Infinitesimal generator for conjugation action

Question

Let $G \circlearrowright G$ be a Lie group acting on itself by conjugation: $\varphi_g(h) \doteq ghg^{-1}$. Denote by $\varphi^h$ the other partial map, that is, $\varphi^h(g) = ghg^{-1}$. I take as the definition of the infinitesimal generator of the action associated to $u \in \mathfrak{g}$ the vector field $(u^\#)_h = {\rm d}(\varphi^h)_e(u)$, where $u$ is the identity of $G$.

I have checked that for the action given by left translation, $u^\#$ is the unique right-invariant extension of $u$, and for the action given by right translation we have that $u^\#$ is minus the unique left-invariant extension of $u$.

I do not know how to compute $u^\#$ in this case, though. If $\alpha$ is a curve on $G$ with $\alpha(0) = e$ and $\alpha'(0) = u$, I get stumped at $$ (u^\#)_h = \frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} \alpha(t)h\alpha(t)^{-1},$$since I can't just "factor" something like $L_h$ or $R_h$ like in the previous cases. Help?

Answer 1

I solved it, but I'd expect a simpler solution. Let $m_3\colon G^3 \to G$ denote the multiplication of three elements, in order. Brute force tells us that $${\rm d}(m_3)_{(g_1,g_2,g_3)}(v_1,v_2,v_3) = {\rm d}(R_{g_2g_3})_{g_1}(v_1)+ {\rm d}(L_{g_1}\circ R_{g_3})_{g_2}(v_2)+{\rm d}(L_{g_1g_2})_{g_3}(v_3).$$ If ${\rm inv}\colon G \to G$ denotes inversion, we observe that the relation $$ g^{-1}(ghg^{-1})g = h$$reads as $m_3\circ ({\rm inv}, \varphi^h, {\rm Id}_G) = {\rm constant}$. Differentiating that with respect to $g$ at the identity gives: (after some simplification, using that ${\rm d}({\rm inv})_e = -{\rm Id}_{\mathfrak{g}}$) $$(u^\#)_h= {\rm d}(\varphi^h)_e(u) = {\rm d}(R_h)_e(u) - {\rm d}(L_h)_e(u).$$

I'll leave the question open for awhile in case someone manages to do it without going through the derivative of $m_3$.