diff --git a/images/milnor_singular.pdf b/images/milnor_singular.pdf new file mode 100644 index 0000000..284c792 Binary files /dev/null and b/images/milnor_singular.pdf differ diff --git a/images/milnor_singular.pdf_tex b/images/milnor_singular.pdf_tex new file mode 100644 index 0000000..9fa2892 --- /dev/null +++ b/images/milnor_singular.pdf_tex @@ -0,0 +1,60 @@ +%% Creator: Inkscape inkscape 0.92.2, www.inkscape.org +%% PDF/EPS/PS + LaTeX output extension by Johan Engelen, 2010 +%% Accompanies image file 'milnor_singular.pdf' (pdf, eps, ps) +%% +%% To include the image in your LaTeX document, write +%% \input{.pdf_tex} +%% instead of +%% \includegraphics{.pdf} +%% To scale the image, write +%% \def\svgwidth{} +%% \input{.pdf_tex} +%% instead of +%% \includegraphics[width=]{.pdf} +%% +%% Images with a different path to the parent latex file can +%% be accessed with the `import' package (which may need to be +%% installed) using +%% \usepackage{import} +%% in the preamble, and then including the image with +%% \import{}{.pdf_tex} +%% Alternatively, one can specify +%% \graphicspath{{/}} +%% +%% For more information, please see info/svg-inkscape on CTAN: +%% http://tug.ctan.org/tex-archive/info/svg-inkscape +%% +\begingroup% + \makeatletter% + \providecommand\color[2][]{% + \errmessage{(Inkscape) Color is used for the text in Inkscape, but the package 'color.sty' is not loaded}% + \renewcommand\color[2][]{}% + }% + \providecommand\transparent[1]{% + \errmessage{(Inkscape) Transparency is used (non-zero) for the text in Inkscape, but the package 'transparent.sty' is not loaded}% + \renewcommand\transparent[1]{}% + }% + \providecommand\rotatebox[2]{#2}% + \ifx\svgwidth\undefined% + \setlength{\unitlength}{595.27559055bp}% + \ifx\svgscale\undefined% + \relax% + \else% + \setlength{\unitlength}{\unitlength * \real{\svgscale}}% + \fi% + \else% + \setlength{\unitlength}{\svgwidth}% + \fi% + \global\let\svgwidth\undefined% + \global\let\svgscale\undefined% + \makeatother% + \begin{picture}(1,1.41428571)% + \put(0,0){\includegraphics[width=\unitlength,page=1]{milnor_singular.pdf}}% + \put(0.65155898,1.03673474){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.41757364\unitlength}\raggedright $F^{-1}(0)$\\ \end{minipage}}}% + \put(0.59036283,1.14112815){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.26998302\unitlength}\raggedright \end{minipage}}}% + \put(0.04886191,0.32141521){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{1.19914935\unitlength}\raggedright $L = F^{-1}(0) \cap S^3$\\ \end{minipage}}}% + \put(0.63715987,1.05113382){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.98633784\unitlength}\raggedright \end{minipage}}}% + \put(0.86754533,0.91794223){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.80274954\unitlength}\raggedright \end{minipage}}}% + \put(0.86394559,0.92514175){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.81714843\unitlength}\raggedright \end{minipage}}}% + \end{picture}% +\endgroup% diff --git a/images/milnor_singular.svg b/images/milnor_singular.svg new file mode 100644 index 0000000..36417de --- /dev/null +++ b/images/milnor_singular.svg @@ -0,0 +1,140 @@ + + + + + + + + + + image/svg+xml + + + + + + + + + + $F^{-1}(0)$ $L = F^{-1}(0) \cap S^3$ + diff --git a/images/torus_lambda.pdf b/images/torus_lambda.pdf new file mode 100644 index 0000000..72f9039 Binary files /dev/null and b/images/torus_lambda.pdf differ diff --git a/images/torus_lambda.pdf_tex b/images/torus_lambda.pdf_tex new file mode 100644 index 0000000..c020368 --- /dev/null +++ b/images/torus_lambda.pdf_tex @@ -0,0 +1,59 @@ +%% Creator: Inkscape inkscape 0.92.2, www.inkscape.org +%% PDF/EPS/PS + LaTeX output extension by Johan Engelen, 2010 +%% Accompanies image file 'torus_lambda.pdf' (pdf, eps, ps) +%% +%% To include the image in your LaTeX document, write +%% \input{.pdf_tex} +%% instead of +%% \includegraphics{.pdf} +%% To scale the image, write +%% \def\svgwidth{} +%% \input{.pdf_tex} +%% instead of +%% \includegraphics[width=]{.pdf} +%% +%% Images with a different path to the parent latex file can +%% be accessed with the `import' package (which may need to be +%% installed) using +%% \usepackage{import} +%% in the preamble, and then including the image with +%% \import{}{.pdf_tex} +%% Alternatively, one can specify +%% \graphicspath{{/}} +%% +%% For more information, please see info/svg-inkscape on CTAN: +%% http://tug.ctan.org/tex-archive/info/svg-inkscape +%% +\begingroup% + \makeatletter% + \providecommand\color[2][]{% + \errmessage{(Inkscape) Color is used for the text in Inkscape, but the package 'color.sty' is not loaded}% + \renewcommand\color[2][]{}% + }% + \providecommand\transparent[1]{% + \errmessage{(Inkscape) Transparency is used (non-zero) for the text in Inkscape, but the package 'transparent.sty' is not loaded}% + \renewcommand\transparent[1]{}% + }% + \providecommand\rotatebox[2]{#2}% + \ifx\svgwidth\undefined% + \setlength{\unitlength}{185.50670233bp}% + \ifx\svgscale\undefined% + \relax% + \else% + \setlength{\unitlength}{\unitlength * \real{\svgscale}}% + \fi% + \else% + \setlength{\unitlength}{\svgwidth}% + \fi% + \global\let\svgwidth\undefined% + \global\let\svgscale\undefined% + \makeatother% + \begin{picture}(1,0.62539361)% + \put(1.05585685,2.21258294){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.51818377\unitlength}\raggedright \end{minipage}}}% + \put(0,0){\includegraphics[width=\unitlength,page=1]{torus_lambda.pdf}}% + \put(0.89358894,0.61469018){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.10667827\unitlength}\raggedright $\lambda$\end{minipage}}}% + \put(0.01954926,0.16801593){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.15987467\unitlength}\raggedright $\mu$\end{minipage}}}% + \put(0,0){\includegraphics[width=\unitlength,page=2]{torus_lambda.pdf}}% + \put(0.8126573,0.13034845){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.24978585\unitlength}\raggedright $K$\end{minipage}}}% + \end{picture}% +\endgroup% diff --git a/images/torus_lambda.svg b/images/torus_lambda.svg new file mode 100644 index 0000000..f8b543e --- /dev/null +++ b/images/torus_lambda.svg @@ -0,0 +1,1096 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + image/svg+xml + + + + + + + $\Sigma$ + + + + $\lambda$ $\mu$ + + $K$ + diff --git a/lec_1.tex b/lec_1.tex index d61cf2a..5a94fae 100644 --- a/lec_1.tex +++ b/lec_1.tex @@ -70,7 +70,7 @@ Borromean link: A link diagram $D_{\pi}$ is a picture over projection $\pi$ of a link $L$ in $\mathbb{R}^3$($S^3$) to $\mathbb{R}^2$ ($S^2$) such that: \begin{enumerate}[label={(\arabic*)}] \item -${D_{\pi}}_{\big|L}$ is non degenerate: \includegraphics[width=0.05\textwidth]{LinkDiagram1.png}, +$D_{\pi |_L}$ is non degenerate: \includegraphics[width=0.05\textwidth]{LinkDiagram1.png}, \item the double points are not degenerate: \includegraphics[width=0.03\textwidth]{LinkDiagram2.png}, \item there are no triple point: \includegraphics[width=0.05\textwidth]{LinkDiagram3.png}. \end{enumerate} @@ -139,6 +139,7 @@ Note: the obtained surface isn't unique and in general doesn't need to be connec \end{figure} \begin{theorem}[Seifert] +\label{theo: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} % @@ -168,13 +169,13 @@ Remark: there are knots that admit non isotopic Seifert surfaces of minimal genu 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 -\\ +\begin{definition} +\label{def:lk_via_homo} 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 $\beta$: \[ \alpha \in H_1(S^3 \setminus \nu(\beta), \mathbb{Z}) \cong \mathbb{Z}.\] - +\end{definition} \begin{example} \begin{itemize} diff --git a/lec_2.tex b/lec_2.tex new file mode 100644 index 0000000..51ea39a --- /dev/null +++ b/lec_2.tex @@ -0,0 +1,269 @@ +\subsection{Existence of Seifert surface - second proof} +%\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}(Theorem \ref{theo:Seifert})\\ +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} +\begin{align*} +N \cong & D^2 \times S^1\\ +\partial N \cong & S^1 \times S^1\\ +H^1(N, \partial N) \cong & \mathbb{Z} \oplus \mathbb{Z} +\end{align*} +\begin{align*} +H^* (S^3, S^3 \setminus N) &\cong H^* (N, \partial N)\\ +\\ +H^ 1 (S^3\setminus N) &\cong H^1(S^3\setminus K) \cong \mathbb{Z} +\end{align*} +\begin{equation*} +\begin{tikzcd}[row sep=huge] +H^1(S^3 \setminus K) \arrow[r,] \arrow[d,"\widetilde{\Theta}"] & +H^1(N \setminus K) \arrow[d,"\Theta"] \\ +{[S^3 \setminus K, S^1]} \arrow[r,]& +{[N \setminus K, S^1]} +\end{tikzcd} +\end{equation*} +\noindent +$\Sigma = \widetilde{\Theta}^{-1}(X)$ is a surface, such that $\partial \Sigma = K$, so it is a Seifert surface. +% +% +% Thom isomorphism, +\end{proof} +\subsection{Alexander polynomial} +\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 +$S = \begin{pmatrix} +-1 & -1 \\ +0 & -1 +\end{pmatrix}$. Then +\[ +\Delta_K(t) = \det +\begin{pmatrix} +-t + 1 & -t\\ +1 & -t +1 +\end{pmatrix} += (t -1)^2 + t = t^2 - t +1 \ne 1 +\Rightarrow \text{trefoil is not trivial.} +\] +\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 + \dots + a_1 t^l )= k - l. +\end{align*} +\end{lemma} +\begin{proof} +If $\Sigma$ is a genus $g$ - Seifert surface for $K$ then $H_1(\Sigma) = \mathbb{Z}^{2g}$, so $S$ is an $2g \times 2g$ matrix. Therefore $\det (tS - S^T)$ is a polynomial of degree at most $2g$. +\end{proof} +\begin{example} +There are not trivial knots with Alexander polynomial equal $1$, for example: +\includegraphics[width=0.3\textwidth]{11n34.png} +$\Delta_{11n34} \equiv 1$. +\end{example} + +\subsection{Decomposition of $3$-sphere} +We know that $3$ - sphere can be obtained by gluing two solid tori: +$S^3 = \partial D^4 = \partial (D^2 \times D^2) = (D^2 \times S^1) \cup (S^1 \times D^2)$. So the complement of solid torus in $S^3$ is another solid torus.\\ +Analytically it can be describes as follow. +Take $(z_1, z_2) \in \mathbb{C}$ such that $max(\mid z_1 \mid, \mid z_2\mid) = 1 +$. Define following sets: $S_1 = \{ (z_1, z_2) \in S^3: \mid z_1 \mid = 0\} \cong S^1 \times D^2 $ and $S_2 = \{(z_1, z_2) \in S ^3: \mid z_2 \mid = 1 \} \cong D^2 \times S^1$. The intersection $S_1 \cap S_2 = \{(z_1, z_2): \mid z_1 \mid = \mid z_2 \mid = 1 \} \cong S^1 \times S^1$ +\begin{figure}[h] +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.3\textwidth}{!}{\includegraphics[width=0.3\textwidth]{sphere_as_torus.png}} +\caption{The complement of solid torus in $S^3$ is another solid torus.} +\label{fig:sphere_as_tori} +} +\end{figure} + +\subsection{Dehn lemma and sphere theorem} +%removing one disk from surface doesn't change $H_1$ (only $H_2$) +% +% +% +\begin{lemma}[Dehn] +Let $M$ be a $3$-manifold and $D^2 \overset{f} \rightarrow M^3$ be a map of a disk such that $f\big|_{\partial D^2}$ is an embedding. Then there exists an embedding +${D^2 \overset{g}\longhookrightarrow M}$ such that: +\[ +g\big|_{\partial D^2} = f\big|_{\partial D^2.} +\] +\end{lemma} +\noindent +Remark: Dehn lemma doesn't hold for dimension four.\\ +Let $M$ be connected, compact three manifold with boundary. +Suppose $\pi_1(\partial M) \longrightarrow \pi_1(M)$ has non-trivial kernel. Then there exists a map $f: (D^2, \partial D^2) \longrightarrow (M, \partial M)$ such that $f\big|_{\partial D^2}$ is non-trivial loop in $\partial M$. +\begin{theorem}[Sphere theorem] +Suppose $\pi_1(M) \ne 0$. Then there exists an embedding $f: S^2 \hookrightarrow M$ that is homotopy non-trivial. +\end{theorem} +\begin{problem} +Prove that $S^3 \ K$ is Eilenberg–MacLane space of type $K(\pi, 1)$. +\end{problem} +\begin{corollary} +Suppose $K \subset S^3$ and $\pi_1(S^3 \setminus K)$ is infinite cyclic ($\mathbb{Z})$. Then $K$ is trivial. +\end{corollary} +\begin{proof} +Let $N$ be a tubular neighbourhood of a knot $K$ and $M = S^3 \setminus N$ its complement. Then $\partial M = S^1 \times S^1$. Let $f : \pi_1(\partial M ) \longrightarrow \pi_1(M)$. +If $\pi_1(M)$ is infinite cyclic group then the map $f$ is non-trivial. Suppose ${\lambda \in \ker (\pi_1(S^1 \times S^1) \longrightarrow \pi_1(M)}$. There is a map $g: (D^2, \partial D^2) \longrightarrow (M, \partial M)$ such that $g(\partial D^2) = \lambda$. By Dehn's lemma there exists an embedding ${h: (D^2, \partial D^2) \longhookrightarrow (M, \partial M)}$ such that +$h\big|_{\partial D^2} = f \big|_{\partial D^2}$ and $h(\partial D^2) = \lambda$. +Let $\Sigma$ be a union of the annulus and the image of $\partial D^2$. +\\???? $g_3$?\\ +If $g(\Sigma) = 0$, then $K$ is trivial. \\ +Now we should proof that: +\[ +H_1(M) \cong \mathbb{Z} \Longrightarrow \lambda \in \ker ( \pi_1(S^1 \times S^1) \longrightarrow \pi_1(M)). +\] +\begin{figure}[h] +\fontsize{40}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.4\textwidth}{!}{\input{images/torus_lambda.pdf_tex}} +} +\caption{$\mu$ is a meridian and $\lambda$ is a longitude.} +\label{fig:meridian_and_longitude} +\end{figure} +Choose a meridian $\mu$ such that $\Lk (\mu, K) = 1$. Recall the definition of linking number via homology group (Definition \ref{def:lk_via_homo}). +$[\mu]$ represents the generator of $H_1(S^3\setminus K, \mathbb{X})$. From definition of $\lambda$ we know that $\lambda$ is trivial in $H_1(M)$ ($\Lk(\lambda, K) =0$, therefore $[\lambda]$ was trivial in $pi_1(M)$). If $K$ is non-trivial then $\lambda$ is non-trivial in $\pi_1(M)$, but it is trivial in $H_1(M)$. +\end{proof} diff --git a/lec_3.tex b/lec_3.tex new file mode 100644 index 0000000..e69de29 diff --git a/lec_4.tex b/lec_4.tex new file mode 100644 index 0000000..60a9465 --- /dev/null +++ b/lec_4.tex @@ -0,0 +1,116 @@ +\begin{definition} +Two knots $K$ and $K^{\prime}$ are called (smoothly) concordant if there exists an annulus $A$ that is smoothly embedded in ${S^3 \times [0, 1]}$ such that +\[ +\partial A = K^{\prime} \times \{1\} \; \sqcup \; K \times \{0\}. +\] +\end{definition} + +\begin{figure}[h] +\fontsize{20}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.8\textwidth}{!}{\input{images/concordance.pdf_tex}} +} +\end{figure} +\begin{definition} +A knot $K$ is called (smoothly) slice if $K$ is smoothly concordant to an unknot. \\ +Put differently: a knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk in $B^4$. +\end{definition} + + + +\noindent +Let $m(K)$ denote a mirror image of a knot $K$. +\begin{fact} +For any $K$, $K \# m(K)$ is slice. +\end{fact} +\begin{fact} +Concordance is an equivalence relation. +\end{fact} +\begin{fact}\label{fact: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{1\textwidth}{!}{\input{images/concordance_sum.pdf_tex}} +} +\caption{Sketch for Fact \ref{fact:concordance_connected}.} +\label{fig:concordance_sum} +\end{figure} + +\end{fact} +\begin{fact} +$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 $-\{K\} = \{mK\}$. +\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. +\\ +\begin{figure}[h] +\fontsize{20}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.5\textwidth}{!}{\input{images/ball_4_pushed_seifert.pdf_tex}} +} +\caption{$Y = F \cup \Sigma$ is a smooth close surface.} +\label{fig:closed_surface} +\end{figure} +\noindent +\\ +Pontryagin-Thom construction tells us that there exists a compact 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 +$\partial B^+ = \beta^+$. +Then +$\Lk(\alpha, \beta^+) = A \cdot B^+$. But $A$ and $B$ are disjoint, so $\Lk(\alpha, \beta^+) = 0$. Then the Seifert form is zero. +\\ +????????????????? +\\ +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} +\[ +\dim \ker (H_1(Y, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z})) = \frac{1}{2} b_1(Y), +\] +where $b_1$ is first Betti number. +\end{proposition} +\begin{proof} +\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_0(Y) \to H_0(\Omega) \to 0 +\end{align*} +\end{proof} + +\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} diff --git a/lectures_on_knot_theory.pdf b/lectures_on_knot_theory.pdf index f97d7be..f36a8dd 100644 Binary files a/lectures_on_knot_theory.pdf and b/lectures_on_knot_theory.pdf differ diff --git a/lectures_on_knot_theory.tex b/lectures_on_knot_theory.tex index bdb5452..038146d 100644 --- a/lectures_on_knot_theory.tex +++ b/lectures_on_knot_theory.tex @@ -90,8 +90,11 @@ \DeclareMathOperator{\Lk}{lk} \DeclareMathOperator{\pt}{\{pt\}} - -\titleformat{\section}{\normalfont \fontsize{12}{15} \bfseries}{% +\titleformat{\subsection}{% + \normalfont \fontsize{12}{15}\bfseries}{% + }{.0ex plus .2ex}{} +\titleformat{\section}{% + \normalfont \fontsize{13}{15} \bfseries}{% Lecture\ \thesection}% {2.3ex plus .2ex}{} \titlespacing*{\section} @@ -114,249 +117,12 @@ \section{Basic definitions \hfill\DTMdate{2019-02-25}} \input{lec_1.tex} -\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} -\begin{align*} -N \cong & D^2 \times S^1\\ -\partial N \cong & S^1 \times S^1\\ -H^1(N, \partial N) \cong & \mathbb{Z} \oplus \mathbb{Z} -\end{align*} -\begin{align*} -H^* (S^3, S^3 \setminus N) &\cong H^* (N, \partial N)\\ -\\ -H^ 1 (S^3\setminus N) &\cong H^1(S^3\setminus K) \cong \mathbb{Z} -\end{align*} -\begin{equation*} -\begin{tikzcd}[row sep=huge] -H^1(S^3 \setminus K) \arrow[r,] \arrow[d,"\widetilde{\Theta}"] & -H^1(N \setminus K) \arrow[d,"\Theta"] \\ -{[S^3 \setminus K, S^1]} \arrow[r,]& -{[N \setminus K, S^1]} -\end{tikzcd} -\end{equation*} -\noindent -$\Sigma = \widetilde{\Theta}^{-1}(X)$ is a surface, such that $\partial \Sigma = K$, so it is a Seifert surface. -% -% -% Thom isomorphism, -\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 -$S = \begin{pmatrix} --1 & -1 \\ -0 & -1 -\end{pmatrix}$. Then -\[ -\Delta_K(t) = \det -\begin{pmatrix} --t + 1 & -t\\ -1 & -t +1 -\end{pmatrix} -= (t -1)^2 + t = t^2 - t +1 \ne 1 -\Rightarrow \text{trefoil is not trivial.} -\] -\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 + \dots + a_1 t^l )= k - l. -\end{align*} -\end{lemma} -\begin{proof} -If $\Sigma$ is a genus $g$ - Seifert surface for $K$ then $H_1(\Sigma) = \mathbb{Z}^{2g}$, so $S$ is an $2g \times 2g$ matrix. Therefore $\det (tS - S^T)$ is a polynomial of degree at most $2g$. -\end{proof} -\begin{example} -There are not trivial knots with Alexander polynomial equal $1$, for example: -\includegraphics[width=0.3\textwidth]{11n34.png} -$\Delta_{11n34} \equiv 1$. -\end{example} -%removing one disk from surface doesn't change $H_1$ (only $H_2$) -% -% -% -\begin{lemma}[Dehn] -Let $M$ be a $3$-manifold and $D^2 \overset{f} \rightarrow M^3$ be a map of a disk such that $f_{\big|\partial D^2}$ is an embedding. Then there exists an embedding -${D^2 \overset{g}\longhookrightarrow M}$ such that: -\[ -g_{\big| \partial D^2} = f_{\big| \partial D^2.} -\] -\end{lemma} -\noindent -Remark: Dehn lemma doesn't hold for dimension four.\\ -Let $M$ be connected, compact three manifold with boundary. -Suppose $\pi_1(\partial M) \longrightarrow \pi_1(M)$ has non-trivial kernel. Then there exists a map $f: (D^2, \partial D^2) \longrightarrow (M, \partial M)$ such that $f\big| \partial D^2$ is non-trivial loop in $\partial M$ -\begin{theorem}[Sphere theorem] -Suppose $\pi_1(M) \ne 0$. Then there exists an embedding $f: S^2 \hookrightarrow M$ that is homotopy non-trivial. -\end{theorem} -\begin{problem} -Prove that $S^3 \ K$ is Eilenberg–MacLane space of type $K(\pi, 1)$. -\end{problem} -\begin{corollary} -Suppose $K \subset S^3$ and $\pi_1(S^3 \setminus K)$ is infinite cyclic ($\mathbb{Z})$. Then $K$ is trivial. -\end{corollary} -\subsection*{Construction} -We know that $3$ - sphere can be obtained by gluing two solid tori: -$S^3 = (D^2 \times S^1) \cup (S^1 \times D^2)$. So the complement of solid torus in $S^3$ is another solid torus.\\ -Take $(z_1, z_2) \in \mathbb{C}$ such that $max(\mid z_1 \mid, \mid z_2, \mid) = 1 -$ -\begin{figure}[h] -\centering{ -\def\svgwidth{\linewidth} -\resizebox{0.3\textwidth}{!}{\includegraphics[width=0.3\textwidth]{sphere_as_torus.png}} -\caption{The complement of solid torus in $S^3$ is another solid torus.} -\label{fig:sphere_as_tori} -} -\end{figure} +\section{Alexander polynomial \hfill\DTMdate{2019-03-04}} +\input{lec_2.tex} +%add Hurewicz theorem? +\section{\hfill\DTMdate{2019-03-11}} +\input{lec_3.tex} \begin{example} \begin{align*} @@ -364,6 +130,15 @@ $ &F(0) = 0 \end{align*} \end{example} +\begin{figure}[h] +\fontsize{40}{10}\selectfont +\centering{ +\def\svgwidth{\linewidth} +\resizebox{0.2\textwidth}{!}{\input{images/milnor_singular.pdf_tex}} +} +%\caption{$\mu$ is a meridian and $\lambda$ is a longitude.} +\label{fig:milnor_singular} +\end{figure} ???????????? \\ \noindent @@ -392,125 +167,16 @@ Figure 8 knot is negative amphichiral. % % % +\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{Concordance group \hfill\DTMdate{2019-03-18}} -\begin{definition} -Two knots $K$ and $K^{\prime}$ are called (smoothly) concordant if there exists an annulus $A$ that is smoothly embedded in ${S^3 \times [0, 1]}$ such that -\[ -\partial A = K^{\prime} \times \{1\} \; \sqcup \; K \times \{0\}. -\] -\end{definition} - -\begin{figure}[h] -\fontsize{20}{10}\selectfont -\centering{ -\def\svgwidth{\linewidth} -\resizebox{0.8\textwidth}{!}{\input{images/concordance.pdf_tex}} -} -\end{figure} -\begin{definition} -A knot $K$ is called (smoothly) slice if $K$ is smoothly concordant to an unknot. \\ -Put differently: a knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk in $B^4$. -\end{definition} +\input{lec_4.tex} -\noindent -Let $m(K)$ denote a mirror image of a knot $K$. -\begin{fact} -For any $K$, $K \# m(K)$ is slice. -\end{fact} -\begin{fact} -Concordance is an equivalence relation. -\end{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{1\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. -\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 $-\{K\} = \{mK\}$. -\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. -\\ -\begin{figure}[h] -\fontsize{20}{10}\selectfont -\centering{ -\def\svgwidth{\linewidth} -\resizebox{0.5\textwidth}{!}{\input{images/ball_4_pushed_seifert.pdf_tex}} -} -\caption{$Y = F \cup \Sigma$ is a smooth close surface.} -\label{fig:closed_surface} -\end{figure} -\noindent -\\ -Pontryagin-Thom construction tells us that there exists a compact 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 -$\partial B^+ = \beta^+$. -Then -$\Lk(\alpha, \beta^+) = A \cdot B^+$. But $A$ and $B$ are disjoint, so $\Lk(\alpha, \beta^+) = 0$. Then the Seifert form is zero. -\\ -????????????????? -\\ -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} -\[ -\dim \ker (H_1(Y, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z})) = \frac{1}{2} b_1(Y), -\] -where $b_1$ is first Betti number. -\end{proposition} -\begin{proof} -\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_0(Y) \to H_0(\Omega) \to 0 -\end{align*} -\end{proof} - -\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 Fakt %%\label{fig:concordance_m} -\end{figure} - \section{\hfill\DTMdate{2019-03-25}} \begin{definition} The (smooth) four genus $g_4(K)$ is the minimal genus of the surface $\Sigma \in B^4$ such that $\Sigma$ is compact, orientable and $\partial \Sigma = K$. @@ -587,10 +253,6 @@ has a subspace of dimension $g_{\Sigma}$ on which it is zero: -\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}}