A compact connected solvable Lie group is a torus

Question

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

Answer 1

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.

Answer 2

@Gis: I would like to add to Tim's answer the following:

1. The lie algebra $\mathfrak{g}$ admits a bi-invariant scalar product $<\,,>$, i.e. the adjoint action $\mathrm{ad}_x$ is antisymetric $\forall x\in\mathfrak{g}$ (since $G$ is compact, it admits a bi-invariant riemannian metric).

2. $\mathfrak{g}=\mathfrak{Z}(\mathfrak{g})\oplus D(\mathfrak{g})$, where $\mathfrak{Z}(\mathfrak{g})$ is the center and $D(\mathfrak{g})=\mathfrak{Z}(\mathfrak{g})^\perp$ is the derived ideal.

3. The restriction of the Killing form on $D(\mathfrak{g})$ is definite negative https://math.stackexchange.com/a/59193/14409. In particular it's semi-simple.

4. Since $\mathfrak{g}$ is solvable, its semi-simple part $D(\mathfrak{g})$ must be zero. So $\mathfrak{g}$ is abelian and also $G$.

5. The exponontial map $\exp : \mathfrak{g}\to G$ is surjective and since it's a local diffeomorphism its kernel is discrete (isomorphic to $\mathbb{Z}^n$) so $G\simeq \mathfrak{g}/\mathbb{Z}^n\simeq\mathbb{R}^n/\mathbb{Z}^n$.

P.S: Sorry for my bad English!