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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!

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If you assume that it's compact, then it is true. This is a standard theorem. It's non-surjective, and $G$ is not abelian, in the case $G=\text{SL}_2(\mathbb{R})$. – Alex Youcis Mar 4 '14 at 9:15
But, $\mathrm{sl}_2 (\Bbb{R})$ is not trivial right? – Singhal Mar 4 '14 at 12:52
up vote 3 down vote accepted

Another solution without using the universal cover is as follows If $X,Y$ are two vector fields then $[X,Y]=0$ is equivalent to $\phi_t \circ \psi_s=\psi_s\circ\phi_t $ where $\phi,\psi$ are the flows of $X,Y$. So if the lie algebra of $G$ is commutative that means $\exp(tX)$ commutes with $\exp(sY)$ for all $s,t\in\mathbb R$ and $X,Y\in g$. Since $G$ is connected that implies that the subgroup generated by any open set containing the identity is equal to $G$ and since $\exp$ is a local diffeomorphism at $e$. It follows that there is an open set containing the identity such that any two elements of it commute but then the elements in the subgroup generated by it would also commute hence $G$ is abelian. To show that $\exp$ is surjective first show that it is a group homomorphism (using the fact that $G$ is abelian) and so its image is a subgroup of $G$ containing a neighborhood of the identity (by being a local diffeomorphism at $e$) hence it is equal to $G$

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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

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For commutative Lie group $G$, there exist a bi-invariant Riemannian metric on $G$, so it is Riemannian symmetric space. Thus, it is complete. By hopf-Rinow theorem, we know exponential map is surjective.

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