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.


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.