $\tilde{p}'$ and so defines an element of $A'$.  We extend this
%map over generators linearly over $\Lambda$ taking into account the isomorphism $D\rightarrow D'$ defined above
%so that $g\cdot\gamma\mapsto g'\cdot\gamma'$.  This association defines the desired isomorphism of
%$\Lambda$-modules.  It follows that $\Delta_{M'}=1$.
%\end{proof}

%For instance we can study the polynomials wit trace zero and try to prove that they are symmetric.
%Generalize the Symmetry result to higher Betti number, meaning try...
%If we allow the condition ${\rm tr}(\lambda)=0$ in the Generalized Levine's Theorem,  we produce an $M'$ with $b_{1}M'=b_{1}M+1$, in which
 %we claim that
%\[  \Delta_{M'}(t_{1},\dots ,t_{n},1) = \lambda (t_{1},\dots ,t_{n})\cdot\Delta_{M} (t_{1},\dots ,t_{n}).  \]
%Thus $\lambda\cdot \Delta_{M} $ would be only a partial trace of $ \Delta_{M'}$.  In any event, 
%\[ {\rm tr} (\Delta_{M'})={\rm tr}(\lambda\cdot\Delta_{M})= {\rm tr}(\lambda)\cdot {\rm tr}(\Delta_{M}) =0.\]




%\begin{color}{blue} THE TABLE \end{color}













\begin{thebibliography}{00}


\bibitem[1]{me} K. Alcaraz, The Alexander Polynomial of Closed 3-manifolds. PhD thesis, University of Oxford (2012). 

\bibitem [2]{Alex} J.W. Alexander, Topological invariants of knots and links.  {\it Trans. AMS}
{\bf 30} (1928), 275--306. 


\bibitem [3] {Blan} R. C. Blanchfield, Intersection theory of manifolds with operators with applications to knot theory. {\it Ann. of Math.}, {\bf 65} (1957), 340--356.

%\bibitem [4]{Bur} G. Burde \& H. Zieschang, {\it Knots.} de Gruyter Studies in Mathematics {\bf 5},
%Walter de Gruyter, Berlin, 1985.

%\bibitem [5]{Fox} R. H. Fox,  A quick trip through knot theory. {\it Proceedings of The University of Georgia Institute}, (1961), 120--167.

\bibitem [4]{FoxI} R. H. Fox, Free differential calculus I.  {\it Annals of Math.}, {\bf 57} (1953), 547--560.


\bibitem [5]{Gordon} C. McA. Gordon, Some aspects of classical knot theory. {\it Lecture Notes in Math.}, {\bf 685} (1978), 1--60.


\bibitem [6]{GrHa} P. Griffiths \&  J. Harris, {\it Principles of Algebraic Geometry.} Wiley Classics Library, Wiley-Interscience, Hoboken, New Jersey, 1994.

%\bibitem[9]{Hill} J. Hillman, {\it Algebraic Invariants of Links}, Series on Knots and Everything, World Scientific, Hackensack, New Jersey, 2002.

\bibitem[7]{Hill1} J. Hillman, The Torres conditions are insufficient. {\it Math. Proc. Cambridge Philos. Soc.} {\bf 89} (1981), 19-22.

\bibitem[8]{Hiro} E. Hironaka, Alexander stratifications of character varieties. {\it Ann. de l'Inst. Four.}, {\bf 47 (2)} (1997), 555--583.

%\bibitem[12]{Ki} R. Kirby, A calculus for framed links in $\SI^{3}$. {\it Inventiones Mathematicae}, {\bf 45} (1978), 35--56.

\bibitem [9]{La} M. Lackenby, New lower bounds on subgroups growth and homology growth. {\it
Proc. London Math. Soc.}, {\bf 98 (2)} (2009), 271--297.

%\bibitem [14]{Lan} S. Lang, {\it Algebra.} Revised Third Edition, Graduate Texts in Mathematics {\bf 73},  Springer-Verlag, New York, 2005.

%\bibitem [15]{Le1} J. Levine, A characterization of knot polynomials. {\it Topology}, {\bf 4} (1965), 135--141.

\bibitem[10]{Le2} J. Levine, A method for generating link polynomials. {\it American Journal of Mathematics}, {\bf 89} (1967), 69--84.

%\bibitem [17]{Li} W. B. R. Lickorish, {\it An Introduction to Knot Theory}. Graduate Texts in Mathematics {\bf 175}, Springer-Verlag, New York, 1997.

\bibitem [11]{Ma} G. Massuyeau, A short introduction to the Alexander polynomial.  www-irma.u-strasbg.fr/\~{}massuyea/talks/Alex.pdf, 2008.

\bibitem [12]{Mc} C. McMullen, The Alexander polynomial of a 3-manifold and the Thurston norm on cohomology. {\it Ann. scient. de \'{E}c. Norm. Sup.} {\bf 35} (2002), 153--172.

%\bibitem [20]{Mi} J. Milnor, Infinite cyclic coverings. in {\it Collected Papers, Volume II: The Fundamental Group}, Publish or Perish, 71--89, 1994.

\bibitem [13]{Pl} Platt, M.L., Insufficiency of Torres' conditions for two-component classical links. {\it Trans. AMS}, {\bf 296} (1986), 125--136.

\bibitem [14]{Ro} D. Rolfsen, {\it Knots and Links}. AMS Chelsea, Providence, RI, 2003.


\bibitem [15]{SW} P. B. Shalen, P. Wagreich,  Growth rates, $Z_{p}$-homology, and volumes of hyperbolic 3-manifolds, Trans. Amer. Math. Soc. {\bf 331} (1992) 895--917.

\bibitem [16]{Sp} E. Spanier, {\it Algebraic Topology}.  Springer-Verlag, New York, 1994.

\bibitem [17]{St} J. Stallings, Homology and central series of groups {\it J. of  Algebra} {\bf 2} (1965) 170--181.
, 2:170?181, 1965.

\bibitem [18]{To} G. Torres,  On the Alexander polynomial. {\it Ann. of Math.}, {\bf 57} (1953), 57--89.

\bibitem [19]{Turaev0}  V. Turaev,  The Alexander polynomial of a three dimensional manifold. {\it Math. USSR Sb.}, {\bf 26} (1975), 313--329.

\bibitem[20]{Turaev} V. Turaev, Reidemeister torsion in knot theory. {\it Russian Math. Surveys}, {\bf 41:1} (1986), 119--182.

\bibitem[21]{Turaevbook} V. Turaev, {\it Torsions of 3-dimensionals manifolds}, volume 208 of {\it Progress in Mathematics}. Birkhauser Verlag, Basel, 2002.




\end{thebibliography}





\end{document}

1406.2042.extracted_bib
[1]{me K. Alcaraz, The Alexander Polynomial of Closed 3-manifolds. PhD thesis, University of Oxford (2012).
1406.2042.extracted_bib
[2]{Alex J.W. Alexander, Topological invariants of knots and links. {Trans. AMS {30 (1928), 275--306.
1406.2042.extracted_bib
[3] {Blan R. C. Blanchfield, Intersection theory of manifolds with operators with applications to knot theory. {Ann. of Math., {65 (1957), 340--356. %
1406.2042.extracted_bib
[4]{Bur G. Burde \& H. Zieschang, {Knots. de Gruyter Studies in Mathematics {5, %Walter de Gruyter, Berlin, 1985. %
1406.2042.extracted_bib
[5]{Fox R. H. Fox, A quick trip through knot theory. {Proceedings of The University of Georgia Institute, (1961), 120--167.
1406.2042.extracted_bib
[4]{FoxI R. H. Fox, Free differential calculus I. {Annals of Math., {57 (1953), 547--560.
1406.2042.extracted_bib
[5]{Gordon C. McA. Gordon, Some aspects of classical knot theory. {Lecture Notes in Math., {685 (1978), 1--60.
1406.2042.extracted_bib
[6]{GrHa P. Griffiths \& J. Harris, {Principles of Algebraic Geometry. Wiley Classics Library, Wiley-Interscience, Hoboken, New Jersey, 1994. %
1406.2042.extracted_bib
[9]{Hill J. Hillman, {Algebraic Invariants of Links, Series on Knots and Everything, World Scientific, Hackensack, New Jersey, 2002.
1406.2042.extracted_bib
[7]{Hill1 J. Hillman, The Torres conditions are insufficient. {Math. Proc. Cambridge Philos. Soc. {89 (1981), 19-22.
1406.2042.extracted_bib
[8]{Hiro E. Hironaka, Alexander stratifications of character varieties. {Ann. de l'Inst. Four., {47 (2) (1997), 555--583. %
1406.2042.extracted_bib
[12]{Ki R. Kirby, A calculus for framed links in $\SI^{3$. {Inventiones Mathematicae, {45 (1978), 35--56.
1406.2042.extracted_bib
[9]{La M. Lackenby, New lower bounds on subgroups growth and homology growth. {Proc. London Math. Soc., {98 (2) (2009), 271--297. %
1406.2042.extracted_bib
[14]{Lan S. Lang, {Algebra. Revised Third Edition, Graduate Texts in Mathematics {73, Springer-Verlag, New York, 2005. %
1406.2042.extracted_bib
[15]{Le1 J. Levine, A characterization of knot polynomials. {Topology, {4 (1965), 135--141.
1406.2042.extracted_bib
[10]{Le2 J. Levine, A method for generating link polynomials. {American Journal of Mathematics, {89 (1967), 69--84. %
1406.2042.extracted_bib
[17]{Li W. B. R. Lickorish, {An Introduction to Knot Theory. Graduate Texts in Mathematics {175, Springer-Verlag, New York, 1997.
1406.2042.extracted_bib
[11]{Ma G. Massuyeau, A short introduction to the Alexander polynomial. www-irma.u-strasbg.fr/\~{massuyea/talks/Alex.pdf, 2008.
1406.2042.extracted_bib
[12]{Mc C. McMullen, The Alexander polynomial of a 3-manifold and the Thurston norm on cohomology. {Ann. scient. de \'{Ec. Norm. Sup. {35 (2002), 153--172. %
1406.2042.extracted_bib
[20]{Mi J. Milnor, Infinite cyclic coverings. in {Collected Papers, Volume II: The Fundamental Group, Publish or Perish, 71--89, 1994.
1406.2042.extracted_bib
[13]{Pl Platt, M.L., Insufficiency of Torres' conditions for two-component classical links. {Trans. AMS, {296 (1986), 125--136.
1406.2042.extracted_bib
[14]{Ro D. Rolfsen, {Knots and Links. AMS Chelsea, Providence, RI, 2003.
1406.2042.extracted_bib
[15]{SW P. B. Shalen, P. Wagreich, Growth rates, $Z_{p$-homology, and volumes of hyperbolic 3-manifolds, Trans. Amer. Math. Soc. {331 (1992) 895--917.
1406.2042.extracted_bib
[16]{Sp E. Spanier, {Algebraic Topology. Springer-Verlag, New York, 1994.
1406.2042.extracted_bib
[17]{St J. Stallings, Homology and central series of groups {J. of Algebra {2 (1965) 170--181. , 2:170?181, 1965.
1406.2042.extracted_bib
[18]{To G. Torres, On the Alexander polynomial. {Ann. of Math., {57 (1953), 57--89.
1406.2042.extracted_bib
[19]{Turaev0 V. Turaev, The Alexander polynomial of a three dimensional manifold. {Math. USSR Sb., {26 (1975), 313--329.
1406.2042.extracted_bib
[20]{Turaev V. Turaev, Reidemeister torsion in knot theory. {Russian Math. Surveys, {41:1 (1986), 119--182.
1406.2042.extracted_bib
[21]{Turaevbook V. Turaev, {Torsions of 3-dimensionals manifolds, volume 208 of {Progress in Mathematics. Birkhauser Verlag, Basel, 2002.