\documentclass[12pt, twoside]{article} \usepackage{amssymb} \usepackage{amsmath} \usepackage{advdate} \usepackage{amsthm} \usepackage[english]{babel} \usepackage{comment} \usepackage{csquotes} \usepackage[useregional]{datetime2} \usepackage{enumitem} \usepackage{fontspec} \usepackage{float} \usepackage{graphicx} \usepackage{hyperref} \usepackage{mathtools} \usepackage{pict2e} \usepackage[pdf]{pstricks} \usepackage{tikz} \usepackage{titlesec} \usepackage{xfrac} \usepackage{unicode-math} \usetikzlibrary{cd} \hypersetup{ colorlinks, citecolor=black, filecolor=black, linkcolor=black, urlcolor=black } \newtheoremstyle{break} {\topsep}{\topsep}% {\itshape}{}% {\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}}} \newcommand{\overbar}[1]{% \mkern 1.5mu=\overline{% \mkern-1.5mu#1\mkern-1.5mu}% \mkern 1.5mu} \newcommand{\sdots}{\smash{\vdots}} \AtBeginDocument{\renewcommand{\setminus}{% \mathbin{\backslash}}} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\Gl}{Gl} \DeclareMathOperator{\Lk}{lk} \titleformat{\section}{\normalfont \fontsize{12}{15} \bfseries}{% Lecture\ \thesection}% {2.3ex plus .2ex}{} \titlespacing*{\section} {0pt}{16.5ex plus 1ex minus .2ex}{4.3ex plus .2ex} \setlist[itemize]{topsep=0pt,before=% \leavevmode\vspace{0.5em}} \input{knots_macros} \graphicspath{ {images/} } \begin{document} \tableofcontents %\newpage %\input{myNotes} \section{Basic definitions \hfill\DTMdate{2019-02-25}} \begin{definition} A knot $K$ in $S^3$ is a smooth (PL - smooth) embedding of a circle $S^1$ in $S^3$: \begin{align*} \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 ${D_{\pi}}_{\big|L}$ 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] 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} \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 $\beta$: \[ \alpha \in H_1(S^3 \setminus \nu(\beta), \mathbb{Z}) \cong \mathbb{Z}.\] \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} \section{\hfill\DTMdate{2019-03-04}} \begin{theorem} For any knot $K \subset S^3$ there exists a connected, compact and orientable surface $\Sigma(K)$ such that $\partial \Sigma(K) = K$ \end{theorem} \begin{proof}("joke")\\ Let $K \in S^3$ be a knot and $N = \nu(K)$ be its tubular neighbourhood. Because $K$ and $N$ are homotopy equivalent, we get: \begin{align*} H^1(S^3 \setminus N ) \cong H^1(S^3 \setminus K). \end{align*} Let us consider a long exact sequence of cohomology of a pair $(S^3, S^3 \setminus N)$ with integer coefficients: \begin{center} \begin{tikzcd} [ column sep=0cm, fill=none, row sep=small, ar symbol/.style =% {draw=none,"\textstyle#1" description,sloped}, isomorphic/.style = {ar symbol={\cong}}, ] &\mathbb{Z} \\ & H^0(S^3) \ar[u,isomorphic] \to &H^0(S^3 \setminus N) \to \\ \to H^1(S^3, S^3 \setminus N) \to & H^1(S^3) \to & H^1(S^3\setminus N) \to \\ & 0 \ar[u,isomorphic]& \\ \to H^2(S^3, S^3 \setminus N) \to & H^2(S^3) \ar[u,isomorphic] \to & H^2(S^3\setminus N) \to \\ \to H^3(S^3, S^3\setminus N)\to & H^3(S) \to & 0 \\ & \mathbb{Z} \ar[u,isomorphic] &\\ \end{tikzcd} \end{center} \[ H^* (S^3, S^3 \setminus N) \cong H^* (N, \partial N) \] \\ ?????????????? \\ \end{proof} \begin{definition} Let $S$ be a Seifert matrix for a knot $K$. The Alexander polynomial $\Delta_K(t)$ is a Laurent polynomial: \[ \Delta_K(t) := \det (tS - S^T) \in \mathbb{Z}[t, t^{-1}] \cong \mathbb{Z}[\mathbb{Z}] \] \end{definition} \begin{theorem} $\Delta_K(t)$ is well defined up to multiplication by $\pm t^k$, for $k \in \mathbb{Z}$. \end{theorem} \begin{proof} We need to show that $\Delta_K(t)$ doesn't depend on $S$-equivalence relation. \begin{enumerate}[label={(\arabic*)}] \item Suppose $S\prime = CSC^T$, $C \in \Gl(n, \mathbb{Z})$ (matrices invertible over $\mathbb{Z}$). Then $\det C = 1$ and: \begin{align*} &\det(tS\prime - S\prime^T) = \det(tCSC^T - (CSC^T)^T) =\\ &\det(tCSC^T - CS^TC^T) = \det C(tS - S^T)C^T = \det(tS - S^T) \end{align*} \item Let \\ $ A := t \begin{pmatrix} \begin{array}{c|c} S & \begin{matrix} \ast & 0 \\ \sdots & \sdots\\ \ast & 0 \end{matrix} \\ \hline \begin{matrix} \ast & \dots & \ast\\ 0 & \dots & 0 \end{matrix} & \begin{matrix} 0 & 0\\ 1 & 0 \end{matrix} \end{array} \end{pmatrix} - \begin{pmatrix} \begin{array}{c|c} S^T & \begin{matrix} \ast & 0 \\ \sdots & \sdots\\ \ast & 0 \end{matrix} \\ \hline \begin{matrix} \ast & \dots & \ast\\ 0 & \dots & 0 \end{matrix} & \begin{matrix} 0 & 1\\ 0 & 0 \end{matrix} \end{array} \end{pmatrix} = \begin{pmatrix} \begin{array}{c|c} tS - S^T & \begin{matrix} \ast & 0 \\ \sdots & \sdots\\ \ast & 0 \end{matrix} \\ \hline \begin{matrix} \ast & \dots & \ast\\ 0 & \dots & 0 \end{matrix} & \begin{matrix} 0 & -1\\ t & 0 \end{matrix} \end{array} \end{pmatrix} $ \\ \\ Using the Laplace expansion we get $\det A = \pm t \det(tS - S^T)$. \end{enumerate} \end{proof} % % % \begin{example} If $K$ is a trefoil then we can take $S = \begin{pmatrix} -1 & -1 \\ 0 & -1 \end{pmatrix}$. 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 + \cdots + 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 ${D^2 \overset{g}\hookrightarrow M}$ such that: \[ g_{\big| \partial D^2} = f_{\big| \partial D^2.} \] \end{lemma} \section{} \begin{example} \begin{align*} &F: \mathbb{C}^2 \rightarrow \mathbb{C} \text{ a polynomial} \\ &F(0) = 0 \end{align*} \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{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 $B^4$. \end{definition} \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} \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. \\ \\ \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{Poincar\'e duality}} H^2(X, \mathbb{Z} ) \xrightarrow{\text{evaluation}}\Hom(H_2(X, \mathbb{Z}), \mathbb{Z}) \end{align*} Intersection form: $H_2(X, \mathbb{Z}) \times H_2(X, \mathbb{Z}) \longrightarrow \mathbb{Z}$ - symmetric, non singular. \\ Let $A$ and $B$ be closed, oriented surfaces in $X$. \begin{proposition} $A \cdot B$ doesn't depend of choice of $A$ and $B$ in their homology classes. %$A \cdot B$ gives the pairing as ?? \end{proposition} \section{\hfill\DTMdate{2019-03-11}} \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}} In other words:\\ Choose a basis $(b_1, ..., b_i)$ \\ ???\\ of $H_2(Y, \mathbb{Z}$, then $A = (b_i, b_y)$ \\??\\ is a matrix of intersection form: \begin{align*} \quot{\mathbb{Z}^n}{A\mathbb{Z}^n} \cong H_1(Y, \mathbb{Z}). \end{align*} In particular $\mid \det A\mid = \# H_1(Y, \mathbb{Z}$.\\ That means - what is happening on boundary is a measure of degeneracy. \begin{center} \begin{tikzcd} [ column sep=tiny, row sep=small, ar symbol/.style =% {draw=none,"\textstyle#1" description,sloped}, isomorphic/.style = {ar symbol={\cong}}, ] H_1(Y, \mathbb{Z}) & \times \quad H_1(Y, \mathbb{Z})& \longrightarrow & \quot{\mathbb{Q}}{\mathbb{Z}} \text{ - a linking form} \\ \quot{\mathbb{Z}^n}{A\mathbb{Z}} \ar[u,isomorphic] & \quot{\mathbb{Z}^n}{A\mathbb{Z}} \ar[u,isomorphic] &\\ \end{tikzcd} $(a, b) \mapsto aA^{-1}b^T$ \end{center} The intersection form on a four-manifold determines the linking on the boundary. \\ \noindent Let $K \in S^1$ be a knot, $\Sigma(K)$ its double branched cover. If $V$ is a Seifert matrix for $K$, then $H_1(\Sigma(K), \mathbb{Z}) \cong \quot{\mathbb{Z}^n}{A\mathbb{Z}}$ where $A = V \times V^T$, where $n = \rank V$. %\input{ink_diag} \begin{figure}[h] \fontsize{40}{10}\selectfont \centering{ \def\svgwidth{\linewidth} \resizebox{0.5\textwidth}{!}{\input{images/ball_4.pdf_tex}} \caption{Pushing the Seifert surface in 4-ball.} \label{fig:pushSeifert} } \end{figure} \noindent Let $X$ be the four-manifold obtained via the double branched cover of $B^4$ branched along $\widetilde{\Sigma}$. \begin{fact} \begin{itemize} \item $X$ is a smooth four-manifold, \item $H_1(X, \mathbb{Z}) =0$, \item $H_2(X, \mathbb{Z}) \cong \mathbb{Z}^n$ \item The intersection form on $X$ is $V + V^T$. \end{itemize} \end{fact} \noindent Let $Y = \Sigma(K)$. Then: \begin{flalign*} H_1(Y, \mathbb{Z}) \times H_1(Y, \mathbb{Z}) \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}}& \\ (a,b) \mapsto a A^{-1} b^{T},\qquad A = V + V^T& \\ H_1(Y, \mathbb{Z}) \cong \quot{\mathbb{Z}^n}{A\mathbb{Z}}&\\ A \longrightarrow BAC^T \quad \text{Smith normal form}& \end{flalign*} ???????????????????????\\ 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 bilinear form - the intersection form on $M$: \begin{center} \begin{tikzcd} [ column sep=tiny, row sep=small, ar symbol/.style = {draw=none,"\textstyle#1" description,sloped}, isomorphic/.style = {ar symbol={\cong}}, ] H_2(M, \mathbb{Z})& \times & H_2(M, \mathbb{Z}) \longrightarrow & \mathbb{Z} \\ \ar[u,isomorphic] \mathbb{Z}^n && &\\ \end{tikzcd} \end{center} \noindent Let us consider a specific case: $M$ has a boundary $Y = \partial M$. Betti number $b_1(Y) = 0$, $H_1(Y, \mathbb{Z})$ is finite. Then the intersection form can be degenerated in the sense that: \begin{align*} H_2(M, \mathbb{Z}) \times H_2(M, \mathbb{Z}) &\longrightarrow \mathbb{Z} \quad& H_2(M, \mathbb{Z}) &\longrightarrow \Hom (H_2(M, \mathbb{Z}), \mathbb{Z})\\ (a, b) &\mapsto \mathbb{Z} \quad& a &\mapsto (a, \_) H_2(M, \mathbb{Z}) \end{align*} has coker precisely $H_1(Y, \mathbb{Z})$. \\???????????????\\ Let $K \subset S^3$ be a knot, \\ $X = S^3 \setminus K$ - a knot complement, \\ $\widetilde{X} \xrightarrow{\enspace \rho \enspace} X$ - an infinite cyclic cover (universal abelian cover). \begin{align*} \pi_1(X) \longrightarrow \quot{\pi_1(X)}{[\pi_1(X), \pi_1(X)]} = H_1(X, \mathbb{Z} ) \cong \mathbb{Z} \end{align*} $C_{*}(\widetilde{X})$ has a structure of a $\mathbb{Z}[t, t^{-1}] \cong \mathbb{Z}[\mathbb{Z}]$ module. \\ $H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}])$ - Alexander module, \\ \begin{align*} H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \times H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}[t, t^{-1}]} \end{align*} \begin{fact} \begin{align*} &H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \cong \quot{\mathbb{Z}{[t, t^{-1}]}^n}{(tV - V^T)\mathbb{Z}[t, t^{-1}]^n}\;, \\ &\text{where $V$ is a Seifert matrix.} \end{align*} \end{fact} \begin{fact} \begin{align*} H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \times H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) &\longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}[t, t^{-1}]}\\ (\alpha, \beta) &\mapsto \alpha^{-1}(t -1)(tV - V^T)^{-1}\beta \end{align*} \end{fact} \noindent Note that $\mathbb{Z}$ is not PID. Therefore we don't have primer decomposition of this moduli. We can simplify this problem by replacing $\mathbb{Z}$ by $\mathbb{R}$. We lose some date by doing this transition. \begin{align*} &\xi \in S^1 \setminus \{ \pm 1\} \quad p_{\xi} = (t - \xi)(t - \xi^{-1}) t^{-1} \\ &\xi \in \mathbb{R} \setminus \{ \pm 1\} \quad q_{\xi} = (t - \xi)(t - \xi^{-1}) t^{-1} \\ & \xi \notin \mathbb{R} \cup S^1 \quad q_{\xi} = (t - \xi)(t - \overbar{\xi})(t - \xi^{-1})(t - \overbar{\xi}^{-1}) t^{-2}\\ & \Lambda = \mathbb{R}[t, t^{-1}]\\ &\text{Then: } H_1(\widetilde{X}, \Lambda) \cong \bigoplus_{\substack{\xi \in S^1 \setminus \{\pm 1 \}\\ k\geq 0}} ( \quot{\Lambda}{p_{\xi}^k })^{n_k, \xi} \oplus \bigoplus_{\substack{\xi \notin S^1 \\ l\geq 0}} (\quot{\Lambda}{q_{\xi}^l})^{n_l, \xi}& \end{align*} We can make this composition orthogonal with respect to the Blanchfield paring. \vspace{0.5cm}\\ Historical remark: \begin{itemize} \item John Milnor, \textit{On isometries of inner product spaces}, 1969, \item Walter Neumann, \textit{Invariants of plane curve singularities} %in: Knots, braids and singulari- ties (Plans-sur-Bex, 1982), 223–232, Monogr. Enseign. Math., 31, Enseignement Math., Geneva , 1983, \item András Némethi, \textit{The real Seifert form and the spectral pairs of isolated hypersurfaceenumerate singularities}, 1995, %Compositio Mathematica, Volume 98 (1995) no. 1, p. 23-41 \item Maciej Borodzik, Stefan Friedl \textit{The unknotting number and classical invariants II}, 2014. \end{itemize} \vspace{0.5cm} Let $p = p_{\xi}$, $k\geq 0$. \begin{align*} \quot{\Lambda}{p^k \Lambda} \times \quot{\Lambda}{p^k \Lambda} &\longrightarrow \quot{\mathbb{Q}(t)}{\Lambda}\\ (1, 1) &\mapsto \kappa\\ \text{Now: } (p^k \cdot 1, 1) &\mapsto 0\\ p^k \kappa = 0 &\in \quot{\mathbb{Q}(t)}{\Lambda}\\ \text{therfore } p^k \kappa &\in \Lambda\\ \text{we have } (1, 1) &\mapsto \frac{h}{p^k}\\ \end{align*} $h$ is not uniquely defined: $h \rightarrow h + g p^k$ doesn't affect paring. \\ Let $h = p^k \kappa$. \begin{example} \begin{align*} \phi_0 ((1, 1))=\frac{+1}{p}\\ \phi_1 ((1, 1)) = \frac{-1}{p} \end{align*} $\phi_0$ and $\phi_1$ are not isomorphic. \end{example} \begin{proof} Let $\Phi: \quot{\Lambda}{p^k \Lambda} \longrightarrow \quot{\Lambda}{p^k \Lambda}$ be an isomorphism. \\ Let: $\Phi(1) = g \in \lambda$ \begin{align*} \quot{\Lambda}{p^k \Lambda} \xrightarrow{\enspace \Phi \enspace}& \quot{\Lambda}{p^k \Lambda}\\ \phi_0((1, 1)) = \frac{1}{p^k} \qquad&\qquad \phi_1((g, g)) = \frac{1}{p^k} \quad \text{($\Phi$ is an isometry).} \end{align*} Suppose for the paring $\phi_1((g, g))=\frac{1}{p^k}$ we have $\phi_1((1, 1)) = \frac{-1}{p^k}$. Then: \begin{align*} \frac{-g\overbar{g}}{p^k} = \frac{1}{p^k} &\in \quot{\mathbb{Q}(t)}{\Lambda}\\ \frac{-g\overbar{g}}{p^k} - \frac{1}{p^k} &\in \Lambda \\ -g\overbar{g} &\equiv 1\pmod{p} \text{ in } \Lambda\\ -g\overbar{g} - 1 &= p^k \omega \text{ for some } \omega \in \Lambda\\ \text{evalueting at $\xi$: }\\ \overbrace{-g(\xi)g(\xi^{-1})}^{>0} - 1 = 0 \quad \contradiction \end{align*} \end{proof} ????????????????????\\ \begin{align*} g &= \sum{g_i t^i}\\ \overbar{g} &= \sum{g_i t^{-i}}\\ \overbar{g}(\xi) &= \sum g_i \xi^i \quad \xi \in S^1\\ \overbar{g}(\xi) &=\overbar{g(\xi)} \end{align*} Suppose $g = (t - \xi)^{\alpha} g^{\prime}$. Then $(t - \xi)^{k - \alpha}$ goes to $0$ in $\quot{\Lambda}{p^k \Lambda}$. \begin{theorem} Every sesquilinear non-degenerate pairing \begin{align*} \quot{\Lambda}{p^k} \times \quot{\Lambda}{p} \longleftrightarrow \frac{h}{p^k} \end{align*} is isomorphic either to the pairing wit $h=1$ or to the paring with $h=-1$ depending on sign of $h(\xi)$ (which is a real number). \end{theorem} \begin{proof} There are two steps of the proof: \begin{enumerate} \item Reduce to the case when $h$ has a constant sign on $S^1$. \item Prove in the case, when $h$ has a constant sign on $S^1$. \end{enumerate} \begin{lemma} If $P$ is a symmetric polynomial such that $P(\eta)\geq 0$ for all $\eta \in S^1$, then $P$ can be written as a product $P = g \overbar{g}$ for some polynomial $g$. \end{lemma} \begin{proof}[Sketch of proof] Induction over $\deg P$.\\ Let $\zeta \notin S^1$ be a root of $P$, $P \in \mathbb{R}[t, t^{-1}]$. Assume $\zeta \notin \mathbb{R}$. We know that polynomial $P$ is divisible by $(t - \zeta)$, $(t - \overbar{\zeta})$, $(t^{-1} - \zeta)$ and $(t^{-1} - \overbar{\zeta})$. Therefore: \begin{align*} &P^{\prime} = \frac{P}{(t - \zeta)(t - \overbar{\zeta})(t^{-1} - \zeta)(t^{-1} - \overbar{\zeta})}\\ &P^{\prime} = g^{\prime}\overbar{g} \end{align*} We set $g = g^{\prime}(t - \zeta)(t - \overbar{\zeta})$ and $P = g \overbar{g}$. Suppose $\zeta \in S^1$. Then $(t - \zeta)^2 \mid P$ (at least - otherwise it would change sign). Therefore: \begin{align*} &P^{\prime} = \frac{P}{(t - \zeta)^2(t^{-1} - \zeta)^2}\\ &g = (t - \zeta)(t^{-1} - \zeta) g^{\prime} \quad \text{etc.} \end{align*} The map $(1, 1) \mapsto \frac{h}{p^k} = \frac{g\overbar{g}h}{p^k}$ is isometric whenever $g$ is coprime with $P$. \end{proof} \begin{lemma}\label{L:coprime polynomials} Suppose $A$ and $B$ are two symmetric polynomials that are coprime and that $\forall z \in S^1$ either $A(z) > 0$ or $B(z) > 0$. Then there exist symmetric polynomials $P$, $Q$ such that $P(z), Q(z) > 0$ for $z \in S^1$ and $PA + QB \equiv 1$. \end{lemma} \begin{proof}[Idea of proof] For any $z$ find an interval $(a_z, b_z)$ such that if $P(z) \in (a_z, b_z)$ and $P(z)A(z) + Q(z)B(z) = 1$, then $Q(z) > 0$, $x(z) = \frac{az + bz}{i}$ is a continues function on $S^1$ approximating $z$ by a polynomial . \\??????????????????????????\\ \begin{flalign*} (1, 1) \mapsto \frac{h}{p^k} \mapsto \frac{g\overbar{g}h}{p^k}&\\ g\overbar{g} h + p^k\omega = 1& \end{flalign*} Apply Lemma \ref{L:coprime polynomials} for $A=h$, $B=p^{2k}$. Then, if the assumptions are satisfied, \begin{align*} Ph + Qp^{2k} = 1\\ p>0 \Rightarrow p = g \overbar{g}\\ p = (t - \xi)(t - \overbar{\xi})t^{-1}\\ \text{so } p \geq 0 \text{ on } S^1\\ p(t) = 0 \Leftrightarrow t = \xi or t = \overbar{\xi}\\ h(\xi) > 0\\ h(\overbar{\xi})>0\\ g\overbar{g}h + Qp^{2k} = 1\\ g\overbar{g}h \equiv 1 \mod{p^{2k}}\\ g\overbar{g} \equiv 1 \mod{p^k} \end{align*} ???????????????????????????????\\ If $P$ has no roots on $S^1$ then $B(z) > 0$ for all $z$, so the assumptions of Lemma \ref{L:coprime polynomials} are satisfied no matter what $A$ is. \end{proof} ?????????????????\\ \begin{align*} (\quot{\Lambda}{p_{\xi}^k} \times \quot{\Lambda}{p_{\xi}^k}) &\longrightarrow \frac{\epsilon}{p_{\xi}^k}, \quad \xi \in S^1 \setminus\{\pm 1\}\\ (\quot{\Lambda}{q_{\xi}^k} \times \quot{\Lambda}{q_{\xi}^k}) &\longrightarrow \frac{1}{q_{\xi}^k}, \quad \xi \notin S^1\\ \end{align*} ??????????????????? 1 ?? epsilon?\\ \begin{theorem}(Matumoto, Conway-Borodzik-Politarczyk) Let $K$ be a knot, \begin{align*} &H_1(\widetilde{X}, \Lambda) \times H_1(\widetilde{X}, \Lambda) = \bigoplus_{\substack{k, \xi, \epsilon\\ \xi in S^1}} (\quot{\Lambda}{p_{\xi}^k}, \epsilon)^{n_k, \xi, \epsilon} \oplus \bigoplus_{k, \eta} (\quot{\Lambda}{p_{\xi}^k})^{m_k} \end{align*} \begin{align*} \text{Let } \delta_{\sigma}(\xi) = \lim_{\varepsilon \rightarrow 0^{+}} \sigma(e^{2\pi i \varepsilon} \xi) - \sigma(e^{-2\pi i \varepsilon} \xi),\\ \text{then } \sigma_j(\xi) = \sigma(\xi) - \frac{1}{2} \lim_{\varepsilon \rightarrow 0} \sigma(e^{2\pi i \varepsilon}\xi) + \sigma(e^{-2 \pi i \varepsilon}\xi) \end{align*} The jump at $\xi$ is equal to $2 \sum\limits_{k_i \text{ odd}} \epsilon_i$. The peak of the signature function is equal to $\sum\limits_{k_i \text{even}} \epsilon_i$. %$(\eta_{k, \xi_l^{+}} -\eta_{k, \xi_l^{-}}$ \end{theorem} \end{proof} \section{\hfill\DTMdate{2019-05-27}} .... \begin{definition} A square hermitian matrix $A$ of size $n$. \end{definition} field of fractions \section{balagan} \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{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{document}