196 lines
6.3 KiB
TeX
196 lines
6.3 KiB
TeX
\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}
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\begin{figure}[h]
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\fontsize{20}{10}\selectfont
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\centering{
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\def\svgwidth{\linewidth}
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\resizebox{0.8\textwidth}{!}{\input{images/concordance.pdf_tex}}
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}
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\end{figure}
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\begin{definition}
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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 $B^4$.
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\end{definition}
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\noindent
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Let $m(K)$ denote a mirror image of a knot $K$.
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\begin{fact}
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For any $K$, $K \# m(K)$ is slice.
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\end{fact}
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\begin{fact}
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Concordance is an equivalence relation.
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\end{fact}
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\begin{fact}\label{fact:concordance_connected}
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If $K_1 \sim {K_1}^{\prime}$ and $K_2 \sim {K_2}^{\prime}$, then
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$K_1 \# K_2 \sim {K_1}^{\prime} \# {K_2}^{\prime}$.
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\begin{figure}[h]
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\fontsize{10}{10}\selectfont
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\centering{
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\def\svgwidth{\linewidth}
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\resizebox{1\textwidth}{!}{\input{images/concordance_sum.pdf_tex}}
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}
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\caption{Sketch for Fact \ref{fact:concordance_connected}.}
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\label{fig:concordance_sum}
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\end{figure}
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\end{fact}
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\begin{fact}
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$K \# m(K) \sim $ the unknot.
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\end{fact}
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\noindent
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\begin{theorem}
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Let $\mathscr{C}$ denote a set of all equivalent classes for knots and $[0]$ denote class of all knots concordant to a trivial knot.
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$\mathscr{C}$ 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 $-[K] = [mK]$.
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\end{theorem}
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\begin{fact}
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The figure eight knot is a torsion element in $\mathscr{C}$ ($2K \sim $ the unknot).
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\end{fact}
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\begin{problem}[open]
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Are there in concordance group torsion elements that are not $2$ torsion elements?
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\end{problem}
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\noindent
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Remark: $K \sim K^{\prime} \Leftrightarrow K \# -K^{\prime}$ is slice.
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\\
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\begin{figure}[h]
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\fontsize{20}{10}\selectfont
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\centering{
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\def\svgwidth{\linewidth}
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\resizebox{0.5\textwidth}{!}{\input{images/ball_4_pushed_seifert.pdf_tex}}
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}
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\caption{$Y = F \cup \Sigma$ is a smooth closed surface.}
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\label{fig:closed_surface}
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\end{figure}
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\noindent
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\\
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Pontryagin-Thom construction tells us that there exists a compact oriented three - manifold $\Omega \subset B^4$ such that $\partial \Omega = Y$.\\
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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
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\[
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\alpha, \beta \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z})).\] Then there are two cycles $A, B \in \Omega$ such that $\partial A = \alpha$ and $\partial B = \beta$.
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Let $B^+$ be a push off of $B$ in the positive normal direction such that
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$\partial B^+ = \beta^+$.
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Then
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$\Lk(\alpha, \beta^+) = A \cdot B^+$. But $A$ and $B$ are disjoint, so $\Lk(\alpha, \beta^+) = 0$. Then the Seifert form is zero.
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\\
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\noindent
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Let us consider following maps:
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\[
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\Sigma \overset{\phi} \longhookrightarrow Y \overset{\psi} \longhookrightarrow \Omega.
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\]
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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 $\gamma \in \ker \psi_*$.
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%
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%
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%
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\begin{proposition}
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\[
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\dim \ker (H_1(Y, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z})) = \frac{1}{2} b_1(Y),
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\]
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where $b_1$ is first Betti number.
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\end{proposition}
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\begin{proof}
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Consider the following long exact sequence for a pair $(\Omega, Y)$:
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\begin{align*}
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& 0 \to H_3(\Omega) \to H_3(\Omega, Y) \to
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\\
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\to & H_2(Y) \to H_2(\Omega) \to H_2(\Omega, Y) \to \\
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\to & H_1(Y) \to H_1(\Omega) \to H_1(\Omega, Y) \to \\
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\to & H_0(Y) \to H_0(\Omega) \to 0
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\end{align*}
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By Poincar\'e duality we know that:
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\begin{align*}
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H_3(\Omega, Y) &\cong H^0(\Omega),\\
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H_2(Y) &\cong H^0(Y),\\
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H_2(\Omega) &\cong H^1(\Omega, Y),\\
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H_1(\Omega, Y) &\cong H^1(\Omega).
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\end{align*}
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Therefore $\dim_{\mathbb{Q}} \quot{H_1(Y)}{V}
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= \dim_{\mathbb{Q}} V
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$.\\
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\noindent
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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 $K$
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has a subspace of dimension $g_{\Sigma}$ on which it is zero:
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\begin{align*}
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\newcommand\coolover[2]%
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{\mathrlap{\smash{\overbrace{\phantom{%
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\begin{matrix} #2 \end{matrix}}}^{\mbox{$#1$}}}}#2}
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\newcommand\coolunder[2]{\mathrlap{\smash{\underbrace{\phantom{%
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\begin{matrix} #2 \end{matrix}}}_{\mbox{$#1$}}}}#2}
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\newcommand\coolleftbrace[2]{%
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#1\left\{\vphantom{\begin{matrix} #2 \end{matrix}}\right.}
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\newcommand\coolrightbrace[2]{%
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\left.\vphantom{\begin{matrix} #1 \end{matrix}}\right\}#2}
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\vphantom{% phantom stuff for correct box dimensions
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\begin{matrix}
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\overbrace{XYZ}^{\mbox{$R$}}\\ \\ \\ \\ \\ \\
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\underbrace{pqr}_{\mbox{$S$}}
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\end{matrix}}%
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V =
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\begin{matrix}% matrix for left braces
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\coolleftbrace{g_{\Sigma}}{ \\ \\ \\}
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\\ \\ \\ \\
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\end{matrix}%
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\begin{pmatrix}
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\coolover{g_{\Sigma}}{0 & \dots & 0 } & * & \dots & *\\
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\sdots & & \sdots & \sdots & & \sdots \\
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0 & \dots & 0 & * & \dots & *\\
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* & \dots & * & * & \dots & *\\
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\sdots & & \sdots & \sdots & & \sdots \\
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* & \dots & * & * & \dots & *
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\end{pmatrix}_{2g_{\Sigma} \times 2g_{\Sigma}}
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\end{align*}
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\end{proof}
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\noindent
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Let $V =
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\begin{pmatrix}
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0 & A\\
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B & C
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\end{pmatrix}$. Then
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\begin{align*}
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tV - V^T =
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\begin{pmatrix}
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0 & tA\\
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tB & tC
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\end{pmatrix}
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-
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\begin{pmatrix}
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0 & B^T\\
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A^T & C^T
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\end{pmatrix}
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=
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\begin{pmatrix}
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0 & tA - B^T\\
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tB - A^T & tC - C^T
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\end{pmatrix}
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\\
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\det (tV - V^T) = \det (tA - B^T) - \det(tB - A^T)
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\end{align*}
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\begin{corollary}
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\label{cor:slice_alex}
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If $K$ is a slice knot then there exists $f \in \mathbb{Z}[t, t^{-1}]$ such that
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\[\Delta_K(t) = f(t) \cdot f(t^{-1}).\]
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\end{corollary}
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\begin{example}
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Figure eight knot is not slice.
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\end{example}
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\begin{fact}
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If $K$ is slice, then the signature $\sigma(K) \equiv 0$:
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\[V + V^T =
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\begin{pmatrix}
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0 & A + B^T\\
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B + A^T & C + C^T
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\end{pmatrix}
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\Rightarrow \sigma = 0
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.\]
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\end{fact}
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