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本节解释如何将子流形Poincaré对偶与子流形上向量丛的Thom类联系起来.
商丛
称$F$是秩$n$向量丛$E$的秩$k$子丛, 若$\,\forall\,x\in M$,
$F_x$为$E_x$子空间.
由此可定义商丛$Q=E/F$,
它是一个向量丛使得$0\rightarrow F\rightarrow E\rightarrow Q\rightarrow 0$正合.
即$\,\forall\,x\in M$, $Q_x=E_x/F_x$为商空间.
首先, 先取$F$中自然标架, 再扩张为$E$的一组标架,
这样它们的局部平凡化相容. 设$E$的局部平凡化转移函数为$g_{\alpha\beta},$
$F$为$h_{\alpha\beta}.$ 则对于$F$上的点, 只有$h_{\alpha\beta}$起作用,
从而有$g_{\alpha\beta}=\begin{pmatrix}
h_{\alpha\beta} & *\\
0 & q_{\alpha\beta}
\end{pmatrix}$. 那么$q_{\alpha\beta}$便是$Q$的转移函数.
易见若$E,F$可定向, $Q$也可定向, 称$Q$的定向使得$E$的定向是直和定向.
因为我们有内积, 所以纤维上有正交补空间$F_x^\perp\subset E_x$.
那么$F^\perp=\bigsqcup F_x^\perp$是$E$的子丛且$E=F\oplus F^\perp$,
其上还有继承于$E$的内积. 显然$Q\cong F^\perp$,
从而$Q/E$上也有诱导的内积. $F^\perp$上也有定向使得$E$的定向为直和定向.
子流形法丛与管状邻域引理
$k$维子流形$S\subset M$, $S$上有切丛$TS$与$i^{-1}TM=TM|_S$,
$i$为包含映射.
显然$TS\subset TM|_S$. 定义$N_{S/M}=TM|_S/TS$, 简记为$N$. 若$M,S$可定向,
则切丛可定向, 从而$N$可定向.
$\,\forall\,x\in S$, 存在$x$点坐标卡$(U,\phi)$,
使得$\phi|_{S\cap U}$为$x$在$S$中坐标卡, 且只有前$k$个坐标有值.
由此诱导切丛上的标架. 限制在$U_\alpha$或$U_\alpha\cap S$上,
记$TM$标架为$\{\frac{\partial {} }{\partial {}x_A^\alpha}\}_{1\le A\le n}$,
$TS$标架为$\{\frac{\partial {} }{\partial {}x_i^\alpha}\}_{1\le i\le k}$,
$N$标架为$\{\frac{\partial {} }{\partial {}x_a^\alpha}\}_{k+1\le a\le n}$.
在$S\cap U_{\alpha\beta}$上,
$\frac{\partial {}x_a^\alpha}{\partial {}x_i^\beta}=0$,
因此$TS$转移函数为$h_{\alpha\beta}=(\frac{\partial {}x_j^\alpha}{\partial {}x_i^\beta})_{k\times k}$.
然而,
$\frac{\partial {}x_j^\alpha}{\partial {}x_a^\beta}$显然不一定为$0$,
但我们在做商丛时模掉了这部分,
因此$N$的转移函数为$q_{\alpha\beta}=(\frac{\partial {}x_b^\alpha}{\partial {}x_a^\beta})_{(n-k)\times (n-k)}$.
$TM|_S$的转移函数即为$g_{\alpha\beta}=\begin{pmatrix}
h_{\alpha\beta} & *\\
0 & q_{\alpha\beta}
\end{pmatrix}=(\frac{\partial {}x_A^\alpha}{\partial {}x_B^\beta})_{n\times n}$.
我们有如下重要定理, 证明可参考其它微分流形教材.
定理 1.1 (管状邻域定理). 存在$j$将$N$嵌入$M$, 且$j\circ s=i$, 即将零截面映到$S$.
称开集$j(N)\subset M$为管状邻域, 记为$T$.
Poincaré对偶与Thom类
记$\Phi$为$S$法丛$N$的Thom类, 则由Thom同构,
$H^\ast (S)\cong H_{cv}^{\ast +n-k}(N)\cong H_{cv}^{\ast +n-k}(T)\rightarrow H^{\ast +n-k}(M)$,
最后一个映射为零延拓.
引理 1.1. $S$的闭Poincaré对偶$[\eta_S]=j_\ast \Phi\in H^{n-k}(M)$,
证: 只需证明$j_\ast \Phi$满足对偶性质. 记$l:S\rightarrow T$为包含,
其为$\pi:T\rightarrow S$为零截面. 它们在上同调的层面为互逆的,
从而$\omega|_T=\pi^\ast l^\ast (\omega|_T)+d\tau$, $\tau\in \Omega^{k-1}(T)$
$$\begin{aligned}
\int_M\omega\wedge j_*\Phi&=\int_T\omega|_T\wedge \Phi & j_*\text{为零延拓}\\
&=\int_T \pi^*l^*(\omega|_T)\wedge \Phi+\int_T d\tau \wedge \Phi&\\
&=\int_S l^*(\omega|_T)\wedge \pi_*\Phi&\text{Stokes定理+投射公式}\\
&=\int_S i^*\omega.&\pi_*\Phi=1\quad+\quad{\omega|_T}|_S=\omega|_S
\end{aligned}$$
又因为子流形$S$在$M$中的管状邻域可充分小, 我们得到:
命题 1.1 (局部化原理). 闭子流形$S$的Poincaré对偶的支集可缩小到$S$任意给定的管状邻域中.
应用
横截相交子流形的Poincaré对偶
称$M$中两个子流形$R,S$为横截相交的, 若$\,\forall\,x\in R\cap S$,
$T_xR+T_xS=T_xM$.
设$R,S$为流形$M$中横截相交的闭子流形,
则$R\cap S$也是闭子流形且$\dim R\cap S=\dim R+\dim S-\dim M$.
进一步$\operatorname{codim}_M R\cap S=\operatorname{codim}_M R+\operatorname{codim}_M S$.
$R\cap S$在$M$中法丛$N_{(R\cap S)/M}=(N_{R/M}\oplus N_{S/M}) |_{R\cap S}$.
将法丛嵌入到$M$中为管状邻域,
则$\Phi(N_{(R\cap S)/M})=\Phi(N_{R/M})\wedge \Phi(N_{S/M})$,
从而$\eta_{R\cap S}=\eta_R\wedge \eta_S$.
命题 1.2. 两个横截相交闭子流形的交的Poincaré对偶等于这两个子流形的Poincaré对偶的外积.
Poincaré对偶的拉回
设$f:M'\rightarrow M$是光滑映射, $S$是$M$的闭子流形.
若$\,\forall\,x\in f^{-1}(S)$, $f_\ast (T_x M')+T_{f(x)}S=T_{f(x)}M$,
则称映射$f$横截于子流形$S$.
即$f(M')$要么与子流形$S$不相交($f^{-1}(S)=\varnothing$自动成立),
要么横截相交($f_\ast (T_x M')=T_{f(x)}f(M')$).
这时, $f^{-1}(S)$是$M'$的光滑子流形, 且任意$S$在$M$中的管状邻域$T$,
$f^{-1}T$是$f^{-1}(S)$在$M'$中的管状邻域. 由Thom类的函子性,
$f^\ast $与$\mathcal{T}$, $j_\ast $均可换.
这时我们有$f^\ast \eta_S=f^\ast j_\ast \Phi(T)=j_\ast \mathcal{T}f^\ast (1)=j_\ast \Phi(f^{-1}T)=\eta_{f^{-1}(S)}$.
即:
命题 1.3. 设$f:M'\rightarrow M$是光滑映射, 横截于$M$的闭子流形$S$. 若$[\eta_S]$是$S$在$M$中的Poincaré对偶, 则$f^\ast [\eta_S]$是$f^{-1}(S)$在$M'$中的Poincaré对偶.
文章最后更新于 2021-10-10 21:10:53
This section explains how to relate the Poincaré duality of a submanifold to the Thom class of vector bundles on the submanifold.
Quotient Bundle
We say $F$ is the rank $k$ of the vector bundle $E$ of rank $n$. subcong, if $\,\forall\,x\in M$,
$F_x$ is the $E_x$ subspace.
From this it can be defined Quotient Bundle $Q=E/F$,
It is a vector bundle such that $0\rightarrow F\rightarrow E\rightarrow Q\rightarrow 0$ is exact.
That is, $\,\forall\,x\in M$ and $Q_x=E_x/F_x$ are quotient spaces.
First, take the natural frame in $F$, and then expand it to a set of frames in $E$,
In this way, their local trivializations are compatible. Let the local trivialization transfer function of $E$ be $g_{\alpha\beta},$
$F$ is $h_{\alpha\beta}.$, then for the point on $F$, only $h_{\alpha\beta}$ works,
So there is $g_{\alpha\beta}=\begin{pmatrix}
h_{\alpha\beta} & *\\
0 & q_{\alpha\beta}
\end{pmatrix}$. Then $q_{\alpha\beta}$ is the transfer function of $Q$.
It is easy to see that if $E,F$ can be oriented, $Q$ can also be oriented, and it is said that the orientation of $Q$ makes the orientation of $E$ straight and oriented.
Because we have an inner product, we have an orthogonal complementary space $F_x^\perp\subset E_x$ on the fiber.
Then $F^\perp=\bigsqcup F_x^\perp$ is a subbundle of $E$ and $E=F\oplus F^\perp$,
There is also an inner product inherited from $E$. Obviously $Q\cong F^\perp$,
Therefore, there is also an induced inner product on $Q/E$. There is also an orientation on $F^\perp$ such that the orientation of $E$ is straight and oriented.
Submanifold normal bundle and tubular neighborhood lemma
There are tangent bundles $TS$ and $i^{-1}TM=TM|_S$ on $k$-dimensional submanifolds $S\subset M$ and $S$,
$i$ is an inclusion mapping.
Obviously $TS\subset TM|_S$. Define $N_{S/M}=TM|_S/TS$, abbreviated as $N$. If $M,S$ can be oriented,
Then the tangent bundle can be oriented, so $N$ can be oriented.
$\,\forall\,x\in S$, there is $x$ point coordinate card $(U,\phi)$,
Make $\phi|_{S\cap U}$ be the coordinate card of $x$ in $S$, and only the first $k$ coordinates have values.
This induces a frame on the tangent bundle. Constrained to $U_\alpha$ or $U_\alpha\cap S$,
Record $TM$ frame as $\{\frac{\partial {} }{\partial {}x_A^\alpha}\}_{1\le A\le n}$,
The $TS$ frame is $\{\frac{\partial {} }{\partial {}x_i^\alpha}\}_{1\le i\le k}$,
The $N$ frame is $\{\frac{\partial {} }{\partial {}x_a^\alpha}\}_{k+1\le a\le n}$.
On $S\cap U_{\alpha\beta}$,
$\frac{\partial {}x_a^\alpha}{\partial {}x_i^\beta}=0$,
Therefore the $TS$ transfer function is $h_{\alpha\beta}=(\frac{\partial {}x_j^\alpha}{\partial {}x_i^\beta})_{k\times k}$.
However,
$\frac{\partial {}x_j^\alpha}{\partial {}x_a^\beta}$ is obviously not necessarily $0$,
But we modeled out this part when we were doing the business cluster.
Therefore, the transfer function of $N$ is $q_{\alpha\beta}=(\frac{\partial {}x_b^\alpha}{\partial {}x_a^\beta})_{(n-k)\times (n-k)}$.
The transfer function of $TM|_S$ is $g_{\alpha\beta}=\begin{pmatrix}
h_{\alpha\beta} & *\\
0 & q_{\alpha\beta}
\end{pmatrix}=(\frac{\partial {}x_A^\alpha}{\partial {}x_B^\beta})_{n\times n}$.
We have the following important theorem. For proof, please refer to other differential manifold textbooks.
Theorem 1.1 (Tubular neighborhood theorem). There is $j$ that embeds $N$ into $M$, and $j\circ s=i$, that is, the zero section is mapped to $S$.
The open set $j(N)\subset M$ is called a tubular neighborhood, denoted as $T$.
Poincaré duality and Thom class
Let $\Phi$ be the Thom class of $S$ method bundle $N$, then it is isomorphic to Thom,
$H^\ast (S)\cong H_{cv}^{\ast +n-k}(N)\cong H_{cv}^{\ast +n-k}(T)\rightarrow H^{\ast +n-k}(M)$,
The last mapping is zero continuation.
Lemma 1.1. The closed Poincaré dual of $S$ $[\eta_S]=j_\ast \Phi\in H^{n-k}(M)$,
Certificate: It is only necessary to prove that $j_\ast \Phi$ satisfies the duality property. Let $l:S\rightarrow T$ be included,
It is $\pi:T\rightarrow S$ with zero cross section. They are mutually inverse at the level of cohomology,
Thus $\omega|_T=\pi^\ast l^\ast (\omega|_T)+d\tau$, $\tau\in \Omega^{k-1}(T)$
$$\begin{aligned}
\int_M\omega\wedge j_*\Phi&=\int_T\omega|_T\wedge \Phi & j_*\text{为零延拓}\\
&=\int_T \pi^*l^*(\omega|_T)\wedge \Phi+\int_T d\tau \wedge \Phi&\\
&=\int_S l^*(\omega|_T)\wedge \pi_*\Phi&\text{Stokes定理+投射公式}\\
&=\int_S i^*\omega.&\pi_*\Phi=1\quad+\quad{\omega|_T}|_S=\omega|_S
\end{aligned}$$
And because the tubular neighborhood of submanifold $S$ in $M$ can be sufficiently small, we get:
Proposition 1.1 (Principle of localization). The support of the Poincaré dual of the closed submanifold $S$ can be reduced to any given tubular neighborhood of $S$.
Application
Poincaré duality of transversal intersecting submanifolds
Two submanifolds $R,S$ in $M$ are said to intersect transversely, if $\,\forall\,x\in R\cap S$,
$T_xR+T_xS=T_xM$.
Let $R,S$ be a closed submanifold that intersects transversely in the manifold $M$,
Then $R\cap S$ is also a closed submanifold and $\dim R\cap S=\dim R+\dim S-\dim M$.
Further$\operatorname{codim}_M R\cap S=\operatorname{codim}_M R+\operatorname{codim}_M S$.
$R\cap S$ is in $M$ with $N_{(R\cap S)/M}=(N_{R/M}\oplus N_{S/M}) |_{R\cap S}$.
Embedding the normal bundle into $M$ is a tubular neighborhood,
Then $\Phi(N_{(R\cap S)/M})=\Phi(N_{R/M})\wedge \Phi(N_{S/M})$,
Thus $\eta_{R\cap S}=\eta_R\wedge \eta_S$.
Proposition 1.2. The Poincaré duality of the intersection of two transversally intersecting closed submanifolds is equal to the outer product of the Poincaré duality of the two submanifolds.
Pullback of the Poincaré Duality
Assume $f:M'\rightarrow M$ is a smooth map and $S$ is the closed submanifold of $M$.
If $\,\forall\,x\in f^{-1}(S)$, $f_\ast (T_x M')+T_{f(x)}S=T_{f(x)}M$,
Then the map $f$ is said to be transverse to the submanifold $S$.
That is, $f(M')$ either does not intersect with the submanifold $S$ ($f^{-1}(S)=\varnothing$ is automatically established),
Or transverse intersection ($f_\ast (T_x M')=T_{f(x)}f(M')$).
At this time, $f^{-1}(S)$ is the smooth submanifold of $M'$, and any tubular neighborhood $T$ of $S$ in $M$,
$f^{-1}T$ is the tubular neighborhood of $f^{-1}(S)$ in $M'$. According to the functority of Thom class,
$f^\ast $ can be replaced with $\mathcal{T}$, $j_\ast $.
At this point we have $f^\ast \eta_S=f^\ast j_\ast \Phi(T)=j_\ast \mathcal{T}f^\ast (1)=j_\ast \Phi(f^{-1}T)=\eta_{f^{-1}(S)}$.
That is:
Proposition 1.3. Let $f:M'\rightarrow M$ be a smooth map, a closed submanifold $S$ transverse to $M$. If $[\eta_S]$ is the Poincaré dual of $S$ in $M$, then $f^\ast [\eta_S]$ is the Poincaré dual of $f^{-1}(S)$ in $M'$.
The article was last updated on 2021-10-10 21:10:53
