Horrocks correspondence

Transcription

1 Horrocks correspondence Department of Mathematics and Computer Science University of Missouri-St.Louis May, 2016 (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

2 P n over a field k. For any sheaf B on P n, H i (B) := k Z H i (X, B(k)), a graded module over the polynomial ring S. H 0 (F), then Free resolution for F : F = H 0 (F ). 0 P n 1 P 3 P 2 P 1 P 0 F 0. (1) (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

3 Dualize this: get a complex, with cohomology modules of finite length 0 F P 0 P 1 P 2 P n 2 P n 1 0. (2) or sheafified: get an exact sequence of bundles 0 F P 0 P 1 P 2 P n 2 P n 1 0. (3) (Notation: P i is a free module, P i is a free bundle/sheaf.) From this it becomes clear that Next resolve F and splice it to P. H i (F) = H i (P ), 0 i n 1 (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

4 Get enlarged P, Properties: 0 P n P 0 P n 1 0 H i (P ) = 0 if i / {1,..., n 1}, H i (P ) = finite length module if i {1,..., n 1} Horrocks correspondence is to match a bundle F to such a complex of free modules. Horrocks calls it a Z-complex. rabhakar Rao (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

5 Get enlarged P, Properties: 0 P n P 0 P n 1 0 H i (P ) = 0 if i / {1,..., n 1}, H i (P ) = finite length module if i {1,..., n 1} Horrocks correspondence is to match a bundle F to such a complex of free modules. Horrocks calls it a Z-complex. Given another such complex Q quasi-isomorphic to P, let G = ker(q 0 Q 1 ), let G be the sheafification of the module G. Then G and F are stably isomorphic: ie. There exist free bundles R, S such that F R = G S. rabhakar Rao (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

6 Call such a Z-complex a Horrocks data complex since it contains as data all the intermediate cohomology of a bundle as well as data of a bundle F with that cohomology. Horrocks data complexes modulo quasi-isomorphisms vector bundles modulo stable equivalence Fact: There are non-isomorphic bundles F 1, F 2 on P 3 with identical intermediate cohomology modules: for null-correlation bundles, H 1 (F i ( 1) = k, H 2 (F i ( 3) = k are the only non-zero cohomologies. (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

7 Horrocks Correspondence Theorem: Let F, G be two vector bundles on P n with no free summands. If φ : F G induces isomorphisms then φ is an isomorphism. H i (φ) : H i (F) H i (G), for all 1 i n 1, Horrocks Splitting Criterion: If F is a vector bundle on P n and for all 1 i n 1, H i (F) = 0, then F is free (F = j O P n(a j)). (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

8 ACM varieties X n a smooth ACM variety with respect to the polarization O X (1). This implies that H i (O X ) = 0 for all 1 i n 1. A = H 0 (O X ) is a graded noetherian ring. E a vector bundle on X. H i (E) is an A-module of finite length for any 1 i n 1. A vector bundle B on X is called an ACM bundle if H i (B) = 0 for all 1 i n 1. Examples: O X (a), ω X (b), free bundles j O X (a j ). E bundle on X. E := H 0 (E) an A-module. E = H 0 (E ). Repeat Horrocks steps. (Joint work with F. Malaspina) (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

9 Free resolution for E : but stop after C n 2 0 K C n 2 C 3 C 2 C 1 C 0 E 0. (4) K sheafified gives K an ACM bundle. Dualize. Sheafify: 0 E C 0 C 1 C n 2 K 0. (5) 0 E C 0 C 1 C n 2 K 0. (6) H i (E) = H i (C ) for all i n 1. Next resolve E (maybe infinite free resolution) and splice it to C. (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

10 Get enlarged C... C m C m+1 C 0 C 1 C n 2 K 0 with cohomology supported in degrees 1, 2,..., n 1. Choose a free resolution P of C. P has the same cohomology, supported in degrees 1, 2,..., n 1. Call P a Horrocks data complex on X. Let F equal kernel P 0 P 1. F is F sheafified. Call F a Horrocks data bundle on X. We get a diagram (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

11 0 F P 0 δ 1 P P 1 δ 2 P... δn 2 P P n 2 P n 1 0 β 0 E C 0 δ 1 C C 1 δ 2 C and β : F E induces H i (F) = H i (E). Properties of F... δn 2 C C n 2 K 0 F is determined by E up to direct sum with a free bundle. The dual gives a finite free resolution for F. F has no summands of the form B or B, where B is a non-free ACM bundle. The complex can be replaced by a minimal complex giving β : F m E, where F m has no free summand. For any two minimal Horrocks data bundles F, F, if φ : F F induces an isomorphism on intermediate cohomology modules, then φ is an isomorphism. (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

12 History: Old ideas going back to Auslander. β : E F can be modified to 0 N E L F 0 where N is a maximal Cohen Macaulay module when X is arithmetically Gorenstein and where L is free. Auslander s approximation theorem. Buchweitz extends this to non-commutative rings. (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

13 Constructing (F, β) given E will be called finding Horrocks data for E. Choosing a minimal complex 0 F m P 0 m P 1 m... gives minimal Horrocks data (F m, β) for E. Push out the short exact sequence 0 F m P 0 m G m 0, using β, getting 0 E A(E) G m 0. A(E) is an ACM bundle determined by E. (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

14 Correspondence Theorem I Let σ : E E where E and E have no ACM summands and H i σ is an isomorphism for 1 i n 1. Suppose that one of the following holds: H 1 (E A(E ) ) = H 1 (E A(E ) ). H 1 (E B ) = H 1 (E B ) for all irreducible ACM bundles B. Then σ is an isomorphism. Correspondence Theorem II Given bundles E, E with no ACM summands and suppose they have Horrocks data (F m, β), (F m, β ) with the same F m. For any ACM bundle B, let V B denote the kernel of H 1 (F m B ) H 1 (E B ). Likewise define V B for E. If one of the following holds: V B = V B for each irreducible non-free summand B appearing in A(E) or A(E ). V B = V B for all irreducible non-free ACM bundles B Then E = E. (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

15 Let B be an ACM bundle, j O X (a j ) B a surjection onto global sections. F m gives rise to H 1 (F m B ) H 1 (F m j O X ( a j )). H 1 (F m B ) soc will denote the kernel of this map and will be called the sub-module of H 1 (F m B ) consisting of B-socle elements. Given E, Horrocks data (F m, β) and an ACM bundle B, V B H 1 (F m B ) soc. Reason: H 1 (F m B ) H 1 (F m j O X ( a j )) H 1 (E B ) H 1 (E j O X ( a j )) (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

16 Conversely, given a minimal Horrocks data bundle F m, any element α H 1 (F m (t) B ) soc gives rise the the extension 0 F m β D B( t) 0. Pullback by any section O X (a) B is split hence (F m, β) gives Horrocks data for D. Question Given F m, what are all the submodules V B H 1 (F m B ) soc that arise from bundles E with F m as Horrocks data bundle? (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

17 Splitting Criterion? Mainly a tautology: Requirements for E to be free of the form j O X (a j ): H i (E) = 0, 1 i n 1. For all ACM bundles B, H 1 (E B) = 0. But if E is a non-free ACM bundle, then a short resolution of E 0 C j O X (e j ) E 0 yields C is ACM and H 1 (E C) 0. (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

18 Example 1 Q n quadric hypersurface of odd dimension. Knörrer proved Q has (up to twists) a unique non-free irreducible ACM bundle: Σ the spinor bundle. Knowing this the splitting criterion becomes Ottaviani s theorem. Correspondence Theorem I simplifies considerably. Correspondence Theorem II can be pursued further. (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

19 H 1 (F m Σ ) soc as an A-module is annihilated by the ideal (X 0, X 1,..., X n+2 ), hence is just a graded vector space. Theorem For any graded subspace V of H 1 (F m Σ ) soc, there exists a bundle E with invariants (F m, V ). Proof: A basis of V yields an element in H 1 (F m V ) where V = V k Σ. Hence get E where 0 F m E V 0 More work with Σ is needed to see that now V Σ is not larger than V. Conclusion Up to sums of ACM bundles, { E } { (F m, V ) }. (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

20 Example 2 X equals Veronese surface in P 5, O X (1) the restriction of O P 5(1). The only non-free indecomposable ACM bundles (up to twists) are L = O P 2(1) and U = Ω 1 X L. H 1 (F m L ) soc ) and H 1 (F m U ) soc ) again have A-module structure of a graded vector space. Given E with Horrocks data (F m, β), get kernels V L, V U. The natural map 3L( 1) U gives H 1 (F m U ) ψ Fm 3H 1 (F m L (1). Extra requirement: ψ Fm (V U ) 3V L (1). Conclusion Up to sums of ACM bundles, { E } { (Fm, V, W ) ψ Fm (V ) 3W (1) }. (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

21 Example 3: in progress X is P 1 P 1 embedded in P 5 by O X (1) = O(1, 2). Non-trivial ACM line bundles up to twist are L = O(0, 1), M = O( 1, 1), L = O(0, 1) and O( 1, 0). Ext 1 (M, L) = H 1 (O(1, 2) = k k hence there is a 1-parameter family of ACM bundles {D λ } 0 L D λ M 0. Faenzi & Malaspina show that every irreducible ACM bundle D after twists, appears in a short exact sequence 0 L a D M b 0. So the splitting criterion reduces to requiring H 1 (E B) = 0 for B = L, M, L, O( 1, 0). (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

22 For Correspondence Theorem I, assume E, E have no ACM bundle summands, and φ : E E satisfies H 1 (φ) is an isomorphism, H 1 (E B ) = H 1 (E B ) are isomorphisms for B = L, M, L, O( 1, 0). Does this imply that H 1 (E D ) = H 1 (E D ) for any ACM bundle D? (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

23 Possible framework for future: Costa,Miro-Roig have a definition of Cohen-Macaulay exceptional block collections on a smooth projective X n. A sheaf/bounded complex of sheaves F is exceptional if Ext 0 (F, F) = k, Ext i (F, F) = 0, i 0. (F 1, F 2,..., F m ) is an exceptional collection if each F i is exceptional and for j < k, Ext (F k, F j ) = 0. (F 1, F 2,..., F m ) is a block of length m if each F i is exceptional and for j k, Ext (F k, F j ) = 0. An m-block collection of type (α 0, α 1,..., α m ) is a sequence (F 0, F 1,..., F m ) of blocks F i of length α i such that the totality of members in the blocks (in that order) forms an exceptional collection. An m-block collection is full if the totality of members in the blocks generates the derived category of X. On (X n, O X (1)), an n-block collection of sheaves is Cohen-Macaulay if for each F n p F n and F k q F k, k < n, F n p F k q is ACM. (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23

24 This definition of a full Cohen-Macaulay n-block collection on (X n, O X (1)) includes the cases of quadrics, Grassmannians. In many of these instances, F n will consist of a single sheaf O X. So all F k q will be ACM bundles on X. In this context, their theorem is Costa, Miro-Roig Let X be a smooth ACM variety of dimension n with a full Cohen-Macaulay n-block collection, with F n = {O X }. Let E be a vector bundle such that E F k q is ACM for all k, q. Then E is free. Question A similar correspondence theorem? (Department of Mathematics and Computer Horrocks Science correspondence University of Missouri-St.Louis) May, / 23