I would like to get references or answers, for the following. How do I show that the anti-canonical line bundle (i.e. dual to top wedge power of cotangent bundle) on a flag variety (of a semisimple algebraic group in char. 0) is ample?

Thank you, Sasha

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    $\begingroup$ If you know already that the anticanonical divisor class is represented by the sum of all (closures of) codimension 1 Bruhat cells, then you can check as follows. First the linear system is base-point-free: indeed, by homogeneity for every point there is a translation which moves the representing point into the open Bruhat cell, hence the inverse translation moves the representing divisor to a linear equivalent divisor which "misses" the point. Thus the complete anticanonical linear system defines a morphism to projective space ... $\endgroup$ – Jason Starr Jun 12 '12 at 13:05
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    $\begingroup$ ... To prove this morphism is finite, and hence the anticanonical invertible sheaf is ample, it suffices to prove that no curve is contracted, i.e., every curve has positive intersection number with the representing divisor. For this curve, consider a specialization under the action of the Borel group (e.g., consider a Borel fixed point in the same connected component of the Hilbert scheme / Chow variety). The Borel fixed curves are precisely unions of (closures of) dimension 1 Bruhat cells. ... $\endgroup$ – Jason Starr Jun 12 '12 at 13:09
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    $\begingroup$ ... So to prove the original curve has positive intersection number with the representing divisor, it suffices to check for each dimension 1 Bruhat cell. This follows by self-duality of the basis of (closures of) Bruhat cells (for the Chow ring). $\endgroup$ – Jason Starr Jun 12 '12 at 13:10
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    $\begingroup$ Another reference for this (and generalizations) can be found in the nice survey of Michel Brion, Lectures on the geometry of flag varieties, Proposition 2.2.8. See explicitly www-fourier.ujf-grenoble.fr/~mbrion/notes.html $\endgroup$ – Karl Schwede Jun 12 '12 at 19:57
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    $\begingroup$ Jason's argument shows that $\rho$ is already ample. As Jim Humphreys' answer explains, the anticanonical class is $2\rho$. $\endgroup$ – Allen Knutson Jun 16 '12 at 14:36

It's worth pointing out that the criterion for ampleness applies uniformly for reductive groups and flag varieties in any characteristic. From the algebraic group viewpoint, the only subtle point is that the flag variety $G/B$ should involve a Borel subgroup $B$ corresponding to negative rather than positive roots. Here you have a fixed maximal torus $T\subset B$ whose character group $X(T)$ contains the root lattice.

Classical ideas in algebraic geometry related to this set-up in characteristic 0 are used by Demazure in his 1976 Invent. Math. paper "A very short proof of Bott's theorem". Choosing $B$ in this nonstandard way avoids a sign change in the weight which determines a line bundle. With this convention (opposite to yours), ample line bundles correspond precisely to dominant weights which are also regular (so that some large enough positive multiple is very ample). In particular, you are concerned about the weight $2\rho$: here $2\rho$ is the sum of all positive roots, corresponding to the highest wedge product of the positive part of the Lie algebra. It's an elementary observation about roots and weights that $\rho$ itself is both dominant and regular (the smallest such weight in the usual partial ordering). So it already defines an ample line bundle whenever $\rho \in X(T)$, e.g., when the derived group of the reductive group is simply connected.

All of this machinery carries over intact to prime characteristic, where it's exploited by Kempf and then by Henning Andersen to explore the cohomology of line bundles on a flag variety. One textbook reference worth consulting is Section II.4.4 of Jantzen's large AMS book (second edition, 2003): Representations of Algebraic Groups. There are many earlier references I could point out in the work I've mentioned. In any case, the "right" answer to the question asked depends partly on what language is most comfortable in this interface of algebraic geometry and algebraic groups. But the question itself is fairly narrow and doesn't require the detailed study of Schubert varieties.

ADDED: Here are a few (incomplete) remarks about history and sources. Papers by Demazure in the 1960s shifted the study of cohomology of line bundles on flag varieties into the realm of algebraic geometry, though he stayed mostly in characteristic 0. Andersen's many papers from the 1970s onward exploit creatively the ampleness of a line bundle attached to a regular dominant weight in prime characteristic. He was inspired by Demazure's proof of Bott's theorem and also by Iversen's foundational paper in Advances in Mathematics (1976), which showed how to formulate many ideas about algebraic geometry and algebraic groups over an arbitrary algebraically closed field. Meanwhile Kempf in the 1970s proved some deep results applicable to prime characteristic in this framework. His long 1978 Advances paper "The Grothendieck-Cousin complex of an induced representation" treats line bundles in detail in part I (see especially Lemma 5.3). For me his papers are quite challenging to read, however. The reference I gave to Jantzen's Chapter II.4 (on Kempf's vanishing theorem) deals more concisely with ampleness of line bundles in 4.3 and 4.4. (Here the approaches of Andersen and Haboush to Kempf's theorem are explained.)

CODA: I should return to the short question originally asked. The essential point is to identify the weight $\chi \in X(T)$ which determines the line bundle in question on a flag variety $G/B$ where $T$ is a maximal torus in $B$ and $B$ corresponds (say) to negative roots of $G$ relative to $T$. This requires extracting $\chi$ from the formalism of the cotangent bundle: this bundle involves the subspace $\mathfrak{n}$ of the Lie algebra (= tangent space) which is the sum of positive root spaces; here one knows that $\dim \mathfrak{n} = N = \dim G/B$, so the $N$th exterior power carries the (regular dominant) weight $2\rho$ = sum of positive roots. Note too that duality doesn't affect this particular weight.

  • $\begingroup$ Thank you. I looked into this article, "A very SIMPLE proof of Bott's theorem". I did not see ampleness there. Am I wrong and it can be deduced from there? $\endgroup$ – Sasha Jun 13 '12 at 7:14
  • $\begingroup$ @Sasha: I've added some remarks above, which I intended to do earlier. For me the algebraic groups approach in Jantzen's book is of course most accessible, but the earlier papers mentioned are essential sources. The detailed study of cohomology of line bundles on a flag variety in prime characteristic is still incomplete, though my speculative approach via Kazhdan-Lusztig theory for an affine Weyl group seems to be holding up well. Whether all of this will have further implications remains to be seen, however. $\endgroup$ – Jim Humphreys Jun 13 '12 at 13:29
  • $\begingroup$ Now I looked at Jantzen, II.4.4. This is exactly to the point! Thank you. $\endgroup$ – Sasha Jun 15 '12 at 7:08

Let me try to add to Jim Humphreys's comprehensive answer by pointing out that the (very) ampleness of line bundles coming from regular dominant weights was already observed by Borel and Weil, and essentially appears as Thm.4 in Serre's Bourbaki report

Représentations linéaires et espaces homogènes kählériens des groupes de Lie compacts (d'après Armand Borel et André Weil). Séminaire Bourbaki, Vol. 2, Exp. No. 100, 447–454, Soc. Math. France, Paris, 1995.

(See the example following Thm.4 on p.453.)

Although Borel and Weil work over $\mathbb C$ their argument is probably worth summarizing here. They begin by observing that the line bundle $L_\lambda \to G/B$ coming from a weight $\lambda$ is spanned by global sections iff $\lambda$ is dominant. In this case the image of the map $G/B \to \mathbb P (H^0(G/B,L_\lambda)^\ast)$ defined by sections is of the form $G/P$ where $P\supset B$ is the parabolic subgroup defined by the simple roots perpendicular to $\lambda$. Thus if $L_\lambda$ is not very ample, so that the aforementioned fibration $G/B \to G/P$ is not an embedding, one concludes that $P$ must properly contain $B$ and consequently that $\lambda$ must be perpendicular to some simple root, hence is not regular.

  • $\begingroup$ @Faisal: Thanks for calling attention to this earlier reference for this use of ampleness. (Kempf's 1976 theorem in prime characteristic essentially carries over the conclusion of Borel-Weil without appealing to Kodaira vanishing or the like.) $\endgroup$ – Jim Humphreys Jun 14 '12 at 12:52

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