Covering Maps Abelian Lie groups

Question

I am trying to understand a sentence I found in a paper christened "Tori and Jacobians" by M. Cornalba. The sentence is the following: Let $X$ be a complex connected and compact Lie group. Consider its universal cover $p: \tilde{X} \rightarrow X.$ It is clear that $\tilde{X}$ is an abelian, simply connected Lie group. The article further says that $\tilde{X}$ is therefore isomorphic to a $\mathbb{C}^N$ and I do not understand why. Can someone give a proof or at least an argument of how that works? Thank you in advance

Answer 1

Half an idea:

Well...it has to be more than "abelian simply connected lie groups are all isomorphic to $\Bbb C^N$ for some $N$," because the 3-sphere is a counterexample to that claim.

What does "complex" in the description of $X$ indicate? That $X$ is a complex manifold with a smooth group structure? Perhaps that's a hint. Presumably, whatever it means, it lifts to universal covers. At that point some classification of simply connected even-dimensional Lie groups might be the answer...

Answer 2

Let $\mathfrak x$ be the Lie algebra of $\tilde X$. Since $\tilde X$ is abelian, $\exp\colon\mathfrak{x}\longrightarrow\tilde X$ is a group homomorphism. Since $\tilde X$ is connected, and $\exp(\mathfrak{x})$ is a neighborhood of $e_{\tilde X}$, $\exp$ is surjective. Furthermore, its kernel is a discrete subgroup of $\mathfrak x$. So, $\exp$ is a covering map. Since $\tilde X$ is simply connected, $\exp$ is a diffeomorphism.