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Let $X$ be a (generalised) flag variety over an algebraically closed field $k$ of characteristic zero, that is to say, $X$ is a projective variety which is a homogeneous space for some algebraic group $G$. Any such $X$ can be realised as a quotient $X=G/P$, where $P$ is a parabolic subgroup (please correct me if this is wrong!)

Some basic properties of $X$ are:

  • $X$ is a fano variety
  • $X$ is unirational

The fact that it is fano is not perhaps immediately obvious, but it is clear that it is unirational since it is dominated by $G$, which is a rational variety. There is no reason to expect your average fano variety to be rational, however are flag varieties rational? I am particularly interested in the case where $dim X = 3$.

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Yes, they are, by the Bruhat decomposition. –  Piotr Achinger Apr 12 '11 at 11:18
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You must forgive me, but algebraic groups are not really my area. How does you use this result to show rationality? –  Daniel Loughran Apr 12 '11 at 13:46
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For the general proof, the reference and reasons Jim Humphreys provides is a good source. For intuition, think of the affine cell decomposition of the Grassmannian. (For explicit matrix representatives for this and the type A flag variety, see, e.g., Fulton's Young tableaux.) All $G/P$'s have such a cell decomposition. Another reference is Kumar's book on Kac-Moody groups. –  Dave Anderson Apr 12 '11 at 16:57
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up vote 5 down vote accepted

As indicated by Piotr, the answer to your question is yes. One helpful source is the Springer graduate text by Borel, Linear Algebraic Groups; see AG.13.7 and Chapter 14, especially 14.14. For the study of generalized flag varieties you may as well assume the group $G$ is reductive because the unipotent radical is factored out. Here the study of the Bruhat decomposition in $G$ over any algebraically closed field leads to the fact that $G$ itself is a rational variety. The basic idea is that $G$ has a Zariski-dense open subset which is just a product of copies of the additive and multiplicative groups of the field. Adapting this to $G/P$ is then straightforward though not written down explicitly by Borel. For instance, the usual flag variety $G/B$ has a dense open subset isomorphic to the product of copies of the additive group (one for each positive root); this mirrors the situation in the Lie algebra. In $G/P$ only some of the positive roots play this role. (As Dave Anderson indicates in his comment, the essential feature of Bruhat decomposition for flag varieties is a nice cell decomposition defined in terms of the root system and Weyl group.)

P.S. My convention here is that $B$ involves the negative roots, a convention useful for classical representation theory. But all Borel subgroups are conjugate, so it doesn't matter.

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Thanks for the useful answer! –  Daniel Loughran Apr 13 '11 at 8:51
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