78 lines
2.9 KiB
TeX
78 lines
2.9 KiB
TeX
\begin{theorem}
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If $K$ is slice,
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then $\sigma_K(t)
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= \sign ( (1 - t)S +(1 - \bar{t})S^T)$
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is zero except possibly of finitely many points and $\sigma_K(-1) = \sign(S + S^T) \neq 0$.
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\end{theorem}
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\begin{proof}
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\begin{lemma}
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\label{lem:metabolic}
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If $V$ is a Hermitian matrix ($\bar{V} = V^T$), $V$ is of size $2n \times 2n$ and
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$
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V = \begin{pmatrix}
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0 & A \\
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\bar{A}^T & B
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\end{pmatrix}
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$ and $\det V \neq 0$ then $\sigma(V) = 0$.
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\end{lemma}
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\begin{definition}
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A Hermitian form $V$ is metabolic if $V$ has structure
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$\begin{pmatrix}
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0 & A\\
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\bar{A}^T & B
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\end{pmatrix}$ with half-dimensional null-space.
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\end{definition}
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\noindent
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In other words: non-degenerate metabolic hermitian form has vanishing signature.\\
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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}. \\
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Let $t \in S^1 \setminus \{1\}$. Then:
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\begin{align*}
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&\det((1 - t) S + (1 - \bar{t}) S^T) =
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\det((1 - t) S + (t\bar{t} - \bar{t}) S^T) =\\
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&\det((1 - t) (S - \bar{t} - S^T)) =
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\det((1 -t)(S - \bar{t} S^T)).
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\end{align*}
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As $\det (S + S^T) \neq 0$, so $S - \bar{t}S^T \neq 0$.
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\end{proof}
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?????????????????s\\
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\begin{corollary}
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If $K \sim K^\prime$ then for all but finitely many $t \in S^1 \setminus \{1\}: \sigma_K(t) = \sigma_{K^\prime}(t)$.
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\end{corollary}
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\begin{proof}
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If $ K \sim K^\prime$ then $K \# K^\prime$ is slice.
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\[
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\sigma_{-K^\prime}(t) = -\sigma_{K^\prime}(t)
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\]
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\\??????????????\\
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The signature give a homomorphism from the concordance group to $\mathbb{Z}$.\\
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??????????????????\\
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Remark: if $t \in S^1$ is not algebraic over $\mathbb{Z}$, then $\sigma_K(t) \neq 0$
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(we can is the argument that $\mathscr{C} \longrightarrow \mathbb{Z}$ as well).
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\end{proof}
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\begin{figure}[h]
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\fontsize{20}{10}\selectfont
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\centering{
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\def\svgwidth{\linewidth}
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\resizebox{0.5\textwidth}{!}{\input{images/genus_2_bordism.pdf_tex}}
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}
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\caption{$K$ and $K^\prime$ are connected by a genus $g$ surface of genus.}\label{fig:genus_2_bordism}
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\end{figure}
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???????????????????????\\
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\begin{proposition}[Kawauchi inequality]
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If there exists a genus $g$ surface as in Figure \ref{fig:genus_2_bordism}
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then for almost all $t \in S^1 \setminus \{1\}$ we have $\vert \sigma_K(t) - \sigma_{K^\prime}(t) \vert \leq 2 g$.
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\end{proposition}
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% Kawauchi Chapter 12 ???
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\begin{lemma}
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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 $\begin{pmatrix}
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0 & A\\
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B & C
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\end{pmatrix}$, where $0$ is $(n - g) \times (n - g)$ submatrix.
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\end{lemma}
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\begin{definition}
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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 $\partial \Sigma = K$.
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\end{definition}
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\noindent
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Remark: $3$ - genus is additive under taking connected sum, but $4$ - genus is not.
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