Notes for AKT-140228/0:41:45 - Revision history

Leo algknt at 22:02, 26 July 2018

2018-07-26T22:02:44Z

← Older revision		Revision as of 18:02, 26 July 2018
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	'''<math>(A^g)^h = A^{(gh)}</math> ?'''		'''Showing ''''''<math>(A^g)^h = A^{(gh)}</math> '''

Leo algknt at 21:38, 26 July 2018

2018-07-26T21:38:46Z

← Older revision		Revision as of 17:38, 26 July 2018
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	<center><math>\begin{align}		<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}\mathrm{d}~~(gh~~) + ~~h^{-1}~~\mathrm{d}h\\		&= (gh)^{-1}A(gh) + (gh)^{-1}\Big((\mathrm{d}g)h + g(\mathrm{d}h)\Big)\\
	&= A^{(gh)} + h^{-1}\mathrm{d}h		&= (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}		\end{align}
	</math></center>		</math></center>

			The equality shows that the action is a group action.

Leo algknt at 19:11, 18 July 2018

2018-07-18T19:11:44Z

← Older revision		Revision as of 15:11, 18 July 2018
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	<center><math>\begin{align}		<center><math>\begin{align}
	(A^g)^h &= h^{-1}(g^{-1}Ag + g^{-1}\mathrm{d}g)h + h^{-1}\mathrm{d}h \\		(A^g)^h &= h^{-1}(g^{-1}Ag + g^{-1}\mathrm{d}g)h + h^{-1}\mathrm{d}h \\
			&= (gh)^{-1}A(gh) + (gh)^{-1}\mathrm{d}(gh) + h^{-1}\mathrm{d}h\\
	&=
			&= A^{(gh)} + h^{-1}\mathrm{d}h
	\end{align}		\end{align}
	</math></center>		</math></center>

Leo algknt: Created page with "''' $(A^g)^h = A^{(gh)}$ ?''' $\begin{align} (A^g)^h &= h^{-1}(g^{-1}Ag + g^{-1}\mathrm{d}g)h + h^{-1}\mathrm{d}h \\ &= \end{align}$ "

2018-07-18T19:07:09Z

Created page with "'''<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 \\ &= \end{align} </math></center>"

New page

'''<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 \\
&=
\end{align}
</math></center>

Notes for AKT-140228/0:41:45 - Revision history

Leo algknt at 22:02, 26 July 2018

Leo algknt at 21:38, 26 July 2018

Leo algknt at 19:11, 18 July 2018

Leo algknt: Created page with "'''(A^g)^h = A^{(gh)} ?''' \begin{align} (A^g)^h &= h^{-1}(g^{-1}Ag + g^{-1}\mathrm{d}g)h + h^{-1}\mathrm{d}h \\ &= \end{align} "

Leo algknt: Created page with "''' $(A^g)^h = A^{(gh)}$ ?''' $\begin{align} (A^g)^h &= h^{-1}(g^{-1}Ag + g^{-1}\mathrm{d}g)h + h^{-1}\mathrm{d}h \\ &= \end{align}$ "