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
$$
\boxed{
Y\longmapsto Y+dT
+\frac1{24\pi^2}\operatorname{Tr}[(dg\,g^{-1})^3]}.
$$

Solution

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

Substitute $\rho=\sigma^\alpha T_\alpha$ into the <Maurer-Cartan equation>. Antisymmetry of the wedge product gives
$$
\rho\wedge\rho
=\frac12\sigma^\alpha\wedge\sigma^\beta
[T_\alpha,T_\beta]
=\frac12c^\gamma{}_{\alpha\beta}
\sigma^\alpha\wedge\sigma^\beta T_\gamma.
$$
Equating coefficients of $T_\gamma$ yields
$$
\boxed{d\sigma^\gamma
=-\frac12\sum_{\alpha,\beta}
c^\gamma{}_{\alpha\beta}\sigma^\alpha\wedge\sigma^\beta},
$$
so $\boxed{f^\gamma{}_{\alpha\beta}=-c^\gamma{}_{\alpha\beta}/2}$.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 313 / 3 / b / Solution

{c}

The Maurer-Cartan equation is $d\rho+\rho\wedge\rho=0$ for the left-invariant form $\rho=g^{-1}dg$.


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