# Inﬁnitesimal Generators of Lie Group Actions

If we have a smooth right action $$(p,g)\mapsto p\cdot g$$ of a Lie group $$G$$ on a smooth manifold $$M$$ then each $$X\in \text{Lie} (G)$$ induces a smooth global ﬂow on $$M$$ via $$\theta(t,p): (t,p)\mapsto p\cdot \exp tX.$$ Let $$\hat X$$ be the infinitesimal generator of this flow, so that $$\hat X_p=\theta'(0,p).$$ Finally, define $$\hat \theta$$ to be the map that sends $$X$$ to $$\hat X.\ \hat \theta$$ is called the infinitesimal generator of the action.

The theorem (Lee's proof in his Introduction to Smooth Manifolds) is a converse of this statement. It says that if $$\hat \theta:\text{Lie}(G)\to \mathfrak X(M)$$ is a homomorphism such that $$\hat \theta (X)$$ is complete (its flow exists for all time) for every $$X\in \text{Lie}(G)$$ then there is a unique smooth right $$G$$-action on $$M$$ whose inﬁnitesimal generator is $$\hat \theta.$$

Set $$\hat \theta (X)=\hat X$$ and let $$\eta_{\hat X}$$ be the flow of $$\hat X$$. The conclusion of the theorem is that there is an action as advertised, given by $$p\cdot g=\eta_{\hat X}(1,p)$$ for $$g=\exp X$$ in a neighborhood of $$e$$, which is enough since every element of $$G$$ can be expressed as a ﬁnite product of elements of the form $$\exp X$$ (use the fact that $$\exp$$ is a local diffeomorphism).

So far so good.

The claim now is that $$\hat \theta$$ is the inﬁnitesimal generator of the action.

Lee says it's an "immediate consequence" of a particular line in the proof, namely

$$p\cdot g=\eta_{\hat X}(1,p)$$.

I have to show that $$\hat X_p=\frac{d(p\cdot \exp tX)}{dt}|_{t=0}=\frac{d \eta_{t\hat X}(1,p)}{dt}|_{t=0}$$

How do I calculate this derivative?

It suffices to show that $$\eta_{t\hat X}(1,p)=\eta_{\hat X}(t,p)$$ for all $$t\in \mathbb R.$$ We will appeal to the uniqueness of integral curves through $$p\in G.$$
$$\gamma:\mathbb R\to G$$ defined by $$\gamma(s)=\eta_{t\hat X}(s,p)$$ is an intgral curve of $$t\hat X$$ starting at $$p$$.
Consider $$\sigma:s\mapsto \eta_{\hat X}(ts,p).$$ Then, $$\sigma(0)=p$$ and $$\frac{d\sigma}{ds}=\frac{td(\eta_{\hat X}(ts,p))}{ds}=t\hat X_{\eta_{\hat X}(ts,p)}=t\hat X_{\sigma(s)}$$, which means that $$\sigma$$ is an integral curve of $$t\hat X$$ starting at $$p.$$
The conclusion is then that $$\eta_{t\hat X}(s,p)=\eta_{\hat X}(ts,p)$$ and setting $$s=1$$ finishes the proof.