Theorem \ref{the:sign_slice} can be also express as follow:
non-degenerate metabolic hermitian form has vanishing signature.
\begin{proof}
\noindent
We note that $\det(S + S^T)\neq0$. Hence $\det((1- t) S +(1-\bar{t})S^T)$ is not identically zero on $S^1$, so it is non-zero except possibly at finitely many points. We apply the Lemma \ref{lem:metabolic}.
\\
Let $t \in S^1\setminus\{1\}$.
Then:
\begin{align*}
\det((1 - t) S + (1 - \bar{t}) S^T) =&
\det((1 - t) S + (t\bar{t} - \bar{t}) S^T) =\\
&\det((1 - t) (S - \bar{t} - S^T)) =
\det((1 -t)(S - \bar{t} S^T)).
\end{align*}
As $\det(S + S^T)\neq0$, so $S -\bar{t}S^T \neq0$.
\end{proof}
\begin{corollary}
If $K \sim K^\prime$ then for all but finitely many $t \in S^1\setminus\{1\}: \sigma_K(t)=-\sigma_{K^\prime}(t)$.
\end{corollary}
\begin{proof}
If $ K \sim K^\prime$ then $K \# K^\prime$ is slice.
\[
\sigma_{-K^\prime}(t) = -\sigma_{K^\prime}(t)
\]
The signature gives a homomorphism from the concordance group to $\mathbb{Z}$.
Remark: if $t \in S^1$ is not algebraic over $\mathbb{Z}$, then $\sigma_K(t)\neq0$
(we can use the argument that $\mathscr{C}\longrightarrow\mathbb{Z}$ as well).
\caption{There exists a $3$ - manifold $\Omega$ such that $\partial\Omega= X \cup\Sigma$.}\label{fig:omega_in_B_4}
\end{figure}
\noindent
Let $K$ be a knot and $\Sigma$ its Seifert surface as in Figure \ref{fig:omega_in_B_4}.
There exists a $3$ - submanifold
$\Omega$ such that
$\partial\Omega= Y = X \cup\Sigma$
(by Thom-Pontryagin construction).
If $\alpha, \beta\in\ker(H_1(\Sigma)\longrightarrow H_1(\Omega))$,
then ${\Lk(\alpha, \beta^+)=0}$. Now we have to determine the size of the kernel. We know that
${\dim H_1(\Sigma)=2 n}$. When we glue $\Sigma$ (genus $n$) and $X$ (genus $g$) along a circle we get a surface of genus $n + g$. Therefore $\dim H_1(Y)=2 n +2 g$. Then:
\[
\dim (\ker (H_1(Y) \longrightarrow H_1(\Omega)) = n + g.
\]
So we have $H_1(W)$ of dimension
$2 n +2 g$
- the image of $H_1(Y)$
with a subspace
corresponding to the image of $H_1(\Sigma)$ with dimension $2 n$ and a subspace corresponding to the kernel
of $H_1(Y)\longrightarrow H_1(\Omega)$ of size $n + g$.
We consider minimal possible intersection of this subspaces that corresponds to the kernel of the composition $H_1(\Sigma)\longrightarrow H_1(Y)\longrightarrow H_1(\Omega)$. As the first map is injective, elements of the kernel of the composition have to be in the kernel of the second map.
So we can calculate:
\[
\dim\ker (H_1(\Sigma) \longrightarrow H_1(\Omega)) = 2 n + n + g -2 n - 2 g = n - g.
\]
\end{proof}
\begin{corollary}
If $t$ is not a root of
$\det(tS - S^T)$, then
$\vert\sigma_K(t)\vert\leq2g$.
\end{corollary}
\begin{fact}
If there exists cobordism of genus $g$ between $K$ and $K^\prime$ like shown in Figure \ref{fig:proof_for_bound_disk}, then $K \#-K^\prime$ bounds a surface of genus $g$ in $B^4$.
\caption{If $K$ and $K^\prime$ are connected by a genus $g$ surface, then $K \#-K^\prime$ bounds a genus $g$ surface.}\label{fig:proof_for_bound_disk}
\end{figure}
\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
Remarks:
\begin{enumerate}[label={(\arabic*)}]
\item
$3$ - genus is additive under taking connected sum, but $4$ - genus is not,
\item
for any knot $K$ we have $g_4(K)\leq g_3(K)$.
\end{enumerate}
\begin{example}
\begin{itemize}
\item Let $K = T(2, 3)$. $\sigma(K)=-2$, therefore $T(2, 3)$ isn't a slice knot.
\item Let $K$ be a trefoil and $K^\prime$ a mirror of a trefoil. $g_4(K^\prime)=1$, but $g_4(K \# K^\prime)=0$, so we see that $4$-genus isn't additive,
\item
the equality:
\[
g_4(T(p, q) ) = \frac{1}{2} (p - 1) (g -1)
\]
was conjecture in the '70 and proved by P. Kronheimer and T. Mrówka (1994).
% OZSVATH-SZABO AND RASMUSSEN
\end{itemize}
\end{example}
\begin{proposition}
$g_4(T(p, q)\#-T(r, s))$ is in general hopelessly unknown.
\end{proposition}
\begin{proposition}
Supremum of the signature function of the knot is bounded almost everywhere by two times $4$ - genus:
\[
\ess\sup\vert\sigma_K(t) \vert\leq 2 g_4(K).
\]
\end{proposition}
\subsection{Topological genus}
\begin{definition}
A knot $K$ is called topologically slice if $K$ bounds a topological locally flat disc in $B^4$ (i.e. the disk has tubular neighbourhood).
\end{definition}
\begin{theorem}[Freedman, '82]
If $\Delta_K(t)=1$, then $K$ is topologically slice (but not necessarily smoothly slice).
\end{theorem}
\begin{theorem}[Powell, 2015]
If $K$ is genus $g$
(topologically flat)
cobordant to $K^\prime$,
then
\[
\vert\sigma_K(t) - \sigma_{K^\prime}(t) \vert\leq 2 g
\]
if $g_4^{\mytop}(K)\geq\ess\sup\vert\sigma_K(t)\vert$.
\end{theorem}
\noindent
The proof for smooth category was based on following equality:
For this equality we assumed that there exists a $3$ - dimensional manifold $\Omega$ (as shown in Figure \ref{fig:omega_in_B_4}) which was guaranteed by Pontryagin-Thom Construction.\\
Pontryagin-Thom Construction relays on taking $\Omega$ as preimage of regular value:
\[
H^1 (B^4 \setminus Y, \mathbb{Z}) = [B^4 \setminus Y, S^1],
\]
what relies on Sard's theorem, that the set of regular values has positive measure. But Sard's theorem doesn't work for topologically locally flat category. So there was a gap in the proof for topological locally flat category - the existence of $\Omega$.\\
\noindent
Remark: unless $p=2$ or $p =3\wedge q =4$:
\[
g_4^{\mytop} (T(p, q)) < q_4(T(p, q)).
\]
% Wilczyński '93
%Feller 2014
%Baoder 2017
%Lemark
\\
\noindent
From the category of cobordant knots (or topologically cobordant knots) there exists a map to $\mathbb{Z}$ given by signature function. To any element $K$ we can associate a form
\] This association is not well define because id depends on the choice of Seifert form. However, different choices lead ever to congruent forms ($S \mapsto CSC^T$) or induced the change on the form by adding or subtracting a hyperbolic element.
\begin{definition}
The Witt group $W$ of $\mathbb{Z}[t, t^{-1}]$ elements are classes of non-degenerate
forms over $\mathbb{Z}[t, t^{-1}]$ under the equivalence relation $V \sim W$ if $V \oplus- W$ is metabolic.
\end{definition}
\noindent
If $S$ differs from $S^\prime$ by a row extension, then
$(1- t) S +(1-\bar{t}^{-1}) S^T$ is Witt equivalence to $(1- t) S^\prime+(1- t^{-1})S^T$.
\\
\noindent
A form is meant as hermitian with respect to this involution: $A^T = A: (a, b)=\bar{(a, b)}$.