Lie group, Lie algebra and surjectivity

Question

Let $G$ be a connected Lie group. If the Lie algebra $\mathfrak{g}$ is commutative, is the exponential mapping surjective? If not, do we at least have that $G$ is abelian? Any counter-examples as I have not been able to think of one. Thank you in advance!

Answer 1

First of all, if $G$ is a connected Lie group with the universal cover $p: \tilde G\to G$ and Lie algebra ${\mathfrak g}$, then exponential maps for $G$ and $\tilde G$ commute with the covering map $p$. Therefore, if $\exp: {\mathfrak g}\to \tilde G$ is surjective, so is $\exp: {\mathfrak g}\to G$. Therefore, for your question it suffices to consider the case of simply-connected Lie groups. Such Lie groups are uniquely determined by their Lie algebras. The key now is the assumption that ${\mathfrak g}$ is commutative, i.e., isomorphic to ${\mathbb R}^n$ with zero Lie bracket. (I am assuming you are working with real Lie algebras, the complex case is analogous.) For this Lie algebra and the associated Lie group $\tilde G= ({\mathbb R}_+)^n$ (with multiplicative binary operation) the exponential map is clearly surjective. qed