lectures_on_knot_theory/lec_4.tex

\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}