Under what conditions is the exponential map on a Lie algebra injective?

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$ and let $\exp :\mathfrak{g}\rightarrow G$ be the exponential map.

In his blog, Terrence Tao notes that if a Lie group is not simply-connected, then $\exp$ will not be injective. Conversely, is it true that if a Lie group is simply-connected, then $\exp$ is injective? If not, what is a counter-example?

-

There is a complete characterization, in a large part due to Dixmier and Saito (both independently in 1957):

If $G$ is a real Lie group with Lie algebra $\mathfrak{g}$, then the following are equivalent:

1. $\exp$ is injective
2. $\exp$ is bijective
3. $\exp$ is a real analytic diffeomorphism
4. $G$ is solvable, simply connected, and $\mathfrak{g}$ does not admit $\mathfrak{e}$ as subalgebra of a quotient.
5. $G$ is solvable, simply connected, and $\mathfrak{g}$ does not admit $\mathfrak{e}$ or $\tilde{\mathfrak{e}}$ as subalgebra

Here $\mathfrak{e}$ is the 3-dimensional Lie algebra with basis $(H,X,Y)$ and bracket $[H,X]=Y$, $[H,Y]=-X$, $[X,Y]=0$. It is isomorphic to the Lie algebra of the group of isometries of the plane. Its central extension $\tilde{\mathfrak{e}}$ is defined as the 4-dimensional Lie algebra defined by adding a central generator $Z$ and the additional nonzero bracket $[X,Y]=Z$.

On the proof:

Injectivity of the exponential implies (as mentioned in Qiaochu's post) that there is no closed subgroup isomorphic to the circle, which means that the maximal compact subgroup in $G$ is trivial, that is, $G$ is contractible. A contractible Lie group is always isomorphic to $R\rtimes S^k$ where $R$ is a simply connected solvable Lie group, $k$ is a non-negative integer and $S$ is the universal covering $\widetilde{\mathrm{SL}_2(\mathbf{R})}$. The latter has a non-injective exponential map (as we see by unfolding two distinct circle from $\mathrm{SL}_2(\mathbf{R})$. So if the exponential map is injective we have $k=0$, i.e. $G$ is a simply connected solvable Lie group (for a solvable Lie group, contractible and simply connected are equivalent assumptions).

This is not enough since in the simply connected Lie group associated to $\mathfrak{e}$, the exponential map is not injective (this can been seen concretely, as in can be realized as the group of motions of the 3-dimensional Euclidean space generated by horizontal translations and a given 1-parameter group of vertical screwings).

That (4) implies (2) and (3) is due to Dixmier (Numdam freely available link) (Bull. SMF, 1957, in French). Dixmier also proved that (2), (3) and (4) are equivalent for simply connected solvable Lie groups, which together with the previous paragraph shows the equivalence between (2), (3), and (4) in general.

To complete the proof of the equivalences, one needs to show that for a simply connected solvable Lie group $G$, (1) implies the last (sub-quotient) condition in (4). A careful look at Dixmier's proof seems to show this: if $G$ does not satisfy (4) he even obtains that the exponential map is not locally injective.

That (4) implies (5) is easy, the converse is a bit harder but was done by Saito (M. Saito. Sur certains groupes de Lie résolubles. Scientific Papers of the College of Arts and Sciences. The University of Tokyo, 7:1-11, 1957; available here; in French too). To obtain that (1) implies (5), it is enough to check by hand that the simply connected Lie groups $E$ and $\tilde{E}$ associated to $\mathfrak{e}$ and $\tilde{\mathfrak{e}}$ have a non-injective exponential map, which is easy (not locally injective is a bit harder).

Note that we then have another characterization, it terms of the 4 minimal counterexamples:

1. $G$ has no closed subgroup isomorphic to either the circle $\mathbf{R}/\mathbf{Z}$, the universal covering $\widetilde{SL_2(\mathbf{R})}$, $E$ or $\tilde{E}$.
-

$SU(2)$ is simply connected, but its exponential map is not injective -- it's a double cover of $SO(3)$, so rotating by $4\pi$ around any axis is the identity.

-

The blog post already notes that injectivity fails whenever $G$ contains $S^1$ as a (Lie?) subgroup, in particular whenever $G$ is (positive-dimensional and) compact.

-