Compact connected iff semi-simple for Lie Groups?

Question

Are compact & connected Lie Groups in correspondence with semi-simple Lie groups? I think there is a condition on the center (discrete?) but I'm not sure.

Answer 1

The answer to your title is "no"; lots of semi-simple Lie groups are not compact (for example, $SL_2(\mathbb{R})$). You're getting this mixed up with the fact that a complex semi-simple Lie group has a unique compact real form, and that this is a bijection to semi-simple compact Lie groups. (Complex reductive groups are in bijection with general compact Lie groups; this allows torus factors on both sides).

Answer 2

To amplify Ben's answer, I'd point to an earlier post that has lots more detail: here. The subject of compact groups is old and well-studied, so there are many references to choose from, even Wikipedia perhaps. Anyway, it's good to browse older Lie group entries on MO first.

PS: This supplementary "answer" is really a suggestion that the question is too close to the earlier post I cited to qualify as a fresh question. Textbook material of this kind calls mainly for references rather than discussion.

Answer 3

I would like to add the following:

I think the source of confusion is the fact that the Killing form is nondegenerate (for semi-simple Lie groups) and negative definite (stronger than non-degenerate) for compact Lie groups with discrete center

$SL(2)$ is semi-simple but not compact.

The torus $S^1$ is compact but not semi-simple (abelian).

Compact groups are reductive and semi-simple only when in the case of discrete center.

Answer 4

I hope it will be useful (even after many years) to make a precise statement of the relation between compact Lie groups and semisimple Lie groups.

The classification of connected compact Lie groups is: every connected compact Lie group is a quotient $(G\times T)/A$ where $G=\prod_i G_i$ is a product of finitely many compact, connected and simply connected simple Lie groups, each with center say $Z_i\subseteq G_i$, and $T$ is a torus $T=\mathbb{R}^n/\mathbb{Z}^n$ and we let $Z:=\prod_i Z_i$ and $A\subseteq Z \times T$ is a finite abelian subgroup so isomorphic to a product $A=\prod_a \mathbb{Z}/n_a\mathbb{Z}$ of cyclic groups. Conversely, any choice of compact connected and simply connected simple Lie groups $G_i$, integer $n\ge 0$ and finite subgroup $A\subset Z\times T$ gives rise to a compact connected Lie group $(G\times T)/A$. For any connected Dynkin diagram $\Delta$, there is one compact, connected and simply connected Lie group $G=G_{\Delta}$ whose Lie algebra has root system with that Dynkin diagram. Hence the classification reduces to a finite set of connected Dynkin diagrams, an integer $n$, and the finite subgroup $A$ of a compact abelian group.

For an elegant and complete proof, see Claudio Procesi, Lie groups, Universitext, Springer, New York, 2007, An approach through invariants and representations. MR MR2265844 (2007j:22016)chapter 10, section 7.2, theorem 4, page 380.

Note that, by the explicit construction (due to Lie, Engel, Killing and Cartan) of compact, connected and simply connected Lie groups associated to the Dynkin diagrams, we know that each one is a real linear algebraic group, and complexifies to a unique complex simple Lie group. Every compact Lie group is thus isomorphic to the real points of a unique real linear algebraic group each of whose complex components contains a real point and which contains no linear algebraic Zariski closed $\mathbb{R}^{\times}$ subgroup; moreover the linear algebraic group is semisimple just when the compact Lie group has finite center.