A differential $d\rho$ of a Lie homomorphism $\rho:G\to H$ commutes with the Lie bracket. Some clarification.

Question

Let $G,H$ be Lie groups, $\rho:G\to H$ be a Lie group homomorphism, and $\psi_g:G\to G$ be an automorphism given by $\psi_g(a)=gag^{-1}$. Consider $X,Y\in T_eG$. Then we define a Lie bracket $[X,Y]:=ad(X)(Y)$ where $ad$ is the differential at the identity of the map $$Ad:G\to Aut(T_eG)$$ where $Ad(g)$ is equal to $d(\psi_g)_e$.

I want to show that $(d\rho)_e([X,Y])=[(d\rho)_e(X),(d\rho)_e(Y)]$ i.e. $$(d\rho)_e(ad(X)(Y))=ad((d\rho)_e(X))((d\rho)_e(Y)).$$

Somehow, it follows from the fact that $\rho$ and $d\rho$ respect the adjoint representation, but I don't see it directly.

Any comment/clarification/hints will be greatly appreciated.

Answer 1

The fact $f^*[X,Y]=[f^*X,f^*Y]$ is true for every differentiable map defined on a manifold.

Pushforward of Lie Bracket

Answer 2

As Tsemo Aristide points out, in terms of left-invariant vector field it follows from a very general rule for Lie brackets of $f$-related fields.

If you want to establish this directly from the $\text{ad}$ definition, you can do so in a few steps. I'll use tildes to denote objects in $H$.

First, we note that conjugation maps commute with $\rho$, i.e. $\rho(\psi_ga)=\widetilde{\psi}_{\rho(g)}\rho(a)$, or, alternately $$ \rho\circ\psi=\widetilde{\psi}\circ(\rho\times\rho) $$ This follows directly from the fact that $\rho$ is a group homomorphism. Thinking of both sides of the above equation as maps $G\times G\to H$ and differentiating with respect to the second argument, we have an equality of maps $G\times T_eG\to T_eH$ $$ d_e\rho\circ\text{Ad}(g)=\widetilde{Ad}(\rho(g))\circ d_e\rho $$ We can equivalently interpret these as maps $G\to\text{Hom}(T_eG,T_eH)$ where $\text{Hom}$ here denotes the space of linear maps. Differentiating this expression again, we have $$ d_e\rho\circ\text{ad}(v)=\widetilde{\text{ad}}(d_e\rho(v))\circ d_e\rho $$ Were we use the fact that, under the canonical identification $T_A\text{Hom}(T_eG,T_eH)\cong\text{Hom}(T_eG,T_eH)$ the differential of left/right multiplication by a fixed matrix is left/right multiplication by the same matrix. Adding a dummy vuriable $u\in T_eG$, we arrive at the desired equality. $$ d_e\rho(\text{ad}(v)(u))=\widetilde{\text{ad}}(d_e\rho(v))(d_e\rho(u)) $$