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
\n\\[
\nR^E=(\\nabla^E)^2\\in\\Omega^{\\cdot+2}(M;\\End(E)),
\n\\]
\nit can be viewed as a matrix of 2-forms. What is more, it satisfy the \\emph{Bianchi identity} $[\\nabla^E,R^E]=0$.
\n\u00a0
\nFor 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 \\emph{trace} of $A$ is a smooth function on $M$, denote it by $\\tr[A]$. The function $\\tr[A]$ further induces a map
\n\\begin{align*}
\n\\tr:\\Omega^\\cdot(M;\\End(E))&\\to\\Omega^\\cdot(M),\\\\
\n\\omega\\otimes A&\\mapsto\\tr[A]\\omega,
\n\\end{align*}
\nwhere $\\omega\\in\\Omega^\\cdot(M)$ and $A\\in\\Gamma(\\End(E))$, we still call it the function of trace.<\/p>\n
We also extend the Lie bracket operation on $\\End(E)$ to $\\Omega^\\cdot(M,\\End(E))$ as
\n\\[
\n[A,B]=A\\wedge B-(-1)^{|A||B|}A\\wedge B,
\n\\]
\nwhere $A\\in\\Omega^k(M;\\End(E))$, $B\\in\\Omega^l(M;\\End(E))$ (thus, $|A|=k$, $|B|=l$).<\/p>\n
Now, we turn to proof the two fundamental Lemmas as a preliminary of the Theorem of Chern-Weil.<\/p>\n
\\begin{lem}
\n For any $A,B\\in\\Omega^\\cdot(M;\\End(E))$, we have
\n \\[
\n \\tr[A,B]=0.
\n \\]
\n\\end{lem}
\n\\begin{proof}
\n 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)
\n \\begin{align*}
\n [A,B]&=(\\omega A_0)\\wedge(\\eta B_0)-(-1)^{kl}(\\eta B_0)\\wedge(\\omega A_0)\\\\
\n &=(\\omega\\wedge\\eta)A_0 B_0-(-1)^{kl}(\\eta\\wedge\\omega)(B_0A_0)\\\\
\n &=(\\omega\\wedge\\eta)(A_0B_0-B_0A_0)\\\\
\n &=(\\omega\\wedge\\eta)[A_0,B_0].
\n \\end{align*}
\n\\end{proof}
\n\\begin{lem}\\label{lem:1_3_2}
\n For $A\\in\\Omega^\\cdot(M;\\End(E))$, we have
\n \\[
\n \\tr\\left[[\\nabla^E,A]\\right]=\\rd(\\tr[A]).
\n \\]
\n\\end{lem}
\nBefore we dealing with the proof, Let us recall some facts. Firstly,
\n\\[
\n[\\nabla^E,A]=\\nabla^EA-(-1)^{|A|}A\\nabla^E.
\n\\]
\nIn 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 $s\\in\\Omega^\\cdot(M;\\End(E))$.<\/p>\n
Now, for $\\nabla^E\\mathpunct{:}\\Omega^\\cdot(M;E)\\to\\Omega^{\\cdot+1}(M;E)$, we have
\n\\begin{align*}
\n[\\nabla^E,A](fs)&=(\\nabla^EA-(-1)^{|A|}A\\nabla^E)(fs)\\\\
\n&=\\nabla^E(A(fs))-(-1)^{|A|}A(\\nabla^E(fs))\\\\
\n&=\\nabla^E(f(As))-(-1)^{|A|}A(\\nabla^E(fs))\\\\
\n&=\\rd f\\wedge(As)+f\\nabla^E(As)-(-1)^{|A|}(A(\\rd f\\wedge s+f\\nabla^E s))\\\\
\n&=\\rd f\\wedge(As)-(-1)^{|A|}A(\\rd f\\wedge s)+f\\nabla^E(As)-(-1)^{|A|}fA(\\nabla^E s)\\\\
\n&=f\\cdot(\\nabla^E(As)-(-1)^{|A|}A\\nabla^Es)\\\\
\n&=f\\cdot([\\nabla^E,A]s).
\n\\end{align*}
\nThus, $[\\nabla^E,A]\\in\\Omega^\\cdot(M;\\End(E))$. This show that $\\tr\\left[[\\nabla^E,A]\\right]$ make sense.
\n\\begin{proof}
\n Firstly, if $\\tilde\\nabla^E$ is another connection on $E$, then from the Leibniz rule in the definition of connection, one can verifies that
\n \\[
\n \\nabla^E-\\tilde\\nabla^E\\in\\Omega^\\cdot(M;\\End(E)).
\n \\]
\n Thus, the above Lemma says that
\n \\[
\n \\tr\\left[[\\nabla^E-\\tilde\\nabla^E,A]\\right]=0,
\n \\]
\n that is, the righthand side of the formula in the Lemma is independent on $\\nabla^E$.<\/p>\n
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.<\/p>\n
In fact,
\n \\begin{align*}
\n [\\nabla^E,A]&=[\\rd,A]+[\\omega,A]\\\\
\n &=\\rd\\cdot A-(-1)^{|A|}A\\rd+[\\omega,A],
\n \\end{align*}
\n thus
\n \\[
\n \\tr\\left[[\\nabla^E,A]\\right]=\\tr[\\rd\\cdot A-(-1)^{|A|}A\\rd].
\n \\]
\n Note that
\n \\begin{align*}
\n (\\rd\\cdot A-(-1)^{|A|}A\\rd)s&=\\rd\\cdot(As)-(-1)^{|A|}A(\\rd s)\\\\
\n &=(\\rd A)s+(-1)^{|A|}A\\cdot\\rd s-(-1)^{|A|}A(\\rd s)\\\\
\n &=(\\rd A)s,
\n \\end{align*}
\n thus,
\n \\[
\n \\tr\\left[[\\nabla^E,A]\\right]=\\tr[\\rd A]=\\rd(\\tr[A]).
\n \\]
\n\\end{proof}
\nNow 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 $\\rd(\\tr[f(A)])=0$.<\/p>\n
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 $\\nabla^E$.<\/p>\n
\\emph{The Chern-Weil theory claims that $\\tr\\left[[f(R^E)]\\right]$ is independent on $\\nabla^E$.}<\/p>\n
Before we turn to the proof, let us set some definition
\n\\begin{defn}
\n 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 $f$-\\emph{characteristic classes} of $E$, and $\\tr[f(R^E)]$ the $f$-\\emph{characteristic differential form}, and $\\int_M\\tr[f(R^E)]$ the $f$-\\emph{characteristic number}.
\n\\end{defn}
\nNow, we will prove that the definition is independent on $\\nabla^E$.
\n \\begin{thm}
\n 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
\n \\[
\n \\tr[f(R_0^E)]-\\tr[f(R^E_1)]=\\rd\\omega.
\n \\]
\n \\end{thm}
\n\\begin{quote}
\nthis post is updated, added this proof, since yesterday is too late for me to write it out from my notes.
\n\\end{quote}
\n \\begin{proof}
\n 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
\n \\[
\n \\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,
\n \\]
\n and
\n \\begin{align*}
\n \\int_0^1\\left\\{\\frac{\\rd}{\\rd t}\\tr[f(R_t^E)]\\right\\}\\rd t
\n &=\\int_0^1\\tr\\left[\\frac{\\rd}{\\rd t}\\left(f(R_t^E)\\right)\\right]\\rd t\\\\
\n &=\\int_0^1\\tr\\left[f'(R^E_t)\\cdot\\frac{\\rd R_t^E}{\\rd t}\\right]\\rd t\\\\
\n &=\\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}\\\\
\n &=\\int_0^1\\tr\\left[\\left(\\frac{\\rd \\nabla_t^E}{\\rd t}\\nabla_t^E
\n +\\nabla_t^E\\frac{\\rd\\nabla_t^E}{\\rd t}\\right)f'(R_t^E)\\right]\\rd t\\\\
\n &=\\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}\\\\
\n &\\overset{\\ast}{=}\\int_0^1\\tr\\left[[\\nabla_t^E,\\frac{\\rd\\nabla_t^E}{\\rd t}f'(R_t^E)]\\right]\\rd t\\\\
\n &=\\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}\\\\
\n &=\\rd\\left\\{\\int_0^1\\tr\\left[\\frac{\\rd\\nabla_t^E}{\\rd t}f'(R_t^E)\\right]\\rd t\\right\\},
\n \\end{align*}
\n The stared equality holds, since
\n \\[
\n \\frac{\\rd\\nabla_t^E}{\\rd t}=\\nabla_1^E-\\nabla^E_0\\in\\Omega^\\cdot(M;\\End(E),
\n \\]
\n and
\n \\[
\n [a,bc]=[a,b]c+(-1)^{|a||b|}b[a,c],
\n \\]
\n then apply Bianchi identity.
\n \\end{proof}
\n \\begin{rem}
\n Chern call the integral
\n \\[
\n \\int_0^1\\tr\\left[\\frac{\\rd\\nabla_t^E}{\\rd t}f'(R_t^E)\\right]\\rd t
\n \\]
\n as \\emph{transgressed form}.
\n \\end{rem} <\/p>\n","protected":false},"excerpt":{"rendered":"
Recall that given a vector bundle on $M$ …<\/p>\n