How to prove that a compact, connected, abelian Lie group is a torus? It seems very intuitive.

Any reference?

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    $\begingroup$ In what possible way does this seem intuitive? $\endgroup$ Dec 10, 2017 at 2:03

2 Answers 2


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 $\Gamma \simeq \mathbb{Z}^k$.

  • $\begingroup$ Do you know why it is discrete? $\endgroup$
    – Lemon
    Nov 14, 2018 at 5:04
  • $\begingroup$ @Hawk $\mathrm{exp}$ is a homeomorphism at the identity, so there is a neighbourhood of 0 that doesn't contain any other kernel points. Suitable left-multiplications show it is discrete everywhere. $\endgroup$ Sep 20, 2019 at 7:07

Should be a comment but i haven't got enough reputation :)

This is a classical result in Lie theory and is probably found in any Book about lie groups. Even if you search for scripts of lectures about Lie groups (e.g. on google) you should make a hit. However for example

R. L. Bryant, An Introduction to Lie Groups and Symplectic Geometry


M. R. Sepanski, Compact Lie groups.


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