More Exponential Properties

I’ll just do a quick finish up on proving basic properties about the exponential map for today, since I’ll be moving on to a different topic after this.

The exponential map actually restricts to a diffeomorphism from some neighborhood of 0 in \frak{g} to some neighborhood of e\in G. Well, if you’ve mastered your basic theorems, then this is no surprise. Last time we showed dexp_0(X)=X. The differential is the identity map and hence non-singular. Thus by the Inverse Function Theorem, there exists neighborhoods about 0 and e such that there is a smooth inverse, and hence exp restricted to these is a diffeo.

Probably the most interesting and non-obvious of these facts is the following one. Given another Lie group H (and Lie algebra \frak{h}), and a Lie group homomorphism F:G\to H, we get a commutative diagram (which I’ve yet to figure out how to make on wordpress). Anyway it just says that exp(F_*X)=F(exp(X)). So pushing forward a vector field and then exponentiating it is the same thing as exponentiating first and then doing the group hom.

Let’s do it by showing that exp(tF_*X)=F(exp(tX)) for all t\in\mathbb{R}. Recall that the LHS is just the one-parameter subgroup generated by F_*X. Thus by uniqueness, we just need to show that \gamma(t)=F(exp(tX)) is a hom that satisfies \gamma'(0)=(F_*X)_e. It is a composition of homs, so is itself one. Now we compute the derivative: \gamma'(0)=\frac{d}{dt}\big|_{t=0} F(exp(tX))

The last property for today is used quite a bit as well. It says that flowing by a vector field X is the same thing as right multiplying by exp(tX). In other words, (\theta_X)t=R_{exp(tX)}.

First note that for any g\in G, the map \sigma(t)=L_g(exp(tX)), satisfies \sigma '(0)=dL_g (\frac{d}{dt}\big|_{t=0} exp(tX))=dL_g (X_e)=X_g by left invariance. Also, left multiplication takes integral curves to integral curves, so this is actually the integral curve of X starting at g. Thus L_g(exp(tX))=\theta^g_X(t).

Now we check what we wanted to show. R_{exp(tX)}(g)=g\cdot exp(tX)=L_g(exp(tX))=\theta^g_X(t)=(\theta_X)_t(g). And we’re done!

Next I’ll talk about infinitesimal generators of Lie group actions, so that I can pull in some stuff I did from the Frobenius theorem.

Update: There is some sort of weird bug where when I hit preview it tells me to save the draft. Then when I hit save draft, it posts. So, sorry if you’ve been getting unedited versions of my posts.


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