# One-parameter subgroups

## One-parameter subgroups

### Definition

The first step in proving the theorem about smooth homomorphisms between matrix groups is to understand their one-parameter subgroups.

Definition:

A 1-parameter subgroup of a matrix group $G$ is a smooth homomorphism $F\colon(\mathbf{R},+)\to G$ where $(\mathbf{R},+)$ is the additive group of real numbers. In other words, $F(t_{1}+t_{2})=F(t_{1})F(t_{2})$ and $F(0)=I$ .

It's called a 1-parameter subgroup because it has one parameter, $t$ , and its image is a subgroup. Here's an example.

Example:

For any $X\in\mathfrak{g}$ , $F(t)=\exp(tX)$ is a 1-parameter subgroup. This is because $\exp(t_{1}X)\exp(t_{2}X)=\exp((t_{1}+t_{2})X)$ because $[t_{1}X,t_{2}X]=t_{1}t_{2}[X,X]=0$ . Moreover $\exp(0X)=I$ .

### One-parameter subgroups have the form exp(tX)

It turns out this is the only way to get a 1-parameter subgroup.

Lemma:

Any 1-parameter subgroup $F\colon\mathbf{R}\to G$ has this form, i.e. there exists an $X\in\mathfrak{g}$ such that $F(t)=\exp(tX)$ .

Proof:

We have $F(s+t)=F(s)F(t)$ and $F(0)=I$ . We will differentiate the first equation with respect to $s$ at $s=0$ . Writing the derivative of $F$ with respect to its variable as $\dot{F}$ , we get $\left.\frac{d}{ds}\right|_{s=0}F(s+t)=\dot{F}(t),\quad\left.\frac{d}{ds}\right% |_{s=0}F(s)F(t)=\dot{F}(0)F(t).$ This tells us that $F(t)$ satisfies the ordinary differential equation $\dot{F}(t)=\dot{F}(0)F(t)$ and we can think of $F(0)=I$ as an initial condition. By the existence and uniqueness theorem for ordinary differential equations, this differential equation has a unique solution when the initial condition is specified.

We can check that $\exp(t\dot{F}(0))$ is a solution which satisfies the initial condition, therefore it agrees with the unique solution $F(t)$ . To see that $\exp(t\dot{F}(0))$ solves the equation:

• differentiate it and we get $\frac{d}{dt}\exp(t\dot{F}(0))=\dot{F}(0)\exp(t\dot{F}(0))$ ,

• set $t=0$ and get $\exp(0\dot{F}(0))=I$ .

Remark:

If we hadn't already constructed the exponential map for matrices, this theorem from ordinary differential equations would tell us that it exists. Indeed, that is how the exponential map is constructed for Lie groups in general. I will not prove the existence and uniqueness theorem for ODEs: I assume you either know it already or will go and read about it if you're interested.