# The description of abelian Lie groups

Question. Classify all abelian connected Lie groups.

There is a problem in my problem sheet which asks me to describe all abelian connected Lie groups (moreover this is the first problem so it should be rather easy). I don't understand how this description should look. They mean description up to an isomorphism (of Lie groups), don't they?

I can list some abelian connected (real) Lie groups:

• $$\mathbb{R}^n$$,
• $$\mathbb{C}_{\ne 0}$$ (as a real group under multiplication),
• $$S^1$$ (i.e. $$\big\{z\in\mathbb{C}: |z|=1\big\}$$),
• also some different finite products.

Could you help me to classify them?

• $\mathbb{R}_{\neq 0}$ is not connected, perhaps you meant $\mathbb{R}_{> 0}$ (which is isomorphic to $\mathbb{R}$, of course). Mar 27 '14 at 20:48
• Oh, Really. Sorry. Mar 27 '14 at 20:53
• Hint: show that the exponential map is a homomorphism, where we consider the Lie algebra as an abelian Lie group under addition.
– Nate
Mar 27 '14 at 20:54

Nate's hint does the trick. Let $G$ be an abelian connected Lie group with Lie algebra $\mathfrak g$. The exponential map $\exp:\mathfrak g\to G$ is actually a homomorphism of abelian groups. The image is open in $G$, so $\exp$ is surjective because $G$ is connected. The fact that $\mathfrak g\to G$ is a local homeomorphism means that $\ker(\exp)$ is a discrete subgroup of $\mathfrak g$. It is known that such groups are of the form $\Lambda=\mathbf Z x_1+\cdots + \mathbf Z x_n$, for $x_1,\dots,x_n\in \mathfrak g$ linearly independent over $\mathbf R$. We can extend $x_1,\dots,x_n$ to a basis of $\mathfrak g$ to see that $$G \simeq (S^1)^r \times \mathbf R^s$$ In other words, every connected abelian Lie group is a product of affine space and a torus.

For example, $\mathbf C_{\ne 0} = \mathbf C^\times$ is the product $\mathbf R\times S^1$, via $(r,\theta)\mapsto r e^{i\theta}$.

• Hi @Daniel Miller do you have a reference for that? Could you precize what $r$ and $s$ are and if they are related? Thanks! Jan 5 '16 at 0:14
• @Hamurabi no reference needed: I've included a complete proof. The integers $r$ and $s$ can be found by $r=\mathrm{rank} X_\ast(G)$ and $s=\dim(G)-r$. Jan 5 '16 at 17:43
• Thanks @Daniel Miller. Sorry for another question, but what is $rank X_{*}(G)$? Jan 7 '16 at 13:27
• @DanielMiller: I want to cite the above result in a paper I'm writing. Can you direct me any references/paper where they prove the above fact ? Jan 23 '16 at 17:54

If one knows the fundamental fact that a simply connected connected Lie group is completely determined by its Lie algebra, one can proceed as follows:

Let $G$ be an connected abelian Lie group of dimension $n$. The universal covering space $\tilde G$ is also a connected abelian Lie group of dimension $n$ which is, of course, simply connected; in particular, the Lie algebra of $\tilde G$ is of dimension $n$ and abelian. Since $\mathbb R^n$ is also a connected simply-connected Lie group with abelian Lie algebra of dimension $n$, we must have $\tilde G\cong\mathbb R^n$ as Lie groups. The point here is that there is exactly one^abelian Lie algebra of each dimension.

Now, the covering map $p:\tilde G\to G$ is a group homomorphism, so that $\ker p$ is a discrete subgroup of $\tilde G\cong\mathbb R^n$, so it is isomorphic to the subgroup generated by a linearly independent subset of $\mathbb R^n$. Using this it is easy to conclude what we want.

• Of course, the fundamental theorem I mentioned in the first paragraph is rather more difficult to prove that the result you want :-) Mar 27 '14 at 22:20
• Can you direct me to any reference where it is proved that every connected abelian Lie group is isomorphic to product of affine spaces and torus ? Jan 24 '16 at 4:55
• I'd say that this fact does not need a reference, as it is a basic part of Lie theory. Mine and Daniel's are two essentially complete proofs — you can reference this page, if you want! Jan 24 '16 at 14:37
• Thanks. We work in areas that are not at all relevant to Lie group theory. that's why we wanted to cite the result in view of the target audience. Jan 24 '16 at 15:34
• You should be able to find a classification of abelian Lie groups in standard textbooks on the topic. Jul 17 '20 at 15:01