A Mind for Madness

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One-Parameter Subgroups

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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.

G will always denote a Lie group from here on out. Alright, so the first thing is that we have left invariant vector fields on G. These are elements of the Lie algebra of, that will be denoted \frak{g}=Lie(G). The flow of a left-invariant vector field satisfies a nice group law (\theta_t\circ\theta_s)(p)=\theta_{t+s}(p). 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 F:\mathbb{R}\to G 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 \frak{g}). Proof:

Let X\in \frak{g}. Then since the vector field is left-invariant, it is L_g-related to itself for any left translation. Thus, by naturality, for any integral curve \gamma, L_g\circ \gamma 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 \theta^{(e)}. Let’s translate by \theta^{(e)}(s) for some fixed s\in G.

Then L_{\theta^{(e)}(s)}(\theta^{(e)}(t)) is an integral curve starting at L_{\theta^{(e)}(s)}(e)=\theta^{(e)}(s). But \theta^{(e)}(s+t) is also an integral curve starting at \theta^{(e)}(s). Thus they are equal. Writing out what left translation means, we get \theta^{(e)}(s)\theta^{(e)}(t)=\theta^{(e)}(s+t). Flows are complete on Lie groups, so \theta^{(e)} : \mathbb{R}\to G 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 F:\mathbb{R}\to G is a one-parameter subgroup. Now d/dt is a left-invariant vector field on \mathbb{R}, so let’s let X=F_*(d/dt)\in \frak{g}. The claim is that F is an integral curve of X.

But F_*(d/dt) is the unique left-invariant vector field that is F-related to d/dt. Thus F'(t_0)=dF_{t_0}\left(\frac{d}{dt}\big|_{t_0}\right)=X_{F(t_0)}. Thus F is an integral curve of X.

Now that we have this correspondence, there is no problem in saying “the” one-parameter subgroup generated by X. Our main example for one-parameter subgroups comes from our favorite Lie group GL_n(\mathbb{R}). If we take a left-invariant vector field A\in \frak{gl}_n(\mathbb{R}). Then the one-parameter subgroup generated by this vector field is F(t)=e^{tA}, 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.

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Author: hilbertthm90

I write about math, philosophy, literature, music, science, computer science, gaming or whatever strikes my fancy that day.

One thought on “One-Parameter Subgroups

  1. Pingback: Analytic Theory of Abelian Varieties I « A Mind for Madness

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