151 lines
4.7 KiB
TeX
151 lines
4.7 KiB
TeX
$X$ is a closed orientable four-manifold. For simplicity assume $\pi_1(X) = 0$ (it is not needed to define the intersection form). In particular $H_1(X) = 0$.
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$H_2$ is free (exercise).
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\[
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H_2(X, \mathbb{Z}) \xrightarrow{\text{Poincar\'e duality}} H^2(X, \mathbb{Z} ) \xrightarrow{\text{evaluation}}\Hom(H_2(X, \mathbb{Z}), \mathbb{Z}).
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\]
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\noindent
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Intersection form:
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$H_2(X, \mathbb{Z}) \times
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H_2(X, \mathbb{Z}) \longrightarrow \mathbb{Z}$ is symmetric and non singular.
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\\
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Let $A$ and $B$ be closed, oriented surfaces in $X$.
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\\
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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/intersection_form_A_B.pdf_tex}}
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}
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\caption{$T_X A + T_X B = T_X X$
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}\label{fig:intersection}
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\end{figure}
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???????????????????????
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\begin{align*}
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x \in A \cap B\\
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T_XA \oplus T_X B = T_X X\\
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\{\epsilon_1, \dots , \epsilon_n \} = A \cap C\\
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A \cdot B = \sum^n_{i=1} \epsilon_i
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\end{align*}
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\begin{proposition}
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Intersection form $A \cdot B$ doesn't depend of choice of $A$ and $B$ in their homology classes:
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\[
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[A], [B] \in H_2(X, \mathbb{Z}).
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\]
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\end{proposition}
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\noindent
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\\
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\subsection{Fundamental cycle}
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If $M$ is an $m$ - dimensional close, connected and orientable manifold, then $H_m(M, \mathbb{Z})$ and the orientation of $M$ determined a cycle $[M] \in H_m(M, \mathbb{Z})$, called the fundamental cycle.
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\begin{example}
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If $\omega$ is an $m$ - form then:
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\[
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\int_M \omega = [\omega]([M]), \quad [\omega] \in H^m_\Omega(M), \ [M] \in H_m(M).
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\]
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\end{example}
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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{0.8\textwidth}{!}{\input{images/torus_alpha_beta.pdf_tex}}
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}
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\caption{$\beta$ cross $3$ times the disk bounded by $\alpha$.
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$T_X \alpha + T_X \beta = T_X \Sigma$
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}\label{fig:torus_alpha_beta}
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\end{figure}
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\begin{example}
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K{\"u}nneth
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?????????????????????????\\
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Let $X = S^2 \times S^2$.
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We know that:
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\begin{align*}
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&H_2(S^2, \mathbb{Z}) =\mathbb{Z}\\
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&H_1(S^2, \mathbb{Z}) = 0\\
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&H_0(S^2, \mathbb{Z}) =\mathbb{Z}
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\end{align*}
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We can construct a long exact sequence for a pair:
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\begin{align*}
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&H_2(\partial X) \to H_2(X)
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\to H_2(X, \partial X) \to \\
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\to &H_1(\partial X) \to H_1(X) \to H_1(X, \partial X) \to
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\end{align*}
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????????????????????\\
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Simple case $H_1(\partial X)$ \\????????????\\
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is torsion.
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$H_2(\partial X)$ is torsion free (by universal coefficient theorem),\\
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???????????????????????\\
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therefore it is $0$.
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\\?????????????????????\\
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We know that $b_1(X) = b_2(X)$. Therefore by Poincar\'e duality:
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\begin{align*}
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b_1(X) =
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\dim_{\mathbb{Q}} H_1(X, \mathbb{Q})
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\overset{\mathrm{PD}}{=}
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\dim_{\mathbb{Q}} H^2(X, \mathbb{Q}) =
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\dim_{\mathbb{Q}} H_2(X, \mathbb{Q}) = b_2(X)
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\end{align*}
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???????????????????????????????\\
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$H_2(X, \mathbb{Z})$ is torsion free and
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$H_2(X_1, \mathbb{Q}) = 0$, therefore $H_2(X, \mathbb{Z}) = 0$.
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The map
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$H_2(X, \mathbb{Z}) \longrightarrow H_2(X, \partial X, \mathbb{Z})$ is a monomorphism. \\??????????\\ (because it is an isomorphism after tensoring by $\mathbb{Q}$.
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\\
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Suppose $\alpha_1, \dots, \alpha_n$ is a basis of $H_2(X, \mathbb{Z})$.
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Let $A$ be the intersection matrix in this basis. Then:
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\begin{enumerate}
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\item
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A has integer coefficients,
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\item
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$\det A \neq 0$,
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\item
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$\vert \det A \vert =
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\vert H_1 (\partial X, \mathbb{Z}) \vert =
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\vert \coker H_2(X) \longrightarrow H_2(X, \partial X) \vert$.
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\end{enumerate}
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\end{example}
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???????????????????
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\\
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\\
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If $CUC^T = W$, then for
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$\binom{a}{b} = C^{-1} \binom{1}{0}$ we have:
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\[
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\binom{a}{b} W
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\binom{a}{b} = \binom{1}{0} U \binom{1}{0} = 1 \notin 2 \mathbb{Z}.
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\]
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% if we switch to \mathbb{Q} it will be possible?
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\begin{theorem}[Whitehead]
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Any non-degenerate form
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\[
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A : \mathbb{Z}^4 \times \mathbb{Z}^4 \longrightarrow \mathbb{Z}
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\]
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can be realized as an intersection form of a simple connected $4$-dimensional manifold.
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\end{theorem}
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??????????????????????????
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\begin{theorem}[Donaldson, 1982]
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If $A$ is an even definite intersection form of a smooth $4$-manifold then it is diagonalizable over $\mathbb{Z}$.
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\end{theorem}
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??????????????????????????
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??????????????????????????
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??????????????????????????
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??????????????????????????
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\begin{definition}
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even define
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\end{definition}
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Suppose $X$ us $4$ -manifold with a boundary such that $H_1(X) = 0$.
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%$A \cdot B$ gives the pairing as ??
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\begin{proof}
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Obviously:
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\[H_1(\partial X, \mathbb{Z}) = \coker H_2(X) \longrightarrow H_2(X, \partial X) = \quot{H_2(X, \partial X)}{H_2(X)}.
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\]
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Let $A$ be an $n \times n$ matrix. $A$ determines a \\
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??????????????/\\
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\begin{align*}
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\mathbb{Z}^n \longrightarrow \Hom (\mathbb{Z}^n, \mathbb{Z})\\
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a \mapsto (b \mapsto b^T A a)\\
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\vert \coker A \vert = \vert \det A \vert
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\end{align*}
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all homomorphisms $b = (b_1, \dots, b_n) $???????\\?????????\\
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\end{proof}
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