\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}} \DeclareRobustCommand\longtwoheadrightarrow {\relbar\joinrel\twoheadrightarrow} \newcommand{\longhookrightarrow}{\lhook\joinrel\longrightarrow} \newcommand{\longhookleftarrow}{\longleftarrow\joinrel\rhook} \AtBeginDocument{\renewcommand{\setminus}{% \mathbin{\backslash}}} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\ord}{ord} \DeclareMathOperator{\Gl}{GL} \DeclareMathOperator{\Sl}{SL} \DeclareMathOperator{\Lk}{lk} \DeclareMathOperator{\pt}{\{pt\}} \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}} \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} \[ H^* (S^3, S^3 \setminus N) \cong H^* (N, \partial N) \] \\ ?????????????? \\ \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 + \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} \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} 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{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 $\partial \Sigma = K$. \end{definition} \noindent Remark: $3$ - genus is additive under taking connected sum, but $4$ - genus is not. \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} \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 $.\\ \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} \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} ?????????????????????????????????\\ \noindent 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 $A = V \times V^T$, $n = \rank V$. %\input{ink_diag} \begin{figure}[h] \fontsize{20}{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} \begin{figure}[h] \fontsize{20}{10}\selectfont \centering{ \def\svgwidth{\linewidth} \resizebox{0.5\textwidth}{!}{\input{images/ball_4_pushed_cycle.pdf_tex}} \caption{Cycle pushed in 4-ball.} \label{fig:pushCycle} } \end{figure} \noindent Let $Y = \Sigma(K)$. Then: \begin{align*} 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. \end{align*} ???????????????????????????? \begin{align*} H_1(Y, \mathbb{Z}) \cong \quot{\mathbb{Z}^n}{A\mathbb{Z}}\\ A \longrightarrow BAC^T \quad \text{Smith normal form} \end{align*} ???????????????????????\\ In general %no lecture at 29.04 \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, Borodzik-Conway-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{\hfill\DTMdate{2019-06-03}} \begin{theorem} Let $K$ be a knot and $u(K)$ its unknotting number. Let $g_4$ be a minimal four genus of a smooth surface $S$ in $B^4$ such that $\partial S = K$. Then: \[ u(K) \geq g_4(K) \] \begin{proof} Recall that if $u(K)=u$ then $K$ bounds a disk $\Delta$ with $u$ ordinary double points. \\ \noindent Remove from $\Delta$ the two self intersecting and glue the Seifert surface for the Hopf link. The reality surface $S$ has Euler characteristic $\chi(S) = 1 - 2u$. Therefore $g_4(S) = u$ . \end{proof} ???????????????????\\ \begin{example} The knot $8_{20}$ is slice: $\sigma \equiv 0$ almost everywhere but $\sigma(e^{\frac{ 2\pi i}{6}}) = + 1$. \end{example} %ref Structure in the classical knot concordance group %Tim D. Cochran, Kent E. Orr, Peter Teichner %Journal-ref: Comment. Math. Helv. 79 (2004) 105-123 \subsection*{Surgery} %Rolfsen, geometric group theory, Diffeomorpphism of a torus, Mapping class group Recall that $H_1(S^1 \times S^1, \mathbb{Z}) = \mathbb{Z}^3$. As generators for $H_1$ we can set ${\alpha = [S^1 \times \pt]}$ and ${\beta=[\pt \times S^1]}$. Suppose ${\phi: S^1 \times S^1 \longrightarrow S^1 \times S^1}$ is a diffeomorphism. Consider an induced map on homology group: \begin{align*} H_1(S^1 \times S^1, \mathbb{Z}) \ni \phi_* (\alpha) &= p\alpha + q \beta, \quad p, q \in \mathbb{Z},\\ \phi_*(\beta) &= r \alpha + s \beta, \quad r, s \in \mathbb{Z}, \\ \phi_* &= \begin{pmatrix} p & q\\ r & s \end{pmatrix} \end{align*} As $\phi_*$ is diffeomorphis, it must be invertible over $\mathbb{Z}$. Then for a direction preserving diffeomorphism we have $\det \phi_* = 1$. Therefore $\phi_* \in \Sl(2, \mathbb{Z})$. \end{theorem} \vspace{10cm} \begin{theorem} Every such a matrix can be realized as a torus. \end{theorem} \begin{proof} \begin{enumerate}[label={(\Roman*)}] \item Geometric reason \begin{align*} \phi_t: S^1 \times S^1 &\longrightarrow S^1 \times S^1 \\ S^1 \times \pt &\longrightarrow \pt \times S^1 \\ \pt \times S^1 &\longrightarrow S^1 \times \pt \\ (x, y) & \mapsto (-y, x) \end{align*} \item \end{enumerate} \end{proof} \section{balagan} \noindent \noindent \section{\hfill\DTMdate{2019-05-06}} \begin{definition} Let $X$ be a knot complement. Then $H_1(X, \mathbb{Z}) \cong \mathbb{Z}$ and there exists an epimorphism $\pi_1(X) \overset{\phi}\twoheadrightarrow \mathbb{Z}$.\\ The infinite cyclic cover of a knot complement $X$ is the cover associated with the epimorphism $\phi$. \[ \widetilde{X} \longtwoheadrightarrow X \] \end{definition} %Rolfsen, bachalor thesis of Kamila \begin{figure}[h] \fontsize{10}{10}\selectfont \centering{ \def\svgwidth{\linewidth} \resizebox{1\textwidth}{!}{\input{images/covering.pdf_tex}} \caption{Infinite cyclic cover of a knot complement.} \label{fig:covering} } \end{figure} \begin{figure}[h] \fontsize{10}{10}\selectfont \centering{ \def\svgwidth{\linewidth} \resizebox{0.8\textwidth}{!}{\input{images/knot_complement.pdf_tex}} \caption{A knot complement.} \label{fig:complement} } \end{figure} \noindent Formal sums $\sum \phi_i(t) a_i + \sum \phi_j(t)\alpha_j$ \\ finitely generated as a $\mathbb{Z}[t, t^{-1}]$ module. \\ Let $v_{ij} = \Lk(a_i, a_j^+)$. Then $V = \{ v_ij\}_{i, j = 1}^n$ is the Seifert matrix associated to the surface $\Sigma$ and the basis $a_1, \dots, a_n$. Therefore $a_k^+ = \sum_{j} v_{jk} \alpha_j$. Then $\Lk(a_i, a_k^+)= \Lk(a_k^+, a_i) = \sum_j v_{jk} \Lk(\alpha_j, a_i) = v_{ik}$. We also notice that $\Lk(a_i, a_j^-) = \Lk(a_i^+, a_j)= v_{ij}$ and $a_j^- = \sum_k v_{kj} t^{-1} \alpha_j$. \\ \noindent The homology of $\widetilde{X}$ is generated by $a_1, \dots, a_n$ and relations. \begin{definition} The Nakanishi index of a knot is the minimal number of generators of $H_1(\widetilde{X})$. \end{definition} %see Maciej page \noindent Remark about notation: sometimes one writes $H_1(X; \mathbb{Z}[t, t^{-1}])$ (what is also notation for twisted homology) instead of $H_1(\widetilde{X})$. \\ ????????????????????? \\ \noindent $\Sigma_?(K) \rightarrow S^3$ ?????\\ $H_1(\Sigma_?(K), \mathbb{Z}) = h$\\ $H \times H \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}}$\\ ...\\ Let now $H = H_1(\widetilde{X})$. Can we define a paring? \\ Let $c, d \in H(\widetilde{X})$ (see Figure \ref{fig:covering_pairing}), $\Delta$ an Alexander polynomial. We know that $\Delta c = 0 \in H_1(\widetilde{X})$ (Alexander polynomial annihilates all possible elements). Let consider a surface $F$ such that $\partial F = c$. Now consider intersection points $F \cdot d$. This points can exist in any $N_k$ or $S_k$. \[ \frac{1}{\Delta} \sum_{j\in \mathbb{Z} t^{-j}}(F \cdot t^j d) \in \quot{\mathbb{Q}[t, t^{-1}]}{\mathbb{Z}[t, t^{-1}]} \] \\ ?????????????\\ \begin{figure}[h] \fontsize{10}{10}\selectfont \centering{ \def\svgwidth{\linewidth} \resizebox{1\textwidth}{!}{\input{images/covering_pairing.pdf_tex}} \caption{$c, d \in H_1(\widetilde{X})$.} \label{fig:covering_pairing} } \end{figure} \begin{definition} The $\mathbb{Z}[t, t^{-1}]$ module $H_1(\widetilde{X})$ is called the Alexander module of knot $K$. \end{definition} \noindent Let $R$ be a PID, $M$ a finitely generated $R$ module. Let us consider \[ R^k \overset{A} \longrightarrow R^n \longtwoheadrightarrow M, \] where $A$ is a $k \times n$ matrix, assume $k\ge n$. The order of $M$ is the $\gcd$ of all determinants of the $n \times n$ minors of $A$. If $k = n$ then $\ord M = \det A$. \begin{theorem} Order of $M$ doesn't depend on $A$. \end{theorem} \noindent For knots the order of the Alexander module is the Alexander polynomial. \begin{theorem} \[ \forall x \in M: (\ord M) x = 0. \] \end{theorem} \noindent $M$ is well defined up to a unit in $R$. \subsection*{Blanchfield pairing} \section{balagan} \begin{theorem} Let $H_p$ be a $p$ - torsion part of $H$. There exists an orthogonal decomposition of $H_p$: \[ H_p = H_{p, 1} \oplus \dots \oplus H_{p, r_p}. \] $H_{p, i}$ is a cyclic module: \[ H_{p, i} = \quot{\mathbb{Z}[t, t^{-1}]}{p^{k_i} \mathbb{Z} [t, t^{-1}]} \] \end{theorem} \noindent The proof is the same as over $\mathbb{Z}$. \noindent \end{document}