\begin{definition}
Let $X$ be a knot complement. 
Then $H_1(X, \mathbb{Z}) \cong \mathbb{Z}$ and there exists an epimorphism 
$\pi_1(X) \overset{\phi}\twoheadrightarrow \mathbb{Z}$.\\
The infinite cyclic cover of a knot complement  $X$ is the cover associated with the epimorphism $\phi$. 
\[
\widetilde{X} \longtwoheadrightarrow X
\] 
\end{definition}
%Rolfsen, bachalor thesis of Kamila
\begin{figure}[h]
\fontsize{10}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{1\textwidth}{!}{\input{images/covering.pdf_tex}}
\caption{Infinite cyclic cover of a knot complement.}
\label{fig:covering}
}
\end{figure}
\noindent
\subsection{Double branched cover.}
Let $K \subset S^3$ be a knot and $\Sigma$
its Seifert surface.
Let us consider a knot complement $S^3 \setminus N(K)$. 
\begin{figure}[h]
\fontsize{10}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.8\textwidth}{!}{\input{images/knot_complement.pdf_tex}}
\caption{The double cover of the $3$-sphere branched over a knot $K$.}
\label{fig:complement}
}
\end{figure}
\noindent
Formal sums $\sum \phi_i(t) a_i + \sum \phi_j(t)\alpha_j$ \\
finitely generated as a $\mathbb{Z}[t, t^{-1}]$ module. 
\\
Let $v_{ij} = \Lk(a_i, a_j^+)$. Then
$V = \{ v_ij\}_{i, j = 1}^n$ is the Seifert matrix associated to the surface $\Sigma$ and the basis $a_1, \dots, a_n$. Therefore $a_k^+ = \sum_{j} v_{jk} \alpha_j$. Then
$\Lk(a_i, a_k^+)= \Lk(a_k^+, a_i) = \sum_j v_{jk} \Lk(\alpha_j, a_i) = v_{ik}$. 
We also notice that $\Lk(a_i, a_j^-) = \Lk(a_i^+, a_j)= v_{ij}$ and 
$a_j^- = \sum_k v_{kj} t^{-1} \alpha_j$.
\\
\noindent
The homology of $\widetilde{X}$ is generated by $a_1, \dots, a_n$ and relations. 
Let now $H = H_1(\widetilde{X})$. Can we define a paring? \\
Let $c, d \in H(\widetilde{X})$ (see Figure \ref{fig:covering_pairing}), $\Delta$ an Alexander polynomial. We know that $\Delta c = 0 \in H_1(\widetilde{X})$ (Alexander polynomial annihilates all possible elements). Let consider a surface $F$ such that $\partial F = c$.  Now consider intersection points $F \cdot d$. This points can exist in any $N_k$ or $S_k$. 
\[
\frac{1}{\Delta} \sum_{j\in \mathbb{Z} t^{-j}}(F \cdot t^j d) \in \quot{\mathbb{Q}[t, t^{-1}]}{\mathbb{Z}[t, t^{-1}]}
\] 
\\
?????????????\\
There is at least one paper where the structure of (Alexander module?) is calculated from a specific knot (?minimal number of generators?) 
\\
C. Kearton, S. M. J. Wilson
\\
\begin{fact}
Let $A$ be a matrix over principal ideal domain $R$.  Than there exist matrices $C$, $D$ and $E$ such that $A = CDE$, 
\[D = \begin{bmatrix}
d_1 & 0 & \cdots & \cdots & 0 \\
0 & d_2 & 0 & \cdots & 0 \\
\sdots &  & \ddots &  & \sdots &  \\
0 & \cdots & 0 &  d_{n-1} & 0\\
0 & \cdots & \cdots & 0 & d_n
\end{bmatrix},\] 
where $d_{i + 1} | d_i$, and matrices 
$C$ and $E$ are invertible over $R$.\\
$D$ is called a Smith normal form of the matrix $A$.
\end{fact}

\begin{definition}
The $\mathbb{Z}[t, t^{-1}]$ module $H_1(\widetilde{X})$ is called the Alexander module of a knot $K$.
\end{definition}
\noindent
Let $R$ be a PID, $M$ a finitely generated $R$ module. Let us consider
\[
R^k \overset{A} \longrightarrow R^n \longtwoheadrightarrow M,
\] 
where $A$ is a $k \times n$ matrix, assume $k\ge n$. The order of $M$ is the $\gcd$ of all determinants of the $n \times n$ minors of $A$. If $k = n$ then $\ord M = \det A$. 
\begin{theorem}
Order of $M$ doesn't depend on $A$. 
\end{theorem} 
\noindent
For knots the order of the Alexander module is the Alexander polynomial.
\begin{theorem}
\[
\forall x \in M: (\ord M) x = 0.
\]
\end{theorem}
\noindent
$M$ is well defined up to a unit in $R$.
\\
??????????????????\\
General picture : $K$, $X$ knot complement... 
\begin{eqnarray*}
H_1( X, \mathbb{Z}) = \mathbb{Z} \\
H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \\
\pi_1(X)
\end{eqnarray*}
\begin{definition}
The Nakanishi index of a knot is the minimal number of generators of $H_1(\widetilde{X})$.
\end{definition}
%see Maciej page
\noindent
Remark about notation: sometimes one writes $H_1(X; \mathbb{Z}[t, t^{-1}])$ (what is also notation for twisted homology) instead of $H_1(\widetilde{X})$. 
\\
?????????????????????
\\
\noindent
$\Sigma_?(K) \rightarrow S^3$ ?????\\
$H_1(\Sigma_?(K), \mathbb{Z}) = h$\\
$H \times H \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}}$\\
...\\
\begin{figure}[h]
\fontsize{10}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{1\textwidth}{!}{\input{images/covering_pairing.pdf_tex}}
\caption{$c, d \in H_1(\widetilde{X})$.}
\label{fig:covering_pairing}
}
\end{figure}

\subsection*{Blanchfield pairing}