Is the exponential map ever not injective?

Question

Let $G\subseteq GL_n(\mathbb R)$ and let $\mathfrak g$ denote its Lie algebra.

Let $e: \mathfrak g \to G$ be the map $X \mapsto e^X$.

Does there exist an example of $G$ and $\mathfrak g$ such that $e$ is not injective?

Of course I think the answer is no, there is no such example because $e: \mathbb R \to \mathbb R$ is injective.

What about $G\subseteq GL_n(\mathbb C)$?

Since the map is the same I again think there should not exist such an example.

Over the complex numbers, have you considered the $n=1$ case? — Tobias Kildetoft, Feb 02 '15 at 11:22
@TobiasKildetoft No, I hadn't, good point. If $n=1$ and $G=\mathbb C$ then also $\mathfrak g = \mathbb C$, right? And the complex exponential is not injective. — learner, Feb 02 '15 at 11:34

score 8 · Accepted Answer · answered Feb 02 '15 at 12:00

8

As a simple counterexample: The group $SO(2) \subset GL_2(\mathbb{R})$ is isomorphic to $\mathbb{S}^1$; its Lie algebra is the space of skew-symmetric matricies, and its exponential map is given by $$ e : \left( \begin{array}{cc} 0 & t \\ -t & 0 \end{array} \right) \mapsto \left( \begin{array}{cc} \cos t & \sin t \\ -\sin t & \cos t \end{array} \right) $$ which isn't injective.

answered Feb 02 '15 at 12:00

mollyerin

3,820

That is a nice example! – learner Feb 02 '15 at 23:59

Is the exponential map ever not injective?

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