# Is $A$ in the Lie algebra of a stabiliser subgroup $G_p$ if and only if $\exp(tA) \in G_p$?

Let $$G$$ be a Lie group that acts smoothly and on the right of a smooth manifold $$M$$ by $$\alpha: M \times G \to M$$. Let $$e \in G$$ be the identity of $$G$$. Let $$p \in M$$. Let $$G_p$$ denote the stabiliser subgroup of $$G$$, $$G_p :=$$ $$\{g \in G$$ $$| \alpha(p,g) = \alpha_p(g) = p\}$$, where $$\alpha_p$$ is the smooth map $$\alpha_p: G \to M$$, known as the orbit map, with $$\alpha_p(g) = \alpha(p,g)$$. Observe the image of $$\alpha_p$$ is $$\alpha_p(G)= \alpha(p,G)$$, the orbit of $$p$$.

Observe that $$\alpha^{-1}(p) = G_p$$. The continuity of $$\alpha$$ gives us the following: Since $$M$$ is a T1 space, since $$M$$ is a T2 space, we have $$G_p$$ to be a closed subset of $$G$$.

It can be shown $$G_p$$ is a subgroup of $$G$$. Since $$G_p$$ is a closed subgroup of $$G$$, it follows by the closed subgroup theorem that $$G_p$$ is not merely a Lie group that is also a subset of $$G$$ but an embedded Lie subgroup of $$G$$. (Also, you can show something like $$\alpha_p$$ is equivariant and thus has constant rank and thus $$G_p$$ is embedded.)

For the inclusion map $$i: G_p \to G$$, we have its differential at $$e$$ to be $$i_{\{*,e\}}: T_e(G_p) \to T_eG$$, an injective $$\mathbb R$$-linear map of $$\mathbb R$$-Lie algebras. The image of $$i_{\{*,e\}}$$ is $$i_{\{*,e\}}(T_e(G_p))$$, an $$\mathbb R$$-vector subspace of $$T_eG$$ and is isomorphic to $$T_e(G_p)$$.

Consider the exponential map $$\exp: T_eG \to G$$. Since $$T_eG$$ is an $$\mathbb R$$- vector space, $$tA \in T_eG$$ for all $$A \in T_eG$$ and for all $$t \in \mathbb R$$. Therefore, the expression '$$\exp(tA)$$' is defined.

Question: For all $$A \in T_eG$$, is $$A \in i_{\{*,e\}}(T_e(G_p))$$ (or $$A \in T_e(G_p)$$ under the aforementioned isomorphism) if and only if for each $$t \in \mathbb R$$, $$\exp(tA) \in G_p$$?

Note: That $$\exp(tA) \in G_p$$ for each $$t \in \mathbb R$$ is I think equivalent to that the map $$s_p : \mathbb R \to G$$, with $$s_p = \exp \circ \hat{A}$$ has image as a subset of $$G_p$$, where $$\hat{A}: \mathbb R \to T_eG$$, $$\hat{A}(t) = tA$$. Also, I believe $$s_p$$ and $$\hat{A}$$ are smooth maps.

It seems like $$(\alpha_p \circ \exp)^{-1}p = i_{\{*,e\}}(T_e(G_p))$$ or something, but I really don't know how to begin proving this. This is supposed to be a lemma in proving that for the fundamental vector field $$\xi(A)$$, of $$A$$ under $$\xi: T_eG \to C^{\infty}(M,TM)$$, we have $$\xi(A)_p = Z_p$$ if and only if $$A \in i_{\{*,e\}}(T_e(G_p))$$, where $$Z_p \in T_pM$$ is the zero element of $$T_pM$$. Also, I'm aware that $$c_p := \alpha_p \circ s_p$$ is the integral curve of $$\xi(A)$$ starting at $$p$$.

My answer: Okay I think I discovered the answer, which is affirmative, and I think I can answer without using, for a second time, the fact that $$G_p$$ is closed.

The 'only if' direction is shown under the naturality of the exponential map, which states that for a Lie group homomorphism $$F: G \to B$$, $$F \circ \exp_B = \exp_G \circ F_{\{*,e\}}$$, where $$\exp_B: T_{e_B} \to B$$ and $$\exp_G: T_eG \to G$$ where $$e_B$$ is the identity of $$B$$.

Here, we have '$$F$$' as $$i$$, '$$B$$' as $$G_p$$. For $$A \in i_{\{*,e\}}(T_e(G_p))$$, let $$C = i_{\{*,e\}}^{-1} A \in T_e(G_p)$$. Then $$(F \circ \exp_B)(C) = (i \circ \exp_{G_p})(C) = \exp_{G_p}(C) \in G_p,$$ and $$(F \circ \exp_B)(C) = (\exp_G \circ i_{\{*,e\}})(C) = \exp_G (A).$$

Therefore, $$\exp_G (A) \in G_p$$ if $$A \in i_{\{*,e\}}(T_e(G_p))$$. This applies for any $$A \in i_{\{*,e\}}(T_e(G_p))$$ including its multiples $$tA \in i_{\{*,e\}}(T_e(G_p))$$, where $$tC = t(i_{\{*,e\}}^{-1} A) = i_{\{*,e\}}^{-1} (tA) \in T_e(G_p)$$.

The 'if' direction is also shown by naturality, I think, but I need to think of this a little more.

The fact that $$A$$ is in the Lie algebra of the stabilizer if and only if $$exp(tA)\in G_p$$ is a consequence of the proof of the Cartan theorem (closed group).
To show this theorem, one shows first, if $$H$$ is a closed subgroup of $$G$$, the Lie algebra of $$H$$ is the elements $$A$$ of the Lie algebra of $$G$$ such that $$exp(tA)\in H$$.
• We wouldn't yet know $H$ is a Lie group, so it might not make sense to speak of its Lie algebra (you could say $H$ is Lie subgroup or $H$ is a subset Lie group that is not (yet) necessarily embedded, I guess), but regardless your point is that $T_eH \cong i_{\{*,e\}}(T_eH) = \exp^{-1}H$, correct? Here, $i: H \to G$ is inclusion and $i_{\{*,e\}}$ is its differential at identity. Wait, how do we know $i$ is smooth if we don't yet know $H$ is embedded submanifold? – Ekhin Taylor R. Wilson Nov 16 '19 at 6:37
• if not, then how do you prove that if $H$ is a closed subgroup of $G$, then the Lie algebra of $H$ is the elements $A$ of the Lie algebra of $G$ such that $exp(tA)\in H$? – Ekhin Taylor R. Wilson Nov 17 '19 at 4:06