# The description of abelian 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. $\{z\in\mathbb{C}: |z|=1\}$), 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). – Jan Ladislav Dussek Mar 27 '14 at 20:48
• Oh, Really. Sorry. – Kolyan 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! – Hamurabi 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 $s=\dim(G)-r$. – Daniel Miller Jan 5 '16 at 17:43
• Thanks @Daniel Miller. Sorry for another question, but what is $rank X_{*}(G)$? – Hamurabi 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 ? – pikachuchameleon 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 :-) – Mariano Suárez-Álvarez 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 ? – pikachuchameleon 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! – Mariano Suárez-Álvarez 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. – pikachuchameleon Jan 24 '16 at 15:34