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In Introduction to smooth manifolds Lee says on page 527:

If $\mathfrak{g}$ is an arbitrary finite-dimensional Lie algebra, any Lie algebra homomorphism $\hat{\theta}:\mathfrak{g}\to\mathfrak{X}(M)$ is called a (right) $\mathfrak{g}$-action on $M$. [$M$ is supposed to be a smooth manifold.]

Basically, I want to understand how fundamental vector fields $\hat{X}\in\mathfrak{X}(M)$, given by $$\hat{X}(p)=\left.\frac{\mathrm{d}}{\mathrm{d}t}p\exp(tX)\right|_{t=0}$$ for $X\in\mathfrak{g}$ and $p\in M$, "describe the infinitesimal behaviour of a smooth Lie group action on a smooth manifold" (source:Wikipedia). While I have managed to understand that the map $X\in\mathfrak{g}\mapsto\hat{X}\in\mathfrak{X}(M)$ is a Lie-Algebra homomorphism, I struggle to understand how it is justified to call a Lie algebra homomorphism an action on $M$.

After a short search I haven't found anything helpful concerning this particular question. My questions seems to be related to that question on the Lie-Palais theorem. Since I am pretty new to Lie theory, I would be glad about a clarification and/or any references on this (particular) subject.

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If you have a Lie group action on $M$, then you have a homomorphism $G \to \text{diff}(M)$, the group of diffeomorphism of $M$. The differential of this map is the map $\hat \theta$. In this sense it seems natural to call $\hat\theta$ an infinitesimal action.

On the other hand, if you are given $\hat\theta$, then for any $X\in \mathfrak g$ and $p\in M$, let $Xp = \phi_1(p)$, where $\phi_t$ is the one parameter group of diffeomorphism defined by $\hat \theta(X)$ (assume that $M$ is compact such that $\phi_t$ exists.) Hence there is an action of $\mathfrak g$ on $M$.

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  • $\begingroup$ Thanks.I don't know anything about the group of diffeomorphisms $\operatorname{diff}(M)$ nor about it's Lie algebra. How do you see that the differential of the homomorphism $G\to\operatorname{diff}(M)$ is $\hat{\theta}$? Anyway, shouldn't it be the differential at the identity $e\in G$? $\endgroup$ – gofvonx Nov 27 '13 at 12:06
  • $\begingroup$ Yes, it is the differential at the identity. $\endgroup$ – user99914 Nov 27 '13 at 22:23

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