lectures_on_knot_theory/lec_5.tex

\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}
\noindent
We will use the following lemma.
\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}
\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{align*}
\dim H_1(Z)  = 2 n\\
\dim H_1 (Y) = 2 n + 2 g\\
\dim (\ker (H_1, Y) \longrightarrow H_1(\Omega)) = n + g\\
Y = X \sum \Sigma
\end{align*}
\noindent
If $\alpha, \beta \in \ker(H_1(\Sigma \longrightarrow H_1(\Omega))$, then ${\Lk(\alpha, \beta^+) = 0}$. 
\begin{corollary}
If $t$ is nota ???? of $\det $ ????
then $\vert \sigma_K(t) \vert \leq 2g$.\\
\end{corollary}
\noindent
If there exists cobordism of genus $g$ between $K$ and $K^\prime$ like shown in Figure \ref{fig:genus_2_bordism}, then $K \# -K^\prime$ bounds a surface of genus $g$ in $B^4$.
\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.
\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 $g_(K \# K^\prime) = 0$.
\\?????????????????????\\
\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.
\end{itemize}
\end{example}
\begin{proposition}
$g_4 (T(p, q) \# -T(r, s))$ is in general hopelessly unknown. 
\\???????????????\\
essentially $\sup \vert \sigma_K(t) \vert \leq 2 g_n(K)$
\end{proposition}
\begin{definition}
A knot $K$ is called topologically slice if $K$ bounds a topological locally flat disc in $B^4$ (it has tubular neighbourhood). 
\end{definition}
\begin{theorem}[Freedman, '82]
If $\Delta_K(t) \geq 1$, then $K$ is topologically slice, but not necessarily smoothly slice.  
\end{theorem}
\begin{theorem}[Powell, 2015]
If $K$ is genus g 
\\(top. loc.?????????)\\
cobordant to $K^\prime$, 
then $\vert \sigma_K(t)  - \sigma_{K^\prime}(t) \vert \leq 2 g$. \\
If $g_4^{\mytop}(K) \geq $ ?????ess $\sup \vert \sigma_K(t) \vert$ and ?????????\\
$\dim \ker (H_1 (Y) \longrightarrow H_1(\Omega)) = \frac{1}{2} \dim H_1(Y)$.
\end{theorem}
???????????????
\[
H^1 (B^4 \setminus Y, \mathbb{Z}) = [B^4 \setminus Y, S^1]
\]
\noindent
Remark: unless $p=2$ or $p = 3 \wedge  q = 4$:
\[
g_4^\top (T(p, q)) < q_4(T(p, q))
\]
%??????????????????????
\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: $A^T = A: (a, b) = \bar{(a, b)}$.