1119 lines
36 KiB
TeX
1119 lines
36 KiB
TeX
\documentclass[12pt, twoside]{article}
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\usepackage[useregional]{datetime2}
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\usepackage{mathtools}
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\usepackage[pdf]{pstricks}
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\theoremstyle{break}
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\newtheorem{lemma}{Lemma}[section]
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\newtheorem{fact}{Fact}[section]
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\newtheorem{corollary}{Corollary}[section]
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\newtheorem{proposition}{Proposition}[section]
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\newtheorem{example}{Example}[section]
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\newtheorem{problem}{Problem}[section]
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\newtheorem{definition}{Definition}[section]
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\newtheorem{theorem}{Theorem}[section]
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\newcommand{\contradiction}{%
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\newcommand{\sdots}{\smash{\vdots}}
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\DeclareRobustCommand\longtwoheadrightarrow
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{\relbar\joinrel\twoheadrightarrow}
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\newcommand{\longhookrightarrow}{\lhook\joinrel\longrightarrow}
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\AtBeginDocument{\renewcommand{\setminus}{%
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\mathbin{\backslash}}}
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\DeclareMathOperator{\Hom}{Hom}
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\DeclareMathOperator{\rank}{rank}
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\DeclareMathOperator{\ord}{ord}
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\DeclareMathOperator{\Gl}{GL}
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\DeclareMathOperator{\Sl}{SL}
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\DeclareMathOperator{\Lk}{lk}
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\DeclareMathOperator{\pt}{\{pt\}}
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\titleformat{\section}{\normalfont \fontsize{12}{15} \bfseries}{%
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Lecture\ \thesection}%
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{2.3ex plus .2ex}{}
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\titlespacing*{\section}
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{0pt}{16.5ex plus 1ex minus .2ex}{4.3ex plus .2ex}
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\setlist[itemize]{topsep=0pt,before=%
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\leavevmode\vspace{0.5em}}
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\input{knots_macros}
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\graphicspath{ {images/} }
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\begin{document}
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\tableofcontents
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%\newpage
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%\input{myNotes}
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\section{Basic definitions \hfill\DTMdate{2019-02-25}}
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\input{lec_1.tex}
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\section{\hfill\DTMdate{2019-03-04}}
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\begin{theorem}
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For any knot $K \subset S^3$ there exists a connected, compact and orientable surface $\Sigma(K)$ such that $\partial \Sigma(K) = K$
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\end{theorem}
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\begin{proof}("joke")\\
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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:
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\begin{align*}
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H^1(S^3 \setminus N ) \cong H^1(S^3 \setminus K).
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\end{align*}
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Let us consider a long exact sequence of cohomology of a pair $(S^3, S^3 \setminus N)$ with integer coefficients:
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\begin{center}
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\begin{tikzcd}
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[
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column sep=0cm, fill=none,
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row sep=small,
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ar symbol/.style =%
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{draw=none,"\textstyle#1" description,sloped},
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isomorphic/.style = {ar symbol={\cong}},
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]
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&\mathbb{Z}
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\\
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& H^0(S^3) \ar[u,isomorphic] \to
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&H^0(S^3 \setminus N) \to
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\\
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\to H^1(S^3, S^3 \setminus N) \to
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& H^1(S^3) \to
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& H^1(S^3\setminus N) \to
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\\
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& 0 \ar[u,isomorphic]&
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\\
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\to H^2(S^3, S^3 \setminus N) \to
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& H^2(S^3) \ar[u,isomorphic] \to
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& H^2(S^3\setminus N) \to
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\\
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\to H^3(S^3, S^3\setminus N)\to
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& H^3(S) \to
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& 0
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\\
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& \mathbb{Z} \ar[u,isomorphic] &\\
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\end{tikzcd}
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\end{center}
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\begin{align*}
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N \cong & D^2 \times S^1\\
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\partial N \cong & S^1 \times S^1\\
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H^1(N, \partial N) \cong & \mathbb{Z} \oplus \mathbb{Z}
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\end{align*}
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\begin{align*}
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H^* (S^3, S^3 \setminus N) &\cong H^* (N, \partial N)\\
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\\
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H^ 1 (S^3\setminus N) &\cong H^1(S^3\setminus K) \cong \mathbb{Z}
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\end{align*}
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\begin{equation*}
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\begin{tikzcd}[row sep=huge]
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H^1(S^3 \setminus K) \arrow[r,] \arrow[d,"\widetilde{\Theta}"] &
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H^1(N \setminus K) \arrow[d,"\Theta"] \\
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{[S^3 \setminus K, S^1]} \arrow[r,]&
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{[N \setminus K, S^1]}
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\end{tikzcd}
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\end{equation*}
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\noindent
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$\Sigma = \widetilde{\Theta}^{-1}(X)$ is a surface, such that $\partial \Sigma = K$, so it is a Seifert surface.
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%
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%
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% Thom isomorphism,
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\end{proof}
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\begin{definition}
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Let $S$ be a Seifert matrix for a knot $K$. The Alexander polynomial $\Delta_K(t)$ is a Laurent polynomial:
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\[
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\Delta_K(t) := \det (tS - S^T) \in
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\mathbb{Z}[t, t^{-1}] \cong \mathbb{Z}[\mathbb{Z}]
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\]
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\end{definition}
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\begin{theorem}
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$\Delta_K(t)$ is well defined up to multiplication by $\pm t^k$, for $k \in \mathbb{Z}$.
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\end{theorem}
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\begin{proof}
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We need to show that $\Delta_K(t)$ doesn't depend on $S$-equivalence relation.
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\begin{enumerate}[label={(\arabic*)}]
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\item Suppose $S\prime = CSC^T$, $C \in \Gl(n, \mathbb{Z})$ (matrices invertible over $\mathbb{Z}$). Then $\det C = 1$ and:
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\begin{align*}
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&\det(tS\prime - S\prime^T) =
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\det(tCSC^T - (CSC^T)^T) =\\
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&\det(tCSC^T - CS^TC^T) =
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\det C(tS - S^T)C^T =
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\det(tS - S^T)
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\end{align*}
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\item
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Let \\
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$ A := t
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\begin{pmatrix}
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\begin{array}{c|c}
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S &
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\begin{matrix}
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\ast & 0 \\
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\sdots & \sdots\\
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\ast & 0
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\end{matrix} \\
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\hline
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\begin{matrix}
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\ast & \dots & \ast\\
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0 & \dots & 0
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\end{matrix}
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&
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\begin{matrix}
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0 & 0\\
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1 & 0
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\end{matrix}
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\end{array}
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\end{pmatrix}
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-
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\begin{pmatrix}
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\begin{array}{c|c}
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S^T &
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\begin{matrix}
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\ast & 0 \\
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\sdots & \sdots\\
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\ast & 0
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\end{matrix} \\
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\hline
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\begin{matrix}
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\ast & \dots & \ast\\
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0 & \dots & 0
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\end{matrix}
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&
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\begin{matrix}
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0 & 1\\
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0 & 0
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\end{matrix}
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\end{array}
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\end{pmatrix}
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=
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\begin{pmatrix}
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\begin{array}{c|c}
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tS - S^T &
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\begin{matrix}
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\ast & 0 \\
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\sdots & \sdots\\
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\ast & 0
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\end{matrix} \\
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\hline
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\begin{matrix}
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\ast & \dots & \ast\\
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0 & \dots & 0
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\end{matrix}
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&
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\begin{matrix}
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0 & -1\\
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t & 0
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\end{matrix}
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\end{array}
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\end{pmatrix}
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$
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\\
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\\
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Using the Laplace expansion we get $\det A = \pm t \det(tS - S^T)$.
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\end{enumerate}
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\end{proof}
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%
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%
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%
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\begin{example}
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If $K$ is a trefoil then we can take
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$S = \begin{pmatrix}
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-1 & -1 \\
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0 & -1
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\end{pmatrix}$. Then
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\[
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\Delta_K(t) = \det
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\begin{pmatrix}
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-t + 1 & -t\\
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1 & -t +1
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\end{pmatrix}
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= (t -1)^2 + t = t^2 - t +1 \ne 1
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\Rightarrow \text{trefoil is not trivial.}
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\]
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\end{example}
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\begin{fact}
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$\Delta_K(t)$ is symmetric.
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\end{fact}
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\begin{proof}
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Let $S$ be an $n \times n$ matrix.
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\begin{align*}
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&\Delta_K(t^{-1}) = \det (t^{-1}S - S^T) = (-t)^{-n} \det(tS^T - S) = \\
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&(-t)^{-n} \det (tS - S^T) = (-t)^{-n} \Delta_K(t)
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\end{align*}
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If $K$ is a knot, then $n$ is necessarily even, and so $\Delta_K(t^{-1}) = t^{-n} \Delta_K(t)$.
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\end{proof}
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\begin{lemma}
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\begin{align*}
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\frac{1}{2} \deg \Delta_K(t) \leq g_3(K),
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\text{ where } deg (a_n t^n + \dots + a_1 t^l )= k - l.
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\end{align*}
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\end{lemma}
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\begin{proof}
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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 $2g$.
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\end{proof}
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\begin{example}
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There are not trivial knots with Alexander polynomial equal $1$, for example:
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\includegraphics[width=0.3\textwidth]{11n34.png}
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$\Delta_{11n34} \equiv 1$.
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\end{example}
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%removing one disk from surface doesn't change $H_1$ (only $H_2$)
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%
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%
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%
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\begin{lemma}[Dehn]
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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
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${D^2 \overset{g}\longhookrightarrow M}$ such that:
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\[
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g_{\big| \partial D^2} = f_{\big| \partial D^2.}
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\]
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\end{lemma}
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\noindent
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Remark: Dehn lemma doesn't hold for dimension four.\\
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Let $M$ be connected, compact three manifold with boundary.
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Suppose $\pi_1(\partial M) \longrightarrow \pi_1(M)$ has non-trivial kernel. Then there exists a map $f: (D^2, \partial D^2) \longrightarrow (M, \partial M)$ such that $f\big| \partial D^2$ is non-trivial loop in $\partial M$
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\begin{theorem}[Sphere theorem]
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Suppose $\pi_1(M) \ne 0$. Then there exists an embedding $f: S^2 \hookrightarrow M$ that is homotopy non-trivial.
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\end{theorem}
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\begin{problem}
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Prove that $S^3 \ K$ is Eilenberg–MacLane space of type $K(\pi, 1)$.
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\end{problem}
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\begin{corollary}
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Suppose $K \subset S^3$ and $\pi_1(S^3 \setminus K)$ is infinite cyclic ($\mathbb{Z})$. Then $K$ is trivial.
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\end{corollary}
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\subsection*{Construction}
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We know that $3$ - sphere can be obtained by gluing two solid tori:
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$S^3 = (D^2 \times S^1) \cup (S^1 \times D^2)$. So the complement of solid torus in $S^3$ is another solid torus.\\
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Take $(z_1, z_2) \in \mathbb{C}$ such that $max(\mid z_1 \mid, \mid z_2, \mid) = 1
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$
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\begin{figure}[h]
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\centering{
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\def\svgwidth{\linewidth}
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\resizebox{0.3\textwidth}{!}{\includegraphics[width=0.3\textwidth]{sphere_as_torus.png}}
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\caption{The complement of solid torus in $S^3$ is another solid torus.}
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\label{fig:sphere_as_tori}
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}
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\end{figure}
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\begin{example}
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\begin{align*}
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&F: \mathbb{C}^2 \rightarrow \mathbb{C} \text{ a polynomial} \\
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&F(0) = 0
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\end{align*}
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\end{example}
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????????????
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\\
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\noindent
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as a corollary we see that $K_T^{n, }$ ???? \\
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is not slice unless $m=0$.
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\begin{theorem}
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The map $j: \mathscr{C} \longrightarrow \mathbb{Z}^{\infty}$ is a surjection that maps ${K_n}$ to a linear independent set. Moreover $\mathscr{C} \cong \mathbb{Z}$
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\end{theorem}
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\begin{fact}[Milnor Singular Points of Complex Hypersurfaces]
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\end{fact}
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%\end{comment}
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\noindent
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An oriented knot is called negative amphichiral if the mirror image $m(K)$ of $K$ is equivalent the reverse knot of $K$: $K^r$. \\
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\begin{problem}
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Prove that if $K$ is negative amphichiral, then $K \# K = 0$ in
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$\mathscr{C}$.
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%
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%\\
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%Hint: $ -K = m(K)^r = (K^r)^r = K$
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\end{problem}
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\begin{example}
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Figure 8 knot is negative amphichiral.
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\end{example}
|
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%
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%
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%
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\section{Concordance group \hfill\DTMdate{2019-03-18}}
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\begin{definition}
|
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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
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\[
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\partial A = K^{\prime} \times \{1\} \; \sqcup \; K \times \{0\}.
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\]
|
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\end{definition}
|
||
|
||
\begin{figure}[h]
|
||
\fontsize{20}{10}\selectfont
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||
\centering{
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||
\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. \\
|
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Put differently: a knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk in $B^4$.
|
||
\end{definition}
|
||
|
||
|
||
|
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\noindent
|
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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{
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||
\def\svgwidth{\linewidth}
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||
\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
|
||
%Add NotePrintSaveCiteYour opinionEmailShare
|
||
%Saveliev, Nikolai
|
||
|
||
%Lectures on the Topology of 3-Manifolds
|
||
%An Introduction to the Casson Invariant
|
||
\end{document}
|
||
|