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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?

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1 Answer 1

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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.

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