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.


2 Answers 2


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, 2013 at 12:06
  • $\begingroup$ Yes, it is the differential at the identity. $\endgroup$
    – user99914
    Nov 27, 2013 at 22:23

I think it is just a choice of name to call a Lie algebra morphism $\mathfrak{g}\rightarrow \mathfrak{X}(M)$ to be an action of $\mathfrak{g}$ on $M$. Question is,

Why is this a reasonable name?

Recall : Given a Lie group $G$ and an action of $G$ on $M$ (say by the map $G\times M\rightarrow M$), we have a morphism of Lie algebras associated to this action, $\mathfrak{g}\rightarrow \mathfrak{X}(M)$ given by $A\mapsto A^*$ where $A^*:M\rightarrow TM $ is given by $ m\mapsto (\delta_m)_{*,e}(A)$. So, each action of Lie group $G$ on a manifold $M$ gives a Liegroup homomorphism $\mathfrak{g}\rightarrow \mathfrak{X}(M)$.

Then, author in the book you quoted proves (Fundamental Theorem on Lie Algebra Actions) that, if $\mathfrak{g}=Lie(G)$ for some simply connected Lie group $G$, a special morphism of Lie algebras $\mathfrak{g}\rightarrow \mathfrak{X}(M)$ should come from an action of the Lie group $G$ on $M$ in the sense as mentioned above. Thus, it is reasonable to call an arbitray map $\mathfrak{g}\rightarrow \mathfrak{X}(M)$ as action of $\mathfrak{g}$ on $M$.


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