Gavin.hurd: Created page with "Proof of the claim: $$d(gs) = (dg)s+gds \implies (dg)s = d(gs)-gds$$ So $\begin{align} D_{g^{-1} A g+g^{-1} d g} (s)\\ &=ds+(g^{-1} A g+g^{-1} d g)s\\ &= ds + g..."$

2018-08-23T04:15:59Z

Created page with "Proof of the claim: $$d(gs) = (dg)s+gds \implies (dg)s = d(gs)-gds$$ So <center><math>\begin{align} D_{g^{-1} A g+g^{-1} d g} (s)\\ &=ds+(g^{-1} A g+g^{-1} d g)s\\ &= ds + g..."

New page

Proof of the claim:

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

So
<center><math>\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></center>

Notes for AKT-140228/0:48:42 - Revision history

Gavin.hurd: Created page with "Proof of the claim: $$d(gs) = (dg)s+gds \implies (dg)s = d(gs)-gds$$ So \begin{align} D_{g^{-1} A g+g^{-1} d g} (s)\\ &=ds+(g^{-1} A g+g^{-1} d g)s\\ &= ds + g..."

Gavin.hurd: Created page with "Proof of the claim: $$d(gs) = (dg)s+gds \implies (dg)s = d(gs)-gds$$ So $\begin{align} D_{g^{-1} A g+g^{-1} d g} (s)\\ &=ds+(g^{-1} A g+g^{-1} d g)s\\ &= ds + g..."$