= Solution
For the curvature $F=dA+A\wedge A$, define the <Second Chern form>
$$
\boxed{C_2=\frac1{8\pi^2}\operatorname{Tr}(F\wedge F)}.
$$
Using the graded cyclicity of the trace and $d^2=0$,
$$
d\operatorname{Tr}(A\wedge dA)
=\operatorname{Tr}(dA\wedge dA),
$$
while
$$
d\operatorname{Tr}(A\wedge A\wedge A)
=3\operatorname{Tr}(dA\wedge A\wedge A).
$$
Expanding $\operatorname{Tr}(F\wedge F)$ gives the same two terms with coefficient two on $dA\wedge A\wedge A$; the quartic term has vanishing trace by graded cyclicity. Therefore
$$
C_2=dY,
\qquad
\boxed{Y=\frac1{8\pi^2}\operatorname{Tr}\left(
A\wedge dA+\frac23A\wedge A\wedge A\right)},
$$
so $\boxed{\alpha=2/3}$. This is the <Chern-Simons 3-form>.
Solved by gpt-5.6-sol high.
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