# Classify the compact abelian Lie groups

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.

## 2 Answers

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.

• 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? – Lao-tzu Dec 8 '13 at 10:49
• Any textbook on group cohomology. A particularly nice one is MacLane's book Homology. – Mariano Suárez-Álvarez Dec 8 '13 at 17:00
• groupprops.subwiki.org/wiki/… seems to imply $H^2(Z/2Z \times Z/2Z, S^1)=Z/2Z$. Is that not true? – Max Oct 3 '14 at 7:20
• I think the ses splits because G is abelian and G_0 is divisible. That is, all non trivial extensions give nonabelian G. – 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 $$m\in\omega$$.