% I don't have this first fragent in my notes 
Let $M$ be compact, oriented, connected four-dimensional manifold. If ${H_1(M, \mathbb{Z}) = 0}$ then there exists a 
bilinear form - the intersection form on $M$: 

\begin{center}
\begin{tikzcd}
[
column sep=tiny,
row sep=small,
ar symbol/.style = {draw=none,"\textstyle#1" description,sloped},
isomorphic/.style = {ar symbol={\cong}},
]
H_2(M, \mathbb{Z})&
\times & H_2(M, \mathbb{Z})
\longrightarrow &
\mathbb{Z}
\\
\ar[u,isomorphic] \mathbb{Z}^n  && &\\
\end{tikzcd}
\end{center}
\noindent
Let us consider a specific case: $M$ has a boundary $Y = \partial M$. 
Betti number $b_1(Y) = 0$, $H_1(Y, \mathbb{Z})$ is  finite.
Then the intersection form can be degenerated in the sense that: 
 \begin{align*}
H_2(M, \mathbb{Z})
\times  H_2(M, \mathbb{Z})
&\longrightarrow 
\mathbb{Z} \quad&
H_2(M, \mathbb{Z}) &\longrightarrow \Hom (H_2(M, \mathbb{Z}), \mathbb{Z})\\
(a, b) &\mapsto \mathbb{Z} \quad&
a &\mapsto (a, \_) \in  H_2(M, \mathbb{Z})
\end{align*}
has coker precisely $H_1(Y, \mathbb{Z})$. 
\\???????????????\\
% Here my notes begin: 
Let $K \subset S^3$  be a knot, $X = S^3 \setminus K$  a knot complement and
$\widetilde{X} \xrightarrow{\enspace \rho \enspace} X$  an infinite cyclic cover (universal abelian cover). 

%By Hurewicz theorem we know that:
%\begin{align*}
%\pi_1(X) \longrightarrow \quot{\pi_1(X)}{[\pi_1(X), \pi_1(X)]} = H_1(X, \mathbb{Z} ) \cong \mathbb{Z}
%\end{align*}
\noindent
$C_{*}(\widetilde{X})$ has a structure of a $\mathbb{Z}[t, t^{-1}] \cong  \mathbb{Z}[\mathbb{Z}]$ module. \\
Let $H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}])$ be the Alexander module of the knot $K$ with an intersection form: 
\begin{align*}
H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \times
H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}[t, t^{-1}]}
\end{align*}

\begin{fact}
\begin{align*}
&H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}])  \cong
\quot{\mathbb{Z}{[t, t^{-1}]}^n}{(tV - V^T)\mathbb{Z}[t, t^{-1}]^n}\;, \\
&\text{where $V$ is a Seifert matrix.}
\end{align*}
\end{fact}
\begin{fact}
\begin{align*}
H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \times
H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) &\longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}[t, t^{-1}]}\\
(\alpha, \beta) \quad &\mapsto \alpha^{-1}{(t -1)(tV - V^T)}^{-1}\beta
\end{align*}
\end{fact}
\noindent
Note that $\mathbb{Z}[t, t^{-1}]$ is not PID. 
Therefore we don't have primary decomposition of this module.  
We can simplify this problem by replacing $\mathbb{Z}$ by $\mathbb{R}$. We lose some date by doing this transition, but we can 
\begin{align*}
\xi \in S^1 \setminus \{ \pm 1\} 
&\quad
p_{\xi} = 
(t - \xi)(t - \xi^{-1}) t^{-1}
\\
\xi \in \mathbb{R} \setminus \{ \pm 1\}
&\quad
q_{\xi} = (t - \xi)(t - \xi^{-1}) t^{-1}
\\
\xi \notin \mathbb{R} \cup S^1 
&\quad
q_{\xi} = (t - \xi)(t - \overbar{\xi})(t - \xi^{-1})
(t - \overbar{\xi}^{-1}) t^{-2}
\end{align*}
Let $\Lambda = \mathbb{R}[t, t^{-1}]$. Then: 
\begin{align*}
H_1(\widetilde{X}, \Lambda) \cong  \bigoplus_{\substack{\xi \in S^1 \setminus \{\pm 1 \}\\ k\geq 0}}
( \quot{\Lambda}{p_{\xi}^k })^{n_k, \xi}
\oplus
\bigoplus_{\substack{\xi \notin S^1 \\ l\geq 0}}
(\quot{\Lambda}{q_{\xi}^l})^{n_l, \xi}&
\end{align*}
We can make this composition orthogonal with respect to the Blanchfield paring.
\vspace{0.5cm}\\
Historical remark:
\begin{itemize}
\item John Milnor, \textit{On isometries of inner product spaces}, 1969,
\item Walter Neumann, \textit{Invariants of plane curve singularities} 
%in: Knots, braids and singulari- ties (Plans-sur-Bex, 1982), 223–232, Monogr. Enseign. Math., 31, Enseignement Math., Geneva
, 1983,
\item András Némethi, \textit{The real Seifert form and the spectral pairs of isolated hypersurfaceenumerate singularities}, 1995,
%Compositio Mathematica,  Volume 98 (1995) no. 1,  p. 23-41
\item Maciej Borodzik, Stefan Friedl
\textit{The unknotting number and classical invariants II}, 2014.
\end{itemize}
\vspace{0.5cm}
Let $p = p_{\xi}$, $k\geq 0$.
\begin{align*}
\quot{\Lambda}{p^k \Lambda} \times
\quot{\Lambda}{p^k \Lambda} &\longrightarrow \quot{\mathbb{Q}(t)}{\Lambda}\\
(1, 1) &\mapsto \kappa\\
\text{Now: } (p^k \cdot 1, 1) &\mapsto 0\\
p^k \kappa = 0 &\in \quot{\mathbb{Q}(t)}{\Lambda}\\
\text{therfore } p^k \kappa &\in \Lambda\\
\text{we have } (1, 1) &\mapsto \frac{h}{p^k}\\
\end{align*}
$h$ is not uniquely defined: $h \rightarrow h + g p^k$ doesn't affect paring. \\
Let $h = p^k \kappa$.
\begin{example}
\begin{align*}
\phi_0 ((1, 1))=\frac{+1}{p}\\
\phi_1 ((1, 1)) = \frac{-1}{p} 
\end{align*}
$\phi_0$ and $\phi_1$ are not isomorphic. 
\end{example}
\begin{proof}
Let $\Phi: 
\quot{\Lambda}{p^k \Lambda} \longrightarrow
 \quot{\Lambda}{p^k \Lambda}$
 be an isomorphism. \\
 Let: $\Phi(1) = g \in \lambda$
 \begin{align*}
\quot{\Lambda}{p^k \Lambda} 
\xrightarrow{\enspace \Phi \enspace}&
 \quot{\Lambda}{p^k \Lambda}\\
 \phi_0((1, 1)) = \frac{1}{p^k} \qquad&\qquad
  \phi_1((g, g)) = \frac{1}{p^k} \quad \text{($\Phi$ is an isometry).}
 \end{align*}
 Suppose for the paring $\phi_1((g, g))=\frac{1}{p^k}$ we have $\phi_1((1, 1)) = \frac{-1}{p^k}$. Then:
 \begin{align*}
\frac{-g\overbar{g}}{p^k} = \frac{1}{p^k} &\in \quot{\mathbb{Q}(t)}{\Lambda}\\
\frac{-g\overbar{g}}{p^k} - \frac{1}{p^k} &\in \Lambda \\
-g\overbar{g} &\equiv 1\pmod{p} \text{ in } \Lambda\\
-g\overbar{g} - 1 &= p^k \omega \text{ for some } \omega \in \Lambda\\
\text{evalueting at $\xi$: }\\
\overbrace{-g(\xi)g(\xi^{-1})}^{>0} - 1 = 0 \quad \contradiction
\end{align*}
\end{proof}
????????????????????\\
\begin{align*}
g &= \sum{g_i t^i}\\
\overbar{g} &= \sum{g_i t^{-i}}\\
\overbar{g}(\xi) &= \sum g_i \xi^i \quad \xi \in S^1\\
\overbar{g}(\xi) &=\overbar{g(\xi)}
\end{align*}
Suppose $g = (t - \xi)^{\alpha} g^{\prime}$. Then $(t - \xi)^{k - \alpha}$ goes to $0$ in $\quot{\Lambda}{p^k \Lambda}$.
\begin{theorem}
Every sesquilinear non-degenerate pairing
\begin{align*}
\quot{\Lambda}{p^k} \times \quot{\Lambda}{p}
\longrightarrow \frac{h}{p^k}
\end{align*}
is isomorphic either to the pairing wit $h=1$ or to the paring with $h=-1$ depending on sign of $h(\xi)$ (which is a real number).
\end{theorem}
\begin{proof}
There are two steps of the proof:
\begin{enumerate}
\item
Reduce to the case when $h$ has a constant sign on $S^1$.
\item
Prove in the case, when $h$ has a constant sign on $S^1$.
\end{enumerate}
\begin{lemma}
If $P$ is a symmetric polynomial such that $P(\eta)\geq 0$ for all $\eta \in S^1$, then $P$ can be written as a product $P = g \overbar{g}$ for some polynomial $g$. 
\end{lemma}
\begin{proof}[Sketch of proof]: 
Induction over $\deg P$.\\
Let $\zeta \notin S^1$ be a root of $P$, $P \in \mathbb{R}[t, t^{-1}]$. Assume $\zeta \notin \mathbb{R}$.  We know that polynomial $P$ is divisible by
$(t - \zeta)$, $(t - \overbar{\zeta})$, $(t^{-1} - \zeta)$ and $(t^{-1} - \overbar{\zeta})$.
Therefore:
\begin{align*}
&P^{\prime} = \frac{P}{(t - \zeta)(t - \overbar{\zeta})(t^{-1} - \zeta)(t^{-1} - \overbar{\zeta})}\\
&P^{\prime} = g^{\prime}\overbar{g}
\end{align*}
We set  $g = g^{\prime}(t - \zeta)(t - \overbar{\zeta})$ and
$P = g \overbar{g}$. Suppose $\zeta \in S^1$. Then $(t - \zeta)^2 \vert P$ (at least - otherwise it would change sign). Therefore:
\begin{align*}
&P^{\prime} = \frac{P}{(t - \zeta)^2(t^{-1} - \zeta)^2}\\
&g = (t - \zeta)(t^{-1} - \zeta) g^{\prime} \quad \text{etc.}
\end{align*}
The map $(1, 1) \mapsto \frac{h}{p^k} = \frac{g\overbar{g}h}{p^k}$ is isometric whenever $g$ is coprime with $P$.
\end{proof}
\begin{lemma}\label{L:coprime polynomials}
Suppose $A$ and $B$ are two symmetric polynomials that are coprime and that $\forall z \in S^1$ either $A(z) > 0$ or $B(z) > 0$. Then there exist
 symmetric polynomials $P$, $Q$ such that 
 $P(z), Q(z) > 0$ for $z \in S^1$ and $PA + QB \equiv 1$.
\end{lemma}
\begin{proof}[Idea of proof]
For any $z$ find an interval $(a_z, b_z)$ such that if $P(z) \in (a_z, b_z)$ and $P(z)A(z) + Q(z)B(z) = 1$, then $Q(z) > 0$, $x(z) = \frac{az + bz}{i}$ is a continues function on $S^1$ approximating $z$ by a polynomial .
\\??????????????????????????\\
\begin{flalign*}
(1, 1) \mapsto \frac{h}{p^k} \mapsto \frac{g\overbar{g}h}{p^k}&\\
g\overbar{g} h + p^k\omega = 1&
\end{flalign*}
Apply Lemma \ref{L:coprime polynomials} for $A=h$, $B=p^{2k}$. Then, if the assumptions are satisfied, 
\begin{align*}
Ph + Qp^{2k} = 1\\
p>0 \Rightarrow p = g \overbar{g}\\
p = (t - \xi)(t - \overbar{\xi})t^{-1}\\
\text{so } p \geq 0 \text{ on } S^1\\
p(t) = 0 \Leftrightarrow 
t = \xi or t = \overbar{\xi}\\
h(\xi) > 0\\
h(\overbar{\xi})>0\\
g\overbar{g}h + Qp^{2k} = 1\\
g\overbar{g}h \equiv 1 \mod{p^{2k}}\\
g\overbar{g} \equiv 1 \mod{p^k}
\end{align*}
???????????????????????????????\\
If $P$ has no roots on $S^1$ then $B(z) > 0$ for all $z$, so the assumptions of Lemma \ref{L:coprime polynomials} are satisfied no matter what $A$ is. 
\end{proof}
?????????????????\\
\begin{align*}
\quot{\Lambda}{p_{\xi}^k} \times
\quot{\Lambda}{p_{\xi}^k} &\longrightarrow
\frac{\epsilon}{p_{\xi}^k}, \quad \xi \in S^1 \setminus\{\pm 1\}\\
\quot{\Lambda}{q_{\xi}^k} \times
\quot{\Lambda}{q_{\xi}^k} &\longrightarrow
\frac{1}{q_{\xi}^k}, \quad \xi \notin S^1\\
\end{align*}
??????????????????? 1 ?? epsilon?\\
\begin{theorem}[Matumoto, Borodzik-Conway-Politarczyk]
Let $K$ be a knot, 
\begin{align*}
H_1(\widetilde{X}, \Lambda) \times
H_1(\widetilde{X}, \Lambda) 
= \bigoplus_{\substack{k, \xi, \epsilon\\ \xi \in S^1}}
(\quot{\Lambda}{p_{\xi}^k}, \epsilon)^{n_k, \xi, \epsilon} \oplus \bigoplus_{k, \eta}
(\quot{\Lambda}{p_{\xi}^k})^{m_k} \text{ and} \\
 \delta_{\sigma}(\xi) = \lim_{\varepsilon \rightarrow 0^{+}} 
\sigma(e^{2\pi i \varepsilon} \xi)
- \sigma(e^{-2\pi i \varepsilon} \xi),\\
\text{then } 
\sigma_j(\xi) = \sigma(\xi) - \frac{1}{2} \lim_{\varepsilon \rightarrow 0} 
\sigma(e^{2\pi i \varepsilon}\xi)
+ \sigma(e^{-2 \pi i \varepsilon}\xi)
\end{align*}
The jump at $\xi$ is equal to 
$2\sum\limits_{k_i \odd} \epsilon_i$.\\
The peak of the signature function is equal to ${\sum\limits_{k_i \even}}{\epsilon_i}$.
\\
?????????????????
\\
$(\eta_{k, \xi_l^{+}} -\eta_{k, \xi_l^{-}}$
% Livingston Pacific Jurnal of M. 2012
\end{theorem}
\end{proof}



