Notice from last time that the matrix exponential maps from the Lie algebra to the Lie group. It took an arbitrary matrix (i.e. a tangent vector at the identity) and mapped it to a non-singular matrix (i.e. a matrix in ). Now we want a map that does the same basic idea. Let’s unravel what that idea was.

It was sort of hidden, but we said that the one-parameter subgroup of generated by A, is . So it took a line through the origin: , and sent it to a one-parameter subgroup .

Let’s sort of work backwards. We have a one-parameter subgroup generated by , say from the last post (the integral curve starting at the identity). From our matrix example, we also saw that if we flowed along this for 1 time unit, . Thus we’ll define , where is the one-parameter subgroup generated by . Thus will be mapping from the Lie algebra to the Lie group.

Now we need to check the line through the origin property. i.e. Is the one-parameter subgroup generated by ? Well, it is really a simple matter of rescaling. Let by rescaling. Then is the integral curve of starting at , so .

Now there are lots of properties we should check. I’ll go through a few this time and the rest tomorrow before moving on. First off, we hope that is a smooth map. In other words, in terms of the flow, we need to depend smoothly on .

Define a vector field on by . Note we made the natural identification . Let be a basis for and the coordinates for , . Also, let be smooth coordinates for .

Let . Then in coordinates . Each is a derivative in the direction, so the expression depends smoothly on . Hence is a smooth vector field.

Now the flow of , say , is . Now the fact that is a flow of a smooth map shows that it is smooth. Let be projection. Then and hence is a composition of smooth maps so is itself smooth.

That was sort of exhausting, so I’ll just do one quick other property before quitting: . This is just because is a one-parameter subgroup and hence homomorphism. The group structure on is additive and the Lie group operation is multiplicative: .

Alright, one more quick one, since that one shouldn’t count. This one is a big one, in that it corresponds with a very characteristic property of the complex valued exponential. If we identify and with , then is the identity map!

Let . Then . Thus (the equality comes from the fact that exp is the flow).

I guess this ran a bit long. Oh well, time is running out.

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