# Why is $\mathbb{C}^{g}$ the universal cover of any connected compact complex Lie group of dimension g?

This question came up while I was studying the book "Complex Abelian Varieties" by Lange/Birkenhake.

More precisely, the authors prove in Lemma 1.1 that every connected compact complex Lie group of dimension $g$, $g$ a positive integer, is a complex torus. At one point they are using the fact that the universal cover of any such Lie group is a complex vector space of dimension $g$. They even give a reference (Theorem 18.4.1, "The Structure of Lie Groups" by Gerhard Hochschild). However, Theorem 18.4.1 (which is instead listed as Proposition 18.4.1 in the cited book) reads

"If a semisimple analytic group has a faithful finite-dimensional continuous representation then its center is finite"

I really don't have a clue how this should apply to my problem, because a complex torus is abelian, hence in particular not semisimple. On the other hand my knowledge of Lie groups is very very rudimentary so I might overlook something.

Can anyone show me how Proposition 18.4.1 in Hochschild's book helps me, or instead give me a reference to a proof of the statement in the title of my question? Of course any outline of a proof would also be appreciated, however I'd rather favor a reference. A reference to a direct proof of the above cited Lemma 1.1 would be very helpful too.

Thank you in advance

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My guess would be they're using the theorem (or proposition) exactly to show that a connected compact complex Lie group cannot be semisimple, maybe because they already know the center is not finite. – Marc van Leeuwen Jul 13 '12 at 11:32

## 1 Answer

The following steps lead to a solution. Let me first define standard notation. If $G$ is a Lie group and if $g\in G$, then the map $\phi_{g}:G\to G$ defined by the rule $\phi_g(x)=gxg^{-1}$ for $x\in G$ is referred to as conjugation by $g$. If $f:M\to N$ is a smooth map, then we denote by $T_{p}(f):T_{p}(M)\to T_{f(p)}(N)$ the differential of $f$ at $p\in M$.

Exercise 1: Let $G$ be a complex Lie group. The adjoint representation of $G$ is the map $\rho:G\to \text{GL}(\mathfrak{g})$ defined by the rule $\rho(g)=T_{e}(\phi_{g})\in \text{GL}(\mathfrak{g})$ where $e\in G$ is the identity element. Prove that $\rho:G\to \text{GL}(\mathfrak{g})$ is a complex (holomorphic) representation of $G$.

Exercise 2: In the context of Exercise 1, assume that $G$ is in addition compact and connected. Prove that the adjoint representation $\rho:G\to \text{GL}(\mathfrak{g})$ is trivial. (Hint: use the maximum modulus principle.) Conclude that $G$ is an abelian group.

We have now set the stage to prove that the exponential map $\text{exp}:\mathfrak{g}\to G$ is a covering map.

Exercise 3: Let $G$ be a Lie group. If $X,Y\in \mathfrak{g}$ commute, i.e., $[X,Y]=0$, then prove that $e^{X}e^{Y}=e^{X+Y}=e^{Y}e^{X}$. In the context of Exercise 2, conclude that $\text{exp}:\mathfrak{g}\to G$ is a homomorphism.

Exercise 4: Use the Hopf-Rinow theorem (or look it up if necessary) to prove that $\text{exp}:\mathfrak{g}\to G$ is surjective in the context of Exercise 2.

Exercise 5: Prove that $\text{exp}:\mathfrak{g}\to G$ is a covering map in the context of Exercise 2. (Hint: use Exercise 3 and Exercise 4. Recall that $\text{exp}:\mathfrak{g}\to G$ is a local homeomorphism at the origin of $\mathfrak{g}$ because its differential at the origin of $\mathfrak{g}$ is the identity map.)

I hope this helps!

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Thank you very much! So just to get this straight: now that we know that the universal cover of $G$ is $\mathfrak{g}$, which is a $g$-dimensional vector space ($g$ being the dimension of $G$), and that the covering map (the exponential map) is surjective, we know that $G$ is a quotient of a $g$-dimensional complex vector space by $\ker(\exp)$. And $\ker(\exp)$ is a lattice, because $G$ is compact. So $\exp$ is nothing else than the projection of $\mathfrak{g}$ onto the quotient $\mathfrak{g}/\ker(\exp) \cong G$. So $G$ is a complex torus. Is that right? – Nils Matthes Jul 13 '12 at 13:51
@NilsMatthes Yes. However, $\text{ker}(\text{exp}:\mathfrak{g}\to G)\subseteq \mathfrak{g}$ is a lattice because $\text{exp}:\mathfrak{g}\to G$ is a local homeomorphism (the compactness of $G$ is irrelevant in this deduction). – Amitesh Datta Jul 13 '12 at 15:13
I don't quite get your last comment. How does that follow without compactness of $G$? (maybe I'm just stupid...) – Nils Matthes Jul 14 '12 at 2:15
@NilsMatthes Let $V$ be a (real or complex) vector space. An additive subgroup $\Gamma\subseteq V$ is a lattice if and only if it is a discrete subset of $V$. Furthermore, an additive subgroup $\Gamma\subseteq V$ is discrete if and only if there is an open neighborhood $U\subseteq V$ of $0\in\mathfrak{g}$ such that $U\cap \Gamma=\{0\}$ (prove this). Finally, if $\text{exp}:\mathfrak{g}\to G$ is a local homeomorphism, then there is an open neighborhood $U\subseteq V$ of $0$ such that the restriction of $\text{exp}:\mathfrak{g}\to G$ to $U$ is injective; in particular, $U\cap \Gamma=\{0\}$. – Amitesh Datta Jul 14 '12 at 2:51
Thank you, I will check this out! – Nils Matthes Jul 14 '12 at 11:09