# Chern-Weil Theorem

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.

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

\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*}

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

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.

$\nabla^E-\tilde\nabla^E\in\Omega^\cdot(M;\End(E)).$
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

Now, we will prove that the definition is independent on $\nabla^E$.

$\tr[f(R_0^E)]-\tr[f(R^E_1)]=\rd\omega.$
this post is updated, added this proof, since yesterday is too late for me to write it out from my notes.

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

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