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Let $T_g:=L_g R_{g^{-1}}: G \rightarrow G$ be the standard automorphism of a Lie algebra, then $Ad_g:=DT_g(e): \mathfrak{g} \rightarrow \mathfrak{g} $is called the adjoint representation. Now, I want to show that $[Ad_g \xi, Ad_g \eta] = Ad_g [\xi,\eta].$ Here, $\xi,\eta \in \mathfrak{g}.$

So the goal is to see that $Ad$ respects also the Lie-Bracket. Unfortunately, I don't see how this can be shown. (and I cannot use more properties here, as I only have this basic definitions available).

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  • $\begingroup$ Is your first upper-case $G$ actually meant to be the Lie algebra $\mathfrak g$? And, usually upper-case Ad is $Ad(g):\mathfrak g\to \mathfrak g$ with $g$ in the Lie group. Then the derivative of Ad is lower-case ad, which seems to be what your question is about. Clarify? (At some level, the Jacobi identity is exactly the assertion that lowercase-ad is a Lie algebra hom, without reference to a Lie group.) $\endgroup$ Commented Jul 28, 2015 at 18:42

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By definition $Ad_g=(T_g)_\ast$ and $T_g$ is a diffeomorphism.

Given a diffeomorphism $f$ it's true that $f_\ast[X,Y]=[f_\ast X,f_\ast Y]$ as this answer shows: Pushforward of Lie Bracket

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