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I am trying to see why is it that if we have a transitive action of a lie group $G\rightarrow M$ then if this action is transitive the connected component of the identity $G^0$ also acts transitively on $M$ and that for all $p\in M$, $G/G_0 \cong G_p/(G_p\cap G^0)$.

Since the action is transitive we know that $G/G_p\cong M$ and since the map $G\rightarrow G_p$ is a submersion we get that the map $G\rightarrow M, g\rightarrow g.p$ is open . I don't know how useful this is since we only know that $G^0$ is a closed set I don't think it has to be open since connected components don't necessarily need to be open, and also I am not sure if this map is even closed or if it is how I could try and prove it.

Any enlightment is appreciated. Thanks in advance.

Someone
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  • There are multiple typos in your question, for instance, it should be Lie, not "lie", $G_p$ instead of $G_0$, the map $G\to Gp$, not $G\to G_p$... – Moishe Kohan Jan 16 '21 at 18:12

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Manifolds (and thus also Lie groups) inherit the property of being locally connected from Euclidean space, and connected components of locally connected spaces are both closed and open. Hence $G_0$ is open in every Lie group.

Further, if we know that for every $q \in M$ the map $\phi_q : G \to M, g \mapsto g \cdot q$ is open, then the orbit $\phi_q(G_0) \subset M$ must be open. Then, fixing a single $p \in M$, the union $\bigcup_{q \in M \setminus \phi_p(G_0) } \phi_q(G_0)$ is open as a union of open sets. But note that $$M = \bigcup_{q \in M} \phi_q(G_0)= \phi_p(G_0) \cup \bigcup_{q \in M \setminus \phi_p(G_0) } \phi_q(G_0),$$ because every $m \in M$ is equal to $e \cdot m \in \phi_m(G_0)$ if $e \in G_0$ is the identity.

But then we have written the connected set $M$ as a disjoint union of open sets $\phi_p(G_0) \cup \bigcup_{q \in M \setminus \phi_p(G_0) } \phi_q(G_0)$, and since the first set is nonempty, this must mean that the second one is empty, and hence $\phi_p(G_0) = M$.

(For academic integrity: Pieced this answer together from the comments to the very similar question here.)

Lukas Miaskiwskyi
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    Ah yes thank you , I had forgotten about that propery of locally connected spaces ! Yes I saw that post but these details was what I was missing . – Someone Jan 16 '21 at 15:54
  • Where did you use that $G$ acts transitively? – Invincible Apr 16 '21 at 11:05
  • That assumption sits in the statement "we know that for every $q \in M$ the map $\phi_q: G \to M, g \mapsto g \cdot q$ is open"; that arises from an argument similar to what's in the original question, but I notice there are some typos there. Transitivity implies for every $p \in M$ that $G/G_p \cong M$, via the map $[g] \mapsto g \cdot p$. Further, the quotient map $G \to G/G_p$ is a submersion, hence automatically an open map. Hence the composition $G \to G/G_p \to M$ is an open map. – Lukas Miaskiwskyi Apr 16 '21 at 13:26