Chern-Weil Theorem

2025 年 4 月
一	二	三	四	五	六	日
	1	2	3	4	5	6
7	8	9	10	11	12	13
14	15	16	17	18	19	20
21	22	23	24	25	26	27
28	29	30

Recall that given a vector bundle on $M$ , there exists (many) connections $\nabla^E\mathpunct{:}\Gamma(E)\to\Gamma(T^\ast M\otimes E)$ , which can be extended to $\nabla^E\mathpunct{:}\Omega^\cdot(M;E)\to\Omega^{\cdot+1}(M;E)$ , and we defined the curvature (operator) $R^E$ of $\nabla^E$ as

$R^E=(\nabla^E)^2\in\Omega^{\cdot+2}(M;\End(E)),$
it can be viewed as a matrix of 2-forms. What is more, it satisfy the Bianchi identity

$[\nabla^E,R^E]=0$ .

For any smooth section

$A$ of the bundle of

$\End(E)$ (which is a vector bundle of fiber

$\End(E_p)$ , where

$E_p$ is the fiber of

$E$ at

$p$ ), the fiberwise trace of

$A$ is a smooth function on

$M$ , denote it by

$\tr[A]$ . The function

$\tr[A]$ further induces a map

$\begin{align*} \tr:\Omega^\cdot(M;\End(E))&\to\Omega^\cdot(M),\\ \omega\otimes A&\mapsto\tr[A]\omega, \end{align*}$
where

$\omega\in\Omega^\cdot(M)$ and

$A\in\Gamma(\End(E))$ , we still call it the function of trace.

We also extend the Lie bracket operation on $\End(E)$ to $\Omega^\cdot(M,\End(E))$ as

$[A,B]=A\wedge B-(-1)^{|A||B|}A\wedge B,$
where

$A\in\Omega^k(M;\End(E))$ ,

$B\in\Omega^l(M;\End(E))$ (thus,

$|A|=k$ ,

$|B|=l$ ).

Now, we turn to proof the two fundamental Lemmas as a preliminary of the Theorem of Chern-Weil.

Lemma 1. For any

$A,B\in\Omega^\cdot(M;\End(E))$ , we have

$\tr[A,B]=0.$

Proof . With out loss of generality, assume that

$A=\omega\otimes A_0$ ,

$B=\eta\otimes B_0$ , where

$\omega\in\Omega^k(M)$ ,

$\eta\in\Omega^l(M)$ ,

$A_0$ ,

$B_0\in\Gamma(\End(E))$ (the general case is just a combination of these terms), sometimes, for simplicity, we also omit the tenser product, denote

$\omega\otimes A_0$ as

$\omega A_0$ . Then (note that

$A_0$ ,

$B_0$ are merely matrix)

$\begin{align*} [A,B]&=(\omega A_0)\wedge(\eta B_0)-(-1)^{kl}(\eta B_0)\wedge(\omega A_0)\\ &=(\omega\wedge\eta)A_0 B_0-(-1)^{kl}(\eta\wedge\omega)(B_0A_0)\\ &=(\omega\wedge\eta)(A_0B_0-B_0A_0)\\ &=(\omega\wedge\eta)[A_0,B_0]. \end{align*}$

Lemma 2. For

$A\in\Omega^\cdot(M;\End(E))$ , we have

$\tr\left[[\nabla^E,A]\right]=\rd(\tr[A]).$

Before we dealing with the proof, Let us recall some facts. Firstly,

$[\nabla^E,A]=\nabla^EA-(-1)^{|A|}A\nabla^E.$
In fact, Recall that, if we define a map

$A\mathpunct{:}\Omega^\cdot(M,E)\to\Omega^\cdot(M;E)$ by

$(As)(p)=A(p)s(p)$ , for

$s\in\Omega^\cdot(M;E)$ , then

$A\in\Omega^\cdot(M;\End(E))$ if and only if

$A(fs)=f(As)$ holds for any

$f\in C^\infty(M)$ and

$s\in\Omega^\cdot(M;\End(E))$ .

Now, for $\nabla^E\mathpunct{:}\Omega^\cdot(M;E)\to\Omega^{\cdot+1}(M;E)$ , we have

$\begin{align*} [\nabla^E,A](fs)&=(\nabla^EA-(-1)^{|A|}A\nabla^E)(fs)\\ &=\nabla^E(A(fs))-(-1)^{|A|}A(\nabla^E(fs))\\ &=\nabla^E(f(As))-(-1)^{|A|}A(\nabla^E(fs))\\ &=\rd f\wedge(As)+f\nabla^E(As)-(-1)^{|A|}(A(\rd f\wedge s+f\nabla^E s))\\ &=\rd f\wedge(As)-(-1)^{|A|}A(\rd f\wedge s)+f\nabla^E(As)-(-1)^{|A|}fA(\nabla^E s)\\ &=f\cdot(\nabla^E(As)-(-1)^{|A|}A\nabla^Es)\\ &=f\cdot([\nabla^E,A]s). \end{align*}$
Thus,

$[\nabla^E,A]\in\Omega^\cdot(M;\End(E))$ . This show that

$\tr\left[[\nabla^E,A]\right]$ make sense.

Proof . Firstly, if

$\tilde\nabla^E$ is another connection on

$E$ , then from the Leibniz rule in the definition of connection, one can verifies that

$\nabla^E-\tilde\nabla^E\in\Omega^\cdot(M;\End(E)).$
Thus, the above Lemma says that

$\tr\left[[\nabla^E-\tilde\nabla^E,A]\right]=0,$
that is, the righthand side of the formula in the Lemma is independent on

$\nabla^E$ .

Since the righthand side is a local operator ( $\nabla^E$ is local), we can assume that $E$ is trivial, and take a connection as $\nabla^E=\rd+\omega$ for some $\omega\in\Omega^\cdot(M;\End(E))$ to verify that the formula holds.

In fact,

$\begin{align*} [\nabla^E,A]&=[\rd,A]+[\omega,A]\\ &=\rd\cdot A-(-1)^{|A|}A\rd+[\omega,A], \end{align*}$
thus

$\tr\left[[\nabla^E,A]\right]=\tr[\rd\cdot A-(-1)^{|A|}A\rd].$
Note that

$\begin{align*} (\rd\cdot A-(-1)^{|A|}A\rd)s&=\rd\cdot(As)-(-1)^{|A|}A(\rd s)\\ &=(\rd A)s+(-1)^{|A|}A\cdot\rd s-(-1)^{|A|}A(\rd s)\\ &=(\rd A)s, \end{align*}$
thus,

$\tr\left[[\nabla^E,A]\right]=\tr[\rd A]=\rd(\tr[A]).$

Now we have

$\tr\left[[\nabla^E,A]\right]=\rd(\tr[A])$ , thus, if

$[\nabla^E,A]=0$ , for example, take

$A=R^E$ , then

$\tr[A]$ is closed. This shows that we can find closed forms by this method. Clearly, if

$[\nabla^E,A]=0$ then

$[\nabla^E,(A)^k]=0$ , thus

$\rd(\tr[A^k])=0$ , and similarly, for any power series

$f(x)$ , we have

$[\nabla,f(A)]=0$ , thus

$\rd(\tr[f(A)])=0$ .

The above analysis shows that $[\tr[f(A)]]$ is an element of de Rham cohomology $H_{dR}^\cdot(M;E)$ , it seems depend on $M$ , $E$ and $\nabla^E$ , while our invariant quantity of $E$ should be independent on the connection $\nabla^E$ .

The Chern-Weil theory claims that $\tr\left[[f(R^E)]\right]$ is independent on $\nabla^E$ .

Before we turn to the proof, let us set some definition

Definition 3. Suppose

$R^E$ is the curvature of

$\nabla^E$ , for any power series

$f$ ,

$\tr[f(R^E)]$ is a closed form, thus

$\left[\tr[f(R^E)\right]\in H_{dR}^\cdot(M;\C)$ . We call

$\left[\tr[f(R^E)\right]$ the

$f$ -characteristic classes of

$E$ , and

$\tr[f(R^E)]$ the

$f$ -characteristic differential form, and

$\int_M\tr[f(R^E)]$ the

$f$ -characteristic number.

Now, we will prove that the definition is independent on

$\nabla^E$ .

Theorem 4. If

$\nabla_0^E$ and

$\nabla_1^E$ are two connections on

$E$ and the associated curvature are

$R_0^E$ and

$R^E_1$ , respectively, Then there is a differential form

$\omega\in\Omega^\cdot(M)$ such that

$\tr[f(R_0^E)]-\tr[f(R^E_1)]=\rd\omega.$

this post is updated, added this proof, since yesterday is too late for me to write it out from my notes.

Proof . Define

$\nabla^E_t=(1-t)\nabla_0^E+t\nabla_1^E$ , then you can verify that it is still a connection on

$E$ for

$t\in[0,1]$ . Set

$R_t^E=(\nabla_t^E)^2$ , then

$\tr[f(R_t^E)]$ is a closed form. Note that

$\tr[f(R_1^E)]-\tr[f(R_0^E)]=\int_0^1\left\{\frac{\rd}{\rd t}\tr[f(R_t^E)]\right\}\rd t,$
and

$\begin{align*} \int_0^1\left\{\frac{\rd}{\rd t}\tr[f(R_t^E)]\right\}\rd t &=\int_0^1\tr\left[\frac{\rd}{\rd t}\left(f(R_t^E)\right)\right]\rd t\\ &=\int_0^1\tr\left[f'(R^E_t)\cdot\frac{\rd R_t^E}{\rd t}\right]\rd t\\ &=\int_0^1\tr\left[\frac{\rd R_t^E}{\rd t}\cdot f'(R_t^E)\right]\rd t\quad\text{they are just matrix}\\ &=\int_0^1\tr\left[\left(\frac{\rd \nabla_t^E}{\rd t}\nabla_t^E +\nabla_t^E\frac{\rd\nabla_t^E}{\rd t}\right)f'(R_t^E)\right]\rd t\\ &=\int_0^1\tr\left[[\nabla_t^E,\frac{\rd\nabla_t^E}{\rd t}]f'(R_t^E)\right]\rd t\quad\text{$R_t^E$ is a matrix of 2-forms}\\ &\overset{\ast}{=}\int_0^1\tr\left[[\nabla_t^E,\frac{\rd\nabla_t^E}{\rd t}f'(R_t^E)]\right]\rd t\\ &=\int_0^1\rd\left(\tr\left[\frac{\rd\nabla_t^E}{\rd t}f'(R_t^E)\right]\right)\rd t\quad\text{by Lemma 2}\\ &=\rd\left\{\int_0^1\tr\left[\frac{\rd\nabla_t^E}{\rd t}f'(R_t^E)\right]\rd t\right\}, \end{align*}$
The stared equality holds, since

$\frac{\rd\nabla_t^E}{\rd t}=\nabla_1^E-\nabla^E_0\in\Omega^\cdot(M;\End(E),$
and

$[a,bc]=[a,b]c+(-1)^{|a||b|}b[a,c],$
then apply Bianchi identity.

Remark 1. Chern call the integral

$\int_0^1\tr\left[\frac{\rd\nabla_t^E}{\rd t}f'(R_t^E)\right]\rd t$
as transgressed form.

本作品采用创作共用版权协议, 要求署名、非商业用途和保持一致. 转载本站内容必须也遵循署名-非商业用途-保持一致的创作共用协议.

如果脚本另外用到了python脚本产生的…

如果用vim的话, 结合Ultsnip插…

小节的话可以用这种结构： (bookma…

不要复制粘贴, 我网站上的代码他们用的是…

没法设置唉，老是弹出来string pa…

《 “Chern-Weil Theorem” 》有 2 条评论

游神

12/02/2012

这就是张伟平的《Lectures on Chern-Weil Theory and Witten Deformations》上的内容呀。

回复
1. 艾子
  
  12/03/2012
  
  的确, 这是冯老师讲课的笔记.
  
  回复

回复艾子取消回复

此站点使用Akismet来减少垃圾评论。了解我们如何处理您的评论数据。

小小泪

Latest news

Tags

Comments

Other news

使用Chrome播放本地SWF文件

Chrome下载完成后显示病毒扫描失败的解决办法

C1驾照学习经验