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inkscape:zoom="0.14715637" + inkscape:cx="250.88609" + inkscape:cy="270.29308" inkscape:document-units="px" inkscape:current-layer="layer1" showgrid="false" - inkscape:window-width="1388" + inkscape:window-width="1397" inkscape:window-height="855" inkscape:window-x="0" inkscape:window-y="1" @@ -1858,8 +1858,8 @@ + originx="-390.9762" + originy="80.436871" /> @@ -1877,21 +1877,11 @@ inkscape:label="Layer 1" inkscape:groupmode="layer" id="layer1" - transform="translate(-390.97626,-940.39914)"> + transform="translate(-390.97626,-940.39913)"> $\Sigma$ - - $\Sigma$ $\Sigma$ - - $\Sigma$ \shortstack{$lk(\alpha, \beta) = 3$} \shortstack{$\Lk(\alpha, \beta) = 3$} $\alpha$ $\alpha$  $\beta$ + id="flowPara11317" + style="font-size:40.00024796px;line-height:1.25">$\beta$  diff --git a/lectures_on_knot_theory.tex b/lectures_on_knot_theory.tex index 2328b94..48b7937 100644 --- a/lectures_on_knot_theory.tex +++ b/lectures_on_knot_theory.tex @@ -48,13 +48,16 @@ {\bfseries}{}% {\newline}{}% \theoremstyle{break} + \newtheorem{lemma}{Lemma}[section] \newtheorem{fact}{Fact}[section] \newtheorem{corollary}{Corollary}[section] \newtheorem{proposition}{Proposition}[section] \newtheorem{example}{Example}[section] +\newtheorem{problem}{Problem}[section] \newtheorem{definition}{Definition}[section] \newtheorem{theorem}{Theorem}[section] + \newcommand{\contradiction}{% \ensuremath{{\Rightarrow\mspace{-2mu}\Leftarrow}}} \newcommand*\quot[2]{{^{\textstyle #1}\big/_{\textstyle #2}}} @@ -66,6 +69,10 @@ \newcommand{\sdots}{\smash{\vdots}} + + + + \AtBeginDocument{\renewcommand{\setminus}{% \mathbin{\backslash}}} @@ -73,6 +80,8 @@ \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\Gl}{Gl} +\DeclareMathOperator{\Lk}{lk} + \titleformat{\section}{\normalfont \fontsize{12}{15} \bfseries}{% Lecture\ \thesection}% @@ -263,12 +272,12 @@ Remark: there are knots that admit non isotopic Seifert surfaces of minimal genu \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: $lk(\alpha, \beta) = \frac{1}{2}(N_+ - N_-)$. +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 $\alpha$ 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 $\beta$: +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 $\beta$: \[ \alpha \in H_1(S^3 \setminus \nu(\beta), \mathbb{Z}) \cong \mathbb{Z}.\] @@ -306,7 +315,7 @@ where $b_1$ is first Betti number of $\Sigma$. \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. +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 @@ -581,19 +590,37 @@ g_{\big| \partial D^2} = f_{\big| \partial D^2.} &F: \mathbb{C}^2 \rightarrow \mathbb{C} \text{ a polynomial} \\ &F(0) = 0 \end{align*} -Fact (Milnor Singular Points of Complex Hypersurfaces): -\end{example} -%\end{comment} -An oriented knot is called negative amphichiral if the mirror image $m(K)$ if $K$ is equivalent the reverse knot of $K$. \\ -\begin{example}[Problem] -Prove that if $K$ is negative amphichiral, then $K \# K$ in -$\mathbf{C}$ +\end{example} +???????????? +\\ +\noindent +as a corollary we see that $K_T^{n, }$ ???? \\ +is not slice unless $m=0$. +\begin{theorem} +The map $j: \mathscr{C} \longrightarrow \mathbb{Z}^{\infty}$ is a surjection that maps ${K_n}$ to a linear independent set. Moreover $\mathscr{C} \cong \mathbb{Z}$ +\end{theorem} + + +\begin{fact}[Milnor Singular Points of Complex Hypersurfaces] +\end{fact} +%\end{comment} +\noindent +An oriented knot is called negative amphichiral if the mirror image $m(K)$ of $K$ is equivalent the reverse knot of $K$: $K^r$. \\ +\begin{problem} +Prove that if $K$ is negative amphichiral, then $K \# K = 0$ in +$\mathscr{C}$. +% +%\\ +%Hint: $ -K = m(K)^r = (K^r)^r = K$ +\end{problem} +\begin{example} +Figure 8 knot is negative amphichiral. \end{example} % % % -\section{\hfill\DTMdate{2019-03-18}} +\section{Concordance group \hfill\DTMdate{2019-03-18}} \begin{definition} A knot $K$ is called (smoothly) slice if $K$ is smoothly concordant to an unknot. \\ A knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk in $B^4$. @@ -628,7 +655,7 @@ $K_1 \# K_2 \sim {K_1}^{\prime} \# {K_2}^{\prime}$. \fontsize{10}{10}\selectfont \centering{ \def\svgwidth{\linewidth} -\resizebox{0.8\textwidth}{!}{\input{images/concordance_sum.pdf_tex}} +\resizebox{1\textwidth}{!}{\input{images/concordance_sum.pdf_tex}} } \caption{Sketch for Fakt \ref{fakt:concordance_connected}.} \label{fig:concordance_sum} @@ -639,23 +666,40 @@ $K_1 \# K_2 \sim {K_1}^{\prime} \# {K_2}^{\prime}$. $K \# m(K) \sim $ the unknot. \end{fact} \noindent -Let $\mathscr{C}$ denote all equivalent classes for knots. $\mathscr{C}$ is a group under taking connected sums, with neutral element (the class defined by) an unknot and inverse element (a class defined by) a mirror image.\\ -The figure eight knot is a torsion element in $\mathscr{C}$ ($2K \sim $ the unknot).\\ -\begin{example}[Problem] -Are there in concordance group torsion elements that are not $2$ torsion elements? (open) -\end{example} +\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. - +\\ +\\ +\noindent +Let $\Omega$ be an oriented \\ +???????\\ +Suppose $\Sigma$ is a Seifert matrix with an intersection form ${(\alpha, \beta) \mapsto \Lk(\alpha, \beta^+)}$. Suppose $\alpha, \beta \in H_1(\Sigma, \mathbb{Z}$ (i.e. there are cycles). \\ +??????????????\\ +$\alpha, \beta \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z}))$. 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 +$\Lk(\alpha, \beta^+) = A \cdot B^+$ % % +\\ \section{\hfill\DTMdate{2019-04-08}} % % $X$ is a closed orientable four-manifold. Assume $\pi_1(X) = 0$ (it is not needed to define the intersection form). In particular $H_1(X) = 0$. $H_2$ is free (exercise). \begin{align*} -H_2(X, \mathbb{Z}) \xrightarrow{\text{Poincaré duality}} H^2(X, \mathbb{Z} ) \xrightarrow{\text{evaluation}}\Hom(H_2(X, \mathbb{Z}), \mathbb{Z}) +H_2(X, \mathbb{Z}) \xrightarrow{\text{Poincar\'e duality}} H^2(X, \mathbb{Z} ) \xrightarrow{\text{evaluation}}\Hom(H_2(X, \mathbb{Z}), \mathbb{Z}) \end{align*} Intersection form: $H_2(X, \mathbb{Z}) \times @@ -749,7 +793,7 @@ In general \section{\hfill\DTMdate{2019-05-20}} -Let $M$ be compact, oriented, connected four-dimensional manifold. If $H_1(M, \mathbb{Z}) = 0$ then there exists a +Let $M$ be compact, oriented, connected four-dimensional manifold. If ${H_1(M, \mathbb{Z}) = 0}$ then there exists a bilinear form - the intersection form on $M$: \begin{center} @@ -1006,22 +1050,95 @@ 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.}} +\noindent +\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 $\dim_{\mathbb{Q}} \quot{H_1(Y)}{V} += \dim_{\mathbb{Q}} V +$. +\end{proof} +\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 4\\ +?????\\ +has a subspace of dimension $g_{\Sigma}$ on which it is zero: \[ - \begin{pmatrix} - 0 & 0 & 0 \\ - \sdots & \sdots\makebox[0pt][l]{$\lbrce\left\lceil\frac i2\right\rceil$} & \sdots \\ - 0 & 0 & \\ - \hline - - & & 0 \\ - & & \makebox[0pt][r]{$\left\lfloor\frac i2\right\rfloor\rbrce$}\sdots \\ - 0 & & 0 - \end{pmatrix} +V = +\begin{pmatrix} + \begin{array}{c|c} + 0 & * \\ + \hline + * & * + \end{array} +\end{pmatrix} \] +\begin{align*} + \newcommand*{\AddLeft}[1]{% + \vadjust{% + \vbox to 0pt{% + \vss + \llap{$% + {#1}\left\{ + \vphantom{ + \begin{matrix}1\\\vdots\\0\end{matrix} + } + \right.\kern-\nulldelimiterspace + \kern0.5em + $}% + \kern0pt + }% + }% + } +V = \qquad +\begin{pmatrix} + 0 & \cdots & 0 & * & \cdots & * \\ + \vdots & & \vdots & \vdots & &\vdots \\ + 0 & \cdots & 0 & * & \cdots & * + \AddLeft{g_{\Sigma}}\\ + * & \cdots & * & * & \cdots & * \\ + \vdots & & \vdots & \vdots & &\vdots \\ + * & \cdots & * & * & \cdots & * + \end{pmatrix}_{2g_{\Sigma} \times 2g_{\Sigma}} +\end{align*} + +\begin{align*} +\newcommand\coolover[2]{\mathrlap{\smash{\overbrace{\phantom{% + \begin{matrix} #2 \end{matrix}}}^{\mbox{$#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}}% +\begin{matrix}% matrix for left braces +\vphantom{a}\\ + \coolleftbrace{A}{e \\ y\\ y}\\ + \coolleftbrace{B}{y \\i \\ m} +\end{matrix}% +\begin{bmatrix} +a & \coolover{R}{b & c & d} & x & \coolover{Z}{x & x}\\ + e & f & g & h & x & x & x \\ + y & y & y & y & y & y & y \\ + y & y & y & y & y & y & y \\ + y & y & y & y & y & y & y \\ + i & j & k & l & x & x & x \\ + m & \coolunder{S}{n & o} & \coolunder{W}{p & x & x} & x +\end{bmatrix}% +\begin{matrix}% matrix for right braces + \coolrightbrace{x \\ x \\ y\\ y}{T}\\ + \coolrightbrace{y \\ y \\ x }{U} +\end{matrix} +\end{align*} + -\end{comment} \end{document}