It is really down to the wire with less than 2 weeks until prelim exams. I’m feeling sort of weak in a few areas, so I’ll try to clear them up here. Lie theory is sort of a downfall for me. I like mathematical structures that are simple. Lie things just have too much going on for me.
will always denote a Lie group from here on out. Alright, so the first thing is that we have left invariant vector fields on . These are elements of the Lie algebra of, that will be denoted . The flow of a left-invariant vector field satisfies a nice group law . We want to figure out if this relates at all to the group operation on the Lie group itself (and it should, right?).
We call a Lie group homomorphism a one-parameter subgroup (note that it is a morphism…not a subgroup). The relation we want to establish is that the one-parameter subgroups are the integral curves of left-invariant vector fields starting at the identity (there is a bijective correspondence between the set of one-parameter subgroups and ). Proof:
Let . Then since the vector field is left-invariant, it is -related to itself for any left translation. Thus, by naturality, for any integral curve , is also an integral curve. Let’s pick a particularly nice integral curve to try this fact on: the integral curve starting at the identity . Let’s translate by for some fixed .
Then is an integral curve starting at . But is also an integral curve starting at . Thus they are equal. Writing out what left translation means, we get . Flows are complete on Lie groups, so and the equality shows it is a homomorphism. Thus it is a one-parameter subgroup.
Now we must show every one-parameter subgroup determines an integral curve through the identity. Suppose is a one-parameter subgroup. Now is a left-invariant vector field on , so let’s let . The claim is that is an integral curve of .
But is the unique left-invariant vector field that is -related to . Thus . Thus is an integral curve of .
Now that we have this correspondence, there is no problem in saying “the” one-parameter subgroup generated by . Our main example for one-parameter subgroups comes from our favorite Lie group . If we take a left-invariant vector field . Then the one-parameter subgroup generated by this vector field is , the matrix exponential. I should maybe do the computation to show that it works, but it is sort of annoying.
Next post, we’ll generalize this notion of an exponential map.