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

$\begingroup$ Surely the question is only about complex flag manifolds. The real flag manifolds are not even hermitian. $\endgroup$ – José FigueroaO'Farrill Nov 5 '10 at 23:42

$\begingroup$ Yes, of course. I've put this in the question. $\endgroup$ – Dyke Acland Nov 5 '10 at 23:46

$\begingroup$ KKS metric on complex flag manifolds is Kahler. In fact any flag manifold can be seen as symplectic quotient of Kahler manifold. Since the symplectic quotient of Kahler manifold is Kahler, hence flag manifold is Kahler manifold $\endgroup$ – user21574 Jul 18 '17 at 1:43
Every flag manifold $M=G^{\mathbb{C}}/P=G/C(S)$ where $P$ is a parabolic subgroup and $C(S)=P\cap G$ is the centralizer of a torus $S\subset G$, admits a finite number of invariant Kähler 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) Kähler 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 homogeneous Kähler manifolds corresponding to a compact connected semisimple Lie group.
To be more specific, $M$ admits a finite number of Kähler structures which are parametrized by the wellknown $\frak{t}$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 semisimple part of) the complexification $\frak{k}^{\mathbb{C}}$ (note that $K=C(S)=P\cap G$ is a reductive 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 semisimple part of the reductive complex Lie subalgebra ${\frak{k}}^{\mathbb{C}}$
In particular, there exists a natural 11 correspondence between elements from ${\frak{t}}$ and closed invariant 2forms on $M$. Symplectic 2forms (nondegenerate) correspond to regular elements $t$ of ${\frak{t}}$.
Note that the corresponding symplectic form corresponding to a regular element $t_{0}$ is the KirillovKostantSouriau 2form 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, 1017 October 1993, 335. This article is a very good review on the geometry of flag manifolds.

3$\begingroup$ 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 $J$. $\endgroup$ – Robert Bryant Jun 25 '13 at 16:15

$\begingroup$ The quoted paper of Alekseevsky is online here: elib.mi.sanu.ac.rs/pages/browse_issue.php?db=zr&rbr=14 $\endgroup$ – Francois Ziegler Jul 22 '13 at 3:58

$\begingroup$ Clearly no complex manifold admits a finite number of Kähler structures, unless that number is zero, as you can rescale a Kähler metric. $\endgroup$ – Ben McKay Apr 14 '16 at 14:41
Yes. Use Plucker embedding to embed it into $CP^n$ then restrict FubiniStudy metric.

7$\begingroup$ 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.) $\endgroup$ – 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.