# 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 from R,+ to G where R,+ is the additive group of real numbers. In other words, F of t_1 + t_2 = F of t_1 times F of t_2 and F of 0 equals the identity.

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 little g, F of t equals exp (t X) is a 1-parameter subgroup. This is because exp t_1 X times exp t_2 X equals exp of (t_1 + t_2) X because t_1 X bracket t_2 X equals t_1 t_2 X bracket X. Moreover exp of 0 X equals the identity.

### 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 from R to G has this form, i.e. there exists an X in little g such that F of t equals exp of t X.

Proof:

We have F of s plus t equals F of s times F of t and F of zero equals the identity. We will differentiate the first equation with respect to s at s=0. Writing the derivative of F with respect to its variable as F dot, we get d by d s at s = 0 of F of s plus t equals F dot at t, and d by d s at s = 0 of F of s times F of t equals F dot at zero times F of t. This tells us that F of t satisfies the ordinary differential equation F dot at t equals F dot at zero times F of t and we can think of F of zero equals the identity 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 fo t F dot at 0 is a solution which satisfies the initial condition, therefore it agrees with the unique solution F of t. To see that exp of t F dot at zero solves the equation:

• differentiate it and we get d by d t of exp t F dot at 0 equals F dot at zero times exp t F dot at zero,

• set t=0 and get exp of 0 times F dot at zero equals the identity.

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.