Notes for AKT-140228/0:41:45

'Showing '[math]\displaystyle{ (A^g)^h = A^{(gh)} }[/math]

[math]\displaystyle{ \begin{align} A^{(gh)} &= (gh)^{-1}A(gh) + (gh)^{-1}\mathrm{d}(gh) \\ &= (gh)^{-1}A(gh) + (gh)^{-1}\Big((\mathrm{d}g)h + g(\mathrm{d}h)\Big)\\ &= (gh)^{-1}A(gh) + (gh)^{-1}(\mathrm{d}g)h + (gh)^{-1}g(\mathrm{d}h)\\ &= h^{-1}(g^{-1}Ag)h + h^{-1}(g^{-1}\mathrm{d}g)h + h^{-1}\mathrm{d}h\\ &= h^{-1}\Big(g^{-1}Ag + g^{-1}\mathrm{d}g\Big)h + h^{-1}\mathrm{d}h\\ &= (A^g)^h. \end{align} }[/math]

The equality shows that the action is a group action.