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\put(0.2866221,0.08627117){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.16212197\unitlength}\raggedright genus $1$\\ \end{minipage}}}% + \put(0.87444349,0.06347455){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.1438923\unitlength}\raggedright genus $3$\\ \end{minipage}}}% + \end{picture}% +\endgroup% diff --git a/images/torus_1_2_3.svg b/images/torus_1_2_3.svg new file mode 100644 index 0000000..41561e2 --- /dev/null +++ b/images/torus_1_2_3.svg @@ -0,0 +1,2101 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + image/svg+xml + + + + + + + $\Sigma$ + + + + + + + + + + + + + + + + + + + + + + + + genus $0$ genus $2$ genus $1$ genus $3$ + diff --git a/lectures_on_knot_theory.tex b/lectures_on_knot_theory.tex index 746362f..99b6955 100644 --- a/lectures_on_knot_theory.tex +++ b/lectures_on_knot_theory.tex @@ -48,34 +48,41 @@ {\bfseries}{}% {\newline}{}% \theoremstyle{break} -\newtheorem{lemma}{Lemma} -\newtheorem{fact}{Fact} -\newtheorem{corollary}{Corollary} -\newtheorem{example}{Example} -\newtheorem{definition}{Definition} -\newtheorem{theorem}{Theorem} -\newtheorem{proposition}{Proposition} - +\newtheorem{lemma}{Lemma}[section] +\newtheorem{fact}{Fact}[section] +\newtheorem{corollary}{Corollary}[section] +\newtheorem{proposition}{Proposition}[section] +\newtheorem{example}{Example}[section] +\newtheorem{definition}{Definition}[section] +\newtheorem{theorem}{Theorem}[section] \newcommand{\contradiction}{% \ensuremath{{\Rightarrow\mspace{-2mu}\Leftarrow}}} \newcommand*\quot[2]{{^{\textstyle #1}\big/_{\textstyle #2}}} + \newcommand{\overbar}[1]{% \mkern 1.5mu=\overline{% \mkern-1.5mu#1\mkern-1.5mu}% \mkern 1.5mu} +\newcommand{\sdots}{\smash{\vdots}} + \AtBeginDocument{\renewcommand{\setminus}{% \mathbin{\backslash}}} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\rank}{rank} +\DeclareMathOperator{\Gl}{Gl} + +\titleformat{\section}{\normalfont \fontsize{12}{15} \bfseries}{% + Lecture\ \thesection}% + {2.3ex plus .2ex}{} +\titlespacing*{\section} +{0pt}{16.5ex plus 1ex minus .2ex}{4.3ex plus .2ex} -\titleformat{\section}{\normalfont \large \bfseries}{% - Lecture\ \thesection}{2.3ex plus .2ex}{} - -%\setlist[itemize]{topsep=0pt,before=%\leavevmode\vspace{0.5em}} +\setlist[itemize]{topsep=0pt,before=% +\leavevmode\vspace{0.5em}} \input{knots_macros} @@ -87,7 +94,7 @@ %\newpage %\input{myNotes} -\section{\hfill\DTMdate{2019-02-25}} +\section{Basic definitions \hfill\DTMdate{2019-02-25}} \begin{definition} A knot $K$ in $S^3$ is a smooth (PL - smooth) embedding of a circle $S^1$ in $S^3$: \begin{align*} @@ -96,12 +103,13 @@ A knot $K$ in $S^3$ is a smooth (PL - smooth) embedding of a circle $S^1$ in $S^ \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}, -\includegraphics[width=0.08\textwidth]{trefoil.png}. +\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} @@ -172,7 +180,7 @@ Let $D$ be a diagram of an oriented link (to each component of a link we add an 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. -\section*{Reidemeister moves} +\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\\ @@ -197,7 +205,7 @@ deformed into each other by a finite sequence of Reidemeister moves (and isotopy %Singularities of Differentiable Maps %Authors: Arnold, V.I., Varchenko, Alexander, Gusein-Zade, S.M. -\subsection*{Seifert surface} +\subsection{Seifert surface} \noindent Let $D$ be an oriented diagram of a link $L$. We change the diagram by smoothing each crossing: \begin{align*} @@ -221,7 +229,7 @@ Note: in general the obtained surface doesn't need to be connected, but by takin \begin{figure}[H] \begin{center} -\includegraphics[width=0.6\textwidth]{seifert_connect.png} +\includegraphics[width=0.4\textwidth]{seifert_connect.png} \end{center} \caption{Connecting two surfaces.} \label{fig:SeifertConnect} @@ -259,13 +267,16 @@ On a diagram $L$ consider all crossings between $\alpha$ and $\beta$. Let $N_+$ \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 $S^3$: +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 $\beta$: \[ \alpha \in H_1(S^3 \setminus \nu(\beta), \mathbb{Z}) \cong \mathbb{Z}.\] + + \begin{example} -\begin{itemize}\hfill +\begin{itemize} \item -Hopf link\hfill +Hopf link \begin{figure}[H] \fontsize{20}{10}\selectfont \centering{ @@ -274,7 +285,7 @@ Hopf link\hfill } \end{figure} \item -$T(6, 2)$ link\hfill +$T(6, 2)$ link \begin{figure}[H] \fontsize{20}{10}\selectfont \centering{ @@ -285,94 +296,263 @@ $T(6, 2)$ link\hfill \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 $L$. +\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 -$V \rightarrow AVA^T$ for $A \in $ +$V \rightarrow AVA^T$, where $A$ is a matrix with integer coefficients, + \item + $V \rightarrow \begin{pmatrix} - \alpha & * \\ - \gamma^{*} & \delta -\end{pmatrix} -$\\ -\[ - \begin{pmatrix} \begin{array}{c|c} - \epsilon' [T|_A]\epsilon & \ast \\ - \hline - 0 & _{\overline{B}'} [\overline{T}] - _{\overline{B}\vphantom{\overline{B}'}} + V & + \begin{matrix} + \ast & 0 \\ + \sdots & \sdots\\ + \ast & 0 + \end{matrix} \\ + \hline + \begin{matrix} + \ast & \dots & \ast\\ + 0 & \dots & 0 + \end{matrix} + & + \begin{matrix} + 0 & 0\\ + 1 & 0 + \end{matrix} \end{array} - \end{pmatrix} -\]\\ -\[\left| -\begin{array}{cr} - Q & \begin{matrix} 0 \\ 0 \end{matrix} \\ - \begin{matrix} 2 & 3 \end{matrix} & -1 -\end{array} -\right|\] -\\ -\[ -\left[ -\begin{array}{c@{}c@{}c} - \left[\begin{array}{cc} - a_{11} & a_{12} \\ - a_{21} & a_{22} \\ - \end{array}\right] & \mathbf{0} & \mathbf{0} \\ - \mathbf{0} & \left[\begin{array}{ccc} - b_{11} & b_{12} & b_{13}\\ - b_{21} & b_{22} & b_{23}\\ - b_{31} & b_{32} & b_{33}\\ - \end{array}\right] & \mathbf{0}\\ -\mathbf{0} & \mathbf{0} & \left[ \begin{array}{cc} -c_{11} & c_{12} \\ -c_{21} & c_{22} \\ -\end{array}\right] \\ -\end{array}\right] -\] \\ -\[ -\begin{bmatrix} - \begin{bmatrix} - a_{11} & a_{12}\\ - a_{21} & a_{22}\\ - \end{bmatrix} & \mathbf{0} & \mathbf{0} \\ - \mathbf{0} & \begin{bmatrix} - b_{11} & b_{12} & b_{13}\\ - b_{21} & b_{22} & b_{23}\\ - b_{31} & b_{32} & b_{33}\\ - \end{bmatrix} & \mathbf{0} \\ - \mathbf{0} & \mathbf{0} & \begin{bmatrix} - c_{11} & c_{12}\\ - c_{21} & c_{22}\\ - \end{bmatrix} \\ -\end{bmatrix} -\]\\ -\setlength{\arraycolsep}{2em} -\newcommand{\lbrce}{\smash{\left.\rule{0pt}{25pt}\right\}}} -\newcommand{\rbrce}{\smash{\left\{\rule{0pt}{25pt}\right.}} -\newcommand{\sdots}{\smash{\vdots}} -\[ - \begin{pmatrix} - 0 & 0 & 0 \\ - \sdots & \sdots\makebox[0pt][l]{$\lbrce\left\lceil\frac i2\right\rceil$} & \sdots \\ - 0 & 0 & \\ - & & 0 \\ - & & \makebox[0pt][r]{$\left\lfloor\frac i2\right\rfloor\rbrce$}\sdots \\ - 0 & & 0 - \end{pmatrix} -\] - +\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} +\section{\hfill\DTMdate{2019-03-04}} +\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}("joke")\\ +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} +\[ +H^* (S^3, S^3 \setminus N) \cong H^* (N, \partial N) +\] +\\ +?????????????? +\\ + +\end{proof} + +\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 +$S = \begin{pmatrix} +-1 & -1 \\ +0 & -1 +\end{pmatrix}$. +\[ +\Delta_K(t) = \det +\begin{pmatrix} +-t + 1 & -t\\ +1 & -t +1 +\end{pmatrix} += (t -1)^2 + t = t^2 - t +1 \ne 1 +\Rightarrow \text{trefoil is not trivial} +\] +\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 + \cdots + a_1 t^l )= k - l. +\end{align*} +\end{lemma} +\begin{proof} + +\end{proof} +%removing one disk from surface doesn't change $H_1$ (only $H_2$) \section{} \begin{example} \begin{align*} @@ -389,17 +569,6 @@ Prove that if $K$ is negative amphichiral, then $K \# K$ in $\mathbf{C}$ \end{example} -\section{\hfill\DTMdate{2019-03-04}} -\begin{proof}("joke")\\ -Let $K \in S^3$ be a knot and $N$ be its tubular neighbourhood. -\begin{align*} -H^1(S^3 \setminus N ) \cong H^1(S^3 \setminus K) -\end{align*} -For a pair $(S^3, S^3 \setminus N)$ we have: -\begin{align*} -H^0(S^3) -\end{align*} -\end{proof} \section{\hfill\DTMdate{2019-03-18}} \begin{definition} A knot $K$ is called (smoothly) slice if $K$ is smoothly concordant to an unknot. \\ @@ -447,7 +616,11 @@ $H_2(X, \mathbb{Z}) \times H_2(X, \mathbb{Z}) \longrightarrow \mathbb{Z}$ - symmetric, non singular. \\ 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} \section{\hfill\DTMdate{2019-04-15}} In other words:\\ @@ -459,27 +632,28 @@ of $H_2(Y, \mathbb{Z}$, then $A = (b_i, b_y)$ \\??\\ is a matrix of intersection \end{align*} In particular $\mid \det A\mid = \# H_1(Y, \mathbb{Z}$.\\ That means - what is happening on boundary is a measure of degeneracy. -\\ -\vspace{1cm} + \begin{center} \begin{tikzcd} [ -column sep=tiny, -row sep=small, -ar symbol/.style = {draw=none,"\textstyle#1" description,sloped}, -isomorphic/.style = {ar symbol={\cong}}, + 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} +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] &\\ + \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} + The intersection form on a four-manifold determines the linking on the boundary. \\ \noindent @@ -780,4 +954,22 @@ field of fractions \section{balagan} +\begin{comment} +\setlength{\arraycolsep}{2em} +\newcommand{\lbrce}{\smash{\left.\rule{0pt}{25pt}\right\}}} +\newcommand{\rbrce}{\smash{\left\{\rule{0pt}{25pt}\right.}} +\[ + \begin{pmatrix} + 0 & 0 & 0 \\ + \sdots & \sdots\makebox[0pt][l]{$\lbrce\left\lceil\frac i2\right\rceil$} & \sdots \\ + 0 & 0 & \\ + \hline + + & & 0 \\ + & & \makebox[0pt][r]{$\left\lfloor\frac i2\right\rfloor\rbrce$}\sdots \\ + 0 & & 0 + \end{pmatrix} +\] + +\end{comment} \end{document}