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Does every complex flag manifold have a natural Kähler structure? If so, what is it?

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Surely the question is only about complex flag manifolds. The real flag manifolds are not even hermitian. –  José Figueroa-O'Farrill Nov 5 '10 at 23:42
    
Yes, of course. I've put this in the question. –  Dyke Acland Nov 5 '10 at 23:46

3 Answers 3

up vote 4 down vote accepted

Every flag manifold $M=G^{\mathbb{C}}/P=G/C(S)$ where $P$ is a parabolik subgroup and $C(S)=P\cap G$ is the centralizer of a torus $S\subset G$, admits a finite number of invariant Kahler structures. In particular the complex presentation $G^{\mathbb{C}}/P$ gives rise to an finite number of invariant complex structures (i.e. integrable almost complex structures commuting with the isotropy representation of $M$). Any such complex structure is determined by an invariant ordering $R_{M}^{+}$ on the set of complementary roots $R_{M}=R\backslash R_{K}$ of $M$ and explicitly is given by $$ J_{o}E_{\pm \alpha}=\pm i E_{\pm\alpha}, \quad a\in R_{M}^{+} $$ where $E_{\alpha}$ are root vectors with respect a Weyl basis of $\frak{g}^{\mathbb{C}}$.

On the other hand, the real presentation $G/C(S)$ makes $M$ a (homogeneous) Kahler manifold, as a (co)-adjoint orbit of an element $w\in\frak{g}$ in the Lie algebra $\frak{g}$ of the compact connected (semi)simple Lie group. Flag manifolds exhaust all compact homgeneous Kahler manifolds corresponding to a compact connected semi-simple Lie group.

To be more specific, $M$ admits a finite number of Kahler structures which are parametrized by the well-known $\frak{t}$-chambers (connected components of the set of regular elements of $\frak{t}$) where

$$ {\frak{t}} =( H\in{\frak{h}} : (H, \Pi_{0})=0 ) $$

is a real form of the center ${\frak{z}}$ of the isotropy subgroup $K=C(S)$.

Here $\frak{h}$ is the Cartan subalgebra corresponding to a maximal torus $T$ of $G$ which contains $S$, and $\Pi_{0}\subset\Pi$ is the subgroup of simple roots which define (the semi-simple part of) the complexification $\frak{k}^{\mathbb{C}}$ (note that $K=C(S)=P\cap G$ is areductive Lie group). We have

$$ {\frak{z}}^{\mathbb{C}}={\frak{t}}\oplus i {\frak{t}}, \ \ \ {\frak{k}}^{\mathbb{C}}={\frak{z}}^{\mathbb{C}}\oplus{\frak{k}}_{ss}^{\mathbb{C}} $$

where ${\frak{z}}^{\mathbb{C}}$ is the complexification of the center ${\frak{z}}$ and ${\frak{k}}_{ss}^{\mathbb{C}}$ is the semi-simple part of the reductive complex Lie subalgebra ${\frak{k}}^{\mathbb{C}}$

In particular, there exists a natural 1-1 corespondence between elements from ${\frak{t}}$ and closed invariant 2-forms on $M$. Symplectic 2-forms (non-degenerate) correspond to regular elements $t$ of ${\frak{t}}$.

Note that the corresponding symplectic form corresponding to a regular elemnt $t_{0}$ is the Kirillov-Kostant-Souriau 2-form in the (co)-adjoint orbit $Ad(G)t_{0}$, that is

$$ \omega_{t_{0}}(X, Y)=B(t_{0}, [X, Y]), \ \ X, Y\in T_{t_{0}}M. $$

For more details see: D. Alekseevsky: Flag manifolds (11. Yugoslav Geometrical seminar, Divcibare, 10-17 October 1993, 3-35. This article is a very good review on the geometry of flag manifolds.

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Great answer, thanks a lot! –  Dyke Acland Mar 21 '12 at 20:41
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While the space of $G$-invariant complex structures on the flag manifold is finite, the space of $G$-invariant compatible Kähler structures is only finite dimensional in general, rather than finite. For example, consider the simplest flag variety $F_{1,2} = \mathrm{SU}(3)/\mathbb{T}^2$. The space $S$ of closed, $\mathrm{SU}(3)$-invariant $2$-forms on $F_{1,2}$ has dimension $2$, and, for each of the $\mathrm{SU}(3)$-invariant complex structures $J$, there is an open set $S_J\subset S$ that consists of Kähler forms compatible with $J$. –  Robert Bryant Jun 25 '13 at 16:15
    
The quoted paper of Alekseevsky is online here: elib.mi.sanu.ac.rs/pages/browse_issue.php?db=zr&rbr=14 –  Francois Ziegler Jul 22 '13 at 3:58

Yes. Use Plucker embedding to embed it into $CP^n$ then restrict Fubini-Study metric.

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More concretely, let $\lambda$ be a dominant weight of $G$, and $V_\lambda$ its corresponding irrep. Then the unique closed $G$-orbit on ${\mathbb P}V$ is a flag manifold $G/P$, now projectively embedded. (Which $P$ arises depends on which wall of the Weyl chamber $\lambda$ lies on.) –  Allen Knutson Nov 6 '10 at 4:12

The question has already been answered by Bugs Bunny, but I thought I'd point out that there is a nice paper by H.-C. Wang from the 1950s that discusses the complex structure of homogeneous manifolds in some detail. One of the results proved there is that a compact, simply connected complex homogeneous manifold (such as a complex flag manifold) is Kähler if and only if it has nonzero (ordinary) Euler characteristic. That complex flag manifolds have nonzero Euler characteristics follows, for example, from the Bruhat decomposition.

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