From 94e007761267cf024c5efe47f574365e997caef8 Mon Sep 17 00:00:00 2001 From: Maria Marchwicka Date: Fri, 7 Jun 2019 15:53:13 +0200 Subject: [PATCH] some new text --- lectures_on_knot_theory.tex | 66 +++++++++++++++++++++++-------------- 1 file changed, 41 insertions(+), 25 deletions(-) diff --git a/lectures_on_knot_theory.tex b/lectures_on_knot_theory.tex index 075b2e9..750d8ad 100644 --- a/lectures_on_knot_theory.tex +++ b/lectures_on_knot_theory.tex @@ -157,7 +157,7 @@ a Whitehead link: \includegraphics[width=0.13\textwidth]{WhiteheadLink.png}, \item Borromean link: -\includegraphics[width=0.1\textwidth]{BorromeanRings.png}, +\includegraphics[width=0.1\textwidth]{BorromeanRings.png}. \end{itemize} \end{example} % @@ -214,7 +214,7 @@ Let $D$ be an oriented diagram of a link $L$. We change the diagram by smoothing \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] +\begin{figure}[h] \fontsize{15}{10}\selectfont \centering{ \def\svgwidth{\linewidth} @@ -276,21 +276,21 @@ Let $\nu(\beta)$ be a tubular neighbourhood of $\beta$. The linking number can \begin{example} \begin{itemize} \item -Hopf link +Hopf link: \begin{figure}[h] \fontsize{20}{10}\selectfont \centering{ \def\svgwidth{\linewidth} -\resizebox{0.4\textwidth}{!}{\input{images/linking_hopf.pdf_tex}} +\resizebox{0.4\textwidth}{!}{\input{images/linking_hopf.pdf_tex}}, } \end{figure} \item -$T(6, 2)$ link +$T(6, 2)$ link: \begin{figure}[h] \fontsize{20}{10}\selectfont \centering{ \def\svgwidth{\linewidth} -\resizebox{0.4\textwidth}{!}{\input{images/linking_torus_6_2.pdf_tex}} +\resizebox{0.4\textwidth}{!}{\input{images/linking_torus_6_2.pdf_tex}}. } \end{figure} \end{itemize} @@ -301,7 +301,7 @@ Let $L$ be a link and $\Sigma$ be an oriented Seifert surface for $L$. Choose a 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] +\begin{figure}[h] \fontsize{20}{10}\selectfont \centering{ \def\svgwidth{\linewidth} @@ -521,7 +521,7 @@ If $K$ is a trefoil then we can take $S = \begin{pmatrix} -1 & -1 \\ 0 & -1 -\end{pmatrix}$. +\end{pmatrix}$. Then \[ \Delta_K(t) = \det \begin{pmatrix} @@ -529,7 +529,7 @@ $S = \begin{pmatrix} 1 & -t +1 \end{pmatrix} = (t -1)^2 + t = t^2 - t +1 \ne 1 -\Rightarrow \text{trefoil is not trivial} +\Rightarrow \text{trefoil is not trivial.} \] \end{example} \begin{fact} @@ -554,7 +554,8 @@ If $\Sigma$ is a genus $g$ - Seifert surface for $K$ then $H_1(\Sigma) = \mathbb \end{proof} \begin{example} There are not trivial knots with Alexander polynomial equal $1$, for example: -$\Delta_{11n34} \equiv 1$. +\includegraphics[width=0.3\textwidth]{11n34.png} +$\Delta_{11n34} \equiv 1$. \end{example} %removing one disk from surface doesn't change $H_1$ (only $H_2$) \section{} @@ -582,6 +583,17 @@ A knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk 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} @@ -666,7 +678,7 @@ Let $K \in S^1$ be a knot, $\Sigma(K)$ its double branched cover. If $V$ is a Se $H_1(\Sigma(K), \mathbb{Z}) \cong \quot{\mathbb{Z}^n}{A\mathbb{Z}}$ where $A = V \times V^T$, where $n = \rank V$. %\input{ink_diag} -\begin{figure}[H] +\begin{figure}[h] \fontsize{40}{10}\selectfont \centering{ \def\svgwidth{\linewidth} @@ -727,11 +739,11 @@ 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}\\ -H_2(M, \mathbb{Z}) \longrightarrow \Hom (H_2(M, \mathbb{Z}), \mathbb{Z})\\ -(a, b) \mapsto \mathbb{Z}\\ -a \mapsto (a, \_) 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 $H_1(Y, \mathbb{Z})$. \\???????????????\\ @@ -764,23 +776,27 @@ H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) &\longrightarrow \quot{\mathbb{Q}}{\ma \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{flalign*} -\xi \in S^1 \setminus \{ \pm 1\} +\begin{align*} +&\xi \in S^1 \setminus \{ \pm 1\} \quad p_{\xi} = -(t - \xi)(1 - \xi^{-1}) t^{-1}&\\ -\xi \in \mathbb{R} \setminus \{ \pm 1\} +(t - \xi)(t - \xi^{-1}) t^{-1} +\\ +&\xi \in \mathbb{R} \setminus \{ \pm 1\} \quad -q_{\xi} = (t - \xi)(1 - \xi^{-1}) t^{-1}&\\ +q_{\xi} = (t - \xi)(t - \xi^{-1}) t^{-1} +\\ +& \xi \notin \mathbb{R} \cup S^1 \quad -q_{\xi} = (t - \xi)(t - \overbar{\xi})(1 - \xi^{-1})(1 - \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}} +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{flalign*} +\end{align*} We can make this composition orthogonal with respect to the Blanchfield paring. \vspace{0.5cm}\\ Historical remark: