Notes for AKT-140207/0:42:32

Showing [math]\displaystyle{ \Psi(A) = A\wedge \mathrm{d}A }[/math] is invariant under [math]\displaystyle{ A\mapsto A + \mathrm{d}f }[/math]

[math]\displaystyle{ \begin{align} \Psi(A + \mathrm{d}f) &= (A + \mathrm{d}f)\wedge d(A + \mathrm{d}f)\\ &= \end{align} }[/math]