Simplicity of generic Steiner bundles - Dipartimento di Matematica e ...

Simplicity of generic Steiner bundlesMaria Chiara Brambilla ∗Sunto. – Un fibrato di Steiner E su P n ha una risoluzione lineare della forma0 → O(−1) s → O t → E → 0. In questo lavoro proviamo che il generico fibrato di SteinerE è semplice se e solo se χ(EndE) è minore o uguale a 1. In particolare mostriamo che(E è eccezionale oppure soddisfa la disuguaglianza t ≤n+1+ √ (n+1) 2 −42Abstract. – A Steiner bundle E on P n has a linear resolution of the form0 → O(−1) s →O t → E → 0. In this paper we prove that a generic Steiner bundle Eis simple if and only if χ(EndE) is less or equal(to 1. In particular ) we show that eitherE is exceptional or it satisfies the inequality t ≤s.n+1+ √ (n+1) 2 −42)s.1. – IntroductionAccording to [2] a Steiner bundle E on P(V ) = P N−1 has a linear resolution ofthe form0 → O(−1) s →O t → E → 0.It is well known that Steiner bundles have rank t − s ≥ N − 1 and if equality holdsthen they are stable, in particular they are simple (see [1]). The aim of this paperis to investigate the simplicity of Steiner bundles for higher rank.Main Theorem Let E be a Steiner bundle on P N−1 , with N ≥ 3, defined bythe exact sequence0 → O(−1) s m → O t → E → 0,where m is a generic morphism in Hom(O(−1) s , O t ), then the following statementsare equivalent:(i) E is simple, i.e. h 0 (End E) = 1,(ii) s 2 − Nst + t 2 ≤ 1 i.e. χ(End E) ≤ 1,(iii) either s 2 − Nst + t 2 ≤ 0 i.e. t ≤ ( N+√ N 2 −4)s or (t, s) = (a2 k+1 , a k ), wherea k =„N+√N 2 −42« k−„N−√N 2 −4√N 2 −42« k.∗ The author was supported by the MIUR in the framework of the National Research Projects“ Proprietà geometriche delle varietà reali e complesse” and “Geometria delle varietà algebriche”.2000 Math. Subject Classification: Primary 14F05. Secondary 14J60, 15A54, 15A24.1

The generalized Fibonacci numbers appearing in (iii) satisfy a recurrence relation,as it is clear from the proof of Theorem 2.1.Our result in the case of P 2 is partially contained, although somehow hidden, in[3]. Indeed Drézet and Le Potier find a criterion to check the stability of a genericbundle, given its rank and Chern classes. In the case of a normalized Steiner bundleE on P 2 , it is possible to prove that if E satisfies condition (iii) of the main theorem,then the Drézet-Le Potier condition for stability is satisfied. Hence E is stable and,consequently, simple. On the other hand, when E is not normalized, it is verycomplicated to check the criterion of Drézet-Le Potier, but we can easily prove thesimplicity with other techniques. Anyway the proof that we present in this paper isindependent of [3], is more elementary and works on P n as well.The genericity assumption cannot be dropped, because when rkE = t−s > N −1it is always possible to find a decomposable Steiner bundle, that is in particular nonsimple.Since the equivalence between conditions (ii) and (iii) is an arithmetic statement,our theorem claims that χ(End E) is the responsible for the simplicity of a genericSteiner bundle E. Indeed it is easy to check that if E is simple then χ(End E) ≤ 1(Lemma 3.2) and this is also true for some other bundles, for example for everybundle on P 2 . The converse is not true in general, because it is possible to find anon-simple bundle F on P 2 such that χ(End F) < 1. For example we can considerthe cokernel F of a generic map of the form0 → O(−2) ⊕ O(−1) 4 → O 16 → F → 0,where χ(End F) = −3, but it can be shown that h 0 (End F) = 5 therefore F is notsimple.In the third statement of our theorem we claim that if E is a simple Steiner bundle,then either E is exceptional or it satisfies a numerical inequality (see Theorem2.1). We recall that exceptional bundles have no deformations. The name exceptionalin this setting is justified by the fact that they are the only simple Steinerbundles which violate the numerical inequality. It is remarkable to note that all theexceptional bundles on P 2 can be constructed by the theory of helices, in particularthere exists a correspondence between the exceptional bundles on the projectiveplane and the solutions of the Markov equation x 2 + y 2 + z 2 = 3xyz (see [6]).The plan of the article is as follows: section 2 is devoted to the case of exceptionalbundles and section 3 to the proof of the main theorem. At the end of the paper,Theorem 3.8 is a reformulation in terms of matrices of the main theorem. As a basicreference for bundles on P n see [5].I would like to thank Giorgio Ottaviani, for suggesting me the problem and forhis continuous assistance, and Enrique Arrondo, for many useful discussions. I alsothank very much Jean Vallès, for his helpful comments concerning this work, inparticular for his collaboration in simplifying the proof of Lemma 3.7.2. – Exceptional bundlesIn [7] the theory of helices of exceptional bundles is developed in a general axiomaticpresentation. Here we give the following result as a particular case of this2

theory.Theorem 2.1 [6, 7] Let E k be a generic Steiner bundle on P N−1 , with N ≥ 3,defined by the exact sequence0 → O(−1) a k−1→O a k→ E k → 0,wherea k =(N+ √ N 2 −42) k−(√N2 − 4N− √ ) kN 2 −42,then E k is exceptional (i.e. h 0 (End E) = 1 and h i (End E) = 0 for all i > 0.)On P(V ) = P N−1 we define a sequence of vector bundles as follows:F 0 = O(1), F 1 = O, F n+1 = ker(F n ⊗ Hom(F n , F n−1 ) → ψnF n−1 ), (1)where ψ n is the canonical map.The following lemma can be found in [7]. We underline that it is possible toprove it in a straightforward way only by standard cohomology sequences.Lemma 2.2 Given the definition (1), for all n ≥ 1 the canonical map ψ n is anepimorphism. Moreover the following properties (A n ), (B n ) and (C n ) are satisfiedfor all n ≥ 1:(A n ) Hom(F n , F n ) ∼ = C, Ext i (F n , F n ) = 0, for all i ≥ 1,(B n ) Hom(F n−1 , F n ) = 0, Ext i (F n−1 , F n ) = 0, for all i ≥ 1,(C n ) Hom(F n , F n−1 ) ∼ = V, Ext i (F n , F n−1 ) = 0, for all i ≥ 1.Note that (A n ) means that every F n is an exceptional bundle.Remark 2.3 Following [7] the previous lemma means that (F n , F n−1 ) is a left admissiblepair and (F n+1 , F n ) is the left mutation of (F n , F n−1 ) and that the sequence(F n ) forms an exceptional collection generated by the helix (O(i)) by left mutations.Proof of Theorem 2.1. Lemma 2.2 states that the bundles F n , defined as in(1), are exceptional for all n ≥ 0. Obviously their dual Fn ∗ are exceptional too. Nowwe will prove that, for every n ≥ 1, the bundle Fn ∗ admits the following resolution0 → O(−1) a n−1→O an → F ∗ n→ 0, (2)where {a n } is the sequence defined in the statement. This implies that a genericbundle with this resolution is exceptional. We can prove (2) by induction on n.First of all we notice that the sequence {a n } is also defined recursively by⎧⎨⎩a 0 = 0,a 1 = 1,a n+1 = Na n − a n−1 .3

Therefore if n = 1 the sequence (2) is 0 → O(−1) a 0→O a 1→ F1 ∗ → 0, i.e.0→O→F1 ∗ → 0, and this is true because F 1∼ = O. Now let us suppose that everyFk ∗ admits a resolution (2) for all k ≤ n and we will prove it for F n+1 ∗ . Let usdualize the sequence0 → F n+1 → F n ⊗ Hom(F n , F n−1 ) → F n−1 → 0and by induction hypothesis we have:0 00 F ∗ n−1F ∗ n ⊗ V ∗F ∗ n+1 0O a n−1O an ⊗ V ∗O(−1) a n−2O(−1) a n−1⊗ V ∗00We define the map α : O a n−1→ F ∗ n ⊗ V ∗ as the composition of the known maps.Since Ext 1 (O a n−1, O(−1) a n−1⊗ V ∗ ) ∼ = H 1 (O(−1)) a2 n−1 ⊗ V ∗ = 0, the map α inducesa map ˜α : O a n−1→ O an ⊗ V ∗ such that the following diagram commutes:0 00 F ∗ n−1O a n−1f Fn ∗ ⊗ V ∗α eα O an ⊗ V ∗F ∗ n+1 0O(−1) a n−2O(−1) a n−1⊗ V ∗00We observe that ˜α is injective if and only if H 0 (˜α) is injective and, since H 0 (˜α) =H 0 (f), they are injective. Obviously the cokernel of ˜α is O Nan−a n−1= O a n+1. Let˜β be the restriction of ˜α to O(−1) a n−2. Then we can check that ˜β is injective, its4

cokernel is O(−1) Na n−1−a n−2= O(−1) an and the following diagram commutes:0 0 00 Fn−1∗ Fn ∗ ⊗ V ∗α0 O a eαn−1 O an ⊗ V ∗F ∗ n+1O a n+1000 O(−1) a n−2 eβ O(−1) a n−1⊗ V ∗ O(−1) an 0000It follows that Fn+1 ∗ has the resolution 0 → →O a O(−1)an n+1→ Fn+1 ∗completes the proof of our theorem.→ 0 and this3. – Proof of the main theoremLet E be given by the exact sequence on P N−1 = P(V )0 −→ I ⊗ O(−1)m−→ W ⊗ O −→ E −→ 0, (3)where V , I and W are complex vector spaces of dimension N ≥ 3, s and t respectivelyand m is a generic morphism. If we fix a basis in each of the vector spaces Iand W, the morphism m can be represented by a t × s matrix M whose entries arelinear forms. Let us consider the natural action of GL(I) × GL(W) on the spacei.e. the actionH = Hom(I ⊗ O(−1), W ⊗ O) ∼ = V ⊗ I ∨ ⊗ W,H × GL(I) × GL(W) → H(M, A, B) ↦→ A −1 MB.When the pair (A, B) belongs to the stabilizer of M, it induces a morphism φ : E →E, such that the following diagram commutes:0 I ⊗ O(−1) M W ⊗ OE0(4)A0 I ⊗ O(−1) M W ⊗ O E 0BφI. Now we prove the first part of the theorem, i.e. the fact that (i) implies (ii).5

Remark 3.1 From the sequence (3) it follows that χ(E) = t and χ(E(1)) = (Nt −s). Dualizing (3) and tensoring by E we gettherefore0 −→ End E −→ W ∨ ⊗ E−→I ∨ ⊗ E(1) −→ 0, (5)χ(End E) = tχ(E) − sχ(E(1)) = t 2 − s(Nt − s) = t 2 − Nst + s 2 .Lemma 3.2 If E is a simple Steiner bundle, then χ(End E) ≤ 1.Proof. From the sequences (3) and (5) it is easy to check that H i (End E) = 0,for all i ≥ 2. Moreover h 0 (End E) = 1 because of the simplicity, and consequentlyχ(End E) = 1 − h 1 (End E) ≤ 1.II. Now we prove that statement (ii) is equivalent to (iii).Remark 3.3 Obviously s 2 − Nst + t 2 ≤ 0 is equivalent to ( N−√ N 2 −42)s ≤ t ≤( N+√ N 2 −4)s. Since t > s and N > 2 this inequality is equivalent to t ≤ ( N+√ N 2 −4)s.2 2Thus we have only to prove that s 2 −Nst+t 2 = 1 is equivalent to (t, s) = (a k+1 , a k )where a k has been defined above.Lemma 3.4 All the integer solutions of s 2 − Nst + t 2 = 1, when t > s, are exactlys = a k , t = a k+1 , where a k =„N+√N 2 −42« k−„N−√N 2 −4√N 2 −42« k.Proof. We already know that the sequence {a k } is defined recursively by⎧⎨ a 0 = 0,a 1 = 1,⎩a k+1 = Na k − a k−1 .So we prove by induction on k that (s = a k , t = a k+1 ) is a solution ofs 2 − Nst + t 2 = 1. (6)If k = 0, obviously (s = 0, t = 1) is a solution. Let the pair (a k−1 , a k ) satisfy (6),then, using the recursive definition, we check that (a k , a k+1 ) is a solution too. Hencewe have to prove that there are no other solution. By the change of coordinates{r = 2t − Ns, s = s} our equation becomes the following Pell-Fermat equationr 2 − (N 2 − 4)s 2 = 4. By Number Theory results (see for example [8], page 77, or[4]), we know that all the solutions (r, s) are given by the sequence (r k , s k ) definedbyr k + s k√N2 − 4 = 12 k−1(N + √ N 2 − 4) k ,6

for all k ≥ 0. Now we have only to prove that these solutions are exactly thosealready known. We can easily check that the pair of sequences (s k , t k ) can berecursively defined by⎧⎪⎨⎪⎩r 0 = 2,s 0 = 0,r k+1 = (N2 −4)s k +Nr ks k+1 = Ns k+r k,2.2By a change of coordinates we define t k = Ns k+r kand we check that the pair (s2 k , t k )is exactly (a k , a k+1 ), for all k ≥ 0. In fact (s 0 , t 0 ) = (0, 1) = (a 0 , a 1 ) and, moreover,t k = s k+1 and t k+1 = Ns k+1+r k+1= (N2 −2)s k +Nr k= (N2 −2)s k +N(2t k −Ns k )= Nt2 2 2 k − t k−1 .III. Now we prove the last implication, i.e. (iii) implies (i). In the case (t, s) =(a k+1 , a k ), the generic E is an exceptional bundle by Theorem 2.1, therefore it isin particular simple. So suppose s 2 − Nst + t 2 ≤ 0 and recall that H denotesHom(I ⊗ O(−1), W ⊗ O) ∼ = V ⊗ I ∨ ⊗ W. Let S be the set{A, B, M : A −1 MB = M} ⊂ GL(I) × GL(W) × Hand π 1 and π 2 the projections on GL(I) × GL(W) and on H respectively. Noticethat, for all M ∈ H, π 1 (π2 −1 (M)) is the stabilizer of M with respect to the action ofGL(I) × GL(W). Obviously (λ Id, λ Id) ∈ Stab(M), therefore dim Stab(M) ≥ 1.Lemma 3.5 If E is defined by the sequenceand dim Stab(M) = 1, then E is simple.0 −→ I ⊗ O(−1) M−→ W ⊗ O −→ E −→ 0 (7)Proof. If by contradiction E is not simple, then there exists φ : E → E nontrivial.Applying the functor Hom(−, E) to the sequence (7) we get that φ induces˜φ non-trivial in Hom(W ⊗ O, E). Now applying the functor Hom(W ⊗ O, −) againto the same sequence we get Hom(W ⊗ O, W ⊗ O) ∼ = Hom(W ⊗ O, E) becauseHom(W ⊗O, I ⊗O(−1)) ∼ = W ⊗I ⊗H 0 (O(−1)) = 0 and Ext 1 (W ⊗O, I ⊗O(−1)) ∼ =W ⊗ I ⊗ H 1 (O(−1)) = 0. It follows that there exists ˜φ non-trivial in End(W ⊗ O),i.e. a matrix B ≠ Id in GL(W). Restricting ˜φ to I ⊗ O(−1) and calling A thecorresponding matrix in GL(I), we get the commutative diagram (4). Therefore(A, B) ≠ (λ Id, λ Id) belongs to Stab(M) and consequently dim Stab(M) > 1.Finally it suffices to prove that for all generic M ∈ H, the dimension of thestabilizer is exactly 1. In other words we have to prove the followingProposition 3.6 Let H = V ⊗ I ∨ ⊗ W as above and suppose s 2 − Nst + t 2 ≤ 0.Then the generic orbit in H with respect to the natural action of GL(I) × GL(W)has dimension exactly (s 2 + t 2 − 1).7

Recall that we have defined the following diagramS = {A, B, M : A −1 MB = M} π1π 2GL(I) × GL(W)HLet (A, B) be two fixed Jordan canonical forms in GL(I) × GL(W). We defineG AB ⊂ GL(I)×GL(W) as the set of couples of matrices similar respectively to A andB. Note that π 2 π1 −1 (G AB) = {C −1 MD : A −1 MB = M, C ∈ GL(I), D ∈ GL(W)}.Moreover G Id Id = {(λ Id, λ Id), λ ∈ C} and π 2 π1 −1 (G Id Id ) = H.Lemma 3.7 If s 2 − Nst + t 2 ≤ 0 and (A, B) are Jordan canonical forms differentfrom (λ Id, λ Id) for any λ, then π 2 π1 −1 (G AB) is contained in a Zariski closed subsetstrictly contained in H.Proof. Suppose that the assertion is false. Then there exist two Jordan canonicalforms A and B, different from (λ Id, λ Id), such that π 2 π1 −1 (G AB) is not containedin any closed subset. This implies that we can take a general M ∈ H such thatAM = MB and in particular we can suppose the rank of M maximum.Now we prove that A and B have the same minimal polynomial. First, if p B is theminimal polynomial of B, i.e. p B (B) = 0, then it follows that p B (A)M = Mp B (B) =0 and since M is injective we get p B (A) = 0, hence the minimal polynomial ofB divides that of A. Now if we denote by λ i (1 ≤ i ≤ q) the eigenvalues ofA and by µ j (1 ≤ j ≤ q ′ ) those of B, we obtain that µ j ∈ {λ 1 , . . .,λ q } for all1 ≤ j ≤ q ′ . Let us define A ′ = (A − x Id s ) and B ′ = (B − x Id t ): obviously weobtain A ′ M = MB ′ . We denote by B ′ the matrix of cofactors of B ′ and we knowthat B ′ B ′ = det(B ′ ) Id t = P B (x) Id t , where P B is the characteristic polynomial ofB. ThereforeA ′ MB ′ = P B (x)Mand developing this expression we see that q ′ = q. In fact if there exists a λ i ≠ µ jfor all j = 1, . . ., q ′ , then there is a row of zeroes in M and consequently M isnot generic. Then we get A and B with the same eigenvalues λ i (1 ≤ i ≤ q) withmultiplicity respectively a i ≥ 1 and b i ≥ 1. The hypothesis that (A, B) ≠ (λ Id, λ Id)means that either A and B have more than one eigenvalue or at least one of themis non-diagonal.∑Now consider the first case, i.e. q ≥ 2. Since dim I = s and dim W = t, obviouslyqi=1 a i = s e ∑ qi=1 b i = t. Now we denote M = (M ij ), where M ij has dimensiona i × b j . Since AM = MB, every block M ij is zero for all i ≠ j, i.e. it is possible towrite M with the form ⎛ ⎞∗ 0 · · · 00 ∗ · · · 0M = ⎜⎝.0 0 ..⎟ 0 ⎠ .0 0 · · · ∗8

In particular we can define n 1 = a 1 , n 2 = ∑ pi=2 a i, m 1 = b 1 , m 2 = ∑ pi=2 b i and thusthe matrix M becomes( )(∗)n1 ×mM = 1(0) n1 ×m 2(8)(0) n2 ×m 1(∗) n2 ×m 2where n 1 +n 2 = s and m 1 +m 2 = t and n i , m i ≥ 1 for i = 1, 2. Thus it only sufficesto show that a matrix in the orbitO M = {C −1 MD : C ∈ GL(s), D ∈ GL(t), M with the form (8)},is not generic in H if s 2 − Nst + t 2 ≤ 0. This fact contradicts our assumption andcompletes the proof.In order to show this, we introduce the following diagrams{φ, I 1 , W 1 : φ(I 1 ⊗ V ∨ ) ⊆ W 1 }α1 β 1H = Hom(I ⊗ V ∨ , W) G 1 = G(C n 1, C s ) × G(C m 1, C t ))where G(C k , C h ) denotes the Grassmannian of C k ⊂ C h and{φ, I 2 , W 2 : φ(I 2 ⊗ V ∨ ) ⊆ W 2 }α2 β 2H = Hom(I ⊗ V ∨ , W) G 2 = G(C n 2, C s ) × G(C m 2, C t ))It is easy to check that the matrices of the set O M live in the subvariety˜H = α 1 (β −11 (G 1)) ∩ α 2 (β −12 (G 2)) ⊆ H,then, in order to prove that these matrices are not generic, it suffices to show thatdim ˜H < dim H. Since dim(G i ) = (n 1 n 2 + m 1 m 2 ) for i = 1, 2, we obtainanddim(α 1 (β −11 (G 1))) ≤ dim(β −11 (G 1)) = n 1 n 2 + m 1 m 2 + N(n 1 (m 1 + m 2 ) + n 2 m 2 )dim(α 2 (β −12 (G 2 ))) ≤ dim(β −12 (G 2 )) = n 1 n 2 + m 1 m 2 + N(n 1 m 1 + n 2 (m 1 + m 2 )).Therefore, since dim H = Nst = N(n 1 + n 2 )(m 1 + m 2 ) we only need to show thateither (n 1 n 2 +m 1 m 2 −Nn 2 m 1 ) < 0 or (n 1 n 2 +m 1 m 2 −Nn 1 m 2 ) < 0. In other wordswe have to prove that the system{n1 n 2 + m 1 m 2 − Nn 1 m 2 ≥ 0n 1 n 2 + m 1 m 2 − Nn 2 m 1 ≥ 0has no solutions in our hypotesis s 2 − Nst + t 2 ≤ 0, i.e. ifN − √ N 2 − 42t ≤ s ≤ N + √ N 2 − 4t.29

This is equivalent to prove that the system⎧n 1 n 2 + m 1 m 2 − Nn 1 m 2 ≥ 0⎪⎨ n 1 n 2 + m 1 m 2 − Nn 2 m 1 ≥ 0n 1 + n 2 ≥ ⎪⎩N−√ N 2 −4(m 2 1 + m 2 )n 1 + n 2 ≤ N+√ N 2 −4(m 2 1 + m 2 )has no solutions. In order to do it, consider n 1 and m 1 as parameters and write theprevious system as a system of linear inequalities in two unknowns n 2 and m 2 :⎧⎪⎨⎪⎩n 1 n 2 ≥ (Nn 1 − m 1 )m 2(n 1 − Nm 1 )n 2 ≥ −m 1 m 2n 2 ≥ α − m 2 + (α − m 1 − n 1 )n 2 ≤ α + m 2 + (α + m 1 − n 1 )where we denote α − = N−√ N 2 −4and α2 + = N+√ N 2 −4. Notice that (α2 − + α + ) = Nand α − α + = 1, because they are solutions of the equation s 2 − Nst + t 2 = 0. Nowlet us consider three cases:• if 0 < n 1 − α + m 1 the system{n2 ≥ (Nn 1−m 1 )n 1m 2n 2 ≤ α + m 2 + (α + m 1 − n 1 )has no solutions because (α + m 1 − n 1 ) < 0 and α + < (Nn 1−m 1 )n 1, since (N −α + )n 1 − m 1 = α − n 1 − m 1 = (α + ) −1 (n 1 − α + m 1 ) > 0;• if n 1 − α + m 1 < 0 < n 1 − α − m 1 the system is{n 2 ≥ (Nn 1−m 1 )n 2 ≤n 1m 2m 1m (Nm 1 −n 1 ) 2because Nm 1 − n 1 > α + m 1 − n 1 > 0 and there is no solution becausem 1< (Nn 1−m 1 )(Nm 1 −n 1 ) n 1, since N(Nn 1 m 1 − m 2 1 − n 2 1) > 0;• if n 1 − α − m 1 < 0 then the system{m n2 ≤ 1m (Nm 1 −n 1 ) 2n 2 ≥ α − m 2 + (α − m 1 − n 1 )has no solutions because (α − m 1 − n 1 ) > 0 and α − >(Nm 1 −n 1 )m 1, since (N − α + )m 1 − n 1 = α − m 1 − n 1 > 0.Thus the proof in the case q ≥ 2 is complete.In the second case we consider q = 1 and the two matrices are⎛⎞⎛ ⎞λ 1J 1⎜A = ⎝. ..⎟λ 1⎠ , where J i = ⎜⎝.. ...⎟ . ⎠J hλ10m 1(Nm 1 −n 1 )i.e. α +

and c i denotes the order of J i andB =⎛⎜⎝L 1. ..L k⎞⎟⎠ , where L i =⎛⎜⎝λ 1λ 1. .. . ..λ⎞⎟⎠and d i is the order of L i . We suppose that c 1 ≥ 2 or d 1 ≥ 2 i.e. h < s or k < t. Thena matrix M such that AM = MB has the form M = (M ij ) and M ij is a c i × d jmatrix such that ⎧⎨ T c if c i = d j = cM ij = (0|T c ) if c = c i < d j⎩( T d0) if c i > d j = dand T c is a c × c upper-triangular Toeplitz matrix. It is easy to see that M has atleast k columns in which there are at least (c 1 −1)+(c 2 −1)+. . .+(c h −1) = (s−h)zeroes in such a way that we can order the basis so as to write M in the followingform ( )(∗)h×k (∗) h×(t−k).(0) (s−h)×k (∗) (s−h)×(t−k)Analogously M has at least h rows with at least (t−k) zeroes such that it is possibleto write the matrix in the form( )(∗)h×k (0) h×(t−k).(∗) (s−h)×k (∗) (s−h)×(t−k)Hence there exist non-trivial subspaces I 1 , I 2 , W 1 , W 2 such that M(I i ⊗ V ∨ ) ⊆ W i ,for i = 1, 2, and dim I 1 = s − h, dim W 1 = k, dim I 2 = h, dim W 2 = t − k. Thereforeexactly the same argument used in the first case gives that M is not generic andcompletes the proof.The previous lemma proves Proposition 3.6 and the main theorem follows. Thistheorem can also be reformulated as follows:Theorem 3.8 Let M a (s×t) matrix whose entries are linear forms in N variablesand consider the systemXM = MY, (9)where X ∈ GL(s) and Y ∈ GL(t) are the unknowns. Then if s 2 +t 2 −Nst ≤ 1, thereis a dense subset of the vector space C s ⊗C t ⊗C N , where M lives, such that the onlysolutions of (9) are trivial, i.e. (X, Y ) = (λ Id, λ Id) ∈ GL(s) × GL(t) for λ ∈ C.Conversely if s 2 + t 2 − Nst ≥ 2, then for all M there are non-trivial solutions.11

REFERENCES[1] V. Ancona and G. Ottaviani. Stability of special instanton bundles on P 2n+1 .Trans. Amer. Math. Soc., 341(2): 677–693, 1994.[2] I. Dolgachev and M. Kapranov. Arrangements of hyperplanes and vector bundleson P n . Duke Math. J., 71(3): 633–664, 1993.[3] J.-M. Drézet and J. Le Potier. Fibrés stables et fibrés exceptionnels sur P 2 . Ann.Sci. École Norm. Sup. (4), 18(2): 193–243, 1985.[4] H. W. Lenstra, Jr. Solving the Pell equation. Notices Amer. Math. Soc., 49(2):182–192, 2002.[5] C. Okonek, M. Schneider, and H. Spindler. Vector bundles on complex projectivespaces, volume 3 of Progress in Mathematics. Birkhäuser Boston, Mass., 1980.[6] A. N. Rudakov. Markov numbers and exceptional bundles on P 2 . Izv. Akad.Nauk SSSR Ser. Mat., 52(1): 100–112, 240, 1988.[7] A. N. Rudakov. Helices and vector bundles, volume 148 of London MathematicalSociety Lecture Note Series. Cambridge University Press, Cambridge, 1990.[8] P. Samuel. Théorie algébrique des nombres. Hermann, Paris, 1967.Dipartimento di Matematica “U.Dini”Viale Morgagni 67/A50134 Firenze, Italyemail: brambilla@math.unifi.it12