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?

Solution 1:

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 $(r,\theta)\mapsto r e^{i\theta}$.

Solution 2:

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.