Machine-translated from Chinese.
向量丛上的不变量: 紧垂直上同调, 由沿纤维积分给出.
向量丛(全空间)的紧上同调
若$s:M\rightarrow E$为零截面, 则它把$M$嵌入到$E$. $s(M)$为$E$的形变收缩,
于是$H^\ast (E)\cong H^\ast (M)$. 那么受Poincaré引理启发, 对于紧上同调,
是否有$H_c^\ast (E)\cong H_c^{\ast -n}(M)$? 但对开Möbius带就不成立.
现在我们将具有有限好覆盖的流形称为有限型流形.
引理 1.1. $\pi:E\rightarrow M$为秩$n$向量丛. 若$E,M$是可定向有限型流形, 则公式成立.
由Poincaré对偶转化为上同调的对偶空间, 再由同伦不变性即得.
若$M$为有限型流形, 则$E$也是. 事实上, 若$M$可定向, $\pi$为可定向向量丛,
则$E$也是可定向的. 只需利用$M$上的坐标函数$\psi_\alpha$,
以及$E$的定向平凡化$\phi_\alpha$自然地构造$E$上的坐标函数即可.
即$E$的坐标图册为
$\{(E|_{U_\alpha},(\psi_\alpha\times \mathrm{id})\circ\phi_\alpha)\}$.
容易验证其为定向的, 称为局部乘积定向.
从而有如下命题:
命题 1.1. 若$M$为可定向有限型流形, $\pi:E\rightarrow M$为秩$n$可定向向量丛, 则$H_c^\ast (E)\cong H_c^{\ast -n}(M).$
事实上仿照后文Thom同构证明与MV方法, 可去掉$M$的可定向性假设.
紧垂直上同调与沿纤维积分
记$\Omega^\ast _{cv}(E)$为垂直方向具紧支集的微分形式复形,
即$\omega\in\Omega_{cv}^\ast (E)$当且仅当对任意紧集$K\subset M$,
$\operatorname{supp}\omega\cap \pi^{-1}(K)$紧.
称该微分形式在垂直方向具紧支集. $H_{cv}^\ast (E)$称紧垂直上同调.
对这样的形式, 限制在$E_x$上具紧支集, 因此可沿纤维积分.
记沿纤维积分$\pi_\ast :\Omega_{cv}^\ast (E)\rightarrow \Omega^{\ast -n}(M)$.
先设$E$是平凡丛.
将$\omega\in \Omega_{cv}^\ast (E)$分解为含$dt_1\wedge \cdots\wedge dt_n$的项和其他项的线性组合.
令$\pi_\ast :\pi^\ast \phi\wedge f(x,t)dt_1\wedge \cdots\wedge dt_n\mapsto \phi\int_{\mathbb{R}^n}f(x,t)dt_1\cdots dt_n$,
其它形式映为$0$. 事实上这就是以前记号的高维推广.
对任意定向向量丛, 将$\pi_\ast \omega$在每个局部平凡化邻域做即可.
但需要验证的是, 取两块$U_\alpha,U_\beta$与其上的坐标函数、平凡化,
需要验证其上的$\pi_\ast \omega_\alpha, \pi_\ast \omega_\beta$可以拼起来.
事实上这就是重积分的积分变换公式.
当然, 也可直接在操作层面处理,
利用单位分解将$\omega$分为在每个$U_\alpha$上支撑的$\omega_\alpha$的和,
然后做$\pi_\ast$沿纤维积分即可. 接下来我们有:
命题 1.2. $\pi_\ast$为链映射, 从而诱导紧垂直上同调间的映射.
对乘积丛$E=M\times \mathbb{R}^n$验证即可, 利用定义展开,
和之前证明链同伦差不多.
命题 1.3 (投射公式). $\,\forall\,\tau\in\Omega^\ast (M)$, $\omega\in \Omega_{cv}^\ast (E)$, 则$\pi_\ast (\pi^\ast \tau\wedge \omega)=\tau\wedge \pi_\ast \omega$. 若$M$为$m$维定向流形, 设$\omega\in\Omega^{q}_{cv}(E)$, $\tau \in \Omega_c^{m+n-q}(M)$, $E$取局部乘积定向, 则$\int_E\pi^\ast \tau\wedge\omega=\int_M\tau\wedge\pi_\ast \omega.$
做简单的验证即好, 和之前的证明差不多, 将形式分为两个部分分别验证.
后半命题将$M$单位分解即可, 证明要用到Fubini定理,
就是在$E|_{U_\alpha}$上积分和先沿纤维积分再在$U_\alpha$上积分是一样的.
Thom同构
接下来将紧支集上的Poincaré引理证明照搬即可得到:
命题 1.4 (紧垂直支集的Poincaré引理). $\pi_\ast :H_{cv}^\ast (M\times \mathbb{R}^n)\cong H^{\ast -n}(M).$
接下来我们证明一般情形:
定理 1.1 (Thom同构). 若$\pi:E\rightarrow M$为有限型流形$M$上秩$n$可定向向量丛, 则$H_{cv}^\ast (E)\cong H^{\ast -n}(M).$
证明用到$U\cup V$上的MV序列以及其上纤维$E|_{U\cup V}$的MV序列.
其间由链映射$\pi_\ast$联系. 由五引理,
Thom同构对$E|_U,E|_V,E|_{U\cap V}$成立, 则也对$E|_{U\cup V}$成立.
从而归纳好覆盖开集个数, 由MV方法即得.
记$\mathcal{T}:H^\ast (M)\rightarrow H_{cv}^{\ast +n}(E)$为$\pi_\ast$的逆,
称其为Thom同构. 对$1\in H^0(M)$,
$\Phi=\mathcal{T}(1)\in H_{cv}^n(E)$称为$E$的Thom类.
$\mathcal{T}(\omega)=\mathcal{T}\pi_\ast (\pi^\ast \omega\wedge \Phi)=\pi^\ast \omega\wedge \Phi.$
因为$\mathcal{T}\pi_\ast =\mathrm{id}$, $\pi_\ast \mathcal{T}(1)=1.$
命题 1.5. $\Phi$是$H_{cv}^n(E)$的唯一上同调类, 限制在每个$E_x$上为$H_c^n(E_x)$生成元.
证: $\pi_\ast \Phi=1$, $\pi_\ast \Phi(x)=\int_{E_x}\Phi(x,t)dt=1$,
从而$\Phi|_{E_x}\in H_c^n(E_x)$是生成元.
若$\Phi'$也满足该性质, 则$\pi_\ast \Phi'=1$.
$\pi_\ast (\pi^\ast \omega\wedge\Phi')=\omega$,
$\mathcal{T}(\omega)=\pi^\ast \omega\wedge\Phi'$,
从而$\Phi'=\mathcal{T}(1)=\Phi.$
命题 1.6. 设$\pi_1:E\rightarrow M$, $\pi_2:E'\rightarrow M$为两定向向量丛, 则$\Phi(E\oplus E')=\operatorname{pr}_1^\ast \Phi(E)\wedge \operatorname{pr}_2^\ast \Phi(E')$.
证: 取右式为$\Phi'$, 在$E_x\oplus E'_x$上积分发现为$1$,
由上一命题的唯一性即得.
事实上, $\Phi$也有函子性, 即$f:N\rightarrow M$为光滑映射,
则$\Phi(f^{-1}E)=f^\ast \Phi(E)$, 证明留做习题. (确实是我们的习题)
文章最后更新于 2021-09-23 15:07:12
Invariants on vector bundles: tight vertical cohomology, given by integrating along the fiber.
Compactly Supported Cohomology of the Total Space of a Vector Bundle
If $s:M\rightarrow E$ is zero cross-section, it embeds $M$ into $E$. $s(M)$ is the deformation shrinkage of $E$,
So $H^\ast (E)\cong H^\ast (M)$. Then inspired by Poincaré’s lemma, for compact cohomology,
Is there $H_c^\ast (E)\cong H_c^{\ast -n}(M)$? But this is not true for the Möbius belt.
Now we call a manifold with finite good coverage finite manifold.
Lemma 1.1. $\pi:E\rightarrow M$ is a vector bundle of rank $n$. If $E,M$ is an orientable finite manifold, the formula is established.
The Poincaré duality is transformed into a cohomological duality space, and then it is obtained by homotopy invariance.
If $M$ is a finite manifold, then $E$ is also. In fact, if $M$ is orientable, $\pi$ is an orientable vector bundle,
Then $E$ is also orientable. Just use the coordinate function $\psi_\alpha$ on $M$,
And the directional trivialization $\phi_\alpha$ of $E$ can naturally construct the coordinate function on $E$.
That is, the coordinate atlas of $E$ is
$\{(E|_{U_\alpha},(\psi_\alpha\times \mathrm{id})\circ\phi_\alpha)\}$.
It is easy to verify that it is directional, which is called local product directional.
This leads to the following proposition:
Proposition 1.1. If $M$ is an orientable finite manifold and $\pi:E\rightarrow M$ is a rank $n$ orientable vector bundle, then $H_c^\ast (E)\cong H_c^{\ast -n}(M).$
In fact, following the Thom isomorphism proof and MV method described later, the orientability assumption of $M$ can be removed.
Tight vertical cohomology and integration along fibers
Let $\Omega^\ast _{cv}(E)$ be the differential form complex with compact support in the vertical direction,
That is, $\omega\in\Omega_{cv}^\ast (E)$ iff for any compact set $K\subset M$,
$\operatorname{supp}\omega\cap \pi^{-1}(K)$ Tight.
call this differential form has tight support in the vertical direction. $H_{cv}^\ast (E)$ Scale tight vertical cohomology.
For such a form, the constraint has compact support on $E_x$, so it can be integrated along the fiber.
Record the integral along the fiber $\pi_\ast :\Omega_{cv}^\ast (E)\rightarrow \Omega^{\ast -n}(M)$.
Suppose $E$ is a trivial bundle.
Decompose $\omega\in \Omega_{cv}^\ast (E)$ into linear combinations of terms containing $dt_1\wedge \cdots\wedge dt_n$ and other terms.
Order$\pi_\ast :\pi^\ast \phi\wedge f(x,t)dt_1\wedge \cdots\wedge dt_n\mapsto \phi\int_{\mathbb{R}^n}f(x,t)dt_1\cdots dt_n$,
Other forms map to $0$. In fact, this is a high-dimensional generalization of the previous notation.
For any directed vector bundle, just do $\pi_\ast \omega$ in each local trivialization neighborhood.
But what needs to be verified is to take two blocks $U_\alpha,U_\beta$ and the coordinate functions on them, trivialize them,
It needs to be verified that the $\pi_\ast \omega_\alpha, \pi_\ast \omega_\beta$ on it can be put together.
In fact, this is the integral transformation formula of heavy integral.
Of course, it can also be processed directly at the operational level,
Use partition of unity to divide $\omega$ into the sum of $\omega_\alpha$ supported on each $U_\alpha$,
Then just do $\pi_\ast$ integration along the fiber. Next we have:
Proposition 1.2. $\pi_\ast$ is a chain mapping, thus inducing a mapping between tight vertical cohomologies.
Just verify the product bundle $E=M\times \mathbb{R}^n$, use the definition to expand,
It is almost the same as proving chain homotopy before.
Proposition 1.3 (projection formula). $\,\forall\,\tau\in\Omega^\ast (M)$, $\omega\in \Omega_{cv}^\ast (E)$, then $\pi_\ast (\pi^\ast \tau\wedge \omega)=\tau\wedge \pi_\ast \omega$. If $M$ is a $m$-dimensional oriented manifold, let $\omega\in\Omega^{q}_{cv}(E)$, $\tau \in \Omega_c^{m+n-q}(M)$, $E$ take local product orientation, then $\int_E\pi^\ast \tau\wedge\omega=\int_M\tau\wedge\pi_\ast \omega.$
Just do a simple verification, which is similar to the previous proof. Divide the form into two parts and verify them separately.
For the second half of the proposition, just decompose the $M$ unit. The proof requires the use of Fubini's theorem.
That is, integrating on $E|_{U_\alpha}$ is the same as integrating along the fiber first and then integrating on $U_\alpha$.
Thom isomorphism
Next, copy the proof of Poincaré’s lemma on compact supports to get:
Proposition 1.4 (Poincaré’s lemma for compact vertical supports). $\pi_\ast :H_{cv}^\ast (M\times \mathbb{R}^n)\cong H^{\ast -n}(M).$
Next we prove the general case:
Theorem 1.1 (Thom isomorphism). If $\pi:E\rightarrow M$ is a rank $n$ orientable vector bundle on the finite manifold $M$, then $H_{cv}^\ast (E)\cong H^{\ast -n}(M).$
The proof uses the MV sequence on $U\cup V$ and the MV sequence of the fiber $E|_{U\cup V}$ above it.
They are connected by chain mapping $\pi_\ast$. According to the five lemmas,
If Thom isomorphism holds for $E|_U,E|_V,E|_{U\cap V}$, it also holds for $E|_{U\cup V}$.
Thus, the number of covered open sets can be summarized, which can be obtained by the MV method.
Let $\mathcal{T}:H^\ast (M)\rightarrow H_{cv}^{\ast +n}(E)$ be the inverse of $\pi_\ast$,
Call it Thom isomorphism. For $1\in H^0(M)$,
$\Phi=\mathcal{T}(1)\in H_{cv}^n(E)$ is called the Thom class of $E$.
$\mathcal{T}(\omega)=\mathcal{T}\pi_\ast (\pi^\ast \omega\wedge \Phi)=\pi^\ast \omega\wedge \Phi.$
Because $\mathcal{T}\pi_\ast =\mathrm{id}$, $\pi_\ast \mathcal{T}(1)=1.$
Proposition 1.5. $\Phi$ is the only cohomology class of $H_{cv}^n(E)$, and is restricted to generating elements for $H_c^n(E_x)$ on every $E_x$.
Certificate: $\pi_\ast \Phi=1$, $\pi_\ast \Phi(x)=\int_{E_x}\Phi(x,t)dt=1$,
Therefore $\Phi|_{E_x}\in H_c^n(E_x)$ is a generator.
If $\Phi'$ also satisfies this property, then $\pi_\ast \Phi'=1$.
$\pi_\ast (\pi^\ast \omega\wedge\Phi')=\omega$,
$\mathcal{T}(\omega)=\pi^\ast \omega\wedge\Phi'$,
Thus $\Phi'=\mathcal{T}(1)=\Phi.$
Proposition 1.6. Assume $\pi_1:E\rightarrow M$ and $\pi_2:E'\rightarrow M$ are two directed vector bundles, then $\Phi(E\oplus E')=\operatorname{pr}_1^\ast \Phi(E)\wedge \operatorname{pr}_2^\ast \Phi(E')$.
Certificate: Take the right formula as $\Phi'$, integrate on $E_x\oplus E'_x$ and find that it is $1$,
It is obtained from the uniqueness of the previous proposition.
In fact, $\Phi$ also has functor properties, that is, $f:N\rightarrow M$ is a smooth mapping,
Then $\Phi(f^{-1}E)=f^\ast \Phi(E)$, the proof is left as an exercise. (It is indeed our exercise)
The article was last updated on 2021-09-23 15:07:12
