The Chern-Simons Transgressed Form

In the end of the proof of theorem1.9, we have
\[
tr \left[ f(R^E)\right]-tr \left[ f(\tilde{R}^E)\right]=-d\int_0^1tr \left[ \frac{d\nabla_t^E}{dt}f'(R_t^E)\right]dt.
\]
The transgressed term
\[
-d\int_0^1tr \left[ \frac{d\nabla_t^E}{dt}f'(R_t^E)\right]dt
\]
is usually called a Chern-Simons term. In many interesting cases, it is a closed form. Of course mathematician interests this case.

例子 1. Let $M$ be a compact oriented 3-dimension manifold. Then $TM$ is trivial bundle[Stiefel]. One can choose a fixed global basis $e_1$, $e_2$, $e_3$ of $TM$. Then
\begin{gather*}
\big(\forall X \in \Gamma(TM)\big)\quad \exists f_1,f_2,f_3 \in C^{\infty}(M)\\s.t.\qquad X=f_1e_1+f_2e_2+f_3e_3.
\end{gather*}

Let $d^{TM}$ denote the trivial connection on $TM$ defined by
\[
d^{TM}(f_1e_1+f_2e_2+f_3e_3)=df_1\cdot e_1+df_2\cdot e_2+df_3\cdot e_3.
\]
Then any connection $\nabla^{TM}$ on $TM$ can be written as
\[
\nabla^{TM}= d^{TM}+A, \quad A\in \Omega^1(M; End(M)).
\]
Set
\[
\nabla^{TM}_t=d^{TM}+tA,\quad t\in[0,1].
\]
And, take $f(x)=-x^2$. By $\dim(M)=3$ and $(R^E)^2 \in \Omega^4(M)=\{0\}$, one can obtain $(R^E)^2=0$.
Then
\begin{align*}
0=&tr \left[ f(d^{TM})\right]-tr \left[ f(R^{TM})\right]=-d\int_0^1tr \left[ \frac{d\nabla_t^{TM}}{dt}f'(R_t^{TM})\right]dt\\
=&-d\int_0^1tr\left[A\left(-2(d^{TM}+tA)^2 \right) \right]dt\\
=&2d\int_0^1tr\left[A\wedge\left(t(d^{TM}\circ A+A\circ d^{TM})+t^2A\wedge A \right) \right]dt\\
=&2d\int_0^1tr\left[tA\wedge(d^{TM}A)+t^2A\wedge A\wedge A\right]dt\\
=&d\left\{tr\left[A\wedge (d^{TM}A)+\frac{2}{3}A\wedge A\wedge A\right]\right\}.
\end{align*}
The term ‘$tr\left[A\wedge (d^{TM}A)+\frac{2}{3}A\wedge A\wedge A\right]$’ is (up to a constant) the Chern-simons form.

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