Can we discribe a Lie group action from some local property?

Question

Let G be a Lie group,and it acts on a smooth manifold M.Then can we get that the action is transitive from some local property of the Lie group action.More precisely,Can we get the action is transitive by research the action on a neighbourhood of some point in M or a neighbourhood of some element in G.Is there some similar result?.Thank you.

Answer 1

Just knowing about the action in a neighborhood of one point of the manifold tells you nothing about whether the action is transitive. For example, consider the standard action of $\operatorname{GL}(n,\mathbb R)$ on $\mathbb R^n$. This action is not transitive, because it fixes the origin. On the other hand, we can restrict the action to $\mathbb R^n\smallsetminus\{0\}$, and then it becomes transitive. But these actions are indistinguishable if we look only in a small neighborhood of a particular point of $\mathbb R^n\smallsetminus\{0\}$.

On the other hand, if you know something local about the action in a neighborhood of every point, then sometimes something can be said. For example, given a smooth action of a Lie group $G$ on a smooth manifold $M$, for each point $p\in M$ we get a linear map $\phi_p\colon \operatorname{Lie}(G)\to T_pM$ by $$ \phi_p(\xi) = \left.\frac{d}{dt}\right|_{t=0} (\exp t\xi)\cdot p. $$

Theorem. If $M$ is connected and $\phi_p$ is surjective for each $p\in M$, then the action of $G$ is transitive.

The proof basically boils down to showing that each orbit is open. (This is problem 20-13 in my Introduction to Smooth Manifolds, 2nd ed.)