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: $\gamma'(0)=\frac{d}{dt}\big|_{t=0} F(exp(tX))$
$=dF_0\left(\frac{d}{dt}\big|_{t=0}exp(tX)\right)$
$=dF_0(X_e)$
$=(F_*X)_e$.

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.