In this short note we present a family of counterexamples to the King's
conjecture

Batyrev

Costa

Hille

Mateusz Michałek

English

arXiv

Michałek, Mateusz

Numérisation de Documents Anciens Mathématiques

C. R. Acad. Sci. Paris, Ser. I 349 (2011) 67–69Contents lists available at ScienceDirectC. R. Acad. Sci. Paris, Ser. Iwww.sciencedirect.comAlgebraic GeometryFamily of counterexamples to King’s conjecture ✩Famille de contre-exemples à la conjecture de KingMateusz Michałek a,ba Mathematical Institute of the Polish Academy of Sciences, Św. Tomasza 30, 31-027 Kraków, Polandb Institut Fourier, universite Joseph-Fourier, 100, rue des Maths, BP 74, 38402 St Martin d’Hères, Francea r t i c l e i n f o a b s t r a c tArticle history:Received 4 November 2010Accepted after revision 26 November 2010Available online 18 December 2010Presented by the Editorial BoardIn this short Note we present an infinite family of arbitrary high dimensional counterex-amples to the King’s conjecture.© 2010 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.r é s u m éDans cette Note nous présentons une famille infinie de contre-exemples à la conjecture deA. King, de dimension arbitrairement grande.© 2010 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.1. IntroductionIn [6] King made the following conjecture:Conjecture 1.1. For any smooth, complete toric variety X there exists a full, strongly, exceptional collection of line bundles.The conjecture turned out to be false. The first counterexample was given by Hille and Perling, in [4]. They showeda smooth, complete toric surface that does not have a full strongly exceptional collection of line bundles. Although verynice techniques were used, the proof was burdensome and required large tables of cases. Later counterexamples of higherdimension were found and the case of toric surfaces was classified [5]. The main purpose of this Note is to present a rathercompact, combinatorial proof that an infinite family of higher dimensional varieties does not satisfy King’s conjecture. Thevarieties we consider were suspected to be counterexamples and these are Pn blown up in two points. It turns out that forn sufficiently large the longest strongly exceptional collection of line bundles is (by a factor) shorter then the rank of theGrothendieck group, hence such a collection is far from being full.For more information on this topic the reader is advised to look in [2,3] and references therein.2. NotationThe varieties we consider are of Picard number 3. Such (smooth, complete, toric) varieties have been fully classifiedby Batyrev in [1] in terms of their primitive collections. Using this classification Pn blown up in two points is given by|X0| = |X2| = |X3| = |X4| = 1 and |X1| = n − 1 with all other parameters equal to 0. Choosing the basis of the one parameter✩ The author was supported by a grant of Polish MNiSW (N N201 413539).E-mail address: mateusz.michalek@ujf-grenoble.fr.1631-073X/$ – see front matter © 2010 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.doi:10.1016/j.crma.2010.11.02768 M. Michałek / C. R. Acad. Sci. Paris, Ser. I 349 (2011) 67–69subgroups lattice N equal to v1, y2, . . . , yn−1, z1 the ray generators of the fan are the basis elements and vectors y1, t1, u1satisfying:t1 = −z1, y1 = −y2 − · · · − yn−1 − z1 − v1, u1 = −v1.The rank of the Grothendieck group is equal to the number of maximal cones that is 3n − 1. To each element of Xicorresponds a T -divisor. All divisors in a given Xi are linearly equivalent and are given by D v , D y , Dz , Dt , Du respectivelyfor i = 0,1,2,3,4.In the strongly exceptional collection, differences of divisors cannot have nonzero higher cohomology. Divisors withnonzero higher cohomology will be called forbidden. The following classification of forbidden divisors is very easy to estab-lish. In a general case of Picard number three this has been done in [7], but in this special case one can use arguments ofelementary topology. The forbidden divisors in our case are α1 D v + α2 D y + α3 Dz + α4 Dt + α5 Du , where exactly 2,3 or 5consecutive (in a cyclic way, that is indices are considered modulo 5) α’s are negative and if α2 < 0, then α2  −n + 1.We have Dz = Dt + D y and D v = Du + D y . We choose the basis D y , Dt , Du , what gives us forbidden divisors (α1 +α2 +α3)D y + (α3 + α4)Dt + (α1 + α5)Du with the conditions on α’s as above. A divisor aD y + bDt + cDu will be denoted by(a,b, c) and we reserve precise letters for precise coordinates. A line bundle L1 will be called compatible with L2 iff theycan both appear in a strongly exceptional collection, that is iff L1 − L2 and L2 − L1 = −(L1 − L2) are not forbidden.3. ProofLet us fix a strongly exceptional collection E . We assume without loss of generality that 0 ∈ E and that all other divisorsin E have nonnegative coefficient a.Lemma 3.1. The only divisors with a = 0 compatible with (0,0,0) are:(0,−1,0), (0,0,−1), (0,1,0), (0,0,1), (0,−1,1), (0,1,−1).Proof. If b < −1, then we take α1 = 0, α2 = 1, α3 = −1, α4-negative to obtain b, α5-any to obtain c. Analogously for c < −1,hence −1  b, c  1. Moreover (0,−1,−1) is also bad (so also (0,1,1)). Corollary 3.2. There can be at most 3 distinct line bundles with a = 0 in E. For a fixed a we can have only 3 line bundles in E.Proof. Follows by inspection. Lemma 3.3. For a > 0 the only line bundles (a,b, c) that are not forbidden must satisfy −1  b  a and −1 + a − b  c  a (and bysymmetry −1 + a − c  b  a).Proof. For b < −1 we take α1 = 0, α3 = −1, α2-positive to have a, α4-negative to have b, α5-any to have c. For b > awe look at (−a,−b,−c) and take α3 = −a, α1 = α2 = 0, α4-negative to have −b, α5-any to have −c. In the same way−1  c  a.1 So the only case that we have to exclude is −1  c < −1 + a − b. In such a case we can take α4 = −1,α3 = b + 1, α2 = 0, α1 = a − b − 1, α5 = c − a + b + 1 < 0. Lemma 3.4. For three consecutive parameters a’s there can be at most 8 line bundles in E.Proof. We assume without loss of generality 0  a  2. If the lemma does not hold, then from Corollary 3.2 we would haveto have 3 line bundles for each a. For a = 0 we can have either:Case 1: (0,0,0), (0,−1,0), (0,0,−1) then for a = 1 there is only one compatible from Lemma 3.3 namely (1,0,0).Case 2: (0,0,0), (0,1,0), (0,0,1) then for a = 1 the compatible line bundles are (1,1,1), (1,1,0), (1,0,1). If we chooseall of them then the only one compatible for a = 2 is (2,1,1) from Lemma 3.3.Case 3: (0,0,0), (0,−1,0), (0,−1,1); (0,0,0), (0,0,−1), (0,1,−1); (0,0,0), (0,1,0), (0,1,−1); (0,0,0), (0,0,1), (0,−1,1). All these possibilities are cases 1 or 2 after subtracting a divisor from all three considered divisors. Definition 3.5. Line bundles in the collection E with a > n are called high. Others are called low.Lemma 3.6. A high line bundle is forbidden unless either b = 1 (high bundles of type 1) or c = 1 (high bundles of type 2).Proof. Suppose that b = 0 or b = −1. We show that (−a,−b,−c) is forbidden. Take α1 = −1, α2 = −a +1, α3 = 0, α4 = −b,α5-any to obtain −c. So b  1 and analogously c  1. If both coefficients were strictly greater than 1 we would obtain(−a,−b,−c) by taking all α’s negative. 1 The parameters b and c are in symmetry.M. Michałek / C. R. Acad. Sci. Paris, Ser. I 349 (2011) 67–69 69Lemma 3.7. We cannot have high line bundles of both types in E.Proof. From Lemma 3.3 a high line bundle must have the coordinate different from 1 greater or equal to n − 1. If wesubtract two high line bundles of different types we can assume that the first coordinate is positive and one of the otherswill be less or equal to −n + 2 what contradicts Lemma 3.3 for n > 3. From now on without loss of generality we assume that we only have high line bundles of type 1 in E . Let us projectall high line bundles from E onto the first coordinate obtaining a subset of N. Suppose that this subset has got k elements,that is high line bundles can have k different parameters a. We obtain:Lemma 3.8. There are at most k + 2 high line bundles in E.Proof. We assumed that 0 ∈ E , so the high line bundles in E must not be forbidden. We know that for each high linebundle in E we have b = 1, so from Lemma 3.3 we know that 0  a − c  2. Let us notice that the difference a − c cannotdecrease when a increases for high line bundles in E . Indeed suppose that we have two high line bundles in E of the form(a1,1, c1), (a2,1, c2) with a2 > a1 and a2 − c2 < a1 − c1. By subtracting these two line bundles we obtain (a2 −a1,0, c2 − c1)that is forbidden by Lemma 3.3.Notice that each time we have more than one line bundle for a fixed a then the difference a − c strictly increases. Thismeans that we can have one line bundle for each a plus possibly two more as a − c increases from 0 to 2. This gives us intotal k + 2 line bundles. Proposition 3.9. There are at most 83 (n − 1) + 6 low line bundles (from Lemma 3.4), so k > 0 for n > 13.Remark 3.10. Of course k is at most n + 1. Otherwise we would have two high line bundles in E with the difference that ishigh. By Lemma 3.7 the difference would have b = 0, hence by Lemma 3.6 it would have c = 1 and would be forbidden byLemma 3.3.From the definition of k we know that there is a line bundle L = (a,1, c) in E , with a  n + k. Now we investigate linebundles with a < k, that are called very low.Lemma 3.11. Each very low line bundle in E must have b = 0.Proof. Let B be a very low line bundle. L − B is high, so from Lemma 3.6 either the second or third coordinate is 1. Thethird one is cL − cB  aL − 2 − aB > n + k − 2 − k = n − 2 > 1, for n > 3. We see that bL − bB = 1. As bL = 1 the lemmafollows. For very low line bundles in E the parameter c is either a or a−1 by Lemma 3.3 and Lemma 3.11. Reasoning analogouslyto the proof of Lemma 3.8, we see that there are at most k + 1 very low line bundles (the difference a − c cannot decrease).Theorem 3.12. The sequence E can have at most: k + 1 + 83 (n − k − 1) + 6 + k + 2  83 n − 23 k + 193 < 3n − 1 for n > 20.Remark 3.13. The bounds on n can be easily improved. For example by considering separately the case k = 1 one candecrease the bound to n > 18. We concentrated rather on brevity of the proof than sharp bounds.AcknowledgementsThe author would like to thank very much Rosa Maria Miró-Roig and Laura Costa for inspiring talks, examples and ideas.I also thank Markus Perling for important remarks and careful reading of the article.References[1] V.V. Batyrev, On the classification of smooth projective toric varieties, Tôhoku Math. J. 43 (1991) 569–585.[2] L. Costa, R.M. Miró-Roig, Derived category of toric varieties with small Picard number, preprint.[3] L. Costa, R.M. Miró-Roig, Tilting sheaves on toric varieties, Math. Z. 248 (4) (2004) 849–865.[4] L. Hille, M. Perling, A counterexample to King’s conjecture, Compos. Math. 142 (2006) 1507–1521.[5] L. Hille, M. Perling, Exceptional sequences of invertible sheaves on rational surfaces, Compos. Math., arXiv:0810.1936, in press.[6] A. King, Titling bundles on some rational surfaces, preprint at http://www.maths.bath.ac.uk/~masadk/papers/.[7] M. Lason, M. Michalek, On the derived categories of toric varieties with Picard number three, preprint.

Family of counterexamples to King's conjecture

Comptes Rendus Mathématique

Crossref

http://www.numdam.org/item/10.1016/j.crma.2010.11.027.pdf

Family of counterexamples to King's conjecture

Abstract

Similar works

Full text

Available Versions

Numérisation de Documents Anciens Mathématiques

Comptes Rendus Mathématique

Crossref