\begin{definition} Two knots $K$ and $K^{\prime}$ are called (smoothly) concordant if there exists an annulus $A$ that is smoothly embedded in ${S^3 \times [0, 1]}$ such that \[ \partial A = K^{\prime} \times \{1\} \; \sqcup \; K \times \{0\}. \] \end{definition} \begin{figure}[h] \fontsize{20}{10}\selectfont \centering{ \def\svgwidth{\linewidth} \resizebox{0.8\textwidth}{!}{\input{images/concordance.pdf_tex}} } \end{figure} \begin{definition} A knot $K$ is called (smoothly) slice if $K$ is smoothly concordant to an unknot. \\ Put differently: a knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk in $B^4$. \end{definition} \noindent 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{fact: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 Fact \ref{fact: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{ \def\svgwidth{\linewidth} \resizebox{0.5\textwidth}{!}{\input{images/ball_4_pushed_seifert.pdf_tex}} } \caption{$Y = F \cup \Sigma$ is a smooth closed surface.} \label{fig:closed_surface} \end{figure} \noindent \\ Pontryagin-Thom construction tells us that there exists a compact oriented 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. \\ \noindent 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} Consider the following long exact sequence for a pair $(\Omega, Y)$: \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*} 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_1(\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} \noindent Let $V = \begin{pmatrix} 0 & A\\ B & C \end{pmatrix}$. Then \begin{align*} tV - V^T = \begin{pmatrix} 0 & tA\\ tB & tC \end{pmatrix} - \begin{pmatrix} 0 & B^T\\ A^T & C^T \end{pmatrix} = \begin{pmatrix} 0 & tA - B^T\\ tB - A^T & tC - C^T \end{pmatrix} \\ \det (tV - V^T) = \det (tA - B^T) - \det(tB - A^T) \end{align*} \begin{corollary} \label{cor:slice_alex} If $K$ is a slice knot then there exists $f \in \mathbb{Z}[t, t^{-1}]$ such that \[\Delta_K(t) = f(t) \cdot f(t^{-1}).\] \end{corollary} \begin{example} Figure eight knot is not slice. \end{example} \begin{fact} If $K$ is slice, then the signature $\sigma(K) \equiv 0$: \[V + V^T = \begin{pmatrix} 0 & A + B^T\\ B + A^T & C + C^T \end{pmatrix} \Rightarrow \sigma = 0 .\] \end{fact}