lectures_on_knot_theory/lec_06_05.tex

\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}
\begin{figure}[h]
\fontsize{10}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.8\textwidth}{!}{\input{images/knot_complement.pdf_tex}}
\caption{A knot complement.}
\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. 
\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
$\Sigma_?(K) \rightarrow S^3$ ?????\\
$H_1(\Sigma_?(K), \mathbb{Z}) = h$\\
$H \times H \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}}$\\
...\\

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 $S_k$. 
\[
\frac{1}{\Delta} \sum_{j\in \mathbb{Z} t^{-j}}(F \cdot t^j d) \in \quot{\mathbb{Q}[t, t^{-1}]}{\mathbb{Z}[t, t^{-1}]}
\] 
\\
?????????????\\
\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}


\begin{definition}
The $\mathbb{Z}[t, t^{-1}]$ module $H_1(\widetilde{X})$ is called the Alexander module of 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 $R$.
\subsection*{Blanchfield pairing}
\section{balagan}

\begin{fact}[Milnor Singular Points of Complex Hypersurfaces]
\end{fact}
%\end{comment}
\noindent
An oriented knot is called negative amphichiral if the mirror image $m(K)$ of $K$ is equivalent the reverse knot of $K$: $K^r$. \\
\begin{problem}
Prove that if $K$ is negative amphichiral, then $K \# K = 0$ in 
$\mathscr{C}$.
%
%\\
%Hint: $ -K = m(K)^r = (K^r)^r = K$
\end{problem}
\begin{example}
Figure 8 knot is negative amphichiral. 
\end{example}
%
%
\begin{theorem}
Let $H_p$ be a $p$ - torsion part of $H$. There exists an orthogonal decomposition of $H_p$:
\[
H_p = H_{p, 1} \oplus  \dots \oplus H_{p, r_p}.
\]
$H_{p, i}$ is a cyclic module:
\[
H_{p, i} = \quot{\mathbb{Z}[t, t^{-1}]}{p^{k_i} \mathbb{Z} [t, t^{-1}]}
\]
\end{theorem}
\noindent
The proof is the same as over $\mathbb{Z}$.
\noindent
%Add NotePrintSaveCiteYour opinionEmailShare
%Saveliev, Nikolai

%Lectures on the Topology of 3-Manifolds
%An Introduction to the Casson Invariant

\begin{figure}[h]
\fontsize{10}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/ball_4_alpha_beta.pdf_tex}}
}
%\caption{Sketch for Fact %%\label{fig:concordance_m}
\end{figure}

\end{document}