4 Linear Lie groups and their Lie algebras 4.2 The exponential map 4.4 Lie algebras

4.3 One-parameter subgroups

Lemma 4.15.

The map from $\mathbb{R}$ to $\operatorname{GL}_{n}(\mathbb{C})$ given by

t\mapsto\exp(tX)

is a differentiable group homomorphism.

We have

\frac{d}{dt}\exp(tX)=X\exp(tX)=\exp(tX)X.

In particular,

\left.\frac{d}{dt}\exp(tX)\right|_{t=0}=X.

Proof.

The given map is a group homomorphism by Lemma 4.5 part 4.

By definition,

\exp(tX)=\sum_{k=0}^{\infty}\frac{X^{k}}{k!}t^{k}.

As this power series (and its termwise derivative) are uniformly convergent on any compact subset, we can compute its derivative by differentiating termwise, which gives

\frac{d}{dt}\exp(tX)=\sum_{k=1}^{\infty}\frac{X^{k}}{(k-1)!}t^{k-1}=X\exp(tX).\qed

Definition 4.16.

A one-parameter subgroup of $\operatorname{GL}_{n}(\mathbb{C})$ is a differentiable group homomorphism $f:\mathbb{R}\to\operatorname{GL}_{n}(\mathbb{C})$ . That is, a differentiable map such that

f(s+t)=f(s)f(t)

for all $s,t\in\mathbb{R}$ .

The infinitesimal generator of a one-parameter subgroup $f$ is the element $f^{\prime}(0)\in\mathfrak{gl}_{n,\mathbb{C}}$ .

Remark 4.17.

(non-examinable) For a one-parameter subgroup $f$ , it actually suffices to require that $f$ is continuous. Differentiability then comes for free.

Indeed, if $f$ is continuous, the integral $\int_{0}^{a}f(t)dt$ exists. Moreover,

f(s)\int_{0}^{a}f(t)dt=\int_{0}^{a}f(s+t)dt=\int_{s}^{s+a}f(t)dt.

The RHS is differentiable with respect to $s$ by the fundamental theorem of algebra. Therefore, to prove that $f(s)$ is differentiable, we only need to show that there is an $a>0$ such that $\int_{0}^{a}f(t)dt$ is an invertible matrix. Now consider the function

F(a)=\frac{1}{a}\int_{0}^{a}f(t)dt.

It is well-defined for $a\neq 0$ and $\lim\limits_{a\to 0}F(a)=I$ . Hence, for $0<a\ll 1$ , $F(a)$ is invertible, and therefore so is $aF(a)=\int_{0}^{a}f(t)dt$ .

The following is a very important property of one-parameter subgroups: that they all come from the exponential map.

Proposition 4.18.

Let $f:\mathbb{R}\to\operatorname{GL}_{n}(\mathbb{C})$ be a one-parameter subgroup with infinitesimal generator $X$ .

Then

f(t)=\exp(tX)

for all $t\in\mathbb{R}$ . That is, all one-parameter subgroups arise from the exponential function.

Proof.

From the definition of one-parameter subgroups, we have

f^{\prime}(t)=\lim_{s\to 0}\frac{f(s+t)-f(t)}{s}=f(t)\lim_{s\to 0}\frac{f(s)-f% (0)}{s}=f(t)f^{\prime}(0)=f(t)X.

Now consider the differential equation

g^{\prime}(t)=g(t)X.

We know that both $f(t)$ and $\exp(tX)$ are both solutions with the same initial condition that $g(0)=I$ . Therefore they must be equal. ∎

Example 4.19.

The map $\mathbb{R}\to\mathrm{SO}(3)$ taking $\theta$ to rotation by $\theta$ about a fixed axis is a one-parameter subgroup. Problem 55 asks you to find its infinitesimal generator.