diff --git a/lec_1.tex b/lec_1.tex deleted file mode 100644 index 5a94fae..0000000 --- a/lec_1.tex +++ /dev/null @@ -1,280 +0,0 @@ -\begin{definition} -A knot $K$ in $S^3$ is a smooth (PL - smooth) embedding of a circle $S^1$ in $S^3$: -\begin{align*} -\varphi: S^1 \hookrightarrow S^3 -\end{align*} -\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} (unknot), -\includegraphics[width=0.08\textwidth]{trefoil.png} (trefoil). -\item -Not knots: -\includegraphics[width=0.12\textwidth]{not_injective_knot.png} -(it is not an injection), -\includegraphics[width=0.08\textwidth]{not_smooth_knot.png} -(it is not smooth). -\end{itemize} -\end{example} -\begin{definition} -%\hfill\\ -Two knots $K_0 = \varphi_0(S^1)$, $K_1 = \varphi_1(S^1)$ are equivalent if the embeddings $\varphi_0$ and $\varphi_1$ are isotopic, that is there exists a continues function -\begin{align*} -&\Phi: S^1 \times [0, 1] \hookrightarrow S^3 \\ -&\Phi(x, t) = \Phi_t(x) -\end{align*} - such that $\Phi_t$ is an embedding for any $t \in [0,1]$, $\Phi_0 = \varphi_0$ and -$\Phi_1 = \varphi_1$. -\end{definition} - -\begin{theorem} -Two knots $K_0$ and $K_1$ are isotopic if and only if they are ambient isotopic, i.e. there exists a family of self-diffeomorphisms $\Psi = \{\psi_t: t \in [0, 1]\}$ such that: -\begin{align*} -&\psi(t) = \psi_t \text{ is continius on $t\in [0,1]$}\\ -&\psi_t: S^3 \hookrightarrow S^3,\\ -& \psi_0 = id ,\\ -& \psi_1(K_0) = K_1. -\end{align*} -\end{theorem} -\begin{definition} -A knot is trivial (unknot) if it is equivalent to an embedding $\varphi(t) = (\cos t, \sin t, 0)$, where $t \in [0, 2 \pi] $ is a parametrisation of $S^1$. -\end{definition} -\begin{definition} -A link with k - components is a (smooth) embedding of $\overbrace{S^1 \sqcup \ldots \sqcup S^1}^k$ in $S^3$ -\end{definition} -\begin{example} -Links: -\begin{itemize} -\item -a trivial link with $3$ components: -\includegraphics[width=0.2\textwidth]{3unknots.png}, -\item -a hopf link: \includegraphics[width=0.13\textwidth]{Hopf.png}, -\item -a Whitehead link: -\includegraphics[width=0.13\textwidth]{WhiteheadLink.png}, -\item -Borromean link: -\includegraphics[width=0.1\textwidth]{BorromeanRings.png}. -\end{itemize} -\end{example} -% -% -% -\begin{definition} -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 -$D_{\pi |_L}$ 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} -\end{definition} -\noindent -There are under- and overcrossings (tunnels and bridges) on a link diagrams with an obvious meaning.\\ -Every link admits a link diagram. -\\ -Let $D$ be a diagram of an oriented link (to each component of a link we add an arrow in the diagram).\\ -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. - -\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\\ -\includegraphics[width=0.6\textwidth]{rm1.png}, -\item\hfill\\\includegraphics[width=0.6\textwidth]{rm2.png}, -\item\hfill\\\includegraphics[width=0.4\textwidth]{rm3.png}. -\end{enumerate} - -\begin{theorem} [Reidemeister, 1927 ] -Two diagrams of the same link can be -deformed into each other by a finite sequence of Reidemeister moves (and isotopy of the plane). -\end{theorem} -% -% -% -%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: -\begin{align*} -\PICorientpluscross \mapsto \PICorientLRsplit\\ -\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 $\partial \Sigma = L$.\\ - -\begin{figure}[h] -\fontsize{15}{10}\selectfont -\centering{ -\def\svgwidth{\linewidth} -\resizebox{0.8\textwidth}{!}{\input{images/seifert_alg.pdf_tex}} -\caption{Constructing a Seifert surface.} -\label{fig:SeifertAlg} -} -\end{figure} - -\noindent -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} -\includegraphics[width=0.6\textwidth]{seifert_connect.png} -\end{center} -\caption{Connecting two surfaces.} -\label{fig:SeifertConnect} -\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} -% -\begin{figure}[h] -\fontsize{12}{10}\selectfont -\centering{ -\def\svgwidth{\linewidth} -\resizebox{1\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} -\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} -\item -Hopf link: -\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: -\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} -\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 $\alpha_1, \dots, \alpha_n$. -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 -$V \rightarrow AVA^T$, where $A$ is a matrix with integer coefficients, - -\item - -$V \rightarrow -\begin{pmatrix} - \begin{array}{c|c} - V & - \begin{matrix} - \ast & 0 \\ - \sdots & \sdots\\ - \ast & 0 - \end{matrix} \\ - \hline - \begin{matrix} - \ast & \dots & \ast\\ - 0 & \dots & 0 - \end{matrix} - & - \begin{matrix} - 0 & 0\\ - 1 & 0 - \end{matrix} - \end{array} -\end{pmatrix} \quad$ -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} diff --git a/lec_2.tex b/lec_2.tex deleted file mode 100644 index 1519178..0000000 --- a/lec_2.tex +++ /dev/null @@ -1,281 +0,0 @@ -\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 ${\max(\vert z_1 \vert, \vert z_2\vert) = 1.} -$ -Define following sets: -\begin{align*} -S_1 = \{ (z_1, z_2) \in S^3: \vert z_1 \vert = 0\} \cong S^1 \times D^2 ,\\ -S_2 = \{(z_1, z_2) \in S ^3: \vert z_2 \vert = 1 \} \cong D^2 \times S^1. -\end{align*} -The intersection -$S_1 \cap S_2 = \{(z_1, z_2): \vert z_1 \vert = \vert z_2 \vert = 1 \} \cong S^1 \times S^1$. -\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 $H_1(S^3\setminus K, \mathbb{Z})$. 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 deleted file mode 100644 index ac2c94b..0000000 --- a/lec_3.tex +++ /dev/null @@ -1,176 +0,0 @@ -\subsection{Algebraic knots} -\noindent -Suppose $F: \mathbb{C}^2 \rightarrow \mathbb{C}$ is a polynomial and $F(0) = 0$. Let take small small sphere $S^3$ around zero. This sphere intersect set of roots of $F$ (zero set of $F$) transversally and by the implicit function theorem the intersection is a manifold. -The dimension of sphere is $3$ and $F^{-1}(0)$ has codimension $2$. -So there is a subspace $L$ - compact one dimensional manifold without boundary. -That means that $L$ is a link in $S^3$. -\begin{figure}[h] -\fontsize{40}{10}\selectfont -\centering{ -\def\svgwidth{\linewidth} -\resizebox{0.2\textwidth}{!}{\input{images/milnor_singular.pdf_tex}} -} -\caption{The intersection of a sphere $S^3$ and zero set of polynomial $F$ is a link $L$.} -\label{fig:milnor_singular} -\end{figure} -%ref: Milnor Singular Points of Complex Hypersurfaces -\begin{theorem} - -$L$ is an unknot if and only if -zero is a smooth point, i.e. -$\bigtriangledown F(0) \neq 0$ (provided $S^3$ has a sufficiently small radius). -\end{theorem} -\noindent -Remark: if $S^3$ is large it can happen that $L$ is unlink, but $F^{-1}(0) \cap B^4$ is "complicated". \\ -%Kyle M. Ormsby -\noindent -In other words: if we take sufficiently small sphere, the link is non-trivial if and only if the point $0$ is singular and the isotopy type of the link doesn't depend on the radius of the sphere. -A link obtained is such a way is called an -algebraic link (in older books on knot theory there is another notion of algebraic link with another meaning). -%ref: Eisenbud, D., Neumann, W. -\begin{example} -Let $p$ and $q$ be coprime numbers such that $p1$. \\ -Zero is an isolated singular point ($\bigtriangledown F(0) = 0$). $F$ is quasi - homogeneous polynomial, so the isotopy class of the link doesn't depend on the choice of a sphere. -Consider $S^3 = \{ (z, w) \in \mathbb{C} : \max( \vert z \vert, \vert w \vert )\} = \varepsilon$. -The intersection -$F^{-1}(0) \cap S^3$ is a torus $T(p, q)$. -\\??????????????????? -$F(z, w) = z^p - w^q$\\ -.\\ -$F^{-1}(0) = \{t = t^q, w = t^p\}.$ For unknot $t = \max (\vert t\vert ^p, \vert t \vert^q) = \varepsilon$. -\end{example} -as a corollary we see that $K_T^{n, }$ ???? \\ -is not slice unless $m=0$. \\ -$t = re^{i \Theta}, \Theta \in [0, 2\pi], r = \varepsilon^{\frac{i}{p}}$ - -\begin{figure}[h] -\fontsize{40}{10}\selectfont -\centering{ -\def\svgwidth{\linewidth} -\resizebox{0.2\textwidth}{!}{\input{images/polynomial_and_surface.pdf_tex}} -} -\caption{Sa.} -\label{fig:polynomial_and_surface} -\end{figure} -\begin{theorem} -Suppose $L$ is an algebraic link. $L = F^{-1}(0) \cap S^3$. Let -\begin{align*} -&\varphi : S^3 \setminus L \longrightarrow S^1 \\ -&\varphi(z, w) =\frac{F(z, w)}{\vert F(z, w) \vert}\in S^1, \quad (z, w) \notin F^{-1}(0). -\end{align*} -The map $\varphi$ is a locally trivial fibration. -\end{theorem} -???????\\ -$ rh D \varphi \equiv 1$ -\begin{definition} -A map $\Pi : E \longrightarrow B$ is locally trivial fibration with fiber $F$ if for any $b \in B$, there is a neighbourhood $U \subset B$ such that $\Pi^{-1}(U) \cong U \times $ \\ -????????????\\ $\Gamma$ ?????????????\\ -FIGURES\\ -!!!!!!!!!!!!!!!!!!!!!!!!!!\\ -\end{definition} - -\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} -... -\\ -In general $h$ is defined only up to homotopy, but this means that -\[ -h_* : H_1 (F, \mathbb{Z}) \longrightarrow H_1 (F, \mathbb{Z}) -\] -is well defined \\ -???????????\\ map. -\begin{theorem} -\label{thm:F_as_S} -Suppose $S$ is a Seifert matrix associated with $F$ then $h = S^{-1}S^T$. -\end{theorem} -\begin{proof} -TO WRITE REFERENCE!!!!!!!!!!! -%see Arnold Varchenko vol II -%Picard - Lefschetz formula -%Nemeth (Real Seifert forms -\end{proof} -\noindent -Consequences: -\begin{enumerate}[label={(\arabic*)}] -\item -the Alexander polynomial is the characteristic polynomial of $h$: -\[ -\Delta_L (t) = \det (h - t I d) -\] -In particular $\Delta_L $ is monic (i.e. the top coefficient is $\pm 1$), -???????????????? -\item -S is invertible, -\item -$F$ minimize the genus (i.e. $F$ is minimal genus Seifert surface). -\\??????????????????\\ -\end{enumerate} -% -\begin{definition} -A link $L$ is fibered if there exists a map ${\phi: S^3\setminus L \longrightarrow S^1}$ which is locally trivial fibration. -\end{definition} -\noindent -If $L$ is fibered then Theorem \ref{thm:F_as_S} holds and all its consequences. -\begin{problem} -If $K_1$ and $K_2$ are fibered knots, then also $K_1 \# K_2$ is fibered. -\end{problem} -\noindent -?????????????????????\\ -\begin{problem} -Prove that connected sum is well defined:\\ -$\Delta_{K_1 \# K_2} = -\Delta_{K_1} + \Delta_{K_2}$ and -$g_3(K_1 \# K_2) = g_3(K_1) + g_3(K_2)$. - -\end{problem} -\begin{figure}[h] -\fontsize{12}{10}\selectfont -\centering{ -\def\svgwidth{\linewidth} -\resizebox{1\textwidth}{!}{\input{images/satellite.pdf_tex}}} -\caption{Whitehead double satellite knot.\\ -The pattern knot embedded non-trivially in an unknotted solid torus $T$ (e.i. $K \not\subset S^3\subset T$) on the left and the pattern in a companion knot - trefoil - on the right.} -\label{fig:sattelite} -\end{figure} -\noindent -\subsection{Alternating knot} -\begin{definition} -A knot (link) is called alternating if it admits an alternating diagram. -\end{definition} - -\begin{figure}[h] -\fontsize{12}{10}\selectfont -\centering{ -\def\svgwidth{\linewidth} -\includegraphics[width=0.3\textwidth]{figure8.png} -} -\caption{Example: figure eight knot is an alternating knot.} -\label{fig:fig8} -\end{figure} - -\begin{definition} -A reducible crossing in a knot diagram is a crossing for which we can find a circle such that its intersection with a knot diagram is exactly that crossing. A knot diagram without reducible crossing is called reduced. -\end{definition} -\begin{fact} -Any reduced alternating diagram has minimal number of crossings. -\end{fact} -\begin{definition} -The writhe of the diagram is the difference between the number of positive and negative crossings. -\end{definition} -\begin{fact}[Tait] -Any two diagrams of the same alternating knot have the same writhe. -\end{fact} -\begin{fact} -An alternating knot has Alexander polynomial of the form: -$ -a_1t^{n_1} + a_2t^{n_2} + \dots + a_s t^{n_s} -$, where $n_1 < n_2 < \dots < n_s$ and $a_ia_{i+1} < 0$. -\end{fact} -\begin{problem}[open] -What is the minimal $\alpha \in \mathbb{R}$ such that if $z$ is a root of the Alexander polynomial of an alternating knot, then $\Re(z) > \alpha$.\\ -Remark: alternating knots have very simple knot homologies. -\end{problem} -\begin{proposition} -If $T_{p, q}$ is a torus knot, $p < q$, then it is alternating if and only if $p=2$. -\end{proposition} \ No newline at end of file diff --git a/lec_4.tex b/lec_4.tex deleted file mode 100644 index f2873d7..0000000 --- a/lec_4.tex +++ /dev/null @@ -1,170 +0,0 @@ -\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 closed surface.} -\label{fig:closed_surface} -\end{figure} -\noindent -\\ -Pontryagin-Thom construction tells us that there exists a compact oriented 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. -\\ -\noindent -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} -Consider the following long exact sequence for a pair $(\Omega, Y)$: -\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*} -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_1(\Omega, Y) &\cong H^1(\Omega). -\end{align*} -Therefore $\dim_{\mathbb{Q}} \quot{H_1(Y)}{V} -= \dim_{\mathbb{Q}} V -$.\\ -\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 a $K$ -has a subspace of dimension $g_{\Sigma}$ on which it is zero: - -\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}}% - V = -\begin{matrix}% matrix for left braces - \coolleftbrace{g_{\Sigma}}{ \\ \\ \\} - \\ \\ \\ \\ -\end{matrix}% -\begin{pmatrix} - \coolover{g_{\Sigma}}{0 & \dots & 0 } & * & \dots & *\\ - \sdots & & \sdots & \sdots & & \sdots \\ - 0 & \dots & 0 & * & \dots & *\\ - * & \dots & * & * & \dots & *\\ - \sdots & & \sdots & \sdots & & \sdots \\ - * & \dots & * & * & \dots & * - \end{pmatrix}_{2g_{\Sigma} \times 2g_{\Sigma}} -\end{align*} -\end{proof} -\noindent -Let $V = -\begin{pmatrix} - 0 & A\\ - B & C -\end{pmatrix}$ -\begin{align*} -\det (tV - V^T) = \det (tA - B^T) - \det(tB - A^T) -\end{align*} -\begin{corollary} -\label{cor:slice_alex} -If $K$ is a slice knot then there exists $f \in \mathbb{Z}[t^{\pm 1}]$ such that $\Delta_K(t) = f(t) \cdot f(t^{-1})$. -\end{corollary} -\begin{example} -Figure eight knot is not slice. -\end{example} -\begin{fact} -If $K$ is slice, then the signature $\sigma(K) \equiv 0$. -\end{fact} - - - diff --git a/lec_5.tex b/lec_5.tex deleted file mode 100644 index 9cdf66b..0000000 --- a/lec_5.tex +++ /dev/null @@ -1,234 +0,0 @@ -\subsection{Slice knots and metabolic form} -\begin{theorem} -\label{the:sign_slice} -If $K$ is slice, -then $\sigma_K(t) - = \sign ( (1 - t)S +(1 - \bar{t})S^T)$ -is zero except possibly of finitely many points and $\sigma_K(-1) = \sign(S + S^T) \neq 0$. -\end{theorem} -\begin{lemma} -\label{lem:metabolic} -If $V$ is a Hermitian matrix ($\bar{V} = V^T$), $V$ is of size $2n \times 2n$, -$ -V = \begin{pmatrix} -0 & A \\ -\bar{A}^T & B -\end{pmatrix} -$ and $\det V \neq 0$ then $\sigma(V) = 0$. -\end{lemma} -\begin{definition} -A Hermitian form $V$ is metabolic if $V$ has structure -$\begin{pmatrix} -0 & A\\ -\bar{A}^T & B -\end{pmatrix}$ with half-dimensional null-space. -\end{definition} -\noindent -Theorem \ref{the:sign_slice} can be also express as follow: -non-degenerate metabolic hermitian form has vanishing signature. -\begin{proof} -\noindent -We note that $\det(S + S^T) \neq 0$. Hence $\det ( (1 - t) S + (1 - \bar{t})S^T)$ is not identically zero on $S^1$, so it is non-zero except possibly at finitely many points. We apply the Lemma \ref{lem:metabolic}. -\\ -Let $t \in S^1 \setminus \{1\}$. -Then: -\begin{align*} -\det((1 - t) S + (1 - \bar{t}) S^T) =& -\det((1 - t) S + (t\bar{t} - \bar{t}) S^T) =\\ -&\det((1 - t) (S - \bar{t} - S^T)) = -\det((1 -t)(S - \bar{t} S^T)). -\end{align*} -As $\det (S + S^T) \neq 0$, so $S - \bar{t}S^T \neq 0$. -\end{proof} -\begin{corollary} -If $K \sim K^\prime$ then for all but finitely many $t \in S^1 \setminus \{1\}: \sigma_K(t) = -\sigma_{K^\prime}(t)$. -\end{corollary} -\begin{proof} -If $ K \sim K^\prime$ then $K \# K^\prime$ is slice. -\[ -\sigma_{-K^\prime}(t) = -\sigma_{K^\prime}(t) -\] -The signature gives a homomorphism from the concordance group to $\mathbb{Z}$. -Remark: if $t \in S^1$ is not algebraic over $\mathbb{Z}$, then $\sigma_K(t) \neq 0$ -(we can use the argument that $\mathscr{C} \longrightarrow \mathbb{Z}$ as well). -\end{proof} -\subsection{Four genus} -\begin{figure}[h] -\fontsize{20}{10}\selectfont -\centering{ -\def\svgwidth{\linewidth} -\resizebox{0.7\textwidth}{!}{\input{images/genus_2_bordism.pdf_tex}} -} -\caption{$K$ and $K^\prime$ are connected by a genus $g$ surface.}\label{fig:genus_2_bordism} -\end{figure} - -\begin{proposition}[Kawauchi inequality] -If there exists a genus $g$ surface as in Figure \ref{fig:genus_2_bordism} -then for almost all -$t \in S^1 \setminus \{1\}$ we have -$\vert -\sigma_K(t) - \sigma_{K^\prime}(t) -\vert \leq 2 g$. -\end{proposition} -% Kawauchi Chapter 12 ??? -% Borodzik 2010 Morse theory for plane algebraic curves -\begin{lemma} -If $K$ bounds a genus $g$ surface $X \in B^4$ and $S$ is a Seifert form then ${S \in M_{2n \times 2n}}$ has a block structure $\begin{pmatrix} -0 & A\\ -B & C -\end{pmatrix}$, where $0$ is $(n - g) \times (n - g)$ submatrix. -\end{lemma} - -\begin{proof} -\begin{figure}[h] -\fontsize{20}{10}\selectfont -\centering{ -\def\svgwidth{\linewidth} -\resizebox{0.5\textwidth}{!}{\input{images/genus_bordism_zeros.pdf_tex}} -} -\caption{There exists a $3$ - manifold $\Omega$ such that $\partial \Omega = X \cup \Sigma$.}\label{fig:omega_in_B_4} -\end{figure} -\noindent -Let $K$ be a knot and $\Sigma$ its Seifert surface as in Figure \ref{fig:omega_in_B_4}. -There exists a $3$ - submanifold -$\Omega$ such that -$\partial \Omega = Y = X \cup \Sigma$ -(by Thom-Pontryagin construction). -If $\alpha, \beta \in \ker (H_1(\Sigma) \longrightarrow H_1(\Omega))$, -then ${\Lk(\alpha, \beta^+) = 0}$. Now we have to determine the size of the kernel. We know that -${\dim H_1(\Sigma) = 2 n}$. When we glue $\Sigma$ (genus $n$) and $X$ (genus $g$) along a circle we get a surface of genus $n + g$. Therefore $\dim H_1 (Y) = 2 n + 2 g$. Then: -\[ -\dim (\ker (H_1(Y) \longrightarrow H_1(\Omega)) = n + g. -\] -So we have $H_1(W)$ of dimension - $2 n + 2 g$ -- the image of $H_1(Y)$ -with a subspace -corresponding to the image of $H_1(\Sigma)$ with dimension $2 n$ and a subspace corresponding to the kernel -of $H_1(Y) \longrightarrow H_1(\Omega)$ of size $n + g$. -We consider minimal possible intersection of this subspaces that corresponds to the kernel of the composition $H_1(\Sigma) \longrightarrow H_1(Y) \longrightarrow H_1(\Omega)$. As the first map is injective, elements of the kernel of the composition have to be in the kernel of the second map. -So we can calculate: -\[ -\dim \ker (H_1(\Sigma) \longrightarrow H_1(\Omega)) = 2 n + n + g -2 n - 2 g = n - g. -\] -\end{proof} -\begin{corollary} -If $t$ is not a root of -$\det (tS - S^T) $, then -$\vert \sigma_K(t) \vert \leq 2g$. -\end{corollary} -\begin{fact} -If there exists cobordism of genus $g$ between $K$ and $K^\prime$ like shown in Figure \ref{fig:proof_for_bound_disk}, then $K \# -K^\prime$ bounds a surface of genus $g$ in $B^4$. -\end{fact} -\begin{figure}[H] -\fontsize{20}{10}\selectfont -\centering{ -\def\svgwidth{\linewidth} -\resizebox{0.7\textwidth}{!}{\input{images/genus_bordism_proof.pdf_tex}} -} -\caption{If $K$ and $K^\prime$ are connected by a genus $g$ surface, then $K \# -K^\prime$ bounds a genus $g$ surface.}\label{fig:proof_for_bound_disk} -\end{figure} - -\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 $\partial \Sigma = K$. -\end{definition} -\noindent -Remarks: -\begin{enumerate}[label={(\arabic*)}] -\item -$3$ - genus is additive under taking connected sum, but $4$ - genus is not, -\item -for any knot $K$ we have $g_4(K) \leq g_3(K)$. -\end{enumerate} -\begin{example} -\begin{itemize} -\item Let $K = T(2, 3)$. $\sigma(K) = -2$, therefore $T(2, 3)$ isn't a slice knot. -\item Let $K$ be a trefoil and $K^\prime$ a mirror of a trefoil. $g_4(K^\prime) = 1$, but $g_4(K \# K^\prime) = 0$, so we see that $4$-genus isn't additive, -\item -the equality: -\[ -g_4(T(p, q) ) = \frac{1}{2} (p - 1) (g -1) -\] -was conjecture in the '70 and proved by P. Kronheimer and T. Mrówka (1994). -% OZSVATH-SZABO AND RASMUSSEN -\end{itemize} -\end{example} -\begin{proposition} -$g_4 (T(p, q) \# -T(r, s))$ is in general hopelessly unknown. -\end{proposition} -\begin{proposition} -Supremum of the signature function of the knot is bounded almost everywhere by two times $4$ - genus: -\[ -\ess \sup \vert \sigma_K(t) \vert \leq 2 g_4(K). -\] -\end{proposition} -\subsection{Topological genus} -\begin{definition} -A knot $K$ is called topologically slice if $K$ bounds a topological locally flat disc in $B^4$ (i.e. the disk has tubular neighbourhood). -\end{definition} -\begin{theorem}[Freedman, '82] -If $\Delta_K(t) = 1$, then $K$ is topologically slice (but not necessarily smoothly slice). -\end{theorem} -\begin{theorem}[Powell, 2015] -If $K$ is genus $g$ -(topologically flat) -cobordant to $K^\prime$, -then -\[ -\vert \sigma_K(t) - \sigma_{K^\prime}(t) \vert \leq 2 g -\] -if $g_4^{\mytop}(K) \geq \ess \sup \vert \sigma_K(t) \vert$. -\end{theorem} -\noindent -The proof for smooth category was based on following equality: -\[ -\dim \ker (H_1 (Y) \longrightarrow H_1(\Omega)) = \frac{1}{2} \dim H_1(Y). -\] -For this equality we assumed that there exists a $3$ - dimensional manifold $\Omega$ (as shown in Figure \ref{fig:omega_in_B_4}) which was guaranteed by Pontryagin-Thom Construction.\\ -Pontryagin-Thom Construction relays on taking $\Omega$ as preimage of regular value: -\[ -H^1 (B^4 \setminus Y, \mathbb{Z}) = [B^4 \setminus Y, S^1], -\] -what relies on Sard's theorem, that the set of regular values has positive measure. But Sard's theorem doesn't work for topologically locally flat category. So there was a gap in the proof for topological locally flat category - the existence of $\Omega$.\\ -\noindent -Remark: unless $p=2$ or $p = 3 \wedge q = 4$: -\[ -g_4^{\mytop} (T(p, q)) < q_4(T(p, q)). -\] -% Wilczyński '93 -%Feller 2014 -%Baoder 2017 -%Lemark -\\ -\noindent -From the category of cobordant knots (or topologically cobordant knots) there exists a map to $\mathbb{Z}$ given by signature function. To any element $K$ we can associate a form -\[ -(1 - t)S + (1 - \bar{t})S^T) \in W(\mathbb{Z}[t, t^{-1}]). -\] This association is not well define because id depends on the choice of Seifert form. However, different choices lead ever to congruent forms ($S \mapsto CSC^T$) or induced the change on the form by adding or subtracting a hyperbolic element. -\begin{definition} -The Witt group $W$ of $\mathbb{Z}[t, t^{-1}]$ elements are classes of non-degenerate -forms over $\mathbb{Z}[t, t^{-1}]$ under the equivalence relation $V \sim W$ if $V \oplus - W$ is metabolic. -\end{definition} -\noindent -If $S$ differs from $S^\prime$ by a row extension, then -$(1 - t) S + (1 - \bar{t}^{-1}) S^T$ is Witt equivalence to $(1 - t) S^\prime + (1 - t^{-1})S^T$. -\\ -\noindent -A form is meant as hermitian with respect to this involution: $A^T = A: (a, b) = \bar{(a, b)}$. -\\ -$ -W(\mathbb{Z}_p) = \mathbb{Z}_2 \oplus -\mathbb{Z}_2$ or -$\mathbb{Z}_4$ -\\ -??????????????????????? -\\ -$\sum a_gt^j \longrightarrow \sum a_g t^{-1}$\\ -\begin{theorem}[Levine '68] -\[ -W(\mathbb{Z}[t^{\pm 1}]) -\longrightarrow \mathbb{Z}_2^\infty \oplus -\mathbb{Z}_4^\infty \oplus -\mathbb{Z} -\] -\end{theorem} diff --git a/lec_6.tex b/lec_6.tex deleted file mode 100644 index e803a2a..0000000 --- a/lec_6.tex +++ /dev/null @@ -1,147 +0,0 @@ -$X$ 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{Poincar\'e duality}} H^2(X, \mathbb{Z} ) \xrightarrow{\text{evaluation}}\Hom(H_2(X, \mathbb{Z}), \mathbb{Z}) -\end{align*} - -Intersection form: -$H_2(X, \mathbb{Z}) \times -H_2(X, \mathbb{Z}) \longrightarrow \mathbb{Z}$ is symmetric and non singular. -\\ -Let $A$ and $B$ be closed, oriented surfaces in $X$. -\\ -\begin{figure}[h] -\fontsize{20}{10}\selectfont -\centering{ -\def\svgwidth{\linewidth} -\resizebox{0.5\textwidth}{!}{\input{images/intersection_form_A_B.pdf_tex}} -} -\caption{$T_X A + T_X B = T_X X$ -}\label{fig:torus_alpha_beta} -\end{figure} -??????????????????????? -\begin{align*} -x \in A \cap B\\ -T_XA \oplus T_X B = T_X X\\ -\{\epsilon_1, \dots , \epsilon_n \} = A \cap C\\ -A \cdot B = \sum^n_{i=1} \epsilon_i -\end{align*} -\begin{proposition} -Intersection form $A \cdot B$ doesn't depend of choice of $A$ and $B$ in their homology classes: -\[ -[A], [B] \in H_2(X, \mathbb{Z}). -\] -\end{proposition} -\noindent -\\ - -If $M$ is an $m$ - dimensional close, connected and orientable manifold, then $H_m(M, \mathbb{Z})$ and the orientation if $M$ determined a cycle $[M] \in H_m(M, \mathbb{Z})$, called the fundamental cycle. -\begin{example} -If $\omega$ is an $m$ - form then: -\[ -\int_M \omega = [\omega]([M]), \quad [\omega] \in H^m_\Omega(M), \ [M] \in H_m(M). -\] - -\end{example} -???????????????????????????????????????????????? -\begin{figure}[h] -\fontsize{20}{10}\selectfont -\centering{ -\def\svgwidth{\linewidth} -\resizebox{0.8\textwidth}{!}{\input{images/torus_alpha_beta.pdf_tex}} -} -\caption{$\beta$ cross $3$ times the disk bounded by $\alpha$. -$T_X \alpha + T_X \beta = T_X \Sigma$ -}\label{fig:torus_alpha_beta} -\end{figure} -\begin{example} -?????????????????????????\\ -Let $X = S^2 \times S^2$. -We know that: -\begin{align*} -&H_2(S^2, \mathbb{Z}) =\mathbb{Z}\\ -&H_1(S^2, \mathbb{Z}) = 0\\ -&H_0(S^2, \mathbb{Z}) =\mathbb{Z} -\end{align*} -We can construct a long exact sequence for a pair: -\begin{align*} -&H_2(\partial X) \to H_2(X) -\to H_2(X, \partial X) \to \\ -\to &H_1(\partial X) \to H_1(X) \to H_1(X, \partial X) \to -\end{align*} -????????????????????\\ -Simple case $H_1(\partial X)$ \\????????????\\ - is torsion. -$H_2(\partial X)$ is torsion free (by universal coefficient theorem),\\ -???????????????????????\\ -therefore it is $0$. -\\?????????????????????\\ -We know that $b_1(X) = b_2(X)$. Therefore by Poincar\'e duality: -\begin{align*} -b_1(X) = -\dim_{\mathbb{Q}} H_1(X, \mathbb{Q}) -\overset{\mathrm{PD}}{=} -\dim_{\mathbb{Q}} H^2(X, \mathbb{Q}) = -\dim_{\mathbb{Q}} H_2(X, \mathbb{Q}) = b_2(X) -\end{align*} -???????????????????????????????\\ -$H_2(X, \mathbb{Z})$ is torsion free and -$H_2(X_1, \mathbb{Q}) = 0$, therefore $H_2(X, \mathbb{Z}) = 0$. -The map -$H_2(X, \mathbb{Z}) \longrightarrow H_2(X, \partial X, \mathbb{Z})$ is a monomorphism. \\??????????\\ (because it is an isomorphism after tensoring by $\mathbb{Q}$. -\\ -Suppose $\alpha_1, \dots, \alpha_n$ is a basis of $H_2(X, \mathbb{Z})$. -Let $A$ be the intersection matrix in this basis. Then: -\begin{enumerate} -\item -A has integer coefficients, -\item -$\det A \neq 0$, -\item -$\vert \det A \vert = -\vert H_1 (\partial X, \mathbb{Z}) \vert = -\vert \coker H_2(X) \longrightarrow H_2(X, \partial X) \vert$. -\end{enumerate} -\end{example} -???????????????????\\ -If $CVC^T = W$, then for -$\binom{a}{b} = C^{-1} \binom{1}{0}$ we have $\binom{a}{b} $ \\ -????????????????\\ -$\omega \binom{a}{b} = \binom{1}{0} u \binom{1}{0} = 1$. - -\begin{theorem}[Whitehead] -Any non-degenerate form -\[ -A : \mathbb{Z}^4 \times \mathbb{Z}^4 \longrightarrow \mathbb{Z} -\] -can be realized as an intersection form of a simple connected $4$-dimensional manifold. -\end{theorem} -?????????????????????????? -\begin{theorem}[Donaldson, 1982] -If $A$ is an even definite intersection form of a smooth $4$-manifold then it is diagonalizable over $\mathbb{Z}$. -\end{theorem} -?????????????????????????? -?????????????????????????? -?????????????????????????? -?????????????????????????? -\begin{definition} -even define -\end{definition} -Suppose $X$ us $4$ -manifold with a boundary such that $H_1(X) = 0$. - -%$A \cdot B$ gives the pairing as ?? -\begin{proof} -Obviously: -\[H_1(\partial X, \mathbb{Z}) = \coker H_2(X) \longrightarrow H_2(X, \partial X) = \quot{H_2(X, \partial X)}{H_2(X)}. -\] -Let $A$ be an $n \times n$ matrix. $A$ determines a \\ -??????????????/\\ -\begin{align*} -\mathbb{Z}^n \longrightarrow \Hom (\mathbb{Z}^n, \mathbb{Z})\\ -a \mapsto (b \mapsto b^T A a)\\ -\vert \coker A \vert = \vert \det A \vert -\end{align*} -all homomorphisms $b = (b_1, \dots, b_n) $???????\\?????????\\ - -\end{proof}