Structure of connected abelian Lie groups

Let $G$ be a connected Lie group that is abelian, and let $\mathfrak{g}$ be its Lie algebra.

Theorem (classification). There exist integers $n,m\ge 0$ such that

G \cong \mathbb{R}^n \times \mathbb{T}^m \quad\text{as Lie groups},

where $\mathbb{T}^m$ is an $m$ -torus (isomorphic to $(S^1)^m$ ). In particular, $G$ is determined up to isomorphism by the pair $(n,m)$ with $\dim G = n+m$ .

Remarks

Idea of proof (covering + lattice).

Since $G$ is connected and abelian, its universal cover $\widetilde{G}$ is a connected simply connected abelian Lie group. Such a group is isomorphic to the additive group of its Lie algebra, so $\widetilde{G}\cong (\mathfrak{g},+) \cong \mathbb{R}^{n+m}$ ; compare universal covering groups.
The covering map $\widetilde{G}\to G$ has kernel a discrete subgroup of $\widetilde{G}$ . After a linear change of coordinates, every discrete subgroup of $\mathbb{R}^{n+m}$ is $\mathbb{Z}^m$ in an $m$ -dimensional subspace; it need not be cocompact in all of $\mathbb{R}^{n+m}$ .
Splitting $\mathbb{R}^{n+m}$ into directions along this subgroup and a complementary subspace produces $\mathbb{R}^n\times (\mathbb{R}^m/\mathbb{Z}^m)$ .

Context. This result explains why the basic connected abelian examples are $\mathbb{R}^n$ , circles, and tori; see the circle group and the torus. It also matches the infinitesimal picture: $\mathfrak{g}$ is an abelian Lie algebra, so the exponential map is a local group isomorphism whose global kernel records the lattice.