\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{\mytop}{top} \DeclareMathOperator{\Gl}{GL} \DeclareMathOperator{\Sl}{SL} \DeclareMathOperator{\Lk}{lk} \DeclareMathOperator{\pt}{\{pt\}} \DeclareMathOperator{\sign}{sign} \titleformat{\subsection}{% \normalfont \fontsize{12}{15}\bfseries}{% }{.0ex plus .2ex}{} \titleformat{\section}{% \normalfont \fontsize{13}{15} \bfseries}{% Lecture\ \thesection}% {2.3ex plus .2ex}{} \titlespacing*{\section} {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{Alexander polynomial \hfill\DTMdate{2019-03-04}} \input{lec_2.tex} %add Hurewicz theorem? \section{\hfill\DTMdate{2019-03-11}} \input{lec_3.tex} \section{Concordance group \hfill\DTMdate{2019-03-18}} \input{lec_4.tex} \section{\hfill\DTMdate{2019-03-25}} \input{lec_5.tex} \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: \[ [A], [B] \in H_2(X, \mathbb{Z}). \] \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 $\alpha$. $T_X \alpha + T_X \beta = T_Z \Sigma$ }\label{fig:torus_alpha_beta} \end{figure} \begin{theorem} Any non-degenerate form \[ A : \mathbb{Z}^n \times \mathbb{Z}^n \longrightarrow \mathbb{Z} \] 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 defined 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 ?? \section{\hfill\DTMdate{2019-04-15}} \begin{theorem} Suppose that $K \subset S^3$ is a slice knot (i.e. $K$ bound a disk in $B^4$). Then if $F$ is a Seifert surface of $K$ and $V$ denotes the associated Seifet matrix, then there exists $P \in \Gl_g(\mathbb{Z})$ such that: \\??????????????? T ???????? \begin{align} PVP^{-1} = \begin{pmatrix} 0 & A\\ B & C \end{pmatrix}, \quad A, C, C \in M_{g \times g} (\mathbb{Z}) \end{align} \end{theorem} In other words you can find rank $g$ direct summand $\mathcal{Z}$ of $H_1(F)$ \\ ????????????\\ such that for any $\alpha, \beta \in \mathcal{L}$ the linking number $\Lk (\alpha, \beta^+) = 0$. \begin{definition} An abstract Seifert matrix (i. e. \end{definition} 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 $\vert \det A\vert = \# H_1(Y, \mathbb{Z})$.\\ 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 $(t - \zeta)^2 \vert P$ (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{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} % % \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 %Add NotePrintSaveCiteYour opinionEmailShare %Saveliev, Nikolai %Lectures on the Topology of 3-Manifolds %An Introduction to the Casson Invariant \begin{figure}[h] \fontsize{10}{10}\selectfont \centering{ \def\svgwidth{\linewidth} \resizebox{0.5\textwidth}{!}{\input{images/ball_4_alpha_beta.pdf_tex}} } %\caption{Sketch for Fact %%\label{fig:concordance_m} \end{figure} \end{document}