# When is exponential map from Lie algebra to Lie group a covering map?

Suppose $$G$$ is a Lie group and $$\mathfrak{g}$$ its Lie algebra. It is not so difficult to see that if $$G$$ is abelian and connected then $$\exp:\mathfrak{g}\rightarrow G$$ is a universal covering map. What if $$G$$ is non-abelian? Is there a characterization of when $$\exp$$ is a universal covering map?

• About surjectivity of $\exp$ see here. More can be found in Terry Tao's blog. – Dietrich Burde Apr 17 '19 at 8:05

First of all, you should assume that $$G$$ is connected, since otherwise $$\exp$$ cannot be surjective. Then a complete characterization of Lie connected groups for which $$\exp$$ is a covering map is given by YCor in his answer here:
$$\exp$$ is a covering map if and only if $${\mathfrak g}$$ is solvable and does not contain two particular Lie subalgebras $${\mathfrak e}$$, $$\tilde{\mathfrak e}$$.
The reduction to Yves' answer is easy: $$\exp: {\mathfrak g}\to G$$ is a covering map if and only if $$\exp: {\mathfrak g}\to \tilde{G}$$ is a diffeomorphism, where $$\tilde{G}$$ is the universal covering group of $$G$$.