Does every complex flag manifold have a natural Kähler structure? If so, what is it?
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$\begingroup$ Surely the question is only about complex flag manifolds. The real flag manifolds are not even hermitian. $\endgroup$– José Figueroa-O'FarrillNov 5, 2010 at 23:42
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$\begingroup$ Yes, of course. I've put this in the question. $\endgroup$– Dyke AclandNov 5, 2010 at 23:46
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$\begingroup$ 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 $\endgroup$– user21574Jul 18, 2017 at 1:43
3 Answers
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 semi-simple Lie group.
To be more specific, $M$ admits a finite number of Kähler 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 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 semi-simple part of the reductive complex Lie subalgebra ${\frak{k}}^{\mathbb{C}}$
In particular, there exists a natural 1-1 correspondence 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 element $t_{0}$ 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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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 $J$. $\endgroup$ Jun 25, 2013 at 16:15
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$\begingroup$ The quoted paper of Alekseevsky is online here: elib.mi.sanu.ac.rs/pages/browse_issue.php?db=zr&rbr=14 $\endgroup$ Jul 22, 2013 at 3:58
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$\begingroup$ Clearly no complex manifold admits a finite number of Kähler structures, unless that number is zero, as you can rescale a Kähler metric. $\endgroup$ Apr 14, 2016 at 14:41
Yes. Use Plucker embedding to embed it into $CP^n$ then restrict Fubini-Study metric.
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8$\begingroup$ 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.) $\endgroup$ Nov 6, 2010 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.