0
$\begingroup$

To what extent, the following types of Lie algebras are classified :

Those Lie algebras $L$ such that every Lie Group $G$ with $Li(G)\sim L$, is necessarily compact.

$\endgroup$
2
  • 1
    $\begingroup$ There are completely classified, by the classical theory of semisimple Lie algebras. Such a Lie algebra is necessarily semisimple, and the Killing form is negative definite. $\endgroup$ Commented Jan 23, 2014 at 9:36
  • 1
    $\begingroup$ There is an entire section on the topic entitled "Lie groups with compact Lie algebras" in the very readable book Hilgert-Neeb "Structure and Geometry of Lie Groups" $\endgroup$ Commented Jan 24, 2014 at 19:50

2 Answers 2

6
$\begingroup$

They are exactly the semi-simple Lie algebras which admit a positive definite invariant bilinear form (or equivalently, for which the Killing form is negative definite). An excellent reference for this is the first section of Bourbaki, Lie groups and Lie algebras, Chapter 9 (you need very little background to be able to read it).

$\endgroup$
2
  • $\begingroup$ thanks for the answer. at this time the book is not available to me. do you mean that for such Lie algebras, we are sure that there is no non compact lie group $G$ with $Li(G)\sim L$? $\endgroup$ Commented Jan 23, 2014 at 9:51
  • 1
    $\begingroup$ @Ali: yes. It suffices to show that the unique simply connected Lie group with a given Lie algebra satisfying the above condition is compact and this can be done. $\endgroup$ Commented Jan 23, 2014 at 10:11
2
$\begingroup$

They are the compact Lie algebras. Note that are two different definitions in the literature. One is that a compact Lie algebra is the Lie algebra of a compact Lie group. This includes tori, and the Killing form then is negative semidefinite. The Lie algebra is not necessariliy semisimple, but reductive. The other definition is, that a compact Lie algebra is a real Lie algebra whose Killing form is negative definite; this definition is more restrictive and excludes tori.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .