2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $exp(tA)\in H$.

https://en.wikipedia.org/wiki/Closed-subgroup_theorem

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.