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?



Easy argument for "connected simple real rank zero Lie groups are compact"?
(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](http://books.google.com/books?id=oCO0xOzNLhAC&pg=PA284)):



> 
> **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.*
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> 
> 


(Of *that*, I don't know an "easier" proof than what you'll find in Bourbaki or [Hilgert-Neeb](http://books.google.com/books?id=PYWoqskGw1YC&pg=PA467).)


