\subsection{Existence of a Seifert surface - second proof} %\begin{theorem} %For any knot $K \subset S^3$ there exists a connected, compact and orientable surface $\Sigma(K)$ such that $\partial \Sigma(K) = K$ %\end{theorem} \begin{proof}(\Cref{theo:Seifert})\\ 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: \begin{align*} H^1(S^3 \setminus N ) \cong H^1(S^3 \setminus K). \end{align*} Let us consider a long exact sequence of cohomology of a pair $(S^3, S^3 \setminus N)$ with integer coefficients: \begin{center} \begin{tikzcd} [ column sep=0cm, fill=none, row sep=small, ar symbol/.style =% {draw=none,"\textstyle#1" description,sloped}, isomorphic/.style = {ar symbol={\cong}}, ] &\mathbb{Z} \\ & H^0(S^3) \ar[u,isomorphic] \to &H^0(S^3 \setminus N) \to \\ \to H^1(S^3, S^3 \setminus N) \to & H^1(S^3) \to & H^1(S^3\setminus N) \to \\ & 0 \ar[u,isomorphic]& \\ \to H^2(S^3, S^3 \setminus N) \to & H^2(S^3) \ar[u,isomorphic] \to & H^2(S^3\setminus N) \to \\ \to H^3(S^3, S^3\setminus N)\to & H^3(S) \to & 0 \\ & \mathbb{Z} \ar[u,isomorphic] &\\ \end{tikzcd} \end{center} The tubular neighbourhood of the knot is homomorphic to $D^2 \times S^1$. So its boundary $\partial N \cong \ S^1 \times S^1$ and therefore: $H^1(N, \partial N) \cong \ \mathbb{Z} \oplus \mathbb{Z}$. By excision theorem we have: \begin{align*} H^* (S^3, S^3 \setminus N) &\cong H^* (N, \partial N). \end{align*} Therefore: \begin{align*} H^ 1 (S^3\setminus N) &\cong H^1(S^3\setminus K) \cong \mathbb{Z}. \end{align*} Let us consider the following diagram: \begin{equation*} \begin{tikzcd}[row sep=huge] H^1(S^3 \setminus K) \arrow[r,] \arrow[d,"\widetilde{\Theta}"] & H^1(N \setminus K) \arrow[d,"\Theta"] \\ {[S^3 \setminus K, S^1]} \arrow[r,]& {[N \setminus K, S^1]} \end{tikzcd} \end{equation*} \noindent $\Sigma = \widetilde{\Theta}^{-1}(X)$ is a surface, such that $\partial \Sigma = K$, so it is a Seifert surface. % % picture for excision theorem % Thom isomorphism, \end{proof} %$S$ - equivalence $\Sigma$\\ %simple closed curves $\alpha_1, ... \alpha_n \in H_1(\Sigma, \mathbb{Z})$ basis for $H_1$ \subsection{Alexander polynomial} \begin{definition} Let $S$ be a Seifert matrix for a knot $K$. The Alexander polynomial $\Delta_K(t)$ is a Laurent polynomial: \[ \Delta_K(t) := \det (tS - S^T) \in \mathbb{Z}[t, t^{-1}] \cong \mathbb{Z}[\mathbb{Z}] \] \end{definition} \begin{theorem} $\Delta_K(t)$ is well defined up to multiplication by $\pm t^k$, for $k \in \mathbb{Z}$. \end{theorem} \begin{proof} We need to show that $\Delta_K(t)$ doesn't depend on $S$-equivalence relation. \begin{enumerate}[label={(\arabic*)}] \item Suppose $S\prime = CSC^T$, $C \in \Gl(n, \mathbb{Z})$ (matrices invertible over $\mathbb{Z}$). Then $\det C = 1$ and: \begin{align*} &\det(tS\prime - S\prime^T) = \det(tCSC^T - (CSC^T)^T) =\\ &\det(tCSC^T - CS^TC^T) = \det C(tS - S^T)C^T = \det(tS - S^T) \end{align*} \item Let \\ $ A := t \begin{pmatrix} \begin{array}{c|c} S & \begin{matrix} \ast & 0 \\ \sdots & \sdots\\ \ast & 0 \end{matrix} \\ \hline \begin{matrix} \ast & \dots & \ast\\ 0 & \dots & 0 \end{matrix} & \begin{matrix} 0 & 0\\ 1 & 0 \end{matrix} \end{array} \end{pmatrix} - \begin{pmatrix} \begin{array}{c|c} S^T & \begin{matrix} \ast & 0 \\ \sdots & \sdots\\ \ast & 0 \end{matrix} \\ \hline \begin{matrix} \ast & \dots & \ast\\ 0 & \dots & 0 \end{matrix} & \begin{matrix} 0 & 1\\ 0 & 0 \end{matrix} \end{array} \end{pmatrix} = \begin{pmatrix} \begin{array}{c|c} tS - S^T & \begin{matrix} \ast & 0 \\ \sdots & \sdots\\ \ast & 0 \end{matrix} \\ \hline \begin{matrix} \ast & \dots & \ast\\ 0 & \dots & 0 \end{matrix} & \begin{matrix} 0 & -1\\ t & 0 \end{matrix} \end{array} \end{pmatrix} $ \\ \\ Using the Laplace expansion we get $\det A = \pm t \det(tS - S^T)$. \end{enumerate} \end{proof} % % % \begin{example} If $K$ is a trefoil then we can take $S = \begin{pmatrix} -1 & -1 \\ 0 & -1 \end{pmatrix}$. Then \[ \Delta_K(t) = \det \begin{pmatrix} -t + 1 & -t\\ 1 & -t +1 \end{pmatrix} = (t -1)^2 + t = t^2 - t +1 \ne 1 \Rightarrow \text{trefoil is not trivial.} \] \end{example} \begin{lemma} $\Delta_K(t)$ is symmetric. \end{lemma} \begin{proof} Let $S$ be an $n \times n$ matrix. \begin{align*} &\Delta_K(t^{-1}) = \det (t^{-1}S - S^T) = (-t)^{-n} \det(tS^T - S) = \\ &(-t)^{-n} \det (tS - S^T) = (-t)^{-n} \Delta_K(t) \end{align*} If $K$ is a knot, then $n$ is necessarily even, and so $\Delta_K(t^{-1}) = t^{-n} \Delta_K(t)$. \end{proof} \begin{lemma} \begin{align*} \frac{1}{2} \deg \Delta_K(t) \leq g_3(K), \text{ where } deg (a_n t^n + \dots + a_1 t^l )= k - l. \end{align*} \end{lemma} \begin{proof} 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 $2g$. \end{proof} \begin{example} There are not trivial knots with Alexander polynomial equal $1$, for example: \includegraphics[width=0.3\textwidth]{11n34.png} $\Delta_{11n34} \equiv 1$. \end{example} \subsection{Decomposition of \texorpdfstring{ $3$-sphere}{3-sphere}} We know that $3$ - sphere can be obtained by gluing two solid tori: \[ S^3 = \partial D^4 = \partial (D^2 \times D^2) = (D^2 \times S^1) \cup (S^1 \times D^2). \] So the complement of solid torus in $S^3$ is another solid torus.\\ Analytically it can be describes as follow. \\ Take $(z_1, z_2) \in \mathbb{C}$ such that ${\max(\vert z_1 \vert, \vert z_2\vert) = 1.} $ Define following sets: \begin{align*} S_1 = \{ (z_1, z_2) \in S^3: \vert z_1 \vert = 0\} \cong S^1 \times D^2 ,\\ S_2 = \{(z_1, z_2) \in S ^3: \vert z_2 \vert = 1 \} \cong D^2 \times S^1. \end{align*} The intersection $S_1 \cap S_2 = \{(z_1, z_2): \vert z_1 \vert = \vert z_2 \vert = 1 \} \cong S^1 \times S^1$. \begin{figure}[h] \centering{ \def\svgwidth{\linewidth} \resizebox{0.3\textwidth}{!}{\includegraphics[width=0.3\textwidth]{sphere_as_torus.png}} \caption{The complement of solid torus in $S^3$ is another solid torus.} \label{fig:sphere_as_tori} } \end{figure} \subsection{Dehn lemma and sphere theorem} %removing one disk from surface doesn't change $H_1$ (only $H_2$) % % % \begin{lemma}[Dehn] 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 ${D^2 \overset{g}\longhookrightarrow M}$ such that: \[ g\big|_{\partial D^2} = f\big|_{\partial D^2.} \] \end{lemma} \noindent Remark: Dehn lemma doesn't hold for dimension four.\\ Let $M$ be connected, compact three manifold with boundary. 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 $\partial M$. \begin{theorem}[Sphere theorem] Suppose $\pi_1(M) \ne 0$. Then there exists an embedding $f: S^2 \hookrightarrow M$ that is homotopy non-trivial. \end{theorem} \begin{problem} Prove that $S^3 \ K$ is Eilenberg–MacLane space of type $K(\pi, 1)$. \end{problem} \begin{corollary} Suppose $K \subset S^3$ and $\pi_1(S^3 \setminus K)$ is infinite cyclic ($\mathbb{Z})$. Then $K$ is trivial. \end{corollary} \begin{proof} 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 $f : \pi_1(\partial M ) \longrightarrow \pi_1(M)$. If $\pi_1(M)$ is infinite cyclic group then the map $f$ is non-trivial. Suppose ${\lambda \in \ker (\pi_1(S^1 \times S^1) \longrightarrow \pi_1(M))}$. There is a map $g: (D^2, \partial D^2) \longrightarrow (M, \partial M)$ such that $g(\partial D^2) = \lambda$.\\ By Dehn's lemma there exists an embedding ${h: (D^2, \partial D^2) \longhookrightarrow (M, \partial M)}$ such that $h\big|_{\partial D^2} = f \big|_{\partial D^2}$ and $h(\partial D^2) = \lambda$. Let $\Sigma$ be a union of the annulus and the image of $\partial D^2$. %$g_3$ If $g_3(\Sigma) = 0$, then $K$ is trivial. \\ Now we should proof that: \[ H_1(M) \cong \mathbb{Z} \Longrightarrow \lambda \in \ker ( \pi_1(S^1 \times S^1) \longrightarrow \pi_1(M)). \] \begin{figure}[h] \fontsize{40}{10}\selectfont \centering{ \def\svgwidth{\linewidth} \resizebox{0.4\textwidth}{!}{\input{images/torus_lambda.pdf_tex}} } \caption{$\mu$ is a meridian and $\lambda$ is a longitude.} \label{fig:meridian_and_longitude} \end{figure} 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}). $[\mu]$ 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 $H_1(M)$. \end{proof}