TITLE: A compact, connected, abelian Lie group is a torus?
QUESTION [8 upvotes]: How to prove that a compact, connected, abelian Lie group is a torus? It seems very intuitive. 
Any reference?

REPLY [7 votes]: The hardest part of the problem is showing:
If $G$ is a connected abelian Lie group then the exponential map $$ \exp: \mathfrak{g} \longrightarrow G $$ is a $\textbf{surjective homomorphism}$ with $\textbf{discrete kernel}$. 
This can be done by appealing to the fact that $G$ is generated by the image of $\exp$, the first Lie theorem that states $\exp$ is a local homeomorphism at the identity, 1 parameter subgroups, and general considerations on topological groups - it's a good exercise to see if you've grasped the theory.
Our result then follows by noting that $G \simeq \mathbb{R}^k / \Gamma$ where $k$ is the dimension of $\mathfrak{g}$ as a real vector subspace and $\Gamma$ is the discrete kernel. And by a well known fact about cocompact discrete subgroups of euclidean spaces, it follows $\Gamma \simeq \mathbb{Z}^k$.