# Lie algebra homomorphism and action on a manifold

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 $M$.

• 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 $e\in G$? – gofvonx Nov 27 '13 at 12:06
• Yes, it is the differential at the identity. – user99914 Nov 27 '13 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 $$M$$.