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 $p$.

Thanks in advance!

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 $exp(tA)\in H$.


  • $\begingroup$ 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? $\endgroup$ – Ekhin Taylor R. Wilson Nov 16 '19 at 6:37
  • $\begingroup$ Tsemo Aristide, are you referring to the naturality of the exponential map? Please see my answer in the edit. $\endgroup$ – Ekhin Taylor R. Wilson Nov 16 '19 at 7:19
  • $\begingroup$ 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 $exp(tA)\in H$? $\endgroup$ – Ekhin Taylor R. Wilson Nov 17 '19 at 4:06

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