\begin{theorem}
If $K$ is slice, 
then $\sigma_K(t)
 = \sign ( (1 - t)S +(1 - \bar{t})S^T)$ 
is zero except possibly of finitely many points and $\sigma_K(-1) = \sign(S + S^T) \neq 0$.
\end{theorem}
\begin{proof}
\begin{lemma}
\label{lem:metabolic}
If $V$ is a Hermitian matrix ($\bar{V} = V^T$), $V$ is of size  $2n \times 2n$ and 
$
V = \begin{pmatrix}
0 & A \\
\bar{A}^T & B
\end{pmatrix}
$ and $\det V \neq 0$ then $\sigma(V) = 0$. 
\end{lemma}
\begin{definition}
A Hermitian form $V$ is metabolic if $V$ has structure
$\begin{pmatrix}
0 & A\\
\bar{A}^T & B
\end{pmatrix}$ with half-dimensional null-space. 
\end{definition}
\noindent
In other words: non-degenerate metabolic hermitian form has vanishing signature.\\
We note that $\det(S + S^T) \neq 0$. 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 $\det (S + S^T) \neq 0$, so $S - \bar{t}S^T \neq 0$.  
\end{proof}
?????????????????s\\
\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 give a homomorphism from the concordance group to $\mathbb{Z}$.\\
??????????????????\\
Remark: if $t \in S^1$ is not algebraic over $\mathbb{Z}$, then $\sigma_K(t) \neq 0$
(we can is the argument that $\mathscr{C} \longrightarrow \mathbb{Z}$ as well). 
\end{proof}
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/genus_2_bordism.pdf_tex}}
}
\caption{$K$ and $K^\prime$ are connected by a genus $g$ surface of genus.}\label{fig:genus_2_bordism}
\end{figure}
???????????????????????\\
\begin{proposition}[Kawauchi inequality]
If there exists a genus $g$ surface as in Figure \ref{fig:genus_2_bordism}
then for almost all $t \in S^1 \setminus \{1\}$ we have $\vert \sigma_K(t) - \sigma_{K^\prime}(t) \vert \leq 2 g$. 
\end{proposition}
% Kawauchi  Chapter 12 ???
\begin{lemma}
If $K$ bounds a genus $g$ surface $X \in B^4$ and $S$ is a Seifert form then ${S \in M_{2n \times 2n}}$ has a block structure $\begin{pmatrix}
0 & A\\
B & C
\end{pmatrix}$, where $0$ is $(n - g) \times (n - g)$ submatrix.
\end{lemma}

\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
Remark: $3$ - genus is additive under taking connected sum, but $4$ - genus is not.