Notes for AKT-140228/0:41:45: Difference between revisions

Latest revision as of 17:02, 26 July 2018

'Showing ' $(A^{g})^{h}=A^{(gh)}$

{\begin{aligned}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{aligned}}

The equality shows that the action is a group action.

@@ Line 1: / Line 1: @@
-'''<math>(A^g)^h = A^{(gh)}</math> ?'''
+'''Showing ''''''<math>(A^g)^h = A^{(gh)}</math> '''
 <center><math>\begin{align}
-(A^g)^h &= h^{-1}(g^{-1}Ag + g^{-1}\mathrm{d}g)h + h^{-1}\mathrm{d}h \\
+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></center>
+The equality shows that the action is a group action.