Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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 1

up vote 2 down vote accepted

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!

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.