I am looking for the proof of the following statement.
A compact connected solvable Lie group of dimension $n\geq 1$ is a torus, i.e., it isomorphic to the product of $n$ copies of $S^1$.
A search with Google revealed that pages 51-52 of "Lie Groups and Lie Algebras III: Structure of Lie Groups and Lie Algebras" (A.L. Onishchik) deal with this problem but the proofs are not very detailed and I'm not an expert.
Thanks,
Gis
Best Answer
The Lie algebra, $\frak{g}$, of a compact Lie group, $G$, is reductive, meaning its a direct sum of an abelian Lie algebra and a semisimple Lie algebra. This is not difficult to prove. Here's a proof http://books.google.com/books?id=PAJmnm3DU1QC&pg=RA1-PA249#v=onepage&q&f=false
(The proof there uses an equivalent definition of reductive: every ideal has a complimentary ideal. But proving the equivalence is not difficult.)
If $\frak{g}$ is solvable, the semisimple part must be zero. So $\frak{g}$ is abelian. Therefore $G$ is a torus.
If I've skipping too many details, please let me know.