Given a Lie group $G$ with Lie algebra $\mathfrak{g}$, the adjoint action $Ad:G \rightarrow GL(\mathfrak{g})$ is given as the differential of the conjugation map in the unit element.

Given two curves $\gamma: I \rightarrow G$ and $V:I \rightarrow \mathfrak{g}$.

What is $\frac{d}{dt} Ad (\gamma(t)) V(t)$?

I think the answer should be $$\frac{d}{dt} Ad (\gamma(t)) V(t) = ad \left( dL_{\gamma(t)^{-1}}\gamma'(t) \right) V(t)+Ad(\gamma(t))V'(t),$$ but I am not sure.

$Ad$ is a special case of a map $f:M \times N \rightarrow P$ from a product manifold into some other manifold. For $(m,n) \in M \times N$ and $V \in T_mM$, $W \in T_nN$ we have (remember $T(M\times N) \simeq TM \times TN$): $$ df_{(m,n)}(V,W) = df_{(m,n)}(V,0)+df_{(m,n)}(0,W) \\ = d\left( f \mid_{M \times \{n\}} \right)(W,0)+d\left( f \mid_{\{m\} \times N} \right)(0,W). $$

In the case of $Ad$: $$\frac{d}{dt} Ad (\gamma(t)) V(t) \\ = d Ad_{(\gamma(t),V(t))} (\gamma \times V)'(t) \\ = d Ad_{(\gamma(t),V(t))} (\gamma \times const)'(t) + d Ad_{(\gamma(t),V(t))} (const \times V)'(t) \\ = \frac{d}{ds} \mid_{s=t} Ad (\gamma (s)) V(t) + \frac{d}{ds} \mid_{s=t} Ad (\gamma (t)) V(s) \\ = \frac{d}{ds} \mid_{s=t} Ad (\gamma (s)) V(t) + Ad (\gamma(t))V'(t). $$ But why is $\frac{d}{ds} \mid_{s=t} Ad (\gamma (s)) V(t)=ad (dL_{\gamma^{-1}(t)} \gamma'(t)) V(t)$? Or is this the case anyway?


My suspected answer was wrong. The correct formulae are $$ \frac{d}{ds} \mid_{s=t} Ad(\gamma(s)) V(t) = Ad(\gamma(t)) ad(dL_{\gamma^{-1}(t)} \gamma'(t)) V(t) $$ and $$ \frac{d}{dt} Ad(\gamma(t))(V(t)) = Ad(\gamma(t)) ad(dL_{\gamma^{-1}(t)} \gamma'(t)) V(t) + Ad(\gamma(t))V'(t) $$ respectively. The second line follows from the first as mentioned in my initial post. To prove the first line:

$\frac{d}{ds} \mid_{s=t} Ad(\gamma(s)) V(t)$

$=\frac{d}{ds} \mid_{s=t} dL_{\gamma(s)}dR_{\gamma^{-1}(s)} V(t)$

$=\frac{d}{ds} \mid_{s=t} dL_{\gamma(t)} dR_{\gamma^{-1}(t)} dL_{\gamma^{-1}(t)\gamma(s)} dR_{\gamma^{-1}(s) \gamma(t)} V(t)$

$=\frac{d}{ds} \mid_{s=t} dL_{\gamma(t)} dR_{\gamma^{-1}(t)} Ad (L_{\gamma^{-1}(t)} \gamma(s)) V(t)$

$=Ad(\gamma(t)) ad(dL_{\gamma^{-1}(t)} \gamma'(t)) V(t)$.


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