It's a classical theorem of Lie group theory that any compact connected abelian Lie group must be a torus. So it's natural to ask what if we delete the connectedness, i.e. the problem of classification of the compact abelian Lie groups.


The quotient of your Lie group $G$ by the connected component of the identity $G_0$ is a finite discrete abelian group $A$, because the Lie group is compact. It follows that we have an extension $$0\to G_0\to G\to A\to 0$$ Such a thing is classified (forgetting topologies) by an element of $H^2(A,G_0)$. This group is zero, because $A$ is finite and $G_0$ is divisible. It follows that $G\cong A\times G_0$. Since $G_0$ is a compact and connected, it is a torus.

This completely decribes the possible $G$s.

  • $\begingroup$ Thank you! But I don't know how to deduce your statement that such a thing is classified (forgetting topologies) by an element of $H^2(A,G_0)$ and that $G\cong A\times G_0$. Can you give some references? $\endgroup$ – Lao-tzu Dec 8 '13 at 10:49
  • $\begingroup$ Any textbook on group cohomology. A particularly nice one is MacLane's book Homology. $\endgroup$ – Mariano Suárez-Álvarez Dec 8 '13 at 17:00
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    $\begingroup$ groupprops.subwiki.org/wiki/… seems to imply $H^2(Z/2Z \times Z/2Z, S^1)=Z/2Z$. Is that not true? $\endgroup$ – Max Oct 3 '14 at 7:20
  • $\begingroup$ I think the ses splits because G is abelian and G_0 is divisible. That is, all non trivial extensions give nonabelian G. $\endgroup$ – Max Oct 3 '14 at 14:35

Let $G$ be an abelian Lie group. The connected component $G_0$ is a connected abelian Lie group, so it's isomorphic to $\Bbb{R}^n\times\Bbb{T}^m$ for some $n,m\in\omega$ (this is not hard to see: mostly one just needs to observe $\exp$ is a surjective open homomorphism with a discrete kernel). In particular, it is divisible.

Lemma. Suppose $D$ is a divisible subgroup of an abelian group $G$ and $h:D\to H$ is a homomorphism into a divisible abelian group $H$. Then $f$ extends to a homomorphism $\tilde f:G\to H$.

For a proof, see here.

Therefore the identity $\mathrm{id}:G_0\to G_0$ extends to a homomorphism $f:G\to G_0$. Denote $K=\ker(f)$; clearly $K\cdot G_0=G$ and $K\cap G_0={0}$ so (since $G$ is abelian) $G=K\times G_0$. Since $G_0$ is open, we moreover know that $K$ must be discrete (and therefore closed). This shows $G=K\times G_0$ as a topological group and not just an abstract group.

Thus, we have seen abelian Lie groups are exactly the direct products of an abelian discrete group and a connected abelian Lie group.

If we moreover require $G$ to be compact, then $K$ must be finite (since it is compact, as a closed subgroup of $G$, and discrete) and $G_0$ a torus (since all connected compact abelian Lie groups are tori). So a group $G$ is a compact abelian Lie group if and only if it is isomorphic to $F\times \Bbb{T}^m$ for a finite abelian group $F$ and $m\in\omega$.


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