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TRUNCATED MODULES AND

LINEAR PRESENTATIONS OF VECTOR BUNDLES

ADA BORALEVI, DANIELE FAENZI, AND PAOLO LELLA

Abstract. We give a new method to construct linear spaces of matrices of constant rank, basedon truncated graded cohomology modules of certain vector bundles. Our method allows one toproduce several new examples, and provides an alternative point of view on the existing ones.

Introduction

A space of matrices of constant rank is a vector subspace V , say of dimension n + 1, of theset Ma,b(k) of matrices of size a b over a field k, such that any element of V \ {0} has fixedrank r. It is a classical problem, rooted in work of Kronecker and Weierstrass, to look for suchspaces of matrices, and to give relations among the possible values of the parameters a, b, r, n. Themain difficulty here is to construct matrices with small constant corank with respect to a and b,especially so when n grows, since n > 3 is already hard. One can see V as an a b matrix whoseentries are linear forms (a linear matrix), and interpret the cokernel as a vector space varyingsmoothly (i.e. a vector bundle) over Pn. In [EH88, IL99, Syl86] the relation between matrices ofconstant rank and the study of vector bundles on Pn and their invariants was first studied in detail.This interplay was pushed one step further in [BFM13, BM14], where the matrix of constant rankwas interpreted as a 2-extension from the two vector bundles given by its cokernel and its kernel.This allowed to construct skew-symmetric matrices of linear forms in 4 variables of size 14 14and corank 2, beyond the previous record of [Wes96].

In this paper we turn the tide once again and introduce a new effective method to construct linearmatrices of constant rank. This should be seen as a parallel of [BFM13] which complements andexplains the techniques used there, relying on commutative algebra rather than derived categories.

The starting point of our analysis is that linear matrices of relatively small size can be cookedup with two ingredients, namely two finitely generated graded modules E and M over the ringR = k[x0, . . . , xn], admitting a linear resolution up to a certain step. Here, the module M shouldbe thought of as a small modification of E, namely M should be Artinian. Then, under suitableconditions on the Betti numbers of E and M , a surjective map : E M induces a resolutionof ker() which is also linear, and of smaller size, as the presentation matrix of M is subtractedfrom that of E. The key idea here is that, in order for E to fit our purpose, it is necessary totruncate E above a certain range, typically the regularity of E, which ensures linearity of theresolution, while leaving the rank of the matrix presenting E unchanged. All this is stated in amore precise form in Theorem 1.1 and in the auxiliary results of 1.

To connect this result with linear matrices of constant rank, one takes E to be the module ofglobal sections of a vector bundle E over Pn, which guarantees that the presentation matrix of E,

2010 Mathematics Subject Classification. 13D02, 16W50, 15A30, 14J60.Key words and phrases. Graded truncated module, linear presentation, matrix of constant rank, vector bundle,

instanton bundle.Research partially supported by MIUR funds, PRIN 2010-2011 project Geometria delle varieta` algebriche,

FIRB 2012 Moduli spaces and Applications, GEOLMI ANR-11-BS03-0011, J. Draismas Vidi grant from theNetherlands Organization for Scientific Research (NWO) and Universita` degli Studi di Trieste FRA 2011 projectGeometria e topologia delle varieta`. All authors are members of GNSAGA.

1

2 ADA BORALEVI, DANIELE FAENZI, AND PAOLO LELLA

as well as that of any higher truncation of E, actually has constant rank. This is the content ofTheorem 2.1. Then we analyze basic Artinian modules M (especially those of length 1 and 2) inorder to obtain interesting matrices via this method. Note that the choice of where one needs totruncate is arbitrary here; in this sense we define a tree associated with E, rooted in E, whoseedges are the allowed levels where we may truncate E, and whose leaves are the possible forms ofmatrices whose cokernel is E.

One of the goals of this paper is to provide a list of matrices of constant rank arising from vectorbundles over projective spaces. We concentrate here on the most classical constructions (instantons,null-correlation bundles and so forth). To our knowledge however, among the examples of matricesof constant rank constructed from these bundles, very few were known before. Also, the rank andsize of the matrices appearing in many of these examples are in a range that does not seem easyto reach with previously known techniques. All this is developed in 2.

In the last section we examine the conditions needed for a constant rank matrix A con-structed with our method to be skew-symmetric in a suitable basis (we call such matrix skew-symmetrizable). This is tightly related with the results of [BFM13]: there, some skew-symmetriclinear matrices of constant rank were constructed starting with a vector bundle together with apiece of homological data, namely a certain extension class involving the cokernel bundle of A.

This time our approach relies on truncated modules rather than derived categories. Our newtechnique has the advantage of being algorithmic, and can be actually implemented in a veryefficient way, as computations sometimes work even beyond their theoretical scope. This not onlyprovides a detailed explanation of the algorithm appearing in [BFM13, Appendix A], but allowsto construct infinitely many examples of skew-symmetric 10 10 matrices of constant rank 8 in 4variables; up to now, the only example of such being those of [Wes96].

Finally, let us note that all these examples, and many more, can be explicitly constructed thanksto the Macaulay2 package ConstantRankMatrices implementing our algorithms. The interestedreader can find it at the website www.paololella.it/EN/Publications.html.

1. Main results

1.1. Notation and preliminaries. Let R := k[x0, . . . , xn] be a homogeneous polynomial ringover an algebraically closed field k of characteristic other than 2. The ring R comes with a gradingR = R0 R1 . . ., with R0 = k, and is generated by R1 as a k-algebra.

All R-modules here are finitely generated and graded. IfM is such a module, and p, q are integernumbers, we denote by Mp the p-th graded component of M , so that M = pMp, and by M(q)the qth shift ofm, defined by the formulaM(q)p = Mp+q. Finally, the module M>m = p>mMpis the truncation of M at degree m.

A module M is free if M iR(qi) for suitable qi. Given any other finitely generated gradedR-module N , we write Hom(M ,N) for the set of homogeneous maps of all degrees, which is againa graded module, graded by the degrees of the maps. Since graded free resolutions exist, thisconstruction extends to a grading on the modules ExtqR(M ,N). The same holds true for M Nand Tor(M ,N). For any R-module M , we set i,j(M) := dimk Tori(M ,k)j .

For p > 0, the Hilbert function dimk Mp is a polynomial in p. The degree of this polynomial,increased by 1, is dimM , the dimension of M . The degree of M is by definition (dimM !) timesthe leading coefficient of this polynomial.

We say that M hasm-linear resolution over R if the minimal graded free resolution of M reads:

R(m 2)2 R(m 1)1 R(m)0 M 0

for suitable integers i. In other words, M has a m-linear resolution if M r = 0 for r < m, Mis generated by Mm, and M has a resolution where all the maps are represented by matrices oflinear forms. If only the first k maps are matrices of linear forms then M is said by m-linearpresented up to order k, or just linearly presented when k = 1.

3A module M is m-regular if, for all p, the local cohomology groups Hpm(M)r vanish for r =

m p + 1 and also H0m(M )r = 0 for all r > m + 1. The regularity of a module is denoted by

reg(M). Note that M>m always has m-linear resolution if m is large enough.

1.2. Main Theorem. Here comes our main result on the linear presentation of modules.

Theorem 1.1 (The Brunch Theorem). Let E be an R-module, and let m > reg(E) be an in-teger. Let M be an Artinian R-module, m-linearly presented up to order 2, and assume thati,m+i(E>m) > i,m+1(M ) for i = 0, 1, 2. Assume moreover that there is a surjective morphism : E>m M . Then, the kernel F := Ker is m-linearly presented.

Proof. Since is surjective, we have a short exact sequence of graded modules:

0 F E>mM 0

This induces the long exact sequence (see [Pee11, Theorem 38.3]):

(1.1)0Tor0(M ,k)Tor0(E>m,k)Tor0(F ,k)

Tor1(M ,k)Tor1(E>m,k)Tor1(F ,k)

Tor2(M ,k)Tor2(E>m,k)

where the modules Tori(,k) are graded of finite length and the dimensions of the homogeneouspieces are equal to the Betti numbers of the minimal free resolutions of the modules [Pee11,Theorem 11.2].

By assumption, the module E>m has regularity m. By [EG84, Theorem 1.2], M>m has m-linear resolution if and only if M is m-regular. So, there exist integers 0, 1, . . . such that E>mhas the m-linear resolution:

R(m 2)2 R(m 1)1 R(m)0 E>m 0,

so that i,m+i(E>m) = i and i,j(E>m) = 0 for all j 6= m+ i.The first steps of the minimal graded free resolution of M are:

R(m 2)2 R(m 1)1 R(m)0 M 0.

This tells that, for any i = 0, 1, 2:

i,m+i(M) = i, i,j(M ) = 0, if j > m+ i.

The same vanishing holds for all i if j < m+ i.Let us now consider the long exact sequences induced by (1.1) on each degree. Obviously, for

j < m, i,j(E>m) = i,j(M) = 0 implies i,j(F ) = 0. In the case j = m, one has:

0 Tor0(M ,k)m Tor0(E>m,k)m Tor0(F ,k)m 0

i.e. 0,m(F ) = 0,m(E>m) 0,m(M ) = 0 0. The degree m+ 1 part of (1.1) is:

0 Tor1(F ,k)m+1 Tor1(E>m,k)m+1 Tor1(M ,k)m+1 Tor0(F ,k)m+1 0

and thus:

1,m+1(F ) 0,m+1(F ) = 1,m+1(E>m) 1,m+1(M ).

By hypothesis, 1,m+1(E>m) 1,m+1(M ) = 1 1 > 0 and by the minimality of the resolutionof F we have 0,m+1(F ) = 0 and 1,m+1(F ) = 1 1. For degree j = m+ 2, we have:

0 Tor2(F ,k)m+2 Tor2(E>m,k)m+2 Tor2(M ,k)m+2 Tor1(F ,k)m+2 0

and we repeat the same argument as above to conclude 1,m+2(F ) = 0 and 2,m+2(F ) = 2 2.Finally, for j > m + 2, 1,j(F ) = 0 as 1,j(E>m) = 2,j(M) = 0. Hence, the presentation of F


is linear as the modules Tor0(F ,k) and Tor1(F ,k) have length one with Tor0(F ,k)m 6= 0 andTor1(F ,k)m+1 6= 0.

Remark 1.1. The name Brunch Theorem for Theorem 1.1 comes from the fact that a roughdraft of its proof came out exactly during a Saturday morning brunch in Turin, in June 2014.

1.3. A more detailed analysis. Keeping in mind our goal of constructing explicit examples ofconstant rank matrices, we now want to investigate the features of modules satisfying Theorem 1.1shypotheses. We take into consideration Artinian modules of length 1 and 2, and seek numericalranges for Betti numbers satisfying the hypotheses of Theorem 1.1. Note that this does notguarantee the existence of the surjective morphism .

Let us suppose that the module M is generated in degree m and has length + 1. Its Hilbertseries is tm(h0 + h1t+ . . .+ ht

), and it can be expressed as a function of its Betti numbers:

tm(h0 + h1t+ . . .+ ht) =

n+1i=0

((1)i

j ijt

j)

(1 t)n+1.

In our setting one has:

(1 t)n+1tm(h0 + h1t+ . . .+ ht) = 0t

m 1tm+1 + 2t

m+2 + . . .

and from this we deduce:{0 = h0

1 = (n+ 1)h0 h1=

{h0 = 0

h1 = (n+ 1)0 1= (n+ 1)0 1 > 0.

From explicit computations one sees that:

Proposition 1.2. Let M be a finitely generated graded module of length 2 over R = k[x0, . . . , xn],satisfying all hypotheses of Theorem 1.1. Then the last piece of its resolution has the form:

R(r 2)(n+1)1(n+22 )0 R(r 1)1 R(r)0 M 0,

with 2n+43 0 6 1 6 (n+ 1)0.

Remark 1.2. The resolution of Artinian modules of length > 3 is much more involved. In particular,for length > 3, the Hilbert function and the first two Betti numbers do not determine the shapeof the free module of homological position 2.

In what follows we will need the next result:

Lemma 1.3. Let M be a finitely generated graded R-module and let m > reg(M ) be an integernumber. The truncated module M>m is m-regular, and assume that it has linear resolution:

0 R(m n)n,m R(m i)i,m R(m)0,m M>m 0.

Then the truncated module M>m+k, k > 1, has regularity m+ k and linear resolution:

0 R(m k n)n,m+k R(m k)0,m+k M>m+k 0

with

i,m+k = a(i)k 0,m a

(i)k11,m + . . .+ (1)

na(i)knn,m =

nj=0

(1)ja(i)kjj,m,

where for all i = 0, . . . , n, the sequence(a(i)k

)belongs to the set of recursive sequences:

RSn :=

(ak) ak+1 =

nj=0

(1)j(n+ 1

j + 1

)akj

.

5More in detail,(a(i)k

)is defined by the initial values:

(1.2) a(i)1 =

(n+ 1

i+ 1

), a

(i)i = (1)

i and a(i)j = 0 for j 6= 1, i and j < n.

Proof. Looking at the exact sequence 0 M>m+k+1 M>m+k Mm+k 0, one worksrepeatedly with the long exact sequence (1.1).

By induction on n one can also prove:

Lemma 1.4. Any recursive sequence (ak) RSn has a degree n polynomial as its generating func-

tion. In particular, the generating function of the sequence(a(i)k

)defined in (1.2) is the following:

p(i)n (k) =

(n

i

)(k + n 1

n

)k + n

k + i.

Remark that in the case i = 0 one gets that p(0)n (k) =

(k+nn

).

Example 1.1. Consider the Koszul complex of R = k[x0, x1, x2, x3] as resolution of the irrelevantideal m = (x0, x1, x2, x3). It reads:

0 R(4) R(3)4 R(2)6 R(1)4 m 0.

By Lemma 1.3 and 1.4, the resolution of m>1+k for every k > 0 is

0 R(4 k)3(k) R(3 k)2(k) R(2 k)1(k) R(1 k)0(k) m>1+k 0,

where

0(k) =1

6k3 +

3

2k2 +

13

3k + 4, 1(k) =

1

2k3 + 4k2 +

19

2k + 6,

2(k) =1

2k3 +

7

2k2 + 7k + 4, 3(k) =

1

6k3 + k2 +

11

6k + 1.

Therefore, we can predict the resolution of m>10:

0 R(13)220 R(12)715 R(11)780 R(10)286 m>10 0.

2. Linear presentation of vector bundles

We now focus our attention on the case where the graded R-modules are cohomology modulesof vector bundles over the projective space Pn = ProjR.

Theorem 2.1. In the setting of Theorem 1.1, suppose furthermore that E = H0(E) is the graded

module of sections of a rank r vector bundle E on Pn. Set a = 0,m(E>m) 0,m(M ) andb = 1,m+1(E>m) 1,m+1(M ). Then the presentation matrix A of F is linear of size a b and

corank r. Moreover the sheafification F := F of F is isomorphic to E.

Proof. The fact that the presentation matrix of F is linear is consequence of Theorem 1.1.


For i > 0, set i = i,m+i(E>m) and i = i,m+i(M ). Then, from the proof of Theorem 1.1,we extract the next commutative exact diagram.

0 0 0

R(m 1)11 R(m)00 F 0A

R(m 2)2 R(m 1)1 R(m)0 E>m 0

R(m 2)2 R(m 1)1 R(m)0 M 0

0 0 00

Therefore, the matrix A presenting F is linear of size a b, where a = 1 1 and b = 0 0.Note that this holds independently of the fact that E is locally free.

The fact that A has constant corank r follows from the observation that the modules E and Fdiffer by the Artinian module M only, hence their induced coherent sheaves on Pn are isomorphic.

In particular, F = F is a locally free sheaf of rank r, i.e. the restriction of the presentation matrixA of F has constant corank r.

Let us spell out clearly our strategy to construct linear matrices of constant rank based on thisresult. Suppose that, for a given 3-uple of integers (, a, b) with < a 6 b, we want to constructan a b matrix of linear forms of constant rank in the polynomial ring with n + 1 variables.Then we have to search for an a b matrix of linear forms presenting a vector bundle E of rankr = a on Pn. In general, if E is a vector bundle of rank r, its module E of global sectionswill not be linearly presented. But we are free to truncate E in such a way that it is m-linearlypresented. By Theorem 2.1, the presentation matrix will indeed have corank r. However, it isunlikely that its size equals a b. Here is where Theorem 1.1 and 2.1 come into play, as we mayremove a linearly presented Artinian module from our truncation of E in order to reduce the sizeof our presentation matrix, and hopefully arrive at size a b.

Note that, once given E, there are infinitely many truncations of E = H0(E), and for each

truncation E>k one can look at the finitely many Artinian modules of length at most 2 satisfyingthe assumption of Theorem 1.1. We illustrate these possibilities with a 2-level infinite tree structure.The root of the tree is the vector bundle E, the elements at level 1 are all possible truncationsE>regE+k, k N, and the leaves are possible sizes of linear matrices of constant rank producedby the algorithm. (Explicit examples can be found in Figures 1-5.) We emphasize that in order todetermine such a tree, one only needs to compute explicitly the first truncation E>regE with linearresolution, and then apply Lemma 1.3 and Lemma 1.4. The existence of the matrix is guaranteedonly in cases where there are surjective morphisms from E>regE to the Artinian module.

Example 2.1 (Line bundles). To give a first application of Theorem 2.1, let E be a line bundleOPn(l) over Pn. Then, Figure 1 shows how to obtain from Theorem 2.1 matrices of constantrank min{a, b} 1. Clearly these matrices can constitute building blocks to construct bigger sizematrices, such as square matrices of size 10, 12 and 14 of rank 8, 10 and 12, and also skew-symmetricmatrices of size a+ b with rank 2a 2 (assuming a 6 b).

2.1. Steiner bundles, linear resolutions and generalizations. In classical literature a vec-tor bundle E on Pn having a linear resolution of the form:

0 OPn(1)s Os+r

Pn E 0,

with s > 1 and r > n integers, is called a (rank r) Steiner bundle. This motivates the following:

7E = OP2(l)El0 1 2

l 1

k = 0

El+10 1 2

l + 1 3 3 1

k = 1

El+20 1 2

l + 2 6 8 3

k = 25 5 matrix of rank 4

0 1 2 3l + 2 1 3 3 1

El+30 1 2

l + 3 10 15 6


0 1 2 3l + 3 1 3 3 1

8 9 matrix of rank 7

0 1 2 3l + 3 2 6 6 2


0 1 2 3l + 3 3 8 6 l + 4 1

El+40 1 2

l + 4 15 24 10


0 1 2 3l + 4 1 3 3 1


0 1 2 3l + 4 2 6 6 2


0 1 2 3l + 4 3 8 6 l + 5 1


0 1 2 3l + 4 3 9 9 3


0 1 2 3l + 4 4 11 9 1l + 5 1

Figure 1. A piece of the tree associated with a line bundle over P2. Eachtruncation is described by the Betti table of its resolution, and the arrows fromthe truncated modules to possible sizes of constant rank matrices are labeled withthe Betti tables of the Artinian modules satisfying the assumptions of Theorem1.1 and Proposition 1.2. In all cases pictured above one can show the existence ofthe surjective morphism, and thus of the matrix.

Definition 2.1. Let r > n and s > 1 be integer numbers. The cokernel E(m)s,r of a generic

morphism:

OPn(m 1)s Os+r

Pn

is a vector bundle on Pn, that we call generalized Steiner bundle.

Classical Steiner bundles are of the form E(0)s,r . Given a generalized Steiner bundle, the graded

module E(m) := H0(E

(m)s,r ) has the following resolution:

(2.1) 0 R(m 1)s Rs+r E(m) 0,

and its regularity is exactly equal to m. Remark though that the module E(m) does not fit thebill for our purposes: first of all for m 6= 0 the resolution is not linear, and even in the case m = 0


one has Tor2(E(0),k)2 = 0. For this reason we need to truncate it in higher degree, as explained

in the following:

Lemma 2.2. Let E(m)s,r be a rank r generalized Steiner bundles, with graded modules of sections

E(m) = H0

(E

(m)s,r ). The linear resolutions of the truncation E

(m)>m is of the form:

0 R(m i)(m)i,m R(m)

(m)0,m E

(m)>m 0

where:

(2.2) (m)i,m = p

(i)n (m)(s + r) p

(i)n (1)s.

Proof. Looking at the homogeneous piece of degree m+ i of the resolution of E(m)>m

0 R(m i)

(m)i,m

m+i R(m)

(m)0,m

m+i (E

(m)>m

)m+i

0,

we deduce that:

(m)i,m =

ij=1

(1)j1(n+jn

)(m)ij,m + (1)

i dimk(E

(m)>m

)m+i

.

Since(E

(m)>m

)m+i

=(E

(m))m+i

, i > 0, we compute the dimension of the homogeneous piece of

degree m+ i of E(m)>m from the simpler resolution (2.1):

dimk(E

(m)>m

)m+i

= dimk(E

(m))m+i

=(n+m+i

n

)(s+ r)

(n+i1

n

)s.

We now proceed by induction on i; for i = 0, we have:

(m)0,m = dimk

(E

(m)>m

)m=(n+mn

)(s+ r)

(n1n

)s = p(0)n (m)(s+ r) p

(0)n (1)s.

By inductive hypothesis, (2.2) holds for 0, . . . , i 1. We get:

(m)i,m =

ij=1

(1)j1(n+jn

) (p(ij)n (m)(s+ r) p

(ij)n (1)s

)+

(1)i((

n+m+in

)(s+ r)

(n+i1

n

)s)=

=[ ij=1

(1)j1(n+jn

)p(ij)n (m) + (1)

i(n+m+i

n

)](s+ r)

[ ij=1

(1)j1(n+jn

)p(ij)n (1) + (1)

i(n+i1

n

)]s.

The result follows from the observation that the univariate polynomial p(i)n (k) coincides with:

ij=1

(1)j1(n+jn

)p(ij)n (k) + (1)

i(n+k+i

n

),

because both polynomials have degree n and take the value at k = 0,1, . . . , n.

9E(0)1,2

P2

E(0)>1

(7 6, 5) X

E(0)>2

(14 18, 12) X

(13 15, 11) X

(12 13, 10) X

E(0)>3

(23 34, 21) X

(22 31, 20) X

(21 29, 19) X

(21 28, 19) X

(20 26, 18) X

(20 25, 18) X

(19 23, 17) X

(19 22, 17) X

(18 21, 16) X

(18 20, 16) X

(17 18, 15) X

...

(a) E(0)1,2 over P

2.

E(1)2,2

P2

E(1)>1

(11 11, 9) X

E(1)>2

(21 29, 19) X

(20 26, 18) X

(19 24, 17) X

(19 23, 17) X

(18 21, 16) X

(18 20, 16) X

(17 18, 15) X

(16 16, 14) X

...

(b) E(1)2,2 over P

2.

E(2)2,3

P2

E(2)>2

(29 39, 26) X

(28 36, 25) X

(27 34, 24) X

(27 33, 24) X

(26 31, 23) X

(26 30, 23) X

(25 28, 22) X

(25 27, 22) X

(24 26, 21) X

(24 25, 21) X

(23 23, 20) X

...

(c) E(2)2,3 over P

2.

E(1)1,3

P3

E(1)>1

(15 21, 12) X

(14 18, 12) X

(14 17, 11) X

(13 15, 10)

(13 14, 10) X

(12 11, 9)

...

(d) E(1)1,3 over P

3.

E(0)2,3

P3

E(0)>1

(17 26, 14) X

(16 23, 13) X

(16 22, 13) X

(15 20, 12)

(15 19, 12) X

(15 18, 12) X

(14 16, 11) X

(14 15, 11) X

(13 13, 10)

(12 10, 9)

...

(e) E(0)2,3 over P

3.

E(0)1,4

P3

E(0)>1

(18 26, 14) X

(17 23, 13) X

(17 22, 13) X

(16 20, 12)

(16 19, 12) X

(16 18, 12) X

(15 16, 11) X

(15 15, 11) X

(14 13, 10) X

(13 10, 9)

...

(f) E(0)1,4 over P

3.

Figure 2. Examples of possible constant rank matrices arising from generalizedSteiner bundles. The notation (ab, ) stands for an abmatrix of constant rank. Thanks to Lemma 2.2 these trees can be deduced without having to constructthe module E. For every output (a b, ) we take a random Artinian moduleM with the appropriate Betti numbers, and look at the module HomR(E,M)0.The symbols appearing in the rightmost column indicate whether a surjectivemorphism HomR(E,M)0 exists (X), does not exist (), or the computationis too heavy to be performed (?).

2.2. Linear monads and instanton bundles. Mathematical instanton bundles were first intro-duced in [OS86] as rank 2m bundles on P2m+1 satisfying certain cohomological conditions. Theygeneralize particular rank 2 bundles on P3 whose study was motivated by problems from physics,see [AHDM78]. They can also be defined as cohomology of a linear monad; in our work we consideran even more general definition, in the spirit of [Jar06]. First, recall that a monad (on Pn) is acomplex:

Af B

g C

of vector bundles over Pn which is exact everywhere but in the middle, and such that Im f is asubbundle of B. Its cohomology is the vector bundle E = Ker g/ Im f .

Definition 2.2. A generalized instanton bundle is a rank r vector bundle E(r,k) on Pn, r > n 1,

which is the cohomology of a linear monad of type:

(2.3) OPn(1)k f O2k+r

Pn

g OPn(1)

k.


When so, k = dimH1(E(r,k)(1)) and is called the charge of E(r,k). E(r,k) is sometimes called ak-instanton.

According to [Fl00], the condition r > n 1 is equivalent to the existence of a monad of type(2.3). It should be noted that it is not always the case that E(r,k) is a vector bundle; again from[Fl00] we learn that the degeneracy locus of the map f has expected codimension r + 1. Thuswhen dealing with such monads it will be necessary to check that this expected dimension indeedcorresponds to the real dimension.

Figures 3a to 3f are examples of sizes and ranks of matrices that can arise from generalizedinstantons.

E(2,2)

P3

E>2 (11 14, 9) X

(10 11, 8) X

(9 8, 7)

E>3 (29 58, 27) X

(28 55, 26) X

...(18 22, 16)

(18 21, 16)

(17 18, 15)

...

(a) Classical (rank 2) instanton bun-dle over P3 of charge 2.

E(2,4)

P3

E>3 (19 30, 17) X

(18 27, 16) X

(18 26, 16) X

(17 24, 15) X

(17 23, 15) X

(17 22, 15) X

(16 20, 14) X

(16 19, 14) X

(15 17, 13) X

(14 14, 12) X

...

(b) Classical (rank 2) instanton bun-dle over P3 of charge 4.

E(3,2)

P3

E>2 (21 34, 18) X

(20 31, 17) X

(20 30, 17) X

...(18 22, 15) X

(17 21, 14)

(17 20, 14) X

(16 18, 13)

(16 17, 13) X

...

(c) Generalized instanton bundleE(3,2) over P

3.

E(2,2)

P2

E>2 (9 9, 7) X

E>3 (17 23, 15) X

(16 20, 14) X

(15 18, 13) X

(15 17, 13) X

(14 15, 12) X

...

(d) Generalized instanton bundleE(2,2) over P

2.

E(3,2)

P4

E>2 (24 46, 21) X

(24 45, 21) X

(18 22, 15)

...

(18 21, 15)

(17 18, 14)

...

(e) Generalized instanton bundleE(3,2) over P

4.

E(4,1)

P4

E>1 (13 16, 7) X

(13 15, 9) X

(12 12, 8) X

(12 11, 8) X

(11 8, 7)

...

(f) Generalized instanton bundleE(4,1) over P

4.

Figure 3. Examples of possible constant rank matrices that can be obtainedfrom instanton bundles. The notation is the same as in Figure 2.

2.3. Null correlation, Tango, and the Horrocks-Mumford bundle. Null correlation bundlesare examples of rank n 1 bundles on Pn for n odd: they are constructed as kernel of the bundleepimorphism TPn(1) OPn(1). A construction due to Ein [Ein88] generalizes this definition onP3:

Definition 2.3. [Ein88] A rank 2 vector bundle E(e,d,c) on P3 is said to be a generalized null

correlation bundle if it is given as the cohomology of a monad of the form:

OP3(c) OP3(d)OP3(e)OP3(e)OP3(d) OP3(c),

where c > d > e > 0 are given integers.

Figures 4a to 4c show possible sizes of matrices appearing from null correlation bundles.A construction of Tango [Tan76] produces an indecomposable rank n 1 bundle over Pn, for

all Pn, defined as a quotient En of the dual of the kernel of the evaluation map of 1Pn(2), which

is a globally generated bundle. More in detail, one starts by constructing the rank(n2

)bundle En

from the exact sequence:

(2.4) 0 TPn(2) O(n+12 )Pn

En 0,

11

NCP3

E>2 (15 25, 13) X

(14 22, 12) X

(14 21, 12) X

(13 19, 11)

(13 18, 11) X

(13 17, 11) X

(12 15, 10) X

(12 14, 10) X

(11 12, 9)

(10 9, 8)

...

(a) NC bundle over P3.

NCP5

E>1 (13 15, 9) X

(13 14, 9) X

E>2 (63 190, 59) X

...(29 29, 25) ?

...(28 26, 24) ?

(28 25, 24) ?

...

(b) NC bundle over P5.

E(0,1,2)

P3

E>4 (51 113, 49) X

(50 110, 48) X

(50 109, 48) X

(49 107, 47)

...(27 32, 25) ?

(26 30, 24) ?

(26 29, 24) ?

(25 27, 23) ?

...

(c) E(0,1,2) over P3.

Figure 4. Size and rank of matrices of constant rank that can be constructedfrom null correlation bundles. The notation is the same as in Figure 2.

and then take its quotient En:

(2.5) 0 O(n2)nPn

En E

n 0,

that turns out to be a rank n indecomposable vector bundle on Pn containing a trivial subbundleof rank 1. The Tango bundle Fn is defined as the quotient of E

n by its trivial subbundle, and thushas rank n 1.

Indecomposable rank n 2 bundles on Pn are even more difficult to construct; on P4 there isessentially only one example known, whose construction is due to Horrocks and Mumford [HM73].It is an indecomposable rank 2 bundle that can be defined as the cohomology of the monad:

OP4(1)5 (2

P4(2))2 O5

P4.

Figures 5a to 5c show possible examples of constant rank matrices that one can construct startingfrom generalized null correlation, Tango, and the Horrocks-Mumford bundle.

TP4

E>1 (29 65, 26) X

(29 64, 26) X

(28 61, 25)

...(18 21, 15)

(18 20, 15)

(18 19, 15)

(17 17, 14)

(17 16, 14)

...

(a) Tango bundle over P4.

TP5

E>1 (47 129, 43) X

(47 128, 43) X

(46 124, 42)

...(25 26, 21) ?

(25 25, 21) ?

(25 24, 21) ?

(24 22, 20) ?

(24 21, 20) ?

...

(b) Tango bundle over P5.

HMP4

E>5 (99 286, 97) X

(99 285, 97) X

(99 282, 97)

(99 281, 97) X

...(38 38, 36) ?

(37 38, 35) ?

(37 37, 35) ?

(37 36, 35) ?

...

(c) Horrocks-Mumford bundle.

Figure 5. Examples of size and rank of matrices that can be constructed fromTango and Horrocks-Mumford bundles. The notation is the same as in Figure 2.

3. Skew-symmetric matrices

We examine now the situation for linear matrices of constant rank with symmetry. The kerneland cokernel sheaves K and E of such a matrix will be tightly related to one another, and thematrix itself is expressed by an extension class Ext2(E,K). Let us connect this with our map .

Definition 3.1. Let E and K be vector bundles on Pn. For t Z, consider the Yoneda map t:

t : H0(E(t)) Ext2(E,K) H2(K(t)).


Set E = H0(E) and M = H2

(K). Define as the linear map induced by the t:

: Ext2(E,K) HomR(E,M)0.

Theorem 3.1. Assume n > 3 and let A : R(m 1)b R(m)a be skew-symmetrizable ofconstant rank. Set K = kerA and E = CokerA. Then K E(2m1), and there is an element lying in H2(S2E(2m 1)) under the canonical decomposition

Ext2(E,E(2m 1)) H2(S2E(2m 1))H2(2E(2m 1)),

such that A presents ker(). Conversely, if H2(S2E(2m 1)) and = () satisfies theassumptions of Theorem 2.1 and kerA E(2m 1), then A is skew-symmetrizable.

The same holds for symmetrizable A replacing the condition on with H2(2E(2m1)).

Proof. Let us check the first statement. Assume thus that A is skew-symmetric. Then, sheafifyingthe matrix A provided by Theorem 2.1 we get a long exact sequence:

(3.1) 0 K OPn(m 1)a A OPn(m)

a E 0

with K = kerA and E = CokerA. Since A has constant rank, E and K are locally free and hence:

Exti(E,OPn) = Exti(K,OPn) = 0,

for all i > 0. Therefore, dualizing the above sequence and twisting by OPn(2m 1)we get:

0 E(2m 1) OPn(m 1)a A OPn(m)

a K(2m 1) 0

Since the image E of A is obviously the same as the image of A, these exact sequences can beput together to get K E(2m 1). We may thus rewrite them as:

0 E(2m 1) OPn(m 1)a A OPn(m)

a E 0.

This long exact sequence represents an element Ext2(E,E(2m1)) H2(EE(2m1)).By looking at the construction of [BFM13, Lemma 3.1], it is now clear that for A to be skew-symmetric should lie in H2(S2E(2m 1)).

To understand why A presents ker(), let us first expand some details of the definition of .Let again E be the image of A and write for any integer t the exact commutative diagram:

0 K(t) OPn(m 1 + t)a OPn(m+ t)aA

E(t) 0.

E(t)

0 0

Taking cohomology, we get maps:

(3.2) t : H0(E(t)) H2(K(t)).

H1(E(t))

Remark that sequence (3.1) corresponds to Ext2(E,K). Cup product with induces viaYonedas composition the linear maps ts of (3.2). But these maps are obtained from the t bytransposition, so cup product with gives:

= tt = () : H0(E) = E M = H2

(E(2m 1)).

This is obviously a morphism, homogeneous of degree 0. Notice that as soon as n > 3, bothgroups H1(OPn(m+ t)) and H

2(OPn(m 1+ t)) vanish for all values of t, hence is surjective.By construction A appears as presentation matrix of F = ker.

For the converse statement, the element corresponds to a length-2 extension of E(2m 1)by E. Set = (). Theorem 2.1 gives a linear matrix A of constant rank of presentation for

13

F = ker. Note that sheafifying F we get back the bundle E, as M = H2(E(2m 1)) isArtinian. On the other hand, since kerA E(2m 1), the matrix A represents the extensionclass . Therefore, A is skew-symmetrizable by [BFM13, Lemma 3.5 (iii)]. Part (i) of the samelemma says that, when dealing with symmetric matrices, one should replace the condition H2(S2E(2m 1)) with H2(2E(2m 1)). The theorem is thus proved.

In particularly favorable situations, for instance when E is an instanton bundle of charge 2 (cf.[BFM13, Theorem 5.2]), or a generic instanton bundle of charge 4 (cf. [BFM13, Theorem 6.1]),one can use using results from [Dec90] and [Rah97], to check that the map is a surjection. Thismakes considerably easier the search for an element corresponding to a skew-symmetric matrixof the prescribed size and constant rank.

Example 3.1. Let us work out the case of generic instantons of rank 2 to write down an explicit1010 skew-symmetric matrix of constant rank 8. Let E be a rank 2 instanton bundle of charge 2on P3, obtained as cohomology of a monad of type (2.3). We take a special 2-istanton, as describedin [AO95]; its monad has maps:

f =

0 x1x1 x0x0 00 x3

x3 x2x2 0

and g =[x2 x3 0 x0 x1 00 x2 x3 0 x0 x1

].

Once the bundle E is constructed, we call E := H0(E) its module of sections, having resolution:

0 R(4)2 R(3)6 R(2)4 R(1)2 E 0.

We truncate it in degreem = 2 in order to get a 2-linear resolution; then we take the 2nd gradedcohomology module H2

(E), which is a module of length 3, and we set M := (H2

(E)(5))>2. The

truncated moduleM has length 2. By taking a morphismE>2 M we get the following diagram:

R(4)10 R(3)18 R(2)12 E>2 0

R(4)12 R(3)8 R(2)2 M 0

2

Notice that we are in a situation that is very similar to what is described in the hypotheses ofTheorem 1.1, yet here the first column cannot be surjective. Nevertheless, by resolving the kernelof a general element 2, we obtain a 10 10 matrix of constant rank 8, which does not enjoy anyparticular symmetry property.

If we can make sure that the mapE M comes indeed from an element of H2(S2E(5)), thenTheorem 3.1 will guarantee that this matrix is skew-symmetrizable. By [BFM13, Theorem 5.2] weknow that H2(S2E(5)) surjects onto HomR(E,H

2(E)(5))0, and in fact an element there will

have the same kernel as its truncation in degree 2 2, because the map is an isomorphism in degree1. We can thus take a random element in HomR(E,H

2(E)(5))0 and our construction will work


without us having to truncate. This yields the matrix A = A0x0 +A1x1 +A2x2 +A3x3, where:

A0 =

0 108 594 54 36 876 108 18 0 00 0 0 18 192 0 36 0 0

0 0 36 192 0 18 0 00 0 0 0 0 0 0

0 18 18 0 0 00 48 36 0 0

0 36 0 00 0 0

0 00

,

A1 =

0 324 162 0 64 492 324 1934

0 0

0 0 0 16 48 0 412

0 0

0 0 16 264 0 1634

0 0

0 0 24 0 94

0 0

0 16 4 0 0 0

0 48 892

0 0

0 172

0 0

0 272

0

0 00

,

A2 =

0 438 534 108 36 1590 4952

36 324 54

0 300 0 18 0 75 18 0 0

0 54 36 876 7052

36 0 0

0 0 0 272

0 0 0

0 18 18 0 0 00 219 18 0 0

0 18 81 00 0 0

0 00

,

and A3 =

0 498 978 3194

64 10583

438 64 0 0

0 612 23/2 16 3683

48 16 0 0

0 354

16 21163

444 16 0 0

0 0 232

12

0 272

0

0 16 4 0 0 0

0 1283

16 144 24

0 4 0 00 0 0

0 00

.

15

Acknowledgements

We would like to express our gratitude to Emilia Mezzetti for introducing us to this problem.

References

[AHDM78] M. F. Atiyah, N. J. Hitchin, V. G. Drinfeld, and Y. I. Manin, Construction of instantons, Phys. Lett.A 65 (1978), no. 3, 185187.

[AO95] V. Ancona and G. Ottaviani, On moduli of instanton bundles on P2n+1, Pacific J. Math. 171 (1995),no. 2, 343351.

[BFM13] A. Boralevi, D. Faenzi, and E. Mezzetti, Linear spaces of matrices of constant rank and instantonbundles, Adv. Math. 248 (2013), 895920.

[BM14] A. Boralevi and E. Mezzetti, Planes of matrices of constant rank and globally generated vector bundles,to appear in Ann. Inst. Fourier (Grenoble), 2014.

[Dec90] W. Decker, Monads and cohomology modules of rank 2 vector bundles, Compositio Math. 76 (1990),no. 12, 717.

[EG84] D. Eisenbud and S. Goto, Linear free resolutions and minimal multiplicity, J. Algebra 88 (1984), no. 1,89133.

[EH88] D. Eisenbud and J. Harris, Vector spaces of matrices of low rank, Adv. in Math. 70 (1988), no. 2,135155.

[Ein88] L. Ein, Generalized null correlation bundles, Nagoya Math. J. 111 (1988), 1324.

[Fl00] G. Flystad, Monads on projective spaces, Comm. Algebra 28 (2000), no. 12, 55035516, Special issuein honor of Robin Hartshorne.

[HM73] G. Horrocks and D. Mumford, A rank 2 vector bundle on P 4 with 15,000 symmetries, Topology 12(1973), 6381. MR 0382279

[IL99] B. Ilic and J.M. Landsberg, On symmetric degeneracy loci, spaces of symmetric matrices of constantrank and dual varieties, Math. Ann. 314 (1999), no. 1, 159174.

[Jar06] M. Jardim, Instanton sheaves on complex projective spaces, Collect. Math. 57 (2006), no. 1, 6991.[OS86] C. Okonek and H. Spindler, Mathematical instanton bundles on P2n+1, J. Reine Angew. Math. 364

(1986), 3550.[Pee11] I. Peeva, Graded syzygies, Algebra and Applications, vol. 14, Springer-Verlag London, Ltd., London,

2011.[Rah97] O. Rahavandrainy, Resolution des fibres instantons generaux, C. R. Acad. Sci. Paris Ser. I Math. 325

(1997), no. 2, 189192.[Syl86] J. Sylvester, On the dimension of spaces of linear transformations satisfying rank conditions, Linear

Algebra Appl. 78 (1986), 110.[Tan76] H. Tango, An example of indecomposable vector bundle of rank n 1 on Pn, J. Math. Kyoto Univ. 16

(1976), no. 1, 137141.[Wes96] R. Westwick, Spaces of matrices of fixed rank. II, Linear Algebra Appl. 235 (1996), 163169.

TU Eindhoven, Department of Mathematics and Computer Science, PO Box 513, 5600 MB Eindhoven,The Netherlands

E-mail address: [email protected]

Institut de Mathmatiques de Bourgogne, UMR CNRS 5584, Universite de Bourgogne, 9 Avenue AlainSavary, BP 47870, 21078 Dijon Cedex, France

E-mail address: [email protected]

Dipartimento di Matematica, Universita` degli Studi di Trento, Via Sommarive 14, 38123 Trento,Italy

E-mail address: [email protected]: http://www.paololella.it/

Introduction1. Main results1.1. Notation and preliminaries1.2. Main Theorem1.3. A more detailed analysis

2. Linear presentation of vector bundles2.1. Steiner bundles, linear resolutions and generalizations2.2. Linear monads and instanton bundles2.3. Null correlation, Tango, and the Horrocks-Mumford bundle

3. Skew-symmetric matricesAcknowledgementsReferences

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