interaction of left-invariant vector fields and right-translation on a Lie-Group

Question

Given a Lie-Group $G$ denote the set of left-invariant vector fields on $G$ by $LG$ and denote by $R_g$ the right-translation, i. e. for $g \in G$ define $$R_g \colon C^\infty (G) \to C^\infty (M) \colon f(x) \mapsto f(x g) .$$I am interested in how differentiation along a vector field and right translation behave toward each other. So for $X \in LG$, $f \in C^\infty(G)$ and $g \in G$, how can one express $X\big(R_g f\big)$ in terms of $X (f)$ or $R_g \big( X (f ) \big) $ and so on?

Although I am certain there is a general answer to that question, I would also be glad to have an answer for the special case of $G = \operatorname{SL}_2\big( \mathbb{R}\big)$.

Answer 1

I'm not sure if this fits the bill for what you want, but here is what I can show. Recall that for $g\in G$, conjugation $C_g:G\rightarrow G$ preserves the identity element $e\in G$, so induces a map $(C_g)_\ast: T_e G\rightarrow T_e G$. This map is called $Ad_g$. Let $x = X(e)$.

We have $X(R_g f) = (Ad_{g^{-1}} x) f$.

I'll prove this below. To set up some more notation, I'll use $r_g:G\rightarrow G$ for right translation by $g$. That is, $r_g(h) = hg$ for any $h\in G$. Note then, that your map $R_g$ is nothing but $f\mapsto f\circ r_g$.

Thus, you are really asking about computing $X(f\circ r_g)$. Well, by definition, this is the same as the pushforward $((r_g)_\ast X)(f)$.

But the pushforward of a left invariant vector field under right translation is nice.

Proposition: If $X$ is left invariant, then $(r_g)_\ast X$ is also left invariant.

Proof: Let $l_h:G\rightarrow G$ denote left translation by $h$: $l_h(k) = hk$ for any $k\in G$.

We want to show that $(l_h)_\ast((r_g)_\ast X) = (r_g)_\ast X$. The key observation is that associativity of the group implies that $l_h$ and $r_g$ commute: for any $k\in G$, we have $$l_h(r_g(k)) = l_h(kg) = h(kg) = (hk)g = r_g(hk) = r_g(l_h(k)).$$

Then \begin{align*} (l_h)_\ast((r_g)_\ast)X)&= (l_h \circ r_g)_\ast X\\ &= (r_g \circ l_h)_\ast X\\ &= (r_g)_\ast ((l_h)_\ast X)\\ &= (r_g)_\ast X, \end{align*} where the last equality used the fact that $X$ is left invariant. $\square$

Ok, so $(r_g)_\ast X$ is some left invariant vector field, but which one is it?

Proposition: $(r_g)_\ast X = Ad_{g^{-1}}(x)$.

Proof: Since both are left invariant, it's enough to show they agree at $e\in G$.

Since $X$ is left invariant and $x = X(e)$, for any $h\in G$ we have $X(h) = (l_h)_\ast x$. Then \begin{align*} ((r_g)_\ast X)(e) &= (r_g)_\ast(X(g^{-1}))\\ &= (r_g)_\ast ((l_{g^{-1}})_\ast x) \\ &= (l_{g^{-1}} r_g)_\ast x\\ &= Ad_{g^{-1}} x.\end{align*}