Notes for AKT-140228/0:48:42

Proof of the claim:

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

So

[math]\displaystyle{ \begin{align} D_{g^{-1} A g+g^{-1} d g} (s)\\ &=ds+(g^{-1} A g+g^{-1} d g)s\\ &= ds + g^{-1} A g s + g^{-1} (d g) s \\ &=ds + g^{-1} A g s + g^{-1} d(gs)-g^{-1}gds\\ &= g^{-1} A g s + g^{-1} d(gs)\\ &= g^{-1} D_A(gs)\\ &= (D_A)^g(s)\\ \end{align} }[/math]