Notes for AKT-140228/0:48:42

Proof of the claim:

$$d(gs) = (dg)s+gds \implies (dg)s = d(gs)-gds$$

So

${\begin{aligned}D_{g^{-1}Ag+g^{-1}dg}(s)\\&=ds+(g^{-1}Ag+g^{-1}dg)s\\&=ds+g^{-1}Ags+g^{-1}(dg)s\\&=ds+g^{-1}Ags+g^{-1}d(gs)-g^{-1}gds\\&=g^{-1}Ags+g^{-1}d(gs)\\&=g^{-1}D_{A}(gs)\\&=(D_{A})^{g}(s)\\\end{aligned}}$