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, Terence 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?


3 Answers 3


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

If $G$ is a real (finite-dimensional) 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;
  6. $G$ has no closed subgroup isomorphic to either the circle $\mathbf{R}/\mathbf{Z}$, the universal covering $\widetilde{\mathrm{SL}_2(\mathbf{R})}$, $E$ or $\tilde{E}$.

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 $[X,Y]=Z$. And $E$ and $\tilde{E}$ are the 3-dimensional and 4-dimensional simply connected solvable Lie groups associated to $\mathfrak{e}$ and $\tilde{\mathfrak{e}}$ respectively.

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 groups 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).

The equivalence with (6), which is stated in terms of the 4 minimal counterexamples, does not seem to have be stated in printed form, but follows from the proof.

  • $\begingroup$ Excuse me. What's the definition of subalgebra of a quotient? Does it mean that $\mathfrak{e}$ cannot be subalgebra of any quotient algebra of $\mathfrak{g}$ ? $\endgroup$
    – 346699
    Commented Dec 5, 2016 at 13:11
  • 1
    $\begingroup$ @user34669 yes! $\endgroup$
    – YCor
    Commented Dec 5, 2016 at 15:00
  • $\begingroup$ If $\mathfrak{g}$ does not admit $\mathfrak{e}$ it obviously does not admit $\tilde{\mathfrak{e}}$; so why does 5. say "does not admit $\mathfrak{e}$ or $\tilde{\mathfrak{e}}$"? Isn't just "not admitting $\mathfrak{e}$" for a solvable, simply connected $G$ enough for the exponential being a diffeo? $\endgroup$ Commented Mar 7, 2019 at 21:11
  • $\begingroup$ @MohammadrezaBidar why do you you truncate? I wrote in 5. "admit $\mathfrak{e}$ or $\tilde{\mathfrak{e}}$ as a subalgebra". I wrote it because among $\mathfrak{e}$ and $\tilde{\mathfrak{e}}$, neither is isomorphic to a subalgebra of the other one. $\endgroup$
    – YCor
    Commented Mar 7, 2019 at 21:54

$SU(2)$ 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.


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