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 to some neighborhood of . Well, if you’ve mastered your basic theorems, then this is no surprise. Last time we showed . The differential is the identity map and hence non-singular. Thus by the Inverse Function Theorem, there exists neighborhoods about 0 and such that there is a smooth inverse, and hence restricted to these is a diffeo.
Probably the most interesting and non-obvious of these facts is the following one. Given another Lie group (and Lie algebra ), and a Lie group homomorphism , we get a commutative diagram (which I’ve yet to figure out how to make on wordpress). Anyway it just says that . 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 for all . Recall that the LHS is just the one-parameter subgroup generated by . Thus by uniqueness, we just need to show that is a hom that satisfies . It is a composition of homs, so is itself one. Now we compute the derivative:
The last property for today is used quite a bit as well. It says that flowing by a vector field is the same thing as right multiplying by . In other words, .
First note that for any , the map , satisfies by left invariance. Also, left multiplication takes integral curves to integral curves, so this is actually the integral curve of starting at . Thus .
Now we check what we wanted to show. . 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.