Is the application $D(R_p\circ \imath)(e):\mathfrak{h}\rightarrow T_p\mathcal{L}_p$ an isomorphism?

Question

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$ and $H\subseteq G$ a Lie subgroup with Lie subalgebra $\mathfrak{h}$. Consider the right translation $R_p:G\rightarrow G$ given by $R_p(g)=gp$. If $\mathcal{L}_p=R_p(H)$ and $\imath:H\rightarrow G$ is the inclusion we see that $R_p\circ \imath :H\rightarrow R_p(H)$. Is the application $$D(R_p\circ \imath)(e):\mathfrak{h}\rightarrow T_p\mathcal{L}_p,$$ an isomorphism? If that holds, with some aditional hypothesis, I can get a foliation whose leaves will be $\mathcal{L}_p$.

Answer 1

Yes.

Right multiplication by $p$ is a diffeomorphism $r_p\colon G \to G$, so restricts to a diffeomorphism $H \to Hp$. Thus its differential takes any tangent space $T_h H$ isomorphically onto $T_{hp}(Hp)$.

In more detail, right multiplication also bijectively takes curves $\gamma$ in $H$ starting at $e \in G$ to curves $r_p \circ \gamma\colon t \mapsto \gamma(t)p$ in $Hp$ starting at $p$, so to each tangent vector $\gamma'(0) \in \mathfrak{h}$ corresponds the tangent vector $(r_p \circ \gamma)'(0) = (Dr_p)(e)(\gamma'(0))$ in $T_p(Hp)$, and $(Dr_p)(e)$ sends the one to the other.

This happens essentially because of the definition of tangent vectors and the chain rule. I am sweeping your inclusion $\imath$ under the rug, somewhat, however. Closed subgroups of Lie groups turn out to be regular submanifolds, so that questions of smoothness are the same whether or not one composes with the inclusions.