# Generators of compact Lie groups

Suppose $G$ is a compact connected Lie group and let $\{X_i\}$ be a basis for its Lie algebra $\mathfrak g$. We know that the exponential $\exp:\mathfrak g \to G$ is surjective but when is it the case that $G$ is generated by $\{\exp(tX_i) : t\in \mathbb R\}$?

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Ah, whoops... I forgot that the compact + connected implies surjectivity – Dylan Wilson Jan 27 '11 at 5:09

## 1 Answer

The map $\mathbb{R}^n \to G$ sending $(t_1, \dots, t_n)$ to $\mathrm{exp}(t_1 X_1) \dots \mathrm{exp}(t_n X_n)$ has nonsingular Jacobian at $0$, so its image contains a neighborhood of the origin. By a standard argument, a neighborhood of the origin in a connected topological group generates the full group.

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Very nice! Thanks! – Eric O. Korman Jan 27 '11 at 3:04
Can you explain why for the compact connected Lie group the exponential map has to be surjective? (though  there can be cases where the exponential map has a finite injectivity radius) – Anirbit Feb 17 '11 at 8:14
@Anirbit: One way to see this is to use the Hopf-Rinow theorem in Riemannian geometry (since the geodesics with respect to a bi-invariant metric are precisely $\exp(tX)$ for $X$ in the Lie algebra). Sorry for the very delayed response! – Akhil Mathew Feb 25 '12 at 16:00
@Anirbit Akhil's argument is great but if you do not wish to explicitly use Riemannian geometry, then there is an alternative proof if you assume Cartan's theorem on the maximal tori in a compact connected Lie group (Cartan's theorem can be proved using the Lefschetz fixed point theorem in algebraic topology). Indeed, each element of $G$ is an element of some maximal torus $T\subseteq G$ which means that we need only prove the surjectivity of the exponential map $\text{exp}:\mathfrak{t}\to T$ where $T$ is a torus. Of course, this is very easy to do. – Amitesh Datta Jun 27 '12 at 2:22
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