# One-Parameter Subgroups

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.