Chern–Simons form

Remarks

The defining property is $d\, \mathrm{CS}(A) = \tr(F \wedge F)$. Check: $d \tr(A \wedge dA) = \tr(dA \wedge dA)$ and $d \tr(A \wedge A \wedge A) = 3 \tr(dA \wedge A \wedge A)$ by cyclicity of the graded trace; while $\tr(F \wedge F) = \tr(dA \wedge dA) + 2 \tr(dA \wedge A \wedge A) + \tr(A^4)$ and $\tr(A^4) = 0$. Example: if $n = 1$, then $A \wedge A = 0$, $F = dA$ and $\mathrm{CS}(A) = A \wedge dA$; on $M = \R^4$ with coordinates $x_1, \dots, x_4$ and $A = x_1 \, dx_2 + x_3 \, dx_4$, one gets $dA = dx_1 \wedge dx_2 + dx_3 \wedge dx_4$, and both $d(A \wedge dA)$ and $F \wedge F$ equal $2 \, dx_1 \wedge dx_2 \wedge dx_3 \wedge dx_4$. Non-example: for non-abelian $A$, the form $\tr(A \wedge dA)$ alone does not have derivative $\tr(F \wedge F)$ in general, since the term $2 \tr(dA \wedge A \wedge A)$ is missing; the cubic term is needed. More generally, let $f$ be an invariant polynomial of degree $k$ as in the Chern–Weil homomorphism, and let $\omega$ be a connection on a principal bundle $P$. Put $\Omega_t = t\, d\omega + \frac{t^2}{2} [\omega \wedge \omega]$ and $Tf(\omega) = k \int_0^1 f(\omega, \Omega_t, \dots, \Omega_t) \, dt$. Then $Tf(\omega)$ is a $(2k-1)$-form on $P$ with $d \, Tf(\omega) = f(\Omega, \dots, \Omega)$. The $3$-form above is the case $f(X, Y) = \tr(XY)$ in a trivialization. Under a gauge transformation $A \mapsto g^{-1} A g + g^{-1} dg$, $\mathrm{CS}(A)$ changes by an exact form plus $-\frac13 \tr\big((g^{-1}dg)^3\big)$, whose integral over a closed $3$-manifold is an integer multiple of $8\pi^2$ for $G = \mathrm{SU}(n)$.