Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a connected simple Lie group. It is known that if $G$ has real rank zero, then $G$ is compact.

Background: every connected (semi)simple Lie group $G$ (with Lie algebra $\mathfrak{g}$) has a polar decomposition $G=KAK$, where $K$ arises from a Cartan decomposition $\mathfrak{g}=\mathfrak{k} + \mathfrak{p}$ (the group $K$ has Lie algebra $\mathfrak{k}$) and $A$ is an abelian Lie group such that its Lie algebra $\mathfrak{a}$ is a maximal abelian subspace of $\mathfrak{p}$. The real rank of $G$ is defined as the dimension of $\mathfrak{a}$. (Equivalently, the real rank can be defined in a similar way through the Iwasawa decomposition (see http://en.wikipedia.org/wiki/Iwasawa_decomposition)).

A possible way of obtaining the result that "connected simple real rank zero implies compact" is by looking at a list of (connected) simple Lie groups (see for example http://en.wikipedia.org/wiki/List_of_simple_Lie_groups) and check that all such Lie groups with real rank zero are compact. This, however, is not as explanatory as I would like.

Question: Is there an easy and explanatory proof of the fact mentioned above that connected simple real rank zero Lie groups are compact?

Perhaps, one might be able to use that the Lie algebra of a noncompact simple Lie group contains a copy of the real rank one Lie algebra $\mathfrak{sl}(2,\mathbb{R})$?

An easy special case: If $G$ is a connected simple real rank zero Lie group with finite center, then the implication easily follows from the $KAK$-decomposition. Indeed, if $G$ has finite center, the group $K$ in this decomposition is a (maximal) compact subgroup of $G$. This directly implies that $G$ is compact if it has real rank zero.

Using Riemannian symmetric spaces: if $X$ is a Riemannian symmetric space, and $G$ is the connected component of the isometry group of $X$, then the real rank of $G$ is the largest $n \in \mathbb{N}$ such that $X$ contains an $n$-dimensional closed, simply connected, connected, totally geodesic flat submanifold. It follows that $G$ has real rank zero if and only $X$ is compact. This can also be used to prove the fact mentioned above, but as far as I know, one still needs to consider all possible spaces $X$ that might occur. Or is there an easier way here?

share|improve this question
1  
I do not understand what is the problem: If you use the language of symmetric spaces then every noncompact homogeneous Riemannian manifold contains a complete isometrically embedded geodesic (a 1-dimensional flat), which implies that rank is $\ge 1$. You do not "need to consider all possible spaces $X$" to make this conclusion. –  Misha Jul 31 '13 at 9:46
    
You are right. Thanks! However, as I am not so familiar with symmetric spaces, it would still help me to have an easy proof that does not use them at all. –  Tim de Laat Jul 31 '13 at 10:05

1 Answer 1

up vote 7 down vote accepted

Real rank 0 $\Rightarrow \mathfrak a = 0 \Rightarrow \mathfrak p=0 \Rightarrow \mathfrak g$ is a compact Lie algebra.

Therefore this comes down to (e.g. Bourbaki, Lie Groups, Chap IX, §1, no 4):

THEOREM (H. Weyl) Let G be a connected Lie group whose Lie algebra is compact semi-simple. Then G is compact and its centre is finite.

(Of that, I don't know an "easier" proof than what you'll find in Bourbaki or Hilgert-Neeb.)

share|improve this answer
    
Thanks a lot! Intuitively, the reason that $G$ automatically has finite center is what I was not aware of. –  Tim de Laat Jul 31 '13 at 11:04
1  
Since G/center = Ad(G) which is compact for compact $\mathfrak g$, Weyl's theorem $\Leftrightarrow \pi_1(\text{Ad}(G))$ is finite. Helgason quotes a number of proofs of that. Of those I especially like the short and conceptual one by Chevalley-Eilenberg (6 lines: ams.org/mathscinet-getitem?mr=24908, Theorem 16.1). –  Francois Ziegler Jul 31 '13 at 13:07

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.