Surjective geodesic exponential map on a Lie group

Question

Is it possible to put a left invariant connection on an arbitrary Lie group whose associated geodesic exponential map is surjective?

Answer 1

In general, if $G$ acts transitively by isometries on a Riemannian manifold $X$, then $X$ is complete and in particular the geodesic exponential map for the Levi-Civita connection is surjective. In your case you can take $G=X$ acting on itself by left multiplication with any left-invariant metric on $G$. (Assuming of course that $G$ is connected.)

The short version of the proof goes like this: At some point $p \in X$, the exponential map is defined on a ball of radius $r$ centered at $0$ in $T_pX$. Since $X$ is homogeneous, the same holds at every point of $X$. This implies that every geodesic can be extended by some definite amount, say $r/3$, and then by the Hopf-Rinow theorem the exponential map is defined on all of $T_pX$ and moreover is surjective.

Answer 2

Yes, if $G$ is connected. A theorem of Cartan says that a connected Lie group is diffeomorphic to the product of a compact subgroup $C$ and $\mathbb{R}^n$. You can define connections (metric) on $C$ and $\mathbb{R}^n$ whose associated geodesic exponential are surjective, the associated geodesic exponential of the product of these connections is surjective.