288 lines
9.4 KiB
TeX
288 lines
9.4 KiB
TeX
\subsection{Existence of a Seifert surface - second proof}
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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}(\Cref{theo:Seifert})\\
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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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The tubular neighbourhood of the knot is homomorphic to
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$D^2 \times S^1$.
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So its boundary
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$\partial N \cong \ S^1 \times S^1$ and therefore:
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$H^1(N, \partial N) \cong \ \mathbb{Z} \oplus \mathbb{Z}$. By excision theorem we have:
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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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\end{align*}
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Therefore:
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\begin{align*}
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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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Let us consider the following diagram:
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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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% picture for excision theorem
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% Thom isomorphism,
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\end{proof}
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%$S$ - equivalence $\Sigma$\\
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%simple closed curves $\alpha_1, ... \alpha_n \in H_1(\Sigma, \mathbb{Z})$ basis for $H_1$
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\subsection{Alexander polynomial}
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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{lemma}
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$\Delta_K(t)$ is symmetric.
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\end{lemma}
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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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\subsection{Decomposition of \texorpdfstring{
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$3$-sphere}{3-sphere}}
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We know that $3$ - sphere can be obtained by gluing two solid tori:
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\[
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S^3 = \partial D^4 = \partial (D^2 \times D^2) = (D^2 \times S^1) \cup (S^1 \times D^2).
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\]
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So the complement of solid torus in $S^3$ is another solid torus.\\
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Analytically it can be describes as follow. \\
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Take $(z_1, z_2) \in \mathbb{C}$ such that ${\max(\vert z_1 \vert, \vert z_2\vert) = 1.}
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$
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Define following sets:
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\begin{align*}
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S_1 = \{ (z_1, z_2) \in S^3: \vert z_1 \vert = 0\} \cong S^1 \times D^2 ,\\
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S_2 = \{(z_1, z_2) \in S ^3: \vert z_2 \vert = 1 \} \cong D^2 \times S^1.
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\end{align*}
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The intersection
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$S_1 \cap S_2 = \{(z_1, z_2): \vert z_1 \vert = \vert z_2 \vert = 1 \} \cong S^1 \times S^1$.
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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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\subsection{Dehn lemma and sphere theorem}
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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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\begin{proof}
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Let $N$ be a tubular neighbourhood of a knot $K$ and $M = S^3 \setminus N$ its complement. Then $\partial M = S^1 \times S^1$. Let $f : \pi_1(\partial M ) \longrightarrow \pi_1(M)$.
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If $\pi_1(M)$ is infinite cyclic group then the map $f$ is non-trivial. Suppose ${\lambda \in \ker (\pi_1(S^1 \times S^1) \longrightarrow \pi_1(M))}$.
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There is a map $g: (D^2, \partial D^2) \longrightarrow (M, \partial M)$ such that $g(\partial D^2) = \lambda$.\\
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By Dehn's lemma there exists an embedding ${h: (D^2, \partial D^2) \longhookrightarrow (M, \partial M)}$ such that
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$h\big|_{\partial D^2} = f \big|_{\partial D^2}$ and $h(\partial D^2) = \lambda$.
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Let $\Sigma$ be a union of the annulus and the image of $\partial D^2$.
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%$g_3$
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If $g_3(\Sigma) = 0$, then $K$ is trivial. \\
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Now we should proof that:
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\[
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H_1(M) \cong \mathbb{Z} \Longrightarrow \lambda \in \ker ( \pi_1(S^1 \times S^1) \longrightarrow \pi_1(M)).
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\]
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\begin{figure}[h]
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\fontsize{40}{10}\selectfont
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\centering{
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\def\svgwidth{\linewidth}
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\resizebox{0.4\textwidth}{!}{\input{images/torus_lambda.pdf_tex}}
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}
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\caption{$\mu$ is a meridian and $\lambda$ is a longitude.}
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\label{fig:meridian_and_longitude}
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\end{figure}
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Choose a meridian $\mu$ such that $\Lk (\mu, K) = 1$. Recall the definition of linking number via homology group (Definition \ref{def:lk_via_homo}).
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$[\mu]$ represents the generator of $H_1(S^3\setminus K, \mathbb{Z})$. From definition of $\lambda$ we know that $\lambda$ is trivial in $H_1(M)$ ($\Lk(\lambda, K) =0$, therefore $[\lambda]$ was trivial in $pi_1(M)$). If $K$ is non-trivial then $\lambda$ is non-trivial in $\pi_1(M)$, but it is trivial in $H_1(M)$.
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\end{proof}
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