Fundamental Theorem on Lie Algebra Actions

All we’re going to do now is try to take what we did in the last couple of posts and generalize. So instead of working on Lie groups where we have nice left invariance, and we had all our flows were complete, we’re going to try to get things for arbitrary vector fields on a manifold where the flow is not necessarily guaranteed to be complete.

First off, I’m going to want to think of right actions instead of left now. This is because in the last post I showed that flowing is the same thing as right multiplication by exp. From now on, I’m assuming we have a Lie group acting smoothly on a manifold on the right \theta(p,g)=p\cdot g. We want a global flow action of \mathbb{R} on our manifold M, so let X\in Lie(G). Define the action \mathbb{R}\times M \to M by t\cdot p=p\cdot exp(tX) (note: the two dots are two different actions, the one on the right comes from the Lie group).

We have an infinitesimal generator for this flow, say \widehat{X}\in\frak{X}(M). i.e. \widehat{X}_p=\frac{d}{dt}\big|_{t=0}p\cdot exp(tX). Thus we have a map \widehat{\theta}:Lie(G)\to\frak{X}(M) by \widehat{\theta}(X)=\widehat(X).

Let’s break down what this map really is. For any p\in M, examine \theta^p:G\to M by \theta^p(g)=p\cdot g. This is a smooth, since we can identify G\cong \{p\}\times G and then it is just inclusion followed by the smooth action. Thus this is the orbit map of the action. You get everything in the orbit of p by the action of G. Thus \widehat{X}_p=d(\theta^p)_e(X_e).

Let’s go one step further and show that X and \hat{X} are \theta^p related (note now that X, p, and the action are completely arbitrary, so this is really a very general statement).

By the group law p\cdot gh=(p\cdot g)\cdot h, we actually have \theta^p\circ L_g(h)=\theta^{p\cdot g}(h). Now we just check:

\widehat{X}_{p\cdot g}=d(\theta^{p\cdot g})_e(X_e)
= d(\theta^p)_g\circ d(L_g)_e(X_e)
= d(\theta^p)_g(X_g).

Which shows \theta^p related.

Now we easily can get that \widehat{\theta}: Lie(G)\to \frak{X}(M) is a Lie algebra hom. Using linearity of \widehat{\theta} for a fixed p, and the previous statement, we get that [\widehat{X}, \widehat{Y}]_p=d(\theta^p)_e([X, Y]_e)=\widehat{[X, Y]}_p. Thus [\widehat{\theta}, \widehat{\theta}]=\widehat{\theta}([X, Y]).

Now we are ready to state the “Fundamental Theorem on Lie Algebra Actions”. A quick term, we say a \frak{g}-action \widehat{\theta} is complete if \widehat{\theta}(X) is complete for all X.

The FT on LAA says that given any complete Lie(G)-action \widehat{\theta}:Lie(G)\to\frak{X}(M), there is a unique smooth right G-action on M whose infinitesimal generator is \widehat{\theta}.

I won’t prove this, but it was nice to state and a good ending place for the day.


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