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\begin{theorem}
Let $K$ be a knot and $u(K)$ its unknotting number. Let $g_4$ be a minimal four genus of a smooth surface $S$ in $B^4$ such that $\partial S = K$. Then:
\[
u(K) \geq g_4(K)
\]
\begin{proof}
Recall that if $u(K)=u$ then $K$ bounds a disk $\Delta$ with $u$ ordinary double points.
\\
\noindent
Remove from $\Delta$ the two self intersecting disks and glue the Seifert surface for the Hopf link. The reality surface $S$ has Euler characteristic $\chi(S) = 1 - 2u$. Therefore $g_4(S) = u$.
\end{proof}
%Tim D. Cochran and Peter Teichner
\begin{example}
The knot $8_{20}$ is slice: $\sigma \equiv 0$ almost everywhere but $\sigma(e^{\frac{ 2\pi i}{6}}) = + 1$.
\end{example}
%ref Structure in the classical knot concordance group
%Tim D. Cochran, Kent E. Orr, Peter Teichner
%Journal-ref: Comment. Math. Helv. 79 (2004) 105-123
\subsection*{Surgery}
%Rolfsen, geometric group theory, Diffeomorpphism of a torus, Mapping class group
Recall that $H_1(S^1 \times S^1, \mathbb{Z}) = \mathbb{Z}^2$. As generators for $H_1$ we can set ${\alpha = [S^1 \times \pt]}$ and ${\beta=[\pt \times S^1]}$. Suppose ${\phi: S^1 \times S^1 \longrightarrow S^1 \times S^1}$ is a diffeomorphism.
Consider an induced map on the homology group:
\begin{align*}
H_1(S^1 \times S^1, \mathbb{Z}) \ni \phi_* (\alpha) &= p\alpha + q \beta, \quad p, q \in \mathbb{Z},\\
\phi_*(\beta) &= r \alpha + s \beta, \quad r, s \in \mathbb{Z}, \\
\phi_* &=
\begin{pmatrix}
p & q\\
r & s
\end{pmatrix}.
\end{align*}
As $\phi_*$ is diffeomorphis, it must be invertible over $\mathbb{Z}$. Then for a direction preserving diffeomorphism we have $\det \phi_* = 1$. Therefore $\phi_* \in \Sl(2, \mathbb{Z})$.
\end{theorem}
\vspace{10cm}
\begin{theorem}
Every such a matrix can be realized as a torus.
\end{theorem}
\begin{proof}
\begin{enumerate}[label={(\Roman*)}]
\item
Geometric reason
\begin{align*}
\phi_t:
S^1 \times S^1 &\longrightarrow S^1 \times S^1 \\
S^1 \times \pt &\longrightarrow \pt \times S^1 \\
\pt \times S^1 &\longrightarrow S^1 \times \pt \\
(x, y) & \mapsto (-y, x)
\end{align*}
\item
\end{enumerate}
\end{proof}
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/dehn_twist.pdf_tex}}
\caption{Dehn twist.}
\label{fig:dehn_twist}
}
\end{figure}

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\subsection{Existence of Seifert surface - second proof}
%\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}(Theorem \ref{theo:Seifert})\\
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}
\begin{align*}
N \cong & D^2 \times S^1\\
\partial N \cong & S^1 \times S^1\\
H^1(N, \partial N) \cong & \mathbb{Z} \oplus \mathbb{Z}
\end{align*}
\begin{align*}
H^* (S^3, S^3 \setminus N) &\cong H^* (N, \partial N)\\
\\
H^ 1 (S^3\setminus N) &\cong H^1(S^3\setminus K) \cong \mathbb{Z}
\end{align*}
\begin{equation*}
\begin{tikzcd}[row sep=huge]
H^1(S^3 \setminus K) \arrow[r,] \arrow[d,"\widetilde{\Theta}"] &
H^1(N \setminus K) \arrow[d,"\Theta"] \\
{[S^3 \setminus K, S^1]} \arrow[r,]&
{[N \setminus K, S^1]}
\end{tikzcd}
\end{equation*}
\noindent
$\Sigma = \widetilde{\Theta}^{-1}(X)$ is a surface, such that $\partial \Sigma = K$, so it is a Seifert surface.
%
%
% Thom isomorphism,
\end{proof}
\subsection{Alexander polynomial}
\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 + \dots + 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}
\subsection{Decomposition of $3$-sphere}
We know that $3$ - sphere can be obtained by gluing two solid tori:
\[
S^3 = \partial D^4 = \partial (D^2 \times D^2) = (D^2 \times S^1) \cup (S^1 \times D^2).
\]
So the complement of solid torus in $S^3$ is another solid torus.\\
Analytically it can be describes as follow. \\
Take $(z_1, z_2) \in \mathbb{C}$ such that ${\max(\vert z_1 \vert, \vert z_2\vert) = 1.}
$
Define following sets:
\begin{align*}
S_1 = \{ (z_1, z_2) \in S^3: \vert z_1 \vert = 0\} \cong S^1 \times D^2 ,\\
S_2 = \{(z_1, z_2) \in S ^3: \vert z_2 \vert = 1 \} \cong D^2 \times S^1.
\end{align*}
The intersection
$S_1 \cap S_2 = \{(z_1, z_2): \vert z_1 \vert = \vert z_2 \vert = 1 \} \cong S^1 \times S^1$.
\begin{figure}[h]
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.3\textwidth}{!}{\includegraphics[width=0.3\textwidth]{sphere_as_torus.png}}
\caption{The complement of solid torus in $S^3$ is another solid torus.}
\label{fig:sphere_as_tori}
}
\end{figure}
\subsection{Dehn lemma and sphere theorem}
%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
${D^2 \overset{g}\longhookrightarrow M}$ such that:
\[
g\big|_{\partial D^2} = f\big|_{\partial D^2.}
\]
\end{lemma}
\noindent
Remark: Dehn lemma doesn't hold for dimension four.\\
Let $M$ be connected, compact three manifold with boundary.
Suppose $\pi_1(\partial M) \longrightarrow \pi_1(M)$ has non-trivial kernel. Then there exists a map $f: (D^2, \partial D^2) \longrightarrow (M, \partial M)$ such that $f\big|_{\partial D^2}$ is non-trivial loop in $\partial M$.
\begin{theorem}[Sphere theorem]
Suppose $\pi_1(M) \ne 0$. Then there exists an embedding $f: S^2 \hookrightarrow M$ that is homotopy non-trivial.
\end{theorem}
\begin{problem}
Prove that $S^3 \ K$ is Eilenberg–MacLane space of type $K(\pi, 1)$.
\end{problem}
\begin{corollary}
Suppose $K \subset S^3$ and $\pi_1(S^3 \setminus K)$ is infinite cyclic ($\mathbb{Z})$. Then $K$ is trivial.
\end{corollary}
\begin{proof}
Let $N$ be a tubular neighbourhood of a knot $K$ and $M = S^3 \setminus N$ its complement. Then $\partial M = S^1 \times S^1$. Let $f : \pi_1(\partial M ) \longrightarrow \pi_1(M)$.
If $\pi_1(M)$ is infinite cyclic group then the map $f$ is non-trivial. Suppose ${\lambda \in \ker (\pi_1(S^1 \times S^1) \longrightarrow \pi_1(M)}$.
There is a map $g: (D^2, \partial D^2) \longrightarrow (M, \partial M)$ such that $g(\partial D^2) = \lambda$.\\
By Dehn's lemma there exists an embedding ${h: (D^2, \partial D^2) \longhookrightarrow (M, \partial M)}$ such that
$h\big|_{\partial D^2} = f \big|_{\partial D^2}$ and $h(\partial D^2) = \lambda$.
Let $\Sigma$ be a union of the annulus and the image of $\partial D^2$.
\\???? $g_3$?\\
If $g(\Sigma) = 0$, then $K$ is trivial. \\
Now we should proof that:
\[
H_1(M) \cong \mathbb{Z} \Longrightarrow \lambda \in \ker ( \pi_1(S^1 \times S^1) \longrightarrow \pi_1(M)).
\]
\begin{figure}[h]
\fontsize{40}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.4\textwidth}{!}{\input{images/torus_lambda.pdf_tex}}
}
\caption{$\mu$ is a meridian and $\lambda$ is a longitude.}
\label{fig:meridian_and_longitude}
\end{figure}
Choose a meridian $\mu$ such that $\Lk (\mu, K) = 1$. Recall the definition of linking number via homology group (Definition \ref{def:lk_via_homo}).
$[\mu]$ represents the generator of $H_1(S^3\setminus K, \mathbb{Z})$. From definition of $\lambda$ we know that $\lambda$ is trivial in $H_1(M)$ ($\Lk(\lambda, K) =0$, therefore $[\lambda]$ was trivial in $pi_1(M)$). If $K$ is non-trivial then $\lambda$ is non-trivial in $\pi_1(M)$, but it is trivial in $H_1(M)$.
\end{proof}

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\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}

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$X$ 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:
$H_2(X, \mathbb{Z}) \times
H_2(X, \mathbb{Z}) \longrightarrow \mathbb{Z}$ is symmetric and non singular.
\\
Let $A$ and $B$ be closed, oriented surfaces in $X$.
\\
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/intersection_form_A_B.pdf_tex}}
}
\caption{$T_X A + T_X B = T_X X$
}\label{fig:torus_alpha_beta}
\end{figure}
???????????????????????
\begin{align*}
x \in A \cap B\\
T_XA \oplus T_X B = T_X X\\
\{\epsilon_1, \dots , \epsilon_n \} = A \cap C\\
A \cdot B = \sum^n_{i=1} \epsilon_i
\end{align*}
\begin{proposition}
Intersection form $A \cdot B$ doesn't depend of choice of $A$ and $B$ in their homology classes:
\[
[A], [B] \in H_2(X, \mathbb{Z}).
\]
\end{proposition}
\noindent
\\
\subsection{Fundamental cycle}
If $M$ is an $m$ - dimensional close, connected and orientable manifold, then $H_m(M, \mathbb{Z})$ and the orientation of $M$ determined a cycle $[M] \in H_m(M, \mathbb{Z})$, called the fundamental cycle.
\begin{example}
If $\omega$ is an $m$ - form then:
\[
\int_M \omega = [\omega]([M]), \quad [\omega] \in H^m_\Omega(M), \ [M] \in H_m(M).
\]
\end{example}
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.8\textwidth}{!}{\input{images/torus_alpha_beta.pdf_tex}}
}
\caption{$\beta$ cross $3$ times the disk bounded by $\alpha$.
$T_X \alpha + T_X \beta = T_X \Sigma$
}\label{fig:torus_alpha_beta}
\end{figure}
\begin{example}
K{\"u}nneth
?????????????????????????\\
Let $X = S^2 \times S^2$.
We know that:
\begin{align*}
&H_2(S^2, \mathbb{Z}) =\mathbb{Z}\\
&H_1(S^2, \mathbb{Z}) = 0\\
&H_0(S^2, \mathbb{Z}) =\mathbb{Z}
\end{align*}
We can construct a long exact sequence for a pair:
\begin{align*}
&H_2(\partial X) \to H_2(X)
\to H_2(X, \partial X) \to \\
\to &H_1(\partial X) \to H_1(X) \to H_1(X, \partial X) \to
\end{align*}
????????????????????\\
Simple case $H_1(\partial X)$ \\????????????\\
is torsion.
$H_2(\partial X)$ is torsion free (by universal coefficient theorem),\\
???????????????????????\\
therefore it is $0$.
\\?????????????????????\\
We know that $b_1(X) = b_2(X)$. Therefore by Poincar\'e duality:
\begin{align*}
b_1(X) =
\dim_{\mathbb{Q}} H_1(X, \mathbb{Q})
\overset{\mathrm{PD}}{=}
\dim_{\mathbb{Q}} H^2(X, \mathbb{Q}) =
\dim_{\mathbb{Q}} H_2(X, \mathbb{Q}) = b_2(X)
\end{align*}
???????????????????????????????\\
$H_2(X, \mathbb{Z})$ is torsion free and
$H_2(X_1, \mathbb{Q}) = 0$, therefore $H_2(X, \mathbb{Z}) = 0$.
The map
$H_2(X, \mathbb{Z}) \longrightarrow H_2(X, \partial X, \mathbb{Z})$ is a monomorphism. \\??????????\\ (because it is an isomorphism after tensoring by $\mathbb{Q}$.
\\
Suppose $\alpha_1, \dots, \alpha_n$ is a basis of $H_2(X, \mathbb{Z})$.
Let $A$ be the intersection matrix in this basis. Then:
\begin{enumerate}
\item
A has integer coefficients,
\item
$\det A \neq 0$,
\item
$\vert \det A \vert =
\vert H_1 (\partial X, \mathbb{Z}) \vert =
\vert \coker H_2(X) \longrightarrow H_2(X, \partial X) \vert$.
\end{enumerate}
\end{example}
???????????????????
\\
\\
If $CUC^T = W$, then for
$\binom{a}{b} = C^{-1} \binom{1}{0}$ we have:
\[
\binom{a}{b} W
\binom{a}{b} = \binom{1}{0} U \binom{1}{0} = 1.
\]
\begin{theorem}[Whitehead]
Any non-degenerate form
\[
A : \mathbb{Z}^4 \times \mathbb{Z}^4 \longrightarrow \mathbb{Z}
\]
can be realized as an intersection form of a simple connected $4$-dimensional manifold.
\end{theorem}
??????????????????????????
\begin{theorem}[Donaldson, 1982]
If $A$ is an even definite intersection form of a smooth $4$-manifold then it is diagonalizable over $\mathbb{Z}$.
\end{theorem}
??????????????????????????
??????????????????????????
??????????????????????????
??????????????????????????
\begin{definition}
even define
\end{definition}
Suppose $X$ us $4$ -manifold with a boundary such that $H_1(X) = 0$.
%$A \cdot B$ gives the pairing as ??
\begin{proof}
Obviously:
\[H_1(\partial X, \mathbb{Z}) = \coker H_2(X) \longrightarrow H_2(X, \partial X) = \quot{H_2(X, \partial X)}{H_2(X)}.
\]
Let $A$ be an $n \times n$ matrix. $A$ determines a \\
??????????????/\\
\begin{align*}
\mathbb{Z}^n \longrightarrow \Hom (\mathbb{Z}^n, \mathbb{Z})\\
a \mapsto (b \mapsto b^T A a)\\
\vert \coker A \vert = \vert \det A \vert
\end{align*}
all homomorphisms $b = (b_1, \dots, b_n) $???????\\?????????\\
\end{proof}

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\subsection{Algebraic knots}
\noindent
Suppose $F: \mathbb{C}^2 \rightarrow \mathbb{C}$ is a polynomial and $F(0) = 0$. Let take small small sphere $S^3$ around zero. This sphere intersect set of roots of $F$ (zero set of $F$) transversally and by the implicit function theorem the intersection is a manifold.
The dimension of sphere is $3$ and $F^{-1}(0)$ has codimension $2$.
So there is a subspace $L$ - compact one dimensional manifold without boundary.
That means that $L$ is a link in $S^3$.
\begin{figure}[h]
\fontsize{40}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.2\textwidth}{!}{\input{images/milnor_singular.pdf_tex}}
}
\caption{The intersection of a sphere $S^3$ and zero set of polynomial $F$ is a link $L$.}
\label{fig:milnor_singular}
\end{figure}
%ref: Milnor Singular Points of Complex Hypersurfaces
\begin{theorem}
$L$ is an unknot if and only if
zero is a smooth point, i.e.
$\bigtriangledown F(0) \neq 0$ (provided $S^3$ has a sufficiently small radius).
\end{theorem}
\noindent
Remark: if $S^3$ is large it can happen that $L$ is unlink, but $F^{-1}(0) \cap B^4$ is "complicated". \\
%Kyle M. Ormsby
\noindent
In other words: if we take sufficiently small sphere, the link is non-trivial if and only if the point $0$ is singular and the isotopy type of the link doesn't depend on the radius of the sphere.
A link obtained is such a way is called an
algebraic link (in older books on knot theory there is another notion of algebraic link with another meaning).
%ref: Eisenbud, D., Neumann, W.
\begin{example}
Let $p$ and $q$ be coprime numbers such that $p<q$ and $p,q>1$. \\
Zero is an isolated singular point ($\bigtriangledown F(0) = 0$). $F$ is quasi - homogeneous polynomial, so the isotopy class of the link doesn't depend on the choice of a sphere.
Consider $S^3 = \{ (z, w) \in \mathbb{C} : \max( \vert z \vert, \vert w \vert )\} = \varepsilon$.
The intersection
$F^{-1}(0) \cap S^3$ is a torus $T(p, q)$.
\\???????????????????
$F(z, w) = z^p - w^q$\\
.\\
$F^{-1}(0) = \{t = t^q, w = t^p\}.$ For unknot $t = \max (\vert t\vert ^p, \vert t \vert^q) = \varepsilon$.
\end{example}
as a corollary we see that $K_T^{n, }$ ???? \\
is not slice unless $m=0$. \\
$t = re^{i \Theta}, \Theta \in [0, 2\pi], r = \varepsilon^{\frac{i}{p}}$
\begin{figure}[h]
\fontsize{40}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.2\textwidth}{!}{\input{images/polynomial_and_surface.pdf_tex}}
}
\caption{Sa.}
\label{fig:polynomial_and_surface}
\end{figure}
\begin{theorem}
Suppose $L$ is an algebraic link. $L = F^{-1}(0) \cap S^3$. Let
\begin{align*}
&\varphi : S^3 \setminus L \longrightarrow S^1 \\
&\varphi(z, w) =\frac{F(z, w)}{\vert F(z, w) \vert}\in S^1, \quad (z, w) \notin F^{-1}(0).
\end{align*}
The map $\varphi$ is a locally trivial fibration.
\end{theorem}
???????\\
$ rh D \varphi \equiv 1$
\begin{definition}
A map $\Pi : E \longrightarrow B$ is locally trivial fibration with fiber $F$ if for any $b \in B$, there is a neighbourhood $U \subset B$ such that $\Pi^{-1}(U) \cong U \times $ \\
????????????\\ $\Gamma$ ?????????????\\
FIGURES\\
!!!!!!!!!!!!!!!!!!!!!!!!!!\\
\end{definition}
\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}
...
\\
In general $h$ is defined only up to homotopy, but this means that
\[
h_* : H_1 (F, \mathbb{Z}) \longrightarrow H_1 (F, \mathbb{Z})
\]
is well defined \\
???????????\\ map.
\begin{theorem}
\label{thm:F_as_S}
Suppose $S$ is a Seifert matrix associated with $F$ then $h = S^{-1}S^T$.
\end{theorem}
\begin{proof}
TO WRITE REFERENCE!!!!!!!!!!!
%see Arnold Varchenko vol II
%Picard - Lefschetz formula
%Nemeth (Real Seifert forms
\end{proof}
\noindent
Consequences:
\begin{enumerate}[label={(\arabic*)}]
\item
the Alexander polynomial is the characteristic polynomial of $h$:
\[
\Delta_L (t) = \det (h - t I d)
\]
In particular $\Delta_L $ is monic (i.e. the top coefficient is $\pm 1$),
????????????????
\item
S is invertible,
\item
$F$ minimize the genus (i.e. $F$ is minimal genus Seifert surface).
\\??????????????????\\
\end{enumerate}
%
\begin{definition}
A link $L$ is fibered if there exists a map ${\phi: S^3\setminus L \longrightarrow S^1}$ which is locally trivial fibration.
\end{definition}
\noindent
If $L$ is fibered then Theorem \ref{thm:F_as_S} holds and all its consequences.
\begin{problem}
If $K_1$ and $K_2$ are fibered knots, then also $K_1 \# K_2$ is fibered.
\end{problem}
\noindent
?????????????????????\\
\begin{problem}
Prove that connected sum is well defined:\\
$\Delta_{K_1 \# K_2} =
\Delta_{K_1} + \Delta_{K_2}$ and
$g_3(K_1 \# K_2) = g_3(K_1) + g_3(K_2)$.
\end{problem}
\begin{figure}[h]
\fontsize{12}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{1\textwidth}{!}{\input{images/satellite.pdf_tex}}}
\caption{Example for a satellite knot: a Whitehead double of a trefoil.\\
The pattern knot embedded non-trivially in an unknotted solid torus $T$ (e.i. $K \not\subset S^3\subset T$) on the left and the pattern in a companion knot - trefoil - on the right.}
\label{fig:sattelite}
\end{figure}
\noindent
\subsection{Alternating knot}
\begin{definition}
A knot (link) is called alternating if it admits an alternating diagram.
\end{definition}
\begin{figure}[h]
\fontsize{12}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\includegraphics[width=0.3\textwidth]{figure8.png}
}
\caption{Example: figure eight knot is an alternating knot.}
\label{fig:fig8}
\end{figure}
\begin{definition}
A reducible crossing in a knot diagram is a crossing for which we can find a circle such that its intersection with a knot diagram is exactly that crossing. A knot diagram without reducible crossing is called reduced.
\end{definition}
\begin{fact}
Any reduced alternating diagram has minimal number of crossings.
\end{fact}
\begin{definition}
The writhe of the diagram is the difference between the number of positive and negative crossings.
\end{definition}
\begin{fact}[Tait]
Any two diagrams of the same alternating knot have the same writhe.
\end{fact}
\begin{fact}
An alternating knot has Alexander polynomial of the form:
$
a_1t^{n_1} + a_2t^{n_2} + \dots + a_s t^{n_s}
$, where $n_1 < n_2 < \dots < n_s$ and $a_ia_{i+1} < 0$.
\end{fact}
\begin{problem}[open]
What is the minimal $\alpha \in \mathbb{R}$ such that if $z$ is a root of the Alexander polynomial of an alternating knot, then $\Re(z) > \alpha$.\\
Remark: alternating knots have very simple knot homologies.
\end{problem}
\begin{proposition}
If $T_{p, q}$ is a torus knot, $p < q$, then it is alternating if and only if $p=2$.
\end{proposition}

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???????????????????\\
\begin{theorem}
Suppose that $K \subset S^3$ is a slice knot (i.e. $K$ bound a disk in $B^4$).
Then if $F$ is a Seifert surface of $K$ and $V$ denotes the associated Seifet matrix, then there exists $P \in \Gl_g(\mathbb{Z})$ such that:
\\??????????????? T ????????
\begin{align}
PVP^{-1} =
\begin{pmatrix}
0 & A\\
B & C
\end{pmatrix}, \quad A, C, C \in M_{g \times g} (\mathbb{Z})
\end{align}
\end{theorem}
In other words you can find rank $g$ direct summand $\mathcal{Z}$ of $H_1(F)$ \\
????????????\\
such that for any
$\alpha, \beta \in \mathcal{L}$ the linking number $\Lk (\alpha, \beta^+) = 0$.
\begin{definition}
An abstract Seifert matrix (i. e.
\end{definition}
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 $\vert \det A\vert = \# H_1(Y, \mathbb{Z})$.\\
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}
?????????????????????????????????\\
\noindent
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
$A = V \times V^T$, $n = \rank V$.
%\input{ink_diag}
\begin{figure}[h]
\fontsize{20}{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}
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/ball_4_pushed_cycle.pdf_tex}}
\caption{Cycle pushed in 4-ball.}
\label{fig:pushCycle}
}
\end{figure}
\noindent
Let $Y = \Sigma(K)$. Then:
\begin{align*}
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.
\end{align*}
????????????????????????????
\\
We have a primary decomposition of $H_1(Y, \mathbb{Z}) = U$ (as a group). For any $p \in \mathbb{P}$ we define $U_p$ to be the subgroup of elements annihilated by the same power of $p$. We have $U = \bigoplus_p U_p$.
\begin{example}
\begin{align*}
\text{If } U &=
\mathbb{Z}_3 \oplus
\mathbb{Z}_{45} \oplus
\mathbb{Z}_{15} \oplus
\mathbb{Z}_{75}
\text{ then }\\
U_3 &=
\mathbb{Z}_3 \oplus
\mathbb{Z}_9 \oplus
\mathbb{Z}_3 \oplus
\mathbb{Z}_3
\text{ and }\\
U_5 &=
(e) \oplus
\mathbb{Z}_5 \oplus
\mathbb{Z}_5 \oplus
\mathbb{Z}_{25}.
\end{align*}
\end{example}
\begin{lemma}
Suppose $x \in U_{p_1}$, $y \in U_{p_2}$ and $p_1 \neq p_2$. Then $<x, y > = 0$.
\end{lemma}
\begin{proof}
\begin{align*}
x \in U_{p_1}
\end{align*}
\end{proof}
\begin{align*}
H_1(Y, \mathbb{Z}) \cong \quot{\mathbb{Z}^n}{A\mathbb{Z}}\\
A \longrightarrow BAC^T \quad \text{Smith normal form}
\end{align*}
???????????????????????\\
In general
%no lecture at 29.04

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\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}
\begin{definition}
A knot $K$ is called (smoothly) slice if $K$ is smoothly concordant to an unknot. \\
Put differently: a knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk in $B^4$.
\end{definition}
\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{fact: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 Fact \ref{fact: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.
\\
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/ball_4_pushed_seifert.pdf_tex}}
}
\caption{$Y = F \cup \Sigma$ is a smooth closed surface.}
\label{fig:closed_surface}
\end{figure}
\noindent
\\
Pontryagin-Thom construction tells us that there exists a compact oriented three - manifold $\Omega \subset B^4$ such that $\partial \Omega = Y$.
Suppose $\Sigma$ is a Seifert surface and $V$ a Seifert form defined on $\Sigma$: ${(\alpha, \beta) \mapsto \Lk(\alpha, \beta^+)}$. Suppose $\alpha, \beta \in H_1(\Sigma, \mathbb{Z})$, i.e. there are cycles and
$\alpha, \beta \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z}))$. 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
$\Lk(\alpha, \beta^+) = A \cdot B^+$. But $A$ and $B$ are disjoint, so $\Lk(\alpha, \beta^+) = 0$. Then the Seifert form is zero.
\\
\noindent
Let us consider following maps:
\[
\Sigma \overset{\phi} \longhookrightarrow Y \overset{\psi} \longhookrightarrow \Omega.
\]
Let $\phi_*$ and $\psi_*$ be induced maps on the homology group. If an element $\gamma \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z}))$, then $\gamma \in \ker \phi_*$ or $\gamma \in \ker \psi_*$.
%
%
%
\begin{proposition}
\[
\dim \ker (H_1(Y, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z})) = \frac{1}{2} b_1(Y),
\]
where $b_1$ is first Betti number.
\end{proposition}
\begin{proof}
Consider the following long exact sequence for a pair $(\Omega, Y)$:
\begin{align*}
& 0 \to H_3(\Omega) \to H_3(\Omega, Y) \to
\\
\to & H_2(Y) \to H_2(\Omega) \to H_2(\Omega, Y) \to \\
\to & H_1(Y) \to H_1(\Omega) \to H_1(\Omega, Y) \to \\
\to & H_0(Y) \to H_0(\Omega) \to 0
\end{align*}
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_1(\Omega, Y) &\cong H^1(\Omega).
\end{align*}
Therefore $\dim_{\mathbb{Q}} \quot{H_1(Y)}{V}
= \dim_{\mathbb{Q}} V
$.\\
\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 a $K$
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{proof}
\noindent
Let $V =
\begin{pmatrix}
0 & A\\
B & C
\end{pmatrix}$
\begin{align*}
\det (tV - V^T) = \det (tA - B^T) - \det(tB - A^T)
\end{align*}
\begin{corollary}
\label{cor:slice_alex}
If $K$ is a slice knot then there exists $f \in \mathbb{Z}[t^{\pm 1}]$ such that $\Delta_K(t) = f(t) \cdot f(t^{-1})$.
\end{corollary}
\begin{example}
Figure eight knot is not slice.
\end{example}
\begin{fact}
If $K$ is slice, then the signature $\sigma(K) \equiv 0$.
\end{fact}

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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 $H_1(Y, \mathbb{Z})$.
\\???????????????\\
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 $\mathbb{Z}$ 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}
\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 $(t - \zeta)^2 \vert P$ (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}
\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}

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lec_25_02.tex Normal file
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\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
$D_{\pi |_L}$ 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]
\label{theo: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}
\begin{definition}
\label{def:lk_via_homo}
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 $\beta$:
\[
\alpha \in H_1(S^3 \setminus \nu(\beta), \mathbb{Z}) \cong \mathbb{Z}.\]
\end{definition}
\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}

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\subsection{Slice knots and metabolic form}
\begin{theorem}
\label{the:sign_slice}
If $K$ is slice,
then $\sigma_K(t)
= \sign ( (1 - t)S +(1 - \bar{t})S^T)$
is zero except possibly of finitely many points and $\sigma_K(-1) = \sign(S + S^T) \neq 0$.
\end{theorem}
\begin{lemma}
\label{lem:metabolic}
If $V$ is a Hermitian matrix ($\bar{V} = V^T$), $V$ is of size $2n \times 2n$,
$
V = \begin{pmatrix}
0 & A \\
\bar{A}^T & B
\end{pmatrix}
$ and $\det V \neq 0$ then $\sigma(V) = 0$.
\end{lemma}
\begin{definition}
A Hermitian form $V$ is metabolic if $V$ has structure
$\begin{pmatrix}
0 & A\\
\bar{A}^T & B
\end{pmatrix}$ with half-dimensional null-space.
\end{definition}
\noindent
Theorem \ref{the:sign_slice} can be also express as follow:
non-degenerate metabolic hermitian form has vanishing signature.
\begin{proof}
\noindent
We note that $\det(S + S^T) \neq 0$. Hence $\det ( (1 - t) S + (1 - \bar{t})S^T)$ is not identically zero on $S^1$, so it is non-zero except possibly at finitely many points. We apply the Lemma \ref{lem:metabolic}.
\\
Let $t \in S^1 \setminus \{1\}$.
Then:
\begin{align*}
\det((1 - t) S + (1 - \bar{t}) S^T) =&
\det((1 - t) S + (t\bar{t} - \bar{t}) S^T) =\\
&\det((1 - t) (S - \bar{t} - S^T)) =
\det((1 -t)(S - \bar{t} S^T)).
\end{align*}
As $\det (S + S^T) \neq 0$, so $S - \bar{t}S^T \neq 0$.
\end{proof}
\begin{corollary}
If $K \sim K^\prime$ then for all but finitely many $t \in S^1 \setminus \{1\}: \sigma_K(t) = -\sigma_{K^\prime}(t)$.
\end{corollary}
\begin{proof}
If $ K \sim K^\prime$ then $K \# K^\prime$ is slice.
\[
\sigma_{-K^\prime}(t) = -\sigma_{K^\prime}(t)
\]
The signature gives a homomorphism from the concordance group to $\mathbb{Z}$.
Remark: if $t \in S^1$ is not algebraic over $\mathbb{Z}$, then $\sigma_K(t) \neq 0$
(we can use the argument that $\mathscr{C} \longrightarrow \mathbb{Z}$ as well).
\end{proof}
\subsection{Four genus}
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.7\textwidth}{!}{\input{images/genus_2_bordism.pdf_tex}}
}
\caption{$K$ and $K^\prime$ are connected by a genus $g$ surface.}\label{fig:genus_2_bordism}
\end{figure}
\begin{proposition}[Kawauchi inequality]
If there exists a genus $g$ surface as in Figure \ref{fig:genus_2_bordism}
then for almost all
$t \in S^1 \setminus \{1\}$ we have
$\vert
\sigma_K(t) - \sigma_{K^\prime}(t)
\vert \leq 2 g$.
\end{proposition}
% Kawauchi Chapter 12 ???
% Borodzik 2010 Morse theory for plane algebraic curves
\begin{lemma}
If $K$ bounds a genus $g$ surface $X \in B^4$ and $S$ is a Seifert form then ${S \in M_{2n \times 2n}}$ has a block structure $\begin{pmatrix}
0 & A\\
B & C
\end{pmatrix}$, where $0$ is $(n - g) \times (n - g)$ submatrix.
\end{lemma}
\begin{proof}
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/genus_bordism_zeros.pdf_tex}}
}
\caption{There exists a $3$ - manifold $\Omega$ such that $\partial \Omega = X \cup \Sigma$.}\label{fig:omega_in_B_4}
\end{figure}
\noindent
Let $K$ be a knot and $\Sigma$ its Seifert surface as in Figure \ref{fig:omega_in_B_4}.
There exists a $3$ - submanifold
$\Omega$ such that
$\partial \Omega = Y = X \cup \Sigma$
(by Thom-Pontryagin construction).
If $\alpha, \beta \in \ker (H_1(\Sigma) \longrightarrow H_1(\Omega))$,
then ${\Lk(\alpha, \beta^+) = 0}$. Now we have to determine the size of the kernel. We know that
${\dim H_1(\Sigma) = 2 n}$. When we glue $\Sigma$ (genus $n$) and $X$ (genus $g$) along a circle we get a surface of genus $n + g$. Therefore $\dim H_1 (Y) = 2 n + 2 g$. Then:
\[
\dim (\ker (H_1(Y) \longrightarrow H_1(\Omega)) = n + g.
\]
So we have $H_1(W)$ of dimension
$2 n + 2 g$
- the image of $H_1(Y)$
with a subspace
corresponding to the image of $H_1(\Sigma)$ with dimension $2 n$ and a subspace corresponding to the kernel
of $H_1(Y) \longrightarrow H_1(\Omega)$ of size $n + g$.
We consider minimal possible intersection of this subspaces that corresponds to the kernel of the composition $H_1(\Sigma) \longrightarrow H_1(Y) \longrightarrow H_1(\Omega)$. As the first map is injective, elements of the kernel of the composition have to be in the kernel of the second map.
So we can calculate:
\[
\dim \ker (H_1(\Sigma) \longrightarrow H_1(\Omega)) = 2 n + n + g -2 n - 2 g = n - g.
\]
\end{proof}
\begin{corollary}
If $t$ is not a root of
$\det (tS - S^T) $, then
$\vert \sigma_K(t) \vert \leq 2g$.
\end{corollary}
\begin{fact}
If there exists cobordism of genus $g$ between $K$ and $K^\prime$ like shown in Figure \ref{fig:proof_for_bound_disk}, then $K \# -K^\prime$ bounds a surface of genus $g$ in $B^4$.
\end{fact}
\begin{figure}[H]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.7\textwidth}{!}{\input{images/genus_bordism_proof.pdf_tex}}
}
\caption{If $K$ and $K^\prime$ are connected by a genus $g$ surface, then $K \# -K^\prime$ bounds a genus $g$ surface.}\label{fig:proof_for_bound_disk}
\end{figure}
\begin{definition}
The (smooth) four genus $g_4(K)$ is the minimal genus of the surface $\Sigma \in B^4$ such that $\Sigma$ is compact, orientable and $\partial \Sigma = K$.
\end{definition}
\noindent
Remarks:
\begin{enumerate}[label={(\arabic*)}]
\item
$3$ - genus is additive under taking connected sum, but $4$ - genus is not,
\item
for any knot $K$ we have $g_4(K) \leq g_3(K)$.
\end{enumerate}
\begin{example}
\begin{itemize}
\item Let $K = T(2, 3)$. $\sigma(K) = -2$, therefore $T(2, 3)$ isn't a slice knot.
\item Let $K$ be a trefoil and $K^\prime$ a mirror of a trefoil. $g_4(K^\prime) = 1$, but $g_4(K \# K^\prime) = 0$, so we see that $4$-genus isn't additive,
\item
the equality:
\[
g_4(T(p, q) ) = \frac{1}{2} (p - 1) (g -1)
\]
was conjecture in the '70 and proved by P. Kronheimer and T. Mrówka (1994).
% OZSVATH-SZABO AND RASMUSSEN
\end{itemize}
\end{example}
\begin{proposition}
$g_4 (T(p, q) \# -T(r, s))$ is in general hopelessly unknown.
\end{proposition}
\begin{proposition}
Supremum of the signature function of the knot is bounded almost everywhere by two times $4$ - genus:
\[
\ess \sup \vert \sigma_K(t) \vert \leq 2 g_4(K).
\]
\end{proposition}
\subsection{Topological genus}
\begin{definition}
A knot $K$ is called topologically slice if $K$ bounds a topological locally flat disc in $B^4$ (i.e. the disk has tubular neighbourhood).
\end{definition}
\begin{theorem}[Freedman, '82]
If $\Delta_K(t) = 1$, then $K$ is topologically slice (but not necessarily smoothly slice).
\end{theorem}
\begin{theorem}[Powell, 2015]
If $K$ is genus $g$
(topologically flat)
cobordant to $K^\prime$,
then
\[
\vert \sigma_K(t) - \sigma_{K^\prime}(t) \vert \leq 2 g
\]
if $g_4^{\mytop}(K) \geq \ess \sup \vert \sigma_K(t) \vert$.
\end{theorem}
\noindent
The proof for smooth category was based on following equality:
\[
\dim \ker (H_1 (Y) \longrightarrow H_1(\Omega)) = \frac{1}{2} \dim H_1(Y).
\]
For this equality we assumed that there exists a $3$ - dimensional manifold $\Omega$ (as shown in Figure \ref{fig:omega_in_B_4}) which was guaranteed by Pontryagin-Thom Construction.\\
Pontryagin-Thom Construction relays on taking $\Omega$ as preimage of regular value:
\[
H^1 (B^4 \setminus Y, \mathbb{Z}) = [B^4 \setminus Y, S^1],
\]
what relies on Sard's theorem, that the set of regular values has positive measure. But Sard's theorem doesn't work for topologically locally flat category. So there was a gap in the proof for topological locally flat category - the existence of $\Omega$.\\
\noindent
Remark: unless $p=2$ or $p = 3 \wedge q = 4$:
\[
g_4^{\mytop} (T(p, q)) < q_4(T(p, q)).
\]
% Wilczyński '93
%Feller 2014
%Baoder 2017
%Lemark
\\
\noindent
From the category of cobordant knots (or topologically cobordant knots) there exists a map to $\mathbb{Z}$ given by signature function. To any element $K$ we can associate a form
\[
(1 - t)S + (1 - \bar{t})S^T) \in W(\mathbb{Z}[t, t^{-1}]).
\] This association is not well define because id depends on the choice of Seifert form. However, different choices lead ever to congruent forms ($S \mapsto CSC^T$) or induced the change on the form by adding or subtracting a hyperbolic element.
\begin{definition}
The Witt group $W$ of $\mathbb{Z}[t, t^{-1}]$ elements are classes of non-degenerate
forms over $\mathbb{Z}[t, t^{-1}]$ under the equivalence relation $V \sim W$ if $V \oplus - W$ is metabolic.
\end{definition}
\noindent
If $S$ differs from $S^\prime$ by a row extension, then
$(1 - t) S + (1 - \bar{t}^{-1}) S^T$ is Witt equivalence to $(1 - t) S^\prime + (1 - t^{-1})S^T$.
\\
\noindent
A form is meant as hermitian with respect to this involution: $A^T = A: (a, b) = \bar{(a, b)}$.
\\
$
W(\mathbb{Z}_p) = \mathbb{Z}_2 \oplus
\mathbb{Z}_2$ or
$\mathbb{Z}_4$
\\
???????????????????????
\\
$\sum a_gt^j \longrightarrow \sum a_g t^{-1}$\\
\begin{theorem}[Levine '68]
\[
W(\mathbb{Z}[t^{\pm 1}])
\longrightarrow \mathbb{Z}_2^\infty \oplus
\mathbb{Z}_4^\infty \oplus
\mathbb{Z}
\]
\end{theorem}

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\begin{definition}
A square hermitian matrix $A$ of size $n$ with coefficients in \\
the Blanchfield pairing if:
$H_1(\bar{X}$
\end{definition}
field of fractions

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