Notes for AKT-140207/0:42:32

Showing $\Psi (A)=A\wedge \mathrm {d} A$ is invariant under $A\mapsto A+\mathrm {d} f$

${\begin{aligned}\Psi (A+\mathrm {d} f)&=(A+\mathrm {d} f)\wedge d(A+\mathrm {d} f)\\&=\end{aligned}}$