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Maria Marchwicka 2019-06-15 17:23:44 +02:00
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87
lec_4.tex
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@ -47,8 +47,8 @@ $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 $-\{K\} = \{mK\}$.
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).
@ -70,7 +70,7 @@ Remark: $K \sim K^{\prime} \Leftrightarrow K \# -K^{\prime}$ is slice.
\end{figure}
\noindent
\\
Pontryagin-Thom construction tells us that there exists a compact three - manifold $\Omega \subset B^4$ such that $\partial \Omega = Y$.
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
@ -78,16 +78,13 @@ $\partial B^+ = \beta^+$.
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}
@ -97,20 +94,76 @@ Let $\phi_*$ and $\psi_*$ be induced maps on the homology group. If an element $
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_1(Y) \to H_1(\Omega) \to H_1(\Omega, Y) \to \\
\to & H_0(Y) \to H_0(\Omega) \to 0
\end{align*}
\end{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_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}
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}
\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}

62
lectures_on_knot_theory.tex
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@ -200,59 +200,7 @@ 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}
\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
$.\\
\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}
\section{\hfill\DTMdate{2019-04-15}}
@ -776,5 +724,15 @@ The proof is the same as over $\mathbb{Z}$.
%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}