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lectures_on_knot_theory.tex
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@ -50,6 +50,7 @@
\theoremstyle{break}
\newtheorem{lemma}{Lemma}
\newtheorem{fact}{Fact}
\newtheorem{corollary}{Corollary}
\newtheorem{example}{Example}
\newtheorem{definition}{Definition}
\newtheorem{theorem}{Theorem}
@ -190,6 +191,12 @@ deformed into each other by a finite sequence of Reidemeister moves (and isotopy
%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:
@ -198,24 +205,173 @@ Let $D$ be an oriented diagram of a link $L$. We change the diagram by smoothing
\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 $\partial \Sigma = L$.\\
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 $\partial D_2$.
\begin{figure}[H]
\fontsize{15}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.7\textwidth}{!}{\input{images/seifert_surface.pdf_tex}}
\resizebox{0.8\textwidth}{!}{\input{images/seifert_alg.pdf_tex}}
\caption{Constructing a Seifert surface.}
\label{fig:surfaceSeifert}
\label{fig:SeifertAlg}
}
\end{figure}
\includegraphics[width=0.3\textwidth]{seifert3d.png},
\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 $\partial D_2$.
% transversality theorem
%Thom ?
%Singularities of Differentiable Maps
%Authors: Arnold, V.I., Varchenko, Alexander, Gusein-Zade, S.M.
\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{15}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.8\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 $\nu(\beta)$ be a tubular neighbourhood of a closed simple curve $\beta$. 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 in complement of $\beta$ in $S^3$:
\[
\alpha \in H_1(S^3 \setminus \nu(\beta), \mathbb{Z}) \cong \mathbb{Z}.\]
\begin{example}
\begin{itemize}\hfill
\item
Hopf link\hfill
\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\hfill
\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}
Let $L$ be a link and $\Sigma$ be a 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. Let $lk(\alpha_i, \alpha_i^+) = \{a_{ij}\}$. Then the matrix $S = \{a_{ij}\}_{i, j =1}^n$ is called a Seifert matrix for $L$.
\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
$V \rightarrow AVA^T$ for $A \in $
\item
$V \rightarrow
\begin{pmatrix}
\alpha & * \\
\gamma^{*} & \delta
\end{pmatrix}
$\\
\[
\begin{pmatrix}
\begin{array}{c|c}
\epsilon' [T|_A]\epsilon & \ast \\
\hline
0 & _{\overline{B}'} [\overline{T}]
_{\overline{B}\vphantom{\overline{B}'}}
\end{array}
\end{pmatrix}
\]\\
\[\left|
\begin{array}{cr}
Q & \begin{matrix} 0 \\ 0 \end{matrix} \\
\begin{matrix} 2 & 3 \end{matrix} & -1
\end{array}
\right|\]
\\
\[
\left[
\begin{array}{c@{}c@{}c}
\left[\begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{array}\right] & \mathbf{0} & \mathbf{0} \\
\mathbf{0} & \left[\begin{array}{ccc}
b_{11} & b_{12} & b_{13}\\
b_{21} & b_{22} & b_{23}\\
b_{31} & b_{32} & b_{33}\\
\end{array}\right] & \mathbf{0}\\
\mathbf{0} & \mathbf{0} & \left[ \begin{array}{cc}
c_{11} & c_{12} \\
c_{21} & c_{22} \\
\end{array}\right] \\
\end{array}\right]
\] \\
\[
\begin{bmatrix}
\begin{bmatrix}
a_{11} & a_{12}\\
a_{21} & a_{22}\\
\end{bmatrix} & \mathbf{0} & \mathbf{0} \\
\mathbf{0} & \begin{bmatrix}
b_{11} & b_{12} & b_{13}\\
b_{21} & b_{22} & b_{23}\\
b_{31} & b_{32} & b_{33}\\
\end{bmatrix} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \begin{bmatrix}
c_{11} & c_{12}\\
c_{21} & c_{22}\\
\end{bmatrix} \\
\end{bmatrix}
\]\\
\setlength{\arraycolsep}{2em}
\newcommand{\lbrce}{\smash{\left.\rule{0pt}{25pt}\right\}}}
\newcommand{\rbrce}{\smash{\left\{\rule{0pt}{25pt}\right.}}
\newcommand{\sdots}{\smash{\vdots}}
\[
\begin{pmatrix}
0 & 0 & 0 \\
\sdots & \sdots\makebox[0pt][l]{$\lbrce\left\lceil\frac i2\right\rceil$} & \sdots \\
0 & 0 & \\
& & 0 \\
& & \makebox[0pt][r]{$\left\lfloor\frac i2\right\rfloor\rbrce$}\sdots \\
0 & & 0
\end{pmatrix}
\]
\item
inverse of (2)
\end{enumerate}
\end{theorem}
\section{}
\begin{example}