Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I work over an algebraically closed field $k$ of characteristic zero.

Recall that a flag variety is a projective variety which is a homogeneous space for some semisimple algebraic group. Every flag variety is of the form $G/P$, where $G$ is a semisimple algebraic group and $P$ is a parabolic subgroup. It seems to me that this data is "discrete", so I expect flag varieties to have discrete moduli. Moreover flag varieties are Fano varieties, and it is known that there are only finitely many deformation types of Fano varieties of fixed dimension. This leads to my question.

For each $n \in \mathbb{N}$, are there are only finitely flag varieties of dimension $n$, up to isomorphism?

A proof/disproof or a reference for this would be much appreciated. By "up to isomorphism" I mean up to isomorphism as an algebraic variety.

share|improve this question
4  
Yes. Use classification of semisimple algebraic groups in terms of Dynkin diagrams + classification of parabolic subgroups in terms of subsets of Dynkin diagrams. Lower bound dimension of flag variety in terms of dimension of group, giving finiteness. –  Will Sawin Feb 23 at 18:15
    
The thing that is worrying me is something like the possible existence of an infinite sequence $P_i \subset G_i$ ($i=1,\ldots,\infty$) where $\dim G_i \to \infty$, but $\dim G_i/P_i = n$ is fixed. In which case one needs to show that there are only finitely many isomorphism classes of varieties in the set $\{G_i/P_i: i=1,\ldots,\infty\}$. Does your approach achieve this? –  Daniel Loughran Feb 23 at 18:39
7  
Ridiculous observation: in positive characteristic $p$, for every integer $n$, the following smooth, projective, $3$-dimensional variety, $\{([X_0,X_1,X_2],[Y_0,Y_1,Y_2])\in \mathbb{P}^2\times \mathbb{P}^2 : X_0Y_0^{p^n} + X_1Y_1^{p^n} + X_2Y_2^{p^n}\}$, is homogeneous for an action of $\textbf{SL}_3$. –  Jason Starr Feb 23 at 20:11
1  
@Daniel, Ben Webster has already given a slick conceptual proof, but at least in this setting (alg closed, char 0) an easy direct proof comes by treating semisimple $G$'s as a list of examples. You can just use Will's/Jim's suggestion. (After reducing the problem from semisimple to simple there are only three (or four) infinite series of flag varieties: $SL_n/P$, $Sp_{2n}/P$, and $SO_n/P$. By direct inspection, each of these has dimension $\to \infty$ as $n\to\infty$.) –  Dave Anderson Feb 24 at 0:14
1  
@pmath. Flag varieties are examples of Fano varieties. Thus they live in the moduli of Fano varieties, i.e., "anti-canonically polarized varieties". Although the moduli of flag varieties is zero-dimensional, the moduli of Fano's is positive dimensional in general. Consider, for example, Del Pezzo surfaces of degree 1,2,3, or 4. –  Ari Feb 27 at 8:15

1 Answer 1

up vote 8 down vote accepted

Yes, there are only finitely many. One only needs to observe that for a given group, the variety $G/P$ has dimension at least the rank of the group $G$ (the dimension of a maximal torus). You can see this by inspecting cases by hand, but there's also a conceptual reason: the maximal torus of the adjoint group $G$ acts on $G/P$ faithfully with isolated fixed points, so it acts faithfully on the tangent space of some fixed point. Since a faithful module over a torus must have dimension $\geq$ that of the torus, this establishes the desired result.

Thus, for a given dimension $n$, there can only be flag manifolds for groups with rank $\leq n$, there are finitely many of these, and only finitely many flag manifolds for each one. Q.E.D.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.