《代数拓扑与微分形式》笔记(8)-推广的MV原理

这节课要把MV序列从两个开集推广到可数多个开集. 主要想法来自Weil.

可数多个开集的推广

$\{U_\alpha\}$为$M$开覆盖, $I$为可数全序集. 有如下包含序列:

$$M\leftarrow \bigsqcup_{\alpha_0} U_{\alpha_0}\begin{matrix} \xleftarrow{\partial_0}\\\xleftarrow{\partial_1} \end{matrix} \bigsqcup_{\alpha_0<\alpha_1} U_{\alpha_0\alpha_1}\begin{matrix} \xleftarrow{\partial_0}\\\xleftarrow{\partial_1}\\\xleftarrow{\partial_2} \end{matrix}\cdots$$

此处$\partial_i$指忽略第$i$个开集. 那么这样的包含映射$\partial_i$诱导限制映射$\delta_i$. 进而有如下限制序列:

$$\Omega^*(M)\rightarrow \prod_{\alpha_0} \Omega^*(U_{\alpha_0})\begin{matrix} \xrightarrow{\delta_0}\\\xrightarrow{\delta_1} \end{matrix} \prod_{\alpha_0<\alpha_1} \Omega^*(U_{\alpha_0\alpha_1})\begin{matrix} \xrightarrow{\delta_0}\\\xrightarrow{\delta_1}\\\xrightarrow{\delta_2} \end{matrix}\cdots$$

定义差分算子 $$\delta: \prod_{\alpha_0<\cdots<\alpha_p} \Omega^\ast (U_{\alpha_0\cdots\alpha_p})\rightarrow \prod_{\alpha_0<\cdots<\alpha_{p+1} } \Omega^\ast (U_{\alpha_0\cdots\alpha_{p+1} }),$$

$\delta=\sum_{i=0}^{p+1}(-1)^i\delta_i$. $\delta\omega$在$\Omega^\ast (U_{\alpha_0\cdots\alpha_{p+1} })$上的分量即为$\sum_{i=0}^{p+1}(-i)^i \omega_{\alpha_0,...,\hat\alpha_i,...,\alpha_{p+1} }$.

容易证明, $\delta^2=0$. 事实上, 目前为止$\omega_{\alpha_0...\alpha_p}$的下标要求$\alpha_0<...<\alpha_p$, 但可以补充定义去掉该要求, 对任意的$\omega_{\alpha_0...\alpha_p}$, 用$\tau$对下标做置换, 定义它为$\operatorname{sgn}(\tau)\omega_{\alpha_\tau}$. 容易验证这样与之前的系统不矛盾.

命题 1.1 (推广的MV序列). 我们有正合列: $0\rightarrow \Omega^\ast (M)\xrightarrow{r}\prod \Omega^\ast (U_{\alpha_0})\xrightarrow{\delta}\prod\Omega^\ast (U_{\alpha_0\alpha_1})\rightarrow\cdots$

证: 显然限制映射$r$是单射, 只需要证明在$\prod \Omega^\ast (U_{\alpha_0...\alpha_p})$处正合.

$\,\forall\,\alpha_0,\alpha_1$, $(\delta_\omega)_{\alpha_0\alpha_1}=\omega_{\alpha_1}-\omega_{\alpha_0}=0$当且仅当$\omega_{\alpha_1}$, $\omega_{\alpha_0}$在$U_{\alpha_0\alpha_1}$上相容, 当且仅当$\omega_{\alpha}$可以整体定义为$\eta$, 从而$r(\eta)=\omega$.

对于后面的复形, 由于$\delta^2=0$, 只需验证$\ker\delta\subset \operatorname{Im}\delta$. 取$\tau_{\alpha_0\cdots\alpha_{p-1} }=\sum_{\alpha}\rho_\alpha\omega_{\alpha\alpha_0\cdots\alpha_{p-1} }$. 接下来$(\delta\omega)_{\alpha \alpha_0...\alpha_p}=\omega_{\alpha_0\cdots\alpha_p}+\sum_{i=0}^p(-1)^{i+1}\omega_{\alpha\alpha_0\cdots\hat\alpha_i\cdots\alpha_p}=0$, 由此推知$\delta{\tau}=\omega_{\alpha_0\cdots\alpha_p}$.

事实上$\tau$给出了复形上的一个同伦算子$K$, $(K\omega)_{\alpha_0\cdots\alpha_{p-1} }=\sum_\alpha\rho_\alpha\omega_{\alpha\alpha_0\cdots\alpha_{p-1} }$. 满足有$\delta K+K\delta=\mathrm{id}$. 对上闭链, 称$K$是$\delta$的右逆.

双复形

记$K^{p,q}=C^p(\mathcal{U},\Omega^q)=\prod \Omega^q(U_{\alpha_0\cdots\alpha_p})$. 其上的元素称为$q$-形式值的$p$-上链.

将$(p,q)$视为$\mathbb{Z}^2$上的坐标, 则有一张交换图, 行向右的运算为差分算子, 列向上的运算为微分算子. 交换是因为先限制再外微分等于先外微分再限制, 且外微分有线性性.

令$K^n=\oplus_{p+q=n}K^{p,q}$, $D=D'+D''$, $D'=\delta$, $D''=(-1)^pd$. 则现在$D',D''$带符号交换, 因此$D^2=0$, $\{K^\ast ,D\}$成为一个复形.

$C^\ast (\mathcal{U},\Omega^\ast )$称为Čech-de Rham复形, 其上的元简称$D$-上链, 上同调记为$H^\ast _D\{C^\ast (\mathcal{U},\Omega^\ast )\}$.

对链复形进行增广, 在$(-1,q)$位置, 增加$\Omega^p(M)$, 向右有限制映射$r$, 向上是微分算子. 再在左侧再加上$0$向右映射, 则由先前的命题, 所有行正合.

命题 1.2 (推广的MV原理). $r^\ast :H^\ast _{dR}(M)\approx H_D\{C^\ast (\mathcal{U},\Omega^\ast )\}$.

首先$r$是链映射, 所以$r^\ast $定义合理, $r^\ast [\omega]_d=[r\omega]_D$.

接下来证$r^\ast $既满又单即可. 利用行的正合性, 进行基本的图追踪即可. 沿对角线走, 一步步将多余的分量截掉.

应用

可以看出, 若增广双复形每行都是正合的, 那么它的$D$-上同调同构于初始列即增广列的上同调.

取$C^p(\mathcal{U},\mathbb{R})=\prod\{U_{\alpha_0\cdots\alpha_p}\text{上局部常值函数}\}$, 则有包含映射$i:C^p(\mathcal{U},\mathbb{R})\rightarrow C^p(\mathcal{U},\Omega^0)$, 且$\operatorname{Im}i=\ker d$.

有复形$0\rightarrow C^0(\mathcal{U},\mathbb{R})\xrightarrow{\delta}C^1(\mathcal{U},\mathbb{R})\rightarrow\cdots$, 称其上同调为开覆盖$\mathcal{U}$的Čech上同调, 是一个纯组合对象.

与先前证明类似地, 若每列正合, 添加增广行, 则它的上同调也同构于双复形的$D$-上同调.

我们将上面构造的复形作为增广行. 由Poincaré引理, 若$\mathcal{U}$是一个好覆盖, 那么每列确实正合. 从而:

定理 1.3. $H^\ast (M)\cong H^\ast (\mathcal{U},\mathbb{R})$.

所以我们有一个先验来看并不平凡的推论:

推论 1.4. Čech上同调$H^\ast (\mathcal{U},\mathbb{R})$和好覆盖$\mathcal{U}$选取无关.

另一方面, 若流形具有有限好覆盖, $H^\ast (\mathcal{U},\mathbb{R})$有限维. 从而:

推论 1.5. 有限型流形$M$的de Rham上同调$H_{dR}^\ast (M)$是有限维的.

特别地, 紧流形总具有有限好覆盖, 从而de Rham上同调是有限维的.

文章最后更新于 2021-10-19 21:45:05

This lesson will generalize the MV sequence from two open sets to countably multiple open sets. The main idea comes from Weil.

Extension to Countably Many Open Sets

$\{U_\alpha\}$ is an open covering for $M$, and $I$ is a countably ordered set. There are the following inclusive sequences:

Here $\partial_i$ means ignoring the $i$th open set. Then such an inclusion map $\partial_i$ induces a restriction map $\delta_i$. Then there are the following restriction sequences:

Define the difference operator $$\delta: \prod_{\alpha_0<\cdots<\alpha_p} \Omega^\ast (U_{\alpha_0\cdots\alpha_p})\rightarrow \prod_{\alpha_0<\cdots<\alpha_{p+1} } \Omega^\ast (U_{\alpha_0\cdots\alpha_{p+1} }),$$

$\delta=\sum_{i=0}^{p+1}(-1)^i\delta_i$. The component of $\delta\omega$ on $\Omega^\ast (U_{\alpha_0\cdots\alpha_{p+1} })$ is $\sum_{i=0}^{p+1}(-i)^i \omega_{\alpha_0,...,\hat\alpha_i,...,\alpha_{p+1} }$.

It is easy to prove, $\delta^2=0$. In fact, The subscript requirement of $\omega_{\alpha_0...\alpha_p}$ so far is $\alpha_0<...<\alpha_p$, However, you can add a definition to remove this requirement. For any $\omega_{\alpha_0...\alpha_p}$, Use $\tau$ to replace the subscript, Define it as $\operatorname{sgn}(\tau)\omega_{\alpha_\tau}$. It is easy to verify that this does not conflict with the previous system.

Proposition 1.1 (Promoted MV sequence). We have the exact column: $0\rightarrow \Omega^\ast (M)\xrightarrow{r}\prod \Omega^\ast (U_{\alpha_0})\xrightarrow{\delta}\prod\Omega^\ast (U_{\alpha_0\alpha_1})\rightarrow\cdots$

Certificate: Obviously the restriction map $r$ is injective, It is only necessary to prove that it is exact at $\prod \Omega^\ast (U_{\alpha_0...\alpha_p})$.

$\,\forall\,\alpha_0,\alpha_1$, $(\delta_\omega)_{\alpha_0\alpha_1}=\omega_{\alpha_1}-\omega_{\alpha_0}=0$ if and only if $\omega_{\alpha_1}$, $\omega_{\alpha_0}$ is compatible with $U_{\alpha_0\alpha_1}$, If and only if $\omega_{\alpha}$ can be defined as $\eta$ as a whole, then $r(\eta)=\omega$.

For the following complex shape, due to $\delta^2=0$, Just verify $\ker\delta\subset \operatorname{Im}\delta$. Take $\tau_{\alpha_0\cdots\alpha_{p-1} }=\sum_{\alpha}\rho_\alpha\omega_{\alpha\alpha_0\cdots\alpha_{p-1} }$. Next $(\delta\omega)_{\alpha \alpha_0...\alpha_p}=\omega_{\alpha_0\cdots\alpha_p}+\sum_{i=0}^p(-1)^{i+1}\omega_{\alpha\alpha_0\cdots\hat\alpha_i\cdots\alpha_p}=0$, It can be deduced from this $\delta{\tau}=\omega_{\alpha_0\cdots\alpha_p}$.

In fact $\tau$ gives a homotopy operator $K$ on the complex, $(K\omega)_{\alpha_0\cdots\alpha_{p-1} }=\sum_\alpha\rho_\alpha\omega_{\alpha\alpha_0\cdots\alpha_{p-1} }$. It satisfies $\delta K+K\delta=\mathrm{id}$. For the upper closed chain, $K$ is said to be the right inverse of $\delta$.

double complex

Remember $K^{p,q}=C^p(\mathcal{U},\Omega^q)=\prod \Omega^q(U_{\alpha_0\cdots\alpha_p})$. The elements above it are called $q$-form values of the $p$-cochain.

Treat $(p,q)$ as the coordinate on $\mathbb{Z}^2$, then there is a commutation graph, The operations on the right side of the row are differential operators, and the operations on the column direction are differential operators. The exchange is because first limiting and then exterior differentiation is equal to first exterior differentiation and then limiting, and exterior differentiation has linearity.

Let $K^n=\oplus_{p+q=n}K^{p,q}$, $D=D'+D''$, $D'=\delta$, $D''=(-1)^pd$. Then $D',D''$ now has sign exchange, so $D^2=0$ and $\{K^\ast ,D\}$ become a complex.

$C^\ast (\mathcal{U},\Omega^\ast )$ is called Čech-de Rham complex, The element above it is abbreviated as $D$ - on the chain, and the cohomology is recorded as $H^\ast _D\{C^\ast (\mathcal{U},\Omega^\ast )\}$.

Augment the chain complex, add $\Omega^p(M)$ at the $(-1,q)$ position, and limit the mapping to the right $r$, Upward is the differential operator. Then add $0$ on the left to map to the right, and then all the rows are consistent with the previous proposition.

Proposition 1.2 (The MV principle of promotion). $r^\ast :H^\ast _{dR}(M)\approx H_D\{C^\ast (\mathcal{U},\Omega^\ast )\}$.

First of all, $r$ is a chain mapping, so $r^\ast $ is reasonably defined, $r^\ast [\omega]_d=[r\omega]_D$.

Next, prove that $r^\ast $ is both full and single. Just use the consistency of the rows to perform basic graph tracking. Go diagonally and cut off the excess components step by step.

Application

It can be seen that if each row of the augmented bicomplex is exact, Then its $D$-cohomology is isomorphic to the cohomology of the initial sequence, that is, the augmented sequence.

Take $C^p(\mathcal{U},\mathbb{R})=\prod\{U_{\alpha_0\cdots\alpha_p}\text{上局部常值函数}\}$, Then there is an inclusion mapping $i:C^p(\mathcal{U},\mathbb{R})\rightarrow C^p(\mathcal{U},\Omega^0)$, And $\operatorname{Im}i=\ker d$.

There is a complex $0\rightarrow C^0(\mathcal{U},\mathbb{R})\xrightarrow{\delta}C^1(\mathcal{U},\mathbb{R})\rightarrow\cdots$, Call cohomology open covering $\mathcal{U}$ Čech cohomology, is a pure composition object.

Similar to the previous proof, if each column is exact, add augmenting rows, Then its cohomology is also isomorphic to the $D$-cohomology of the double complex.

We take the complex constructed above as an augmented row. According to Poincaré’s lemma, If $\mathcal{U}$ is a good cover, then each column does fit exactly. Thus:

Theorem 1.3. $H^\ast (M)\cong H^\ast (\mathcal{U},\mathbb{R})$.

So we have an inference that is not trivial a priori:

Corollary 1.4. Čech cohomology $H^\ast (\mathcal{U},\mathbb{R})$ and good coverage $\mathcal{U}$ have nothing to do with the selection.

On the other hand, if the manifold has finite good coverage, $H^\ast (\mathcal{U},\mathbb{R})$ has finite dimensions. Thus:

Corollary 1.5. The de Rham cohomology $H_{dR}^\ast (M)$ of a finite manifold $M$ is finite-dimensional.

In particular, compact manifolds always have finite good coverage, so de Rham cohomology is finite-dimensional.

The article was last updated on 2021-10-19 21:45:05