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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 313 / 2 / a / Solution

Set
$$
\omega=dg\,g^{-1},
\qquad
\widetilde A=gAg^{-1}-\omega.
$$
The right <Maurer-Cartan equation> is $d\omega=\omega\wedge\omega$. Direct substitution gives
$$
\widetilde F=gFg^{-1},
$$
so trace invariance proves $\widetilde C_2=C_2$.

For the <Chern-Simons 3-form>, expansion and graded cyclicity give
$$
Y(\widetilde A)-Y(A)
=\frac1{8\pi^2}d\operatorname{Tr}\bigl(
\omega\wedge gAg^{-1}\bigr)
+\frac1{24\pi^2}\operatorname{Tr}(\omega^3).
$$
Since cyclicity also gives
$$
\operatorname{Tr}(\omega\wedge gAg^{-1})
=\operatorname{Tr}(g^{-1}dg\wedge A),
$$
the required two-form may be chosen as
$$
\boxed{T=\frac1{8\pi^2}\operatorname{Tr}(g^{-1}dg\wedge A)}.
$$
Thus
$$
\boxed{
Y\longmapsto Y+dT
+\frac1{24\pi^2}\operatorname{Tr}[(dg\,g^{-1})^3]}.
$$

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 313 / 2 / b / Solution

{c}
{wiki=Chern–Simons_form}

The Chern-Simons 3-form
$$
Y=\frac1{8\pi^2}\operatorname{Tr}\left(A\wedge dA+\frac23A\wedge A\wedge A\right)
$$
satisfies $dY=C_2$.


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 Incoming links: Chern-Simons 3-form