lectures_on_knot_theory/lectures_on_knot_theory.tex

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\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\rank}{rank}
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\DeclareMathOperator{\Lk}{lk}


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\begin{document}
\tableofcontents
%\newpage
%\input{myNotes}

\section{Basic definitions \hfill\DTMdate{2019-02-25}}
\begin{definition}
A knot $K$ in $S^3$ is a smooth (PL - smooth) embedding of a circle $S^1$ in $S^3$:
\begin{align*}
\varphi:  S^1 \hookrightarrow  S^3
\end{align*}   
\end{definition}
\noindent
Usually we think about a knot as an image of an embedding: $K = \varphi(S^1)$.

\begin{example}
\begin{itemize}
\item 
Knots:
\includegraphics[width=0.08\textwidth]{unknot.png} (unknot),
\includegraphics[width=0.08\textwidth]{trefoil.png} (trefoil).
\item 
Not knots: 
\includegraphics[width=0.12\textwidth]{not_injective_knot.png}
(it is not an injection),
\includegraphics[width=0.08\textwidth]{not_smooth_knot.png}
(it is not smooth).
\end{itemize}
\end{example}
\begin{definition}
%\hfill\\
Two knots $K_0 = \varphi_0(S^1)$, $K_1 = \varphi_1(S^1)$ are equivalent if the embeddings $\varphi_0$ and $\varphi_1$ are isotopic, that is there exists a continues function 
\begin{align*}
&\Phi: S^1 \times  [0, 1] \hookrightarrow S^3 \\
&\Phi(x, t) = \Phi_t(x)
\end{align*}
 such that $\Phi_t$ is an embedding for any $t \in [0,1]$, $\Phi_0 = \varphi_0$ and
$\Phi_1 = \varphi_1$.
\end{definition}

\begin{theorem}
Two knots $K_0$ and $K_1$ are isotopic if and only if they are ambient isotopic, i.e. there exists a family of self-diffeomorphisms $\Psi = \{\psi_t: t \in [0, 1]\}$ such that:
\begin{align*}
&\psi(t) = \psi_t \text{ is continius on $t\in [0,1]$}\\
&\psi_t: S^3 \hookrightarrow S^3,\\
& \psi_0 = id ,\\
& \psi_1(K_0) = K_1.
\end{align*}
\end{theorem}
\begin{definition}
A knot is trivial (unknot) if it is equivalent to an embedding $\varphi(t) = (\cos t, \sin t, 0)$, where $t \in [0, 2 \pi] $ is a parametrisation of $S^1$.
\end{definition}
\begin{definition}
A link with k - components is a (smooth) embedding of $\overbrace{S^1 \sqcup \ldots \sqcup S^1}^k$ in $S^3$
\end{definition}
\begin{example}
Links:
\begin{itemize}
\item
a trivial link with $3$ components:
\includegraphics[width=0.2\textwidth]{3unknots.png},
\item
a hopf link: \includegraphics[width=0.13\textwidth]{Hopf.png},
\item
a Whitehead link:
\includegraphics[width=0.13\textwidth]{WhiteheadLink.png},
\item
Borromean link:
\includegraphics[width=0.1\textwidth]{BorromeanRings.png}.
\end{itemize}
\end{example}
%
%
%
\begin{definition}
A link diagram $D_{\pi}$ is a picture over projection $\pi$ of a link $L$ in $\mathbb{R}^3$($S^3$) to $\mathbb{R}^2$ ($S^2$) such that:
\begin{enumerate}[label={(\arabic*)}]
\item 
${D_{\pi}}_{\big|L}$ is non degenerate: \includegraphics[width=0.05\textwidth]{LinkDiagram1.png},
\item the double points are not degenerate: \includegraphics[width=0.03\textwidth]{LinkDiagram2.png},
\item there are no triple point:  \includegraphics[width=0.05\textwidth]{LinkDiagram3.png}.
\end{enumerate}
\end{definition}
\noindent
There are under- and overcrossings  (tunnels and bridges) on a link diagrams with an obvious meaning.\\
Every link admits a link diagram.
\\
Let $D$ be a diagram of an oriented link (to each component of a link we add an arrow in the diagram).\\
We can distinguish two types of crossings: right-handed 
$\left(\PICorientpluscross\right)$, called a positive crossing, and left-handed $\left(\PICorientminuscross\right)$, called a negative crossing.

\subsection{Reidemeister moves}
A Reidemeister move is one of the three types of operation on a link diagram as shown below: 
\begin{enumerate}[label=\Roman*]
\item\hfill\\
\includegraphics[width=0.6\textwidth]{rm1.png},
\item\hfill\\\includegraphics[width=0.6\textwidth]{rm2.png},
\item\hfill\\\includegraphics[width=0.4\textwidth]{rm3.png}.
\end{enumerate}

\begin{theorem} [Reidemeister,  1927 ]
Two diagrams of the same link can be
deformed into each other by a finite sequence of Reidemeister moves (and isotopy of the plane).
\end{theorem}
%
%
%
%The number of Reidemeister Moves Needed for Unknotting
%Joel Hass, Jeffrey C. Lagarias
%(Submitted on 2 Jul 1998)
% Piotr Sumata, praca magisterska
% proof - transversality theorem (Thom)

%Singularities of Differentiable Maps
%Authors: Arnold, V.I., Varchenko, Alexander, Gusein-Zade, S.M.

\subsection{Seifert surface}
\noindent
Let $D$ be an oriented diagram of a link $L$. We change the diagram by smoothing each crossing: 
\begin{align*}
\PICorientpluscross \mapsto \PICorientLRsplit\\ 
\PICorientminuscross \mapsto \PICorientLRsplit
\end{align*}
We smooth all the crossings, so we get a disjoint union of circles on the plane. Each circle bounds a disks in $\mathbb{R}^3$ (we choose disks that don't intersect). For each smoothed crossing we add a twisted band: right-handed for a positive and left-handed for a negative one. We get an orientable surface $\Sigma$ such that $\partial \Sigma = L$.\\ 

\begin{figure}[h]
\fontsize{15}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.8\textwidth}{!}{\input{images/seifert_alg.pdf_tex}}
\caption{Constructing a Seifert surface.}
\label{fig:SeifertAlg}
}
\end{figure}

\noindent
Note: the obtained surface isn't unique and in general doesn't need to be connected, but by taking connected sum of all components we can easily get a connected surface (i.e. we take two disconnected components and cut a disk in each of them: $D_1$ and $D_2$; now we glue both components on the boundaries: $\partial D_1$ and $\partial D_2$. 

\begin{figure}[h]
\begin{center}
\includegraphics[width=0.6\textwidth]{seifert_connect.png}
\end{center}
\caption{Connecting two surfaces.}
\label{fig:SeifertConnect}
\end{figure}

\begin{theorem}[Seifert]
Every link in $S^3$ bounds a surface $\Sigma$ that is compact, connected and orientable. Such a surface is called a Seifert surface.
\end{theorem}
%
\begin{figure}[h]
\fontsize{12}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{1\textwidth}{!}{\input{images/torus_1_2_3.pdf_tex}}
\caption{Genus of an orientable surface.}
\label{fig:genera}
}
\end{figure}
%
%
\begin{definition}
The three genus $g_3(K)$ ($g(K)$) of a knot $K$ is the minimal genus of a Seifert surface $\Sigma$ for $K$.
\end{definition}

\begin{corollary}
A knot $K$ is trivial if and only $g_3(K) = 0$.
\end{corollary}

\noindent
Remark: there are knots that admit non isotopic Seifert surfaces of minimal genus (András Juhász, 2008).

\begin{definition}
Suppose $\alpha$ and $\beta$ are two simple closed curves in $\mathbb{R}^3$. 
On a diagram $L$ consider all crossings between $\alpha$ and $\beta$. Let $N_+$ be the number of positive crossings, $N_-$ - negative. Then the linking number: $\Lk(\alpha, \beta) = \frac{1}{2}(N_+ - N_-)$.
\end{definition}
\hfill
\\
Let $\alpha$ and $\beta$ be two disjoint simple cross curves in $S^3$. 
Let $\nu(\beta)$ be a tubular neighbourhood of $\beta$.  The linking number can be interpreted via first homology group, where $\Lk(\alpha, \beta)$ is equal to evaluation of $\alpha$  as element of first homology group of the complement of $\beta$: 
\[
\alpha \in H_1(S^3 \setminus \nu(\beta), \mathbb{Z}) \cong \mathbb{Z}.\]


\begin{example}
\begin{itemize}
\item
Hopf link:
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.4\textwidth}{!}{\input{images/linking_hopf.pdf_tex}},
}
\end{figure}
\item
$T(6, 2)$ link:
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.4\textwidth}{!}{\input{images/linking_torus_6_2.pdf_tex}}.
}
\end{figure}
\end{itemize}
\end{example}
\begin{fact}
\[
g_3(\Sigma) = \frac{1}{2} b_1 (\Sigma) =
\frac{1}{2} \dim_{\mathbb{R}}H_1(\Sigma, \mathbb{R}),
\]
where $b_1$ is first Betti number of $\Sigma$.
\end{fact}

\subsection{Seifert matrix}
Let $L$ be a link and $\Sigma$ be an oriented Seifert surface for $L$. Choose a basis for $H_1(\Sigma, \mathbb{Z})$ consisting of simple closed $\alpha_1, \dots, \alpha_n$. 
Let $\alpha_1^+, \dots \alpha_n^+$ be copies of $\alpha_i$ lifted up off the surface (push up along a vector field normal to $\Sigma$). Note that elements $\alpha_i$ are contained in the Seifert surface while all $\alpha_i^+$ are don't intersect the surface.
Let $\Lk(\alpha_i, \alpha_j^+) = \{a_{ij}\}$. Then the matrix $S = \{a_{ij}\}_{i, j =1}^n$ is called a Seifert matrix for $L$. Note that by choosing a different basis we get a different matrix. 

\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.8\textwidth}{!}{\input{images/seifert_matrix.pdf_tex}}
}
\end{figure}

\begin{theorem}
The Seifert matrices $S_1$ and $S_2$ for the same link $L$ are S-equivalent, that is, $S_2$ can be obtained from $S_1$ by a sequence of following moves: 
\begin{enumerate}[label={(\arabic*)}]

\item
$V \rightarrow AVA^T$, where $A$ is a matrix with integer coefficients, 

\item

$V \rightarrow 
\begin{pmatrix}
    \begin{array}{c|c}
           V &  
           \begin{matrix}
                 \ast & 0  \\
                 \sdots & \sdots\\
                 \ast & 0
           \end{matrix} \\
    \hline
           \begin{matrix}
                \ast & \dots & \ast\\
                0      & \dots & 0
           \end{matrix}
           &
           \begin{matrix}
               0 & 0\\
               1 & 0
           \end{matrix}
    \end{array}
\end{pmatrix} \quad$ 
or
$\quad
V \rightarrow 
\begin{pmatrix}
    \begin{array}{c|c}
           V &  
           \begin{matrix}
                 \ast & 0  \\
                 \sdots & \sdots\\
                 \ast & 0
           \end{matrix} \\
    \hline
           \begin{matrix}
                \ast & \dots & \ast\\
                0      & \dots & 0
           \end{matrix}
           &
           \begin{matrix}
               0 & 1\\
               0 & 0
           \end{matrix}
    \end{array}
\end{pmatrix}$
\item
inverse of (2)

\end{enumerate} 
\end{theorem}

\section{\hfill\DTMdate{2019-03-04}}
\begin{theorem}
For any knot $K \subset S^3$ there exists a connected, compact and orientable surface $\Sigma(K)$ such that $\partial \Sigma(K) = K$
\end{theorem}
\begin{proof}("joke")\\
Let $K \in S^3$ be a knot and $N = \nu(K)$ be its tubular neighbourhood. Because $K$ and $N$ are homotopy equivalent, we get:
\begin{align*}
H^1(S^3 \setminus N ) \cong H^1(S^3 \setminus K).
\end{align*}
Let us consider a long exact sequence of cohomology of a pair $(S^3, S^3 \setminus N)$  with integer coefficients:

\begin{center}
\begin{tikzcd}
[
    column sep=0cm, fill=none, 
    row sep=small,
    ar symbol/.style =%
        {draw=none,"\textstyle#1" description,sloped},
         isomorphic/.style = {ar symbol={\cong}},
]
&\mathbb{Z}
\\

& H^0(S^3) \ar[u,isomorphic] \to 
&H^0(S^3 \setminus N) \to 
\\
\to H^1(S^3, S^3 \setminus N) \to
 & H^1(S^3) \to
 & H^1(S^3\setminus N) \to
 \\
& 0  \ar[u,isomorphic]&
 \\
 \to H^2(S^3, S^3 \setminus N) \to
 & H^2(S^3) \ar[u,isomorphic] \to
 & H^2(S^3\setminus N) \to
 \\
\to H^3(S^3, S^3\setminus N)\to
& H^3(S) \to
& 0
\\
&  \mathbb{Z} \ar[u,isomorphic] &\\
 \end{tikzcd}
\end{center}
\[
H^* (S^3, S^3 \setminus N) \cong H^* (N, \partial N)
\]
\\
??????????????
\\

\end{proof}

\begin{definition}
Let $S$ be a Seifert matrix for a knot $K$. The Alexander polynomial $\Delta_K(t)$ is a Laurent polynomial:
\[ 
\Delta_K(t) := \det (tS - S^T)  \in 
\mathbb{Z}[t, t^{-1}] \cong \mathbb{Z}[\mathbb{Z}]
\]
\end{definition}

\begin{theorem}
$\Delta_K(t)$ is well defined up to multiplication by $\pm t^k$, for $k \in \mathbb{Z}$.
\end{theorem}
\begin{proof}
We need to show that $\Delta_K(t)$ doesn't depend on $S$-equivalence relation.
\begin{enumerate}[label={(\arabic*)}]
\item Suppose $S\prime = CSC^T$, $C \in \Gl(n, \mathbb{Z})$ (matrices invertible over $\mathbb{Z}$). Then $\det C = 1$ and:
\begin{align*}
&\det(tS\prime - S\prime^T) =
\det(tCSC^T - (CSC^T)^T) =\\
&\det(tCSC^T - CS^TC^T) = 
\det C(tS - S^T)C^T = 
\det(tS - S^T)
\end{align*}
\item
Let \\
$ A := t
\begin{pmatrix}
    \begin{array}{c|c}
           S &  
           \begin{matrix}
                 \ast & 0  \\
                 \sdots & \sdots\\
                 \ast & 0
           \end{matrix} \\
    \hline
           \begin{matrix}
                \ast & \dots & \ast\\
                0      & \dots & 0
           \end{matrix}
           &
           \begin{matrix}
               0 & 0\\
               1 & 0
           \end{matrix}
    \end{array}
\end{pmatrix}
-
\begin{pmatrix}
    \begin{array}{c|c}
           S^T &  
           \begin{matrix}
                 \ast & 0  \\
                 \sdots & \sdots\\
                 \ast & 0
           \end{matrix} \\
    \hline
           \begin{matrix}
                \ast & \dots & \ast\\
                0      & \dots & 0
           \end{matrix}
           &
           \begin{matrix}
               0 & 1\\
               0 & 0
           \end{matrix}
    \end{array}
\end{pmatrix}
=
\begin{pmatrix}
    \begin{array}{c|c}
           tS - S^T &  
           \begin{matrix}
                 \ast & 0  \\
                 \sdots & \sdots\\
                 \ast & 0
           \end{matrix} \\
    \hline
           \begin{matrix}
                \ast & \dots & \ast\\
                0      & \dots & 0
           \end{matrix}
           &
           \begin{matrix}
               0 & -1\\
               t & 0
           \end{matrix}
    \end{array}
\end{pmatrix}
$
\\
\\
Using the Laplace expansion we get $\det A = \pm t \det(tS - S^T)$. 
\end{enumerate} 
\end{proof}
%
%
%
\begin{example}
If $K$ is a trefoil then we can take
$S = \begin{pmatrix}
-1 & -1 \\
0 & -1
\end{pmatrix}$. Then
\[
\Delta_K(t) = \det 
\begin{pmatrix}
-t + 1 & -t\\
1 & -t +1
\end{pmatrix}
= (t -1)^2 + t = t^2 - t +1 \ne 1
\Rightarrow \text{trefoil is not trivial.}
\]
\end{example}
\begin{fact}
$\Delta_K(t)$ is symmetric.
\end{fact}
\begin{proof}
Let $S$ be an $n \times n$ matrix. 
\begin{align*}
&\Delta_K(t^{-1}) = \det (t^{-1}S - S^T) = (-t)^{-n} \det(tS^T - S) = \\
&(-t)^{-n} \det (tS - S^T) = (-t)^{-n} \Delta_K(t)
\end{align*}
If $K$ is a knot, then $n$ is necessarily even, and so $\Delta_K(t^{-1}) = t^{-n} \Delta_K(t)$.
\end{proof}
\begin{lemma}
\begin{align*}
\frac{1}{2} \deg \Delta_K(t) \leq g_3(K),
\text{ where } deg (a_n t^n + \cdots + a_1 t^l )= k - l.
\end{align*}
\end{lemma}
\begin{proof}
If $\Sigma$ is a genus $g$ - Seifert surface for $K$ then $H_1(\Sigma) = \mathbb{Z}^{2g}$, so $S$ is an $2g \times 2g$ matrix. Therefore $\det (tS - S^T)$ is a polynomial of degree at most $2g$.
\end{proof}
\begin{example}
There are not trivial knots with Alexander polynomial equal $1$, for example: 
\includegraphics[width=0.3\textwidth]{11n34.png}
$\Delta_{11n34} \equiv 1$.
\end{example}
%removing one disk from surface doesn't change $H_1$ (only $H_2$)
%
%
%
\begin{lemma}[Dehn]
Let $M$ be a $3$-manifold and $D^2 \overset{f} \rightarrow M^3$ be a map of a disk such that $f_{\big|\partial D^2}$ is an embedding. Then there exists an embedding 
${D^2 \overset{g}\hookrightarrow M}$ such that: 
\[
g_{\big| \partial D^2} = f_{\big| \partial D^2.}
\]
\end{lemma}
\section{}
\begin{example}
\begin{align*}
&F: \mathbb{C}^2 \rightarrow \mathbb{C}  \text{ a polynomial} \\ 
&F(0) = 0
\end{align*}

\end{example}
????????????
\\
\noindent
as a corollary we see that $K_T^{n, }$ ???? \\
is not slice unless $m=0$. 
\begin{theorem}
The map $j: \mathscr{C} \longrightarrow \mathbb{Z}^{\infty}$ is a surjection that maps ${K_n}$ to a linear independent set. Moreover $\mathscr{C} \cong \mathbb{Z}$
\end{theorem}


\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}
%
%
%
\section{Concordance group \hfill\DTMdate{2019-03-18}}
\begin{definition}
A knot $K$ is called (smoothly) slice if $K$ is smoothly concordant to an unknot. \\
A knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk in $B^4$.
\end{definition}
\begin{definition}
Two knots $K$ and $K^{\prime}$ are called (smoothly) concordant if there exists an annulus $A$ that is smoothly embedded in ${S^3 \times [0, 1]}$ such that 
${\partial A = K^{\prime} \times \{1\} \; \sqcup \; K \times \{0\}}$.
\end{definition}

\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.8\textwidth}{!}{\input{images/concordance.pdf_tex}}
}
\end{figure}



\noindent
Let $m(K)$ denote a mirror image of a knot $K$. 
\begin{fact}
For any $K$, $K \# m(K)$ is slice.
\end{fact}
\begin{fact}
Concordance is an equivalence relation.
\end{fact}
\begin{fact}\label{fakt:concordance_connected}
If $K_1 \sim {K_1}^{\prime}$ and $K_2 \sim {K_2}^{\prime}$, then
$K_1 \# K_2 \sim {K_1}^{\prime} \# {K_2}^{\prime}$.
\begin{figure}[h]
\fontsize{10}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{1\textwidth}{!}{\input{images/concordance_sum.pdf_tex}}
}
\caption{Sketch for Fakt \ref{fakt:concordance_connected}.}
\label{fig:concordance_sum}
\end{figure}

\end{fact}
\begin{fact}
$K \# m(K) \sim $ the unknot.
\end{fact}
\noindent
\begin{theorem}
Let $\mathscr{C}$ denote a set of all equivalent classes for knots and $\{0\}$ denote class of all knots concordant to a trivial knot. 
$\mathscr{C}$ is a group under taking connected sums. The neutral element in the group is $\{0\}$ and the inverse element of an element $\{K\} \in \mathscr{C}$ is $-\{K\} = \{mK\}$.
\end{theorem}
\begin{fact}
The figure eight knot is a torsion element in $\mathscr{C}$ ($2K \sim $ the unknot).
\end{fact}
\begin{problem}[open]
Are there in concordance group torsion elements that are not $2$ torsion elements?
\end{problem}
\noindent
Remark: $K \sim K^{\prime} \Leftrightarrow K \# -K^{\prime}$ is slice.
\\
\\
\noindent
Let $\Omega$ be an oriented \\
???????\\
Suppose $\Sigma$ is a Seifert matrix with an intersection form ${(\alpha, \beta) \mapsto \Lk(\alpha, \beta^+)}$. Suppose $\alpha, \beta \in H_1(\Sigma, \mathbb{Z}$ (i.e. there are cycles). \\
??????????????\\
$\alpha, \beta \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z}))$. Then there are two cycles $A, B \in \Omega$ such that $\partial A = \alpha$ and $\partial B = \beta$. 
Let $B^+$ be a push off of $B$ in the positive normal direction such that 
$\partial B^+ = \beta^+$.
Then 
$\Lk(\alpha, \beta^+) = A \cdot B^+$
%
%
\\
\section{\hfill\DTMdate{2019-04-08}}
%
%
$X$ is a closed orientable four-manifold. Assume $\pi_1(X) = 0$ (it is not needed to define the intersection form). In particular $H_1(X) = 0$. 
$H_2$ is free (exercise). 
\begin{align*}
H_2(X, \mathbb{Z}) \xrightarrow{\text{Poincar\'e duality}} H^2(X, \mathbb{Z} ) \xrightarrow{\text{evaluation}}\Hom(H_2(X, \mathbb{Z}), \mathbb{Z})
\end{align*}
Intersection form: 
$H_2(X, \mathbb{Z}) \times 
H_2(X, \mathbb{Z}) \longrightarrow \mathbb{Z}$ - symmetric, non singular. 
\\
Let $A$ and $B$ be closed, oriented surfaces in $X$. 
\begin{proposition}
$A \cdot B$ doesn't depend of choice of $A$ and $B$ in their homology classes. 
%$A \cdot B$  gives the pairing as ??
 
\end{proposition}


\section{\hfill\DTMdate{2019-03-11}}
\begin{definition}
A link $L$ is fibered if there exists a map ${\phi: S^3\setminus L \longleftarrow S^1}$ which is locally trivial fibration. 
\end{definition}


\section{\hfill\DTMdate{2019-04-15}}
In other words:\\
Choose a basis $(b_1, ..., b_i)$ \\
???\\
of $H_2(Y, \mathbb{Z}$, then $A = (b_i, b_y)$ \\??\\ is a matrix of intersection form: 
\begin{align*}
\quot{\mathbb{Z}^n}{A\mathbb{Z}^n} \cong H_1(Y, \mathbb{Z}).
\end{align*}
In particular $\mid \det A\mid = \# H_1(Y, \mathbb{Z}$.\\
That means - what is happening on boundary is a measure of degeneracy.  

\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_1(Y, \mathbb{Z}) &
  \times \quad H_1(Y, \mathbb{Z})&
  \longrightarrow &
  \quot{\mathbb{Q}}{\mathbb{Z}}
  \text{ - a linking form}
\\
   \quot{\mathbb{Z}^n}{A\mathbb{Z}} \ar[u,isomorphic] & 
   \quot{\mathbb{Z}^n}{A\mathbb{Z}} \ar[u,isomorphic] &\\
\end{tikzcd}
$(a, b) \mapsto aA^{-1}b^T$
\end{center}

The intersection form on a four-manifold determines the linking on the boundary. \\

\noindent
Let $K \in S^1$ be a knot, $\Sigma(K)$ its double branched cover. If $V$ is a Seifert matrix for $K$, then 
$H_1(\Sigma(K), \mathbb{Z}) \cong  \quot{\mathbb{Z}^n}{A\mathbb{Z}}$ where 
$A = V \times V^T$, where $n = \rank V$.
%\input{ink_diag}
\begin{figure}[h]
\fontsize{40}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/ball_4.pdf_tex}}
\caption{Pushing the Seifert surface in 4-ball.}
\label{fig:pushSeifert}
}
\end{figure}
\noindent
Let $X$ be the four-manifold obtained via the double branched cover of $B^4$ branched along $\widetilde{\Sigma}$.
\begin{fact}
\begin{itemize}
\item $X$ is a smooth four-manifold, 
\item $H_1(X, \mathbb{Z}) =0$, 
\item $H_2(X, \mathbb{Z}) \cong \mathbb{Z}^n$
\item The intersection form on $X$ is $V + V^T$.
\end{itemize}
\end{fact}
\noindent
Let  $Y = \Sigma(K)$. Then:
\begin{flalign*}
H_1(Y, \mathbb{Z}) \times H_1(Y, \mathbb{Z}) \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}}&
\\
(a,b) \mapsto a A^{-1} b^{T},\qquad
A = V + V^T&
\\
H_1(Y, \mathbb{Z}) \cong \quot{\mathbb{Z}^n}{A\mathbb{Z}}&\\
A \longrightarrow BAC^T \quad \text{Smith normal form}&
\end{flalign*}
???????????????????????\\
In general 

\section{\hfill\DTMdate{2019-05-20}}

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, \_) H_2(M, \mathbb{Z})
\end{align*}
has coker precisely $H_1(Y, \mathbb{Z})$. 
\\???????????????\\
Let $K \subset S^3$  be a knot, \\
$X = S^3 \setminus K$ - a knot complement,  \\
$\widetilde{X} \xrightarrow{\enspace \rho \enspace} X$ - an infinite cyclic cover (universal abelian cover).
\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*}
$C_{*}(\widetilde{X})$ has a structure of a $\mathbb{Z}[t, t^{-1}] \cong  \mathbb{Z}[\mathbb{Z}]$ module. \\
$H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}])$ - Alexander module, \\
\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) &\mapsto \alpha^{-1}(t -1)(tV - V^T)^{-1}\beta
\end{align*}
\end{fact}
\noindent
Note that $\mathbb{Z}$ is not PID. Therefore we don't have primer decomposition of this moduli.  We can simplify this problem by replacing $\mathbb{Z}$ by $\mathbb{R}$. We lose some date by doing this transition. 
\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}\\
&
\Lambda = \mathbb{R}[t, t^{-1}]\\
&\text{Then: } 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}
\longleftrightarrow \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 $(t - \zeta)^2 \mid P$ (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, Conway-Borodzik-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}
\end{align*}
\begin{align*}
\text{Let } \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 
$2 \sum\limits_{k_i \text{ odd}} \epsilon_i$. The peak of the signature function is equal to $\sum\limits_{k_i \text{even}} \epsilon_i$.
%$(\eta_{k, \xi_l^{+}} -\eta_{k, \xi_l^{-}}$
\end{theorem}
\end{proof}
\section{\hfill\DTMdate{2019-05-27}}
....
\begin{definition}
A square hermitian matrix $A$ of size $n$.  
\end{definition}

field of fractions


\section{balagan}

\noindent
\begin{proof}
By Poincar\'e duality we know that:
\begin{align*}
H_3(\Omega, Y) &\cong H^0(\Omega),\\
H_2(Y) &\cong H^0(Y),\\
H_2(\Omega) &\cong H^1(\Omega, Y),\\
H_2(\Omega, Y) &\cong H^1(\Omega).
\end{align*}
Therefore $\dim_{\mathbb{Q}} \quot{H_1(Y)}{V}
= \dim_{\mathbb{Q}} V
$.
\end{proof}
\noindent
Suppose $g(K) = 0$ ($K$ is slice). Then $H_1(\Sigma, \mathbb{Z}) \cong H_1(Y, \mathbb{Z})$. Let $g_{\Sigma}$ be the genus of $\Sigma$, $\dim H_1(Y, \mathbb{Z}) = 2g_{\Sigma}$. Then the Seifert form $V$ on 4\\
?????\\
has a subspace of dimension $g_{\Sigma}$ on which it is zero:

\begin{align*}
\newcommand\coolover[2]%
    {\mathrlap{\smash{\overbrace{\phantom{%
            \begin{matrix} #2 \end{matrix}}}^{\mbox{$#1$}}}}#2}
\newcommand\coolunder[2]{\mathrlap{\smash{\underbrace{\phantom{%
    \begin{matrix} #2 \end{matrix}}}_{\mbox{$#1$}}}}#2}
\newcommand\coolleftbrace[2]{%
    #1\left\{\vphantom{\begin{matrix} #2 \end{matrix}}\right.}
\newcommand\coolrightbrace[2]{%
     \left.\vphantom{\begin{matrix} #1 \end{matrix}}\right\}#2}
 \vphantom{% phantom stuff for correct box dimensions
    \begin{matrix}
    \overbrace{XYZ}^{\mbox{$R$}}\\ \\ \\ \\ \\ \\
    \underbrace{pqr}_{\mbox{$S$}}
    \end{matrix}}%
 V =
\begin{matrix}% matrix for left braces
    \coolleftbrace{g_{\Sigma}}{ \\  \\ \\}
     \\ \\ \\ \\ 
\end{matrix}%
\begin{pmatrix}
    \coolover{g_{\Sigma}}{0 & \dots & 0 } & * & \dots & *\\
    \sdots &  & \sdots & \sdots &  & \sdots \\
    0 & \dots & 0 & * & \dots & *\\
    * & \dots & * & * & \dots & *\\
   \sdots &  & \sdots & \sdots &  & \sdots \\
     * & \dots & * & * & \dots & * 
 \end{pmatrix}_{2g_{\Sigma} \times 2g_{\Sigma}}
\end{align*}
\end{document}