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.


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 $M$.

  • $\begingroup$ 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

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.