Notes for AKT-140207/0:42:32

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

${\begin{aligned}\Psi (A+\mathrm {d} f)&=(A+\mathrm {d} f)\wedge \mathrm {d} (A+\mathrm {d} f)\\&=(A+\mathrm {d} A)\wedge \mathrm {d} A,\;\;\;\;\;\mathrm {since\;d\circ d=0} \\&=A\wedge \mathrm {d} A.\end{aligned}}$