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sum of concordant knots

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Maria Marchwicka 2019-06-07 18:57:28 +02:00
parent 129b2608be
commit 11a97cecc9
4 changed files with 566 additions and 3 deletions
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32
lectures_on_knot_theory.tex
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@ -225,7 +225,7 @@ We smooth all the crossings, so we get a disjoint union of circles on the plane.
\end{figure}
\noindent
Note: in general the obtained surface 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 $\partial D_2$.
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}
@ -295,6 +295,13 @@ $T(6, 2)$ link:
\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 $\alpha_1, \dots, \alpha_n$.
@ -583,7 +590,9 @@ An oriented knot is called negative amphichiral if the mirror image $m(K)$ if $K
Prove that if $K$ is negative amphichiral, then $K \# K$ in
$\mathbf{C}$
\end{example}
%
%
%
\section{\hfill\DTMdate{2019-03-18}}
\begin{definition}
A knot $K$ is called (smoothly) slice if $K$ is smoothly concordant to an unknot. \\
@ -612,9 +621,19 @@ For any $K$, $K \# m(K)$ is slice.
\begin{fact}
Concordance is an equivalence relation.
\end{fact}
\begin{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{0.8\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.
@ -649,6 +668,13 @@ $A \cdot B$ doesn't depend of choice of $A$ and $B$ in their homology classes.
\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 $(b_1, ..., b_i)$ \\