Are compact & connected Lie Groups in correspondence with semisimple Lie groups? I think there is a condition on the center (discrete?) but I'm not sure.
4 Answers
The answer to your title is "no"; lots of semisimple Lie groups are not compact (for example, $SL_2(\mathbb{R})$). You're getting this mixed up with the fact that a complex semisimple Lie group has a unique compact real form, and that this is a bijection to semisimple compact Lie groups. (Complex reductive groups are in bijection with general compact Lie groups; this allows torus factors on both sides).

$\begingroup$ Do you happen to know a source for the bijection between (possibly disconnected) complex reductive groups and (possibly disconnected) compact Lie groups? $\endgroup$– MaximeSep 26, 2013 at 4:49
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 wellstudied, 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.
I would like to add the following:
I think the source of confusion is the fact that the Killing form is nondegenerate (for semisimple Lie groups) and negative definite (stronger than nondegenerate) for compact Lie groups with discrete center
$SL(2)$ is semisimple but not compact.
The torus $S^1$ is compact but not semisimple (abelian).
Compact groups are reductive and semisimple only when in the case of discrete center.

2$\begingroup$ Dear amine, "Trivial center" should perhaps read "finite centre". Regards, Matthew $\endgroup$– EmertonSep 11, 2011 at 3:00

$\begingroup$ @Emerton: Thanks Emerton! Yes, 'trivial center' for Lie algebra which means 'finite center' for Lie groups. $\endgroup$– amineSep 11, 2011 at 4:48
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.