Theta divisors and Ulrich bundles on Geometrically ruled surfaces
M. Aprodu, G. Casnati, L. Costa, R.M. Mir\'o-Roig, M. Teixidor i Bigas

TL;DR
This paper explores the conditions under which geometrically ruled surfaces support Ulrich line bundles, revealing a surprising link with Theta divisors on moduli spaces and providing new existence results for Ulrich bundles.
Contribution
It establishes a novel connection between Theta divisors on moduli spaces and Ulrich line bundles on ruled surfaces, leading to comprehensive existence results.
Findings
Characterization of invariants for Ulrich line bundles on ruled surfaces
Relation between Theta divisors and Ulrich bundle existence
Existence of large families of Ulrich bundles in rank two case
Abstract
We consider the following question: for which invariants and is there a geometrically ruled surface over a curve of genus with invariant such that is the support of an Ulrich line bundle with respect to a very ample line bundle? A surprising relation between the existence of certain proper Theta divisors on some moduli spaces of vector bundles on with the existence of Ulrich line bundles on will be the key to completely solve the above question. The relation is realized by translating the vanishing conditions characterizing Ulrich line bundles to specific geometric conditions on the symmetric powers of the defining vector bundle of a given ruled surface. This general principle leads to some finer existence results of Ulrich line bundles in particular cases. Another focus is on the rank two case where, with very few exceptions, we show…
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Taxonomy
TopicsAlgebraic Geometry and Number Theory · Advanced Algebra and Geometry · Homotopy and Cohomology in Algebraic Topology
Theta divisors and Ulrich bundles
on Geometrically ruled surfaces
Marian Aprodu, Gianfranco Casnati, Laura Costa, Rosa Maria Miró-Roig, Montserrat Teixidor i Bigas
Facultatea de Matematică şi Informatică, Universitatea din Bucureşti, Str. Academiei 14, 010014 Bucureşti, ROMANIA & Institutul de Matematică “Simion Stoilow” al Academiei Române, Calea Griviţei 21, Sector 1, 010702 Bucureşti, ROMANIA
[email protected] & [email protected]
Dipartimento di Scienze Matematiche, Politecnico di Torino, c.so Duca degli Abruzzi 24, 10129 Torino, ITALY
Facultat de Matemàtiques i Informàtica, Departament de Matemàtiques i Informàtica, Gran Via de les Corts Catalanes 585, 08007 Barcelona, SPAIN
Facultat de Matemàtiques i Informàtica, Departament de Matemàtiques i Informàtica, Gran Via de les Corts Catalanes 585, 08007 Barcelona, SPAIN
Mathematics Department, Tufts University, 503 Boston Avenue, Medford MA 02155, USA
Abstract.
We consider the following question: for which invariants and is there a geometrically ruled surface over a curve of genus with invariant such that is the support of an Ulrich line bundle with respect to a very ample line bundle? A surprising relation between the existence of certain proper Theta divisors on some moduli spaces of vector bundles on with the existence of Ulrich line bundles on will be the key to completely solve the above question. The relation is realized by translating the vanishing conditions characterizing Ulrich line bundles to specific geometric conditions on the symmetric powers of the defining vector bundle of a given ruled surface. This general principle leads to some finer existence results of Ulrich line bundles in particular cases. Another focus is on the rank two case where, with very few exceptions, we show the existence of large families of special Ulrich bundles on arbitrary polarized ruled surfaces.
Key words and phrases:
Vector bundle, Ulrich bundle, geometrically ruled surface.
2010 Mathematics Subject Classification:
Primary 14J60; Secondary 14J26
The first author was partly supported by a grant of Ministery of Research and Innovation, CNCS - UEFISCDI, project number PN-III-P4-ID-PCE-2016-0030, within PNCDI III. The second author is a member of GNSAGA group of INdAM and is partially supported by the framework of PRIN 2015 ‘Geometry of Algebraic Varieties’, cofinanced by MIUR. The third and fourth authors have been partially supported by the grant MTM2016-78623-P
1. Introduction
Let be a smooth projective variety of dimension and set . A vector bundle on is an Ulrich bundle with respect to , if
[TABLE]
for each and . For the other equivalent definitions as well as a study of the properties of Ulrich bundles, we refer the interested reader to the papers by D. Eisenbud, F.-O. Schreyer and J. Weyman [8] and by M. Casanellas and R. Hartshorne [4].
Ulrich bundles come in pairs, if is Ulrich then so is its Ulrich dual , see [1]. Special Ulrich bundles are Ulrich self-dual rank-two bundles, [8]. Note that, if is an Ulrich line bundle and is its Ulrich dual, then any extension of by is a special Ulrich bundle.
The existence of Ulrich bundles of low rank on a surface , reflects important geometric properties of . For instance, if supports an Ulrich line bundle, then its associated Cayley-Chow form is linear determinantal and if supports special Ulrich rank two bundles then the Cayley-Chow form of is linear pfaffian (see [3] and [8]).
In this paper we are interested in low rank Ulrich bundles on geometrically ruled surfaces. Recall that if is a smooth curve of genus , then a rank bundle on is called normalized if h^{0}\big{(}C,{\mathcal{E}}\big{)}>0 and h^{0}\big{(}C,{\mathcal{E}}(\mathfrak{v})\big{)}=0 for each divisor on of negative degree. We denote by the geometrically ruled surface defined by a normalized . Note that this is not restrictive as the ruled surface defined by a vector bundle is isomorphic to the surface defined by the vector bundle for any line subbundle of maximal degree. We set and we define the invariant of as the number .
The Picard group of is generated by and by the class of any effective divisor corresponding to a non-zero section in H^{0}\big{(}S,{\mathcal{O}}_{S}(1)\big{)}, which is isomorphic to H^{0}\big{(}C,{\mathcal{E}}\big{)} by the projection formula. Following [10]; Chapter V, Notation 2.8.1, if is a divisor on we will write instead of . Thus the class of each divisor on can be written uniquely as . For instance, the canonical divisor on is in the class , being the canonical divisor on .
The intersection pairing on is given by , , , and we recall that
[TABLE]
M. Nagata proved that in [12]. Moreover, once the curve is fixed, it is well known that each value satisfying such an inequality is actually attained by some geometrically ruled surface on (see [10], Theorem V.2.12, Exercise V.2.5 and the references therein).
In this setting, it is natural to state the following question:
Question 1.1**.**
Let be a geometrically ruled surface and consider a very ample line bundle on .
- (a)
Are there Ulrich line bundles on with respect to ?
- (b)
More generally, what is the minimal rank of Ulrich vector bundles on ?
In [1], a subset of the authors prove that if and is very ample, then the minimal rank of an Ulrich bundle with respect to is if and only if and it is for . Moreover, they prove that if and either or some additional conditions on the numbers , , , are satisfied, then one still has . In particular, for and this gives a negative answer to Question 1.1 (a).
In this paper, we deal with the case . From the properties of ruled surfaces, we know that (see [10]; Chapter V, Exercise 2.5). We will show that is a necessary condition for the existence, see Proposition 2.2.
Our first result is the following statement which summarizes Theorems 2.4 and 2.7 giving a positive answer to Question 1.1 (a).
Theorem A. Let be a curve of genus , a normalised rank bundle on with and a very ample divisor on .
- (1)
If , then there are Ulrich line bundles on with respect to . 2. (2)
If and , then there are Ulrich line bundles on with respect to if and only if is odd. 3. (3)
If and , then there are Ulrich line bundles on with respect to if and only if is even. 4. (4)
If , and , then there are Ulrich line bundles on with respect to . 5. (5)
If and is general in its moduli space, then there are Ulrich line bundles on with respect to .
When and the picture seems to be very intricate. As shown in Proposition 2.2 the description of Ulrich line bundles is strictly related to the existence of suitable generic vanishing results for symmetric powers of rank two bundles on curves. For this reason we formulate the following related question:
Question 1.2**.**
Let , , and be integers such that , and . Is there a geometrically ruled surface over a curve of genus with invariant such that is the support of an Ulrich line bundle with respect to ?
Relating the existence of Ulrich line bundles with the existence of proper Theta divisors on some moduli spaces of vector bundles on we will be able to give a positive answer to Question 1.2 as follows (see Theorem 3.12).
Theorem B. Let , , and be integers such that , and . Then there exist a geometrically ruled surface over a curve of genus with invariant such that is the support of an Ulrich line bundle with respect to .
We then focus on Question 1.1 (b). We conclude the paper by showing the existence of large families of rank two Ulrich vector bundles with respect to . For they can be constructed as an extensions of Ulrich line bundles. For and some mild conditions on we construct them in Theorems 4.1 and 4.3.
Notation: Throughout this note we will work on an algebraically closed field of characteristic [math] and will denote the projective space over of dimension . The words curve and surface will always denote projective smooth connected objects. In several places, we shall mix the multiplicative notation for line bundles and the additive notation for divisors.
2. Ulrich line bundles on ruled surfaces
The goal of this section is to determine the existence of Ulrich line bundles on a geometrically ruled surface with negative invariant and in particular to prove Theorem A stated in the introduction. We start by recalling some useful facts.
If with is a divisor on , then Lemma V.2.4, Exercises III.8.3 and III.8.4 of [10] imply
[TABLE]
where stands for the -th symmetric power of .
On the other hand, since is normalized, there is an everywhere non-zero section in H^{0}\big{(}C,{\mathcal{E}}\big{)} defining the exact sequence
[TABLE]
(see the proof of [10], Theorem V.2.12). Notice that such an extension corresponds to an element \xi\in H^{1}\big{(}C,{\mathcal{O}}_{C}(-\mathfrak{e})\big{)}.
Thus there also exists an exact sequence of the form (see the proof of Lemma 7.6 of [7])
[TABLE]
where . Due to its construction, such an extension depends on the choice of and .
Easy induction on using (1) yields
[TABLE]
for each divisor on .
The following lemma is a particular case of [5]; Corollary 2.2.
Lemma 2.1**.**
Let be a curve of genus , a normalized rank vector bundle on and a very ample divisor on .
The line bundle is Ulrich with respect to if and only if h^{0}\big{(}S,{\mathcal{O}}_{S}(D-h)\big{)}=h^{0}\big{(}S,{\mathcal{O}}_{S}(2h+K_{S}-D)\big{)}=0 and
[TABLE]
The following result is the main part of [1]; Theorem 2.1 where we set
[TABLE]
For the reader’s benefit we repeat here the proof under our current assumptions, mainly with no restrictions on .
Proposition 2.2**.**
Let be a curve of genus , a normalized rank vector bundle on and a very ample divisor on .
There is an Ulrich line bundle on with respect to if and only if and there exist divisors satisfying
[TABLE]
The Ulrich line bundles on are exactly the ones of the form
[TABLE]
and their Ulrich duals
[TABLE]
for each on satisfying condition (5) above. In particular, if is odd, then there are no Ulrich bundles on with respect to .
Proof.
Assume that supports an Ulrich line bundle . Thus its Ulrich dual is also an Ulrich bundle.
In particular, if , then
[TABLE]
Since both and are assumed to be Ulrich with respect to , it follows that . Thus, a direct computation via the Riemann–Roch theorem as in the first part of the proof of [1]; Theorem 2.1 yields the vanishing
[TABLE]
If , for , then the equality yields . It would follow , trivially contradicting the ampleness of . If , then , again a contradiction. Thus we can assume and , whence and
[TABLE]
Let , so that and . The divisor satisfies the equalities (4) and h^{0}\big{(}S,{\mathcal{O}}_{S}(2h+K_{S}-D)\big{)}=h^{0}\big{(}S,\mathcal{L}(-h)\big{)}=0.
On the other hand, according to (1),
[TABLE]
Thus, taking the statement follows from Lemma 2.1. ∎
For the proof of the following result see [1], Theorem 2.1.
Proposition 2.3**.**
Let be a curve of genus , a normalized rank vector bundle on and a very ample divisor on .
If , then there are Ulrich line bundles on with respect to if and only if .
Thanks to Proposition 2.2 we are able to extend the above proposition to the case .
Theorem 2.4**.**
Let be a curve of genus , a normalized rank bundle on and a very ample divisor on .
If , then there exist two families of dimension of Ulrich line bundles with respect to .
Proof.
We have because . Thus, the set of line bundles such that h^{0}\big{(}C,{\mathcal{O}}_{C}(\mathfrak{u})\big{)}=0 is open and non-empty, because it is the complement of the theta divisor . Trivially .
In particular, if Ulrich line bundles on exist, their characterization in Proposition 2.2 means that they form two families according to whether is or . Both these families have the same dimension .
The rest of the proof is devoted to showing that a general divisor in actually satisfies condition (5).
The case is trivial due to the non-emptiness of . So, we can restrict ourselves to the case and . Such case is very easy to handle when (see [5], Example 2.3). Therefore, we will assume .
Notice that for each , we have . Thus, for all and for each with
[TABLE]
Inequality (3) yields h^{0}\big{(}C,(S^{a-1}{\mathcal{E}})(\mathfrak{u})\big{)}=0 for such an . Thus the statement follows from Proposition 2.2. ∎
When and the picture is much more intricate. In order to prove the existence of Ulrich line bundles in this setting, in this section we will relate their existence to the so called Raynaud’s condition and in the next section to the existence of suitable theta divisors. Let us introduce Raynaud’s condition.
Let be a vector bundle of rank on a curve of genus . Riemann–Roch’s Theorem for gives
[TABLE]
If , one has \chi({\mathcal{F}})=\chi({\mathcal{F}}(\mathfrak{v}))=h^{0}\big{(}C,{\mathcal{F}}(\mathfrak{v})\big{)}-h^{1}\big{(}C,{\mathcal{F}}(\mathfrak{v})\big{)}. The integer h^{0}\big{(}C,{\mathcal{F}}(\mathfrak{v})\big{)} is a function of , but there exists a non-empty open subset where it takes a constant value, say .
Assume now . If , then , or, in other words, .
Definition 2.5**.**
Let be a curve of genus and a vector bundle on . We say that satisfies condition if and only if .
Condition is also known in the literature as Raynaud’s condition.
The relation between Raynaud’s condition and Ulrich line bundles is the following
Lemma 2.6**.**
Let be a curve of genus , a normalized rank vector bundle on and a very ample divisor on .
Then there is an Ulrich line bundle on with respect to if and only if and there exist divisors such that satisfies condition .
Proof.
For each divisor (if any, i.e. if is even), we have
[TABLE]
and hence . Thus, the equality is equivalent to condition . ∎
Note that semistable bundles of slope precisely are of special interest in view of condition . Indeed, as pointed out in Raynaud’s work, semistable bundles of smaller slope automatically satisfy this condition, and hence this is a borderline case. For a geometric phenomenon related to the bundles of slope we refer to Proposition 1.8.1 in [14].
The above proposition together with the results proved in [14] allow us to prove the existence of Ulrich line bundles in several cases. In particular, if , then we are able to give a complete answer concerning the existence of Ulrich line bundles.
Theorem 2.7**.**
Let be a curve of genus , a normalized rank bundle on with and a very ample divisor on .
- (1)
If , then there are Ulrich line bundles on with respect to if and only if is odd. 2. (2)
If , then there are Ulrich line bundles on with respect to if and only if is even. 3. (3)
If and , then there are Ulrich line bundles on with respect to . 4. (4)
If and is general in its moduli space, then there are Ulrich line bundles on with respect to .
Proof.
First of all notice that since is a normalized rank two bundle of degre , it is -semistable and the same holds for any of its symmetric powers and their twists. Let us start with the proof of assertion (1). As , if , then the hypothesis forces . Thus if is even there are no Ulrich line bundles on due to Proposition 2.2. If is odd, then is an integer and hence is non-empty. Thus the statement follows from [14], Corollaire 1.7.3 and Lemma 2.6 because is -semistable, as noted above.
Let us prove assertion (2). If is odd, then there are no Ulrich line bundles on due to Proposition 2.2. If is even, then the rank vector bundle is -semistable and the statement follows from Lemma 2.6 and [14], Proposition 1.6.2.
Assertion (3), follows from Lemma 2.6 and [14], Corollaire 1.7.4 due to the fact that is a rank -semistable vector bundle. Finally, since is a rank -semistable vector bundle, (4) follows from [14], section 2.5. ∎
3. Ulrich line bundles and theta divisors
The goal of this section is to determine the existence of geometrically ruled surfaces with negative invariant supporting Ulrich line bundles and in particular to prove Theorem B stated in the introduction.
In some sense Raynaud’s condition is related to the existence of theta divisors on moduli spaces of semistable vector bundles. We will prove the existence of proper theta divisors of some symmetric powers of rank two vector bundles on and this will give us the existence, under some generic conditions, of Ulrich line bundles on geometrically ruled surfaces . In particular, we will be able to give a positive answer to Question 1.2.
Let us recall the definition of theta divisors. Denote by the moduli space of rank semistable vector bundles on of degree .
Definition 3.1**.**
Let be a vector bundle of degree and rank on . Denote by the greatest common divisor of . Then . We define
[TABLE]
If is vector bundle of rank and degree , then . It is expected that for a generic and generic , the space of sections of will be zero and that will be a divisor of the moduli space. If this is the case, is called a theta divisor. For a generic of rank and degree , it is known that is a divisor of (see [14] Prop. 1.8(1)) but this is not true for every . It has been shown that for some values of , there exist vector bundles, sometimes even infinite families of for which
[TABLE]
(see [13]).
In view of Proposition 2.2 we have the following characterization of the existence of Ulrich line bundles in terms of the existence of theta divisors.
Proposition 3.2**.**
Let be a curve of genus , a normalized rank bundle on and a very ample divisor on . Then, there is an Ulrich line bundle on with respect to if and only if and is a proper divisor of .
Proof.
We have if and only if is even, hence
[TABLE]
because and are the rank and the degree of . The statement then follows from Proposition 2.2. ∎
Therefore, our next goal is to study the existence of theta divisors of symmetric powers of rank two normalized vector bundles on .
Lemma 3.3**.**
Let be a curve, a vector bundle of rank two on , a line subbundle of maximal degree on . Then, is normalized.
Proof.
By assumption, there exists an injective map and therefore also a map . Hence, . Assume now that there is a line bundle of negative degree such that , then is a subsheaf of of degree higher than the degree of contradicting the assumption. ∎
Proposition 3.4**.**
Fix integers . For a vector bundle of rank and degree define
[TABLE]
For a fixed with , define
[TABLE]
Then is non-empty, irreducible of dimension and .
Proof.
If there are no integers satisfying the above restrictions. If see [15] Theorem 0.1 and Corollary 1.12. ∎
Corollary 3.5**.**
Let be a curve of genus g, a rank two vector bundle of degree on . Then, if , there exists a line bundle of degree such that is normalized. In particular
[TABLE]
Proof.
The statement follows from Lemma 3.3 and Proposition 3.4. ∎
Proposition 3.6**.**
Let be a curve of genus g, a rank vector bundle of degree and a line bundle of degree . With the notations above is a proper divisor of if and only if is a proper divisor of .
Proof.
The map
[TABLE]
gives a bijection between the two moduli spaces (with inverse ). From the definition of the theta locus, maps to under this map. So one locus is a divisor if and only if the other is. ∎
Corollary 3.7**.**
Let be a curve of genus g, a rank two vector bundle of degree , a line bundle of degree and a positive integer.
- (a)
If is even, is a proper divisor of if and only if is a proper divisor of .
- (b)
If is odd, is a proper divisor of if and only if is a proper divisor of .
Proof.
This follows from Proposition 3.6 as and . ∎
Now we are ready to state the first main result concerning the existence of proper theta divisors. To this end, we first consider a normalized rank two vector bundle of even degree and an integer. In particular, the slope of is . The bundle has rank and slope . Therefore,
[TABLE]
Proposition 3.8**.**
Let be any curve of genus g and fix an even degree and an integer . Let be a normalized vector bundle generic in the stratum . Then, for all , is a proper divisor of .
Proof.
The condition that is a proper divisor is an open condition in the moduli space of vector bundles. Using Proposition 3.4 it suffices to prove the result for the smallest stratum of the moduli space of vector bundles corresponding to those bundles with subbundles of the largest degree. These bundles are extensions of two line bundles of the same degree. They can be deformed to a direct sum of two (different) generic line bundles each of degree (consider the family of extensions of one of the line bundles by the other). Then is the set of tensors in that are invariant under the action of the symmetric group. The set
[TABLE]
is a theta divisor in the Jacobian. Moreover, is contained in the union of these theta divisors as varies. Thus, it is still a divisor. ∎
Corollary 3.9**.**
Let be a curve of genus , a normalized rank two vector bundle on generic among those of degree and a very ample divisor on . Then, there is an Ulrich line bundle on with respect to .
Proof.
From Proposition 3.2, we need to check that is a proper divisor. From Corollary 3.5, it suffices to do this for a generic point of . From Proposition 3.8, this holds. ∎
Let us now consider a normalized rank two vector bundle of odd degree and an integer.
Proposition 3.10**.**
Let be a generic curve of genus g and be a generic normalized rank two vector bundle of degree on . Then for odd , is a proper divisor of and for even , is a proper divisor of the moduli space .
Proof.
We start by deforming the curve to a chain of elliptic curves as follows. Consider generic elliptic curves. Let be generic points on . We construct a curve of arithmetic genus by identifying with . Let us now determine a generic normalized rank two vector bundle of degree on . To this end, take a generic indecomposable vector bundle of rank 2 and degree 1 on , a direct sum of two generic line bundles of degree one on ; and a direct sum of two generic line bundles of degree zero on the remaining components and . We take the gluing so that glue to each other but the gluings are generic otherwise. One can then find a degree zero line subbundle of on that glues with both and and a degree zero line subbundle of that glues with . Gluing these degree zero line subbundles on each component produces a degree zero line subbundle of the vector bundle on the chain. One can in fact check that the largest degree of a subbundle of on is zero (see [16] for details). Hence, is a generic normalized rank two vector bundle on of degree .
In order to show that the theta divisor of the symmetric power of is an actual divisor, it suffices to find a line bundle on (resp. vector bundle ) of the appropriate degrees such that (resp () do not have any limit linear section.
Notice that the vector bundle that we built on has restriction to a direct sum of two generic line bundles of the same degree and has restriction to a generic indecomposable rank two vector bundle of degree . The -symmetric power of is a subsheaf of . The -symmetric power of is a subsheaf of the -tensor power of . Since is a rank two vector bundle, where . On the other hand, it follows from [2]; Lemma 22 that where the are the elements in of order 2.
Therefore,
[TABLE]
A limit linear series of slope of the tensor product of with an arbitrary line bundle of degree (resp. an arbitrary rank two vector bundle of degree ) would give rise to a section on each component with proper vanishing at the nodes.
Assume first that is odd and therefore is even. According to (6) and the construction of ,
[TABLE]
where denotes the restriction of to .
On the other hand, the restriction has degree on , the restriction has degree on the components and degree on the remaining components and . Moreover, it is generic of the stated degree on each of the components. Notice also that are fixed determined by the generic choice of . In addition, the line bundles form a subgroup, so their product is another element in this subgroup and their degree is zero. Hence, the degree of and of is . By the genericity of , the sum of the orders of vanishing of a section of the line bundles at the two nodes of the elliptic curve is at most . In order to have a limit linear series, the order of vanishing at of the sections on the component and the order of vanishing at of the sections on the component needs to be at least . Therefore, the sum of vanishing orders at the nodes satisfies
[TABLE]
which is impossible.
Consider now the case in which is even and then is an arbitrary rank two vector bundle of degree . We take on to be a generic vector bundle of degree , the direct sum of two generic line bundles of degree on the components and the direct sum of two generic line bundles of degree on the remaining components and on . Any two indecomposable vector bundles of rank two and odd degree differ in product with a line bundle (see [2]; corollary to Theorem 7). Therefore, there exists a line bundle of degree such that Then,
[TABLE]
The same argument as before shows that this cannot have a limit section.
∎
Because our proof uses a deformation argument to a special kind of curve, we cannot conclude that the result is true for every curve. It is likely though that, as in the case of even degree, the result holds for every curve.
Corollary 3.11**.**
Let be a generic curve of genus , a normalized rank two vector bundle on generic among those of degree and a very ample divisor on with odd (i.e. ). Then, there is an Ulrich line bundle on with respect to .
Proof.
It follows from Proposition 3.2 and Proposition 3.10 ∎
Now we are ready to give a positive answer to Question 1.2 stated in the introduction
Theorem 3.12**.**
Let , , and be integers such that , and . Then there exist a geometrically ruled surface over a curve of genus with invariant such that is the support of an Ulrich line bundle with respect to .
Proof.
It follows from Corollary 3.9 and Corollary 3.11 ∎
4. Stable rank Ulrich bundles
In this section, we focus our attention on the existence of special stable rank Ulrich bundles. The existence is known for and see [5], Theorem 1.2. When and there are no such bundles, actually it is known (see [9], Corollary to Theorem B), that each Ulrich bundle of rank at least is in this case strictly -semistable. Finally, when the existence of -stable special Ulrich bundles of rank is proved in [6]; Theorem 1.2.
First of all, we study the existence of -stable rank Ulrich bundles on with respect to for . To this end, we set
[TABLE]
With this notation we have the following result:
Theorem 4.1**.**
Let be a curve of genus , a normalized rank bundle on and a very ample divisor on .
If or and
[TABLE]
then, for each general 0-dimensional subscheme of degree and each general , there exist -stable special Ulrich bundles of rank with respect to fitting into the exact sequence
[TABLE]
Proof.
The same argument used in [1], Proposition 3.3 and Theorem 3.4 can be extended almost verbatim to prove that for any these extensions define rank special Ulrich bundles. The unique point where that proof must be slightly modified is when the ampleness of is needed. In order to infer such an ampleness we need if and if . Both inequalities are true since
[TABLE]
We need to prove that bundles in the statement are -stable. They are certainly -stable in the range and . In fact, Ulrich vector bundles are always -semistable and they can only be destabilized by Ulrich line bundles which, in this range, do not exist by [1]; Theorem 2.1.
In order to prove the statement for and , we compute the dimension of the entire family of bundles defined in the statement and we estimate the dimension of the subfamily of strictly -semistable ones.
First, we check that a general belongs to one single extension. To this end, it suffices to prove that
[TABLE]
for general and . Observe that
[TABLE]
On the other hand, by the assumption . Hence h^{1}\big{(}C,{\mathcal{O}}_{C}(\mathfrak{b}+\mathfrak{k}-2\mathfrak{v}+(i+1)\mathfrak{e})\big{)} will be zero for all with , and for a generic choice of . Thus, applying the Riemann–Roch theorem, we obtain
[TABLE]
Hence, for a generic choice of , we have h^{0}\big{(}S,\mathcal{I}_{Z|S}((a-2)C_{0}+(\mathfrak{b}+\mathfrak{k}-2\mathfrak{v}+\mathfrak{e}))\big{)}=0.
The bundles are parameterized by a projective bundle with typical fibre the projectivization of
[TABLE]
over an open subset of the product of the Hilbert scheme of subschemes of of dimension [math] and degree multiplied by .
Let . In the proof of [1]; Theorem 3.4, the authors prove that
[TABLE]
It follows from the first equality (7) and the hypothesis on that
[TABLE]
for a general choice of .
Let . Trivially h^{0}\big{(}S,{\mathcal{O}}_{S}(-2C_{0}+(\mathfrak{b}+2\mathfrak{k}+2\mathfrak{e}-2\mathfrak{v})f)\big{)}=0. Moreover for
[TABLE]
We have , thus
[TABLE]
In both cases above ( and ), the cohomology of the sequence
[TABLE]
tensored by and the equalities (7), (8), (9) yield
[TABLE]
Therefore, since depends on the schemes , the bundles and the extension classes, it follows that
[TABLE]
From now on we will assume that is strictly -semistable. If this is the case, it contains an Ulrich line bundle and, in particular, by Proposition 2.2 this implies that is even. Recall also that we are under the assumption .
Ulrich line bundles are given by Proposition 2.2. We cannot have
[TABLE]
for in this case there would necessarily be a non-zero morphism from to either or which is a contradiction since and .
Thus, we necessarily have and since their quotient must be also an Ulrich line bundle we get an exact sequence of the form
[TABLE]
By the projection formula,
[TABLE]
The extensions as in the exact sequence (11) are parameterised by a space of dimension
[TABLE]
where the inequality follows from the inequality (3).
We have , hence the Riemann–Roch theorem and the Clifford theorem imply
[TABLE]
for each general choice of .
Since , and then
[TABLE]
Hence the strictly -semistable bundles we are interested in are parameterised by a family of dimension at most
[TABLE]
The dimension of the family of strictly -semistable bundles is smaller than the value given by the equality (10) if
[TABLE]
condition which is automatically satisfied, since , by hypothesis.
Since a general is uniquely determined by an extension in , we conclude that there are -stable Ulrich bundles. ∎
Finally, we consider slightly different extensions to remove the restriction on for . To this purpose, we will make use of the following characterization of special Ulrich bundles of rank in our particular setup.
Lemma 4.2**.**
Let be a curve of genus , a normalized rank vector bundle on and a very ample divisor on .
A rank vector bundle on is a special Ulrich bundle with respect to if and only if h^{0}\big{(}S,{\mathcal{F}}(-h)\big{)}=0 and
[TABLE]
Proof.
For the proof we refer the interested reader to [5], Corollary 2.2 or [1], Lemma 3.2. The proof therein is given under the apparently more restrictive hypothesis that is initialized, i.e. h^{0}\big{(}S,{\mathcal{F}}\big{)}\neq 0 and h^{0}\big{(}S,{\mathcal{F}}(-h)\big{)}=0. Only the second vanishing is actually necessary. Indeed the condition h^{0}\big{(}S,{\mathcal{F}}\big{)}\neq 0 follows easily from the Riemann–Roch theorem applied to . ∎
We will end by proving the existence of rank two Ulrich bundles for without the restriction on .
Theorem 4.3**.**
Let be a curve of genus , a normalized rank bundle of degree zero on and a very ample divisor on with . Let and .
Then for each general subscheme of dimension [math] of degree and each general , there exist -stable special Ulrich bundles of rank with respect to fitting into the exact sequence
[TABLE]
Proof.
Let be a non-effective line bundle on .
We define the following two divisors on
[TABLE]
We trivially have .
We observe that . In fact, the restriction of to is very ample: since is an isomorphism on and it follows that
[TABLE]
It is clear that h^{0}\big{(}S,{\mathcal{O}}_{S}(A+K_{S})\big{)}=0. Thus, by [11]; Theorem 5.1.1 there exists a rank vector bundle fitting into the exact sequence
[TABLE]
Simple computations show that the equalities (12) are satisfied. We will show that h^{0}\big{(}S,{\mathcal{F}}(-h)\big{)}=0 and that is -stable for a general choice of and .
We now prove that h^{0}\big{(}S,{\mathcal{F}}(-h)\big{)}=0. To this purpose we will check that
[TABLE]
for a general choice of and .
We have . For a general the inequality (3) yields
[TABLE]
Since , choosing such that for all , we obtain h^{0}\big{(}S,{\mathcal{O}}_{S}(D-h)\big{)}=0.
On the other hand, . Hence again the inequality (3) yields
[TABLE]
Since (see the inequality (13)) and , it follows that for all , which implies
[TABLE]
for general . Thus h^{0}\big{(}C,{\mathcal{O}}_{C}(\mathfrak{b}+\mathfrak{k}-{\mathfrak{v}}+i\mathfrak{e})\big{)}=\deg(\mathfrak{b}) and an easy computation shows
[TABLE]
Thus for a general choice of , h^{0}\big{(}S,{\mathcal{I}}_{Z|S}(D+A-h)\big{)}=0. In particular, by Lemma 4.2, is a special Ulrich bundle of rank .
We now show that a general is -stable. If an Ulrich bundle of rank is not -stable, then it contains a proper Ulrich subbundle of rank . We will show that for each Ulrich line bundle on we have h^{0}\big{(}S,{\mathcal{F}}\otimes\mathcal{L}^{\vee}\big{)}=0. Since fits into the exact sequence (14), this will follow if we prove that
[TABLE]
The Ulrich line bundles on are described in Proposition 2.2. Notice that because is a bundle of degree .
If
[TABLE]
then the required vanishings follow by looking at the coefficient of in and respectively. Now let
[TABLE]
The inequality (3) implies
[TABLE]
Trivially
[TABLE]
hence h^{0}\big{(}C,{\mathcal{O}}_{C}(-\mathfrak{b}+\mathfrak{u}+{\mathfrak{v}}-\mathfrak{k}+i\mathfrak{e})\big{)}=0 which implies that h^{0}\big{(}S,{\mathcal{O}}_{S}(D)\otimes\mathcal{L}^{\vee}\big{)}=0. Finally hence the inequality (3) yields
[TABLE]
We have , so that . If , then
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If , then . If, moreover then
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Finally, if and then
[TABLE]
(recall that when ). Thus, if we start our construction of from a set of points not lying on any divisor in the classes of , then again
[TABLE]
∎
The reference list from the paper itself. Each links out to its DOI / PubMed record.
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- 6[6] G. Casnati: Ulrich bundles on non-special surfaces with p g = 0 subscript 𝑝 𝑔 0 p_{g}=0 and q = 1 𝑞 1 q=1 . Rev. Mat. Complut. (2017). https://doi.org/10.1007/s 13163-017-0248-z
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