0
$\begingroup$

In this comprehensive answer to an old question, it is stated that

Flag manifolds exhaust all compact homogeneous Kähler manifolds corresponding to a compact connected semi-simple Lie group.

Firstly, where can I find a proof of this? Secondly, if I replace compact in the assumptions with complex is the result still true? Finally if one removes connected in the assumptions on the Lie group what happens.

Improve this question
$\endgroup$
7
  • $\begingroup$ Does it really matter, since (real) Lie groups are automatically (real-) analytic? $\endgroup$
    – M.G.
    Commented Oct 17, 2015 at 15:58
  • $\begingroup$ I've edited the question following your comment. $\endgroup$
    – Falertu Vatilski
    Commented Oct 17, 2015 at 16:03
  • $\begingroup$ You meant to link to this question (and accepted answer), not the one you did. $\endgroup$
    – Francois Ziegler
    Commented Oct 17, 2015 at 16:04
  • 5
    $\begingroup$ Also, the flag manifolds are not really "homogeneous Kähler" with respect to the complexified group, as the latter doesn't preserve the metric and 2-form. $\endgroup$
    – Francois Ziegler
    Commented Oct 17, 2015 at 16:09
  • 2
    $\begingroup$ @Francois As for "the flag manifolds are not really "homogeneous Kähler" with respect to the complexified group", that's interesting I did not know this! $\endgroup$
    – Falertu Vatilski
    Commented Oct 17, 2015 at 16:20

2 Answers 2

Reset to default
2
$\begingroup$

The comments make your second question moot unless reformulated, right?

For the first, this goes back to Borel (1954, Thms 1 & 2); more details in e.g. Serre (1954, Thms 1,2,3 and remark following Thm 1), Matsushima (1957, Thm 2), or Besse (1987, Thm 8.89).

Third question: a bit too vague, but you may be interested in what Duistermaat-Kolk (2000) say on p. 316.

Improve this answer
$\endgroup$
2
$\begingroup$

For the first question, there is another (paerhaps more conceptual) proof using the so called moment map of symplectic geometry. This proof, probably due to Souriau, is to consider the so-called moment map of this action. This map transform your manifold into a co-adjoint orbit of your semi-simple Lie group. One proves that this map is an isomorphism, not only for the symplectic, but also the complex structure. Souriau (Structure des systèmes dynamiques, Dunod, Paris, 1970, p 117. unfortunately written in a very high degree of generality).

For the second question, the proof I just sketched gives that your (non compact) manifold is a covering of a co-adjoint orbit. It appears that the fundamental group of the co-adjoint orbit, i.e. the orbit of a point under the co-adjoint action) is the set of connected components of the isotropy group is this point. Sometimes these objects are not simply connected. The proof that in the compact case, co-adjoint orbits are simply connected is not so easy.

For the last assertion, it is easy to make complicated non Kähler (hence non flag) compact complex examples. Consider the action of Sl(2,C) on C^2-(0,0), a plane without its origin. This action commute with homotheties, and therefore pass to the quotient in a transitive action of Sl(2,C) on the (family of) compact complex manifolds obtained by quotienting C^2-(0,O) by a given homothety. These manifolds are certainly not projective, as its first Betti number is 1.

Improve this answer
$\endgroup$

You must log in to answer this question.

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