Maps between moduli spaces of vector bundles and the base locus of the theta divisor

Given a vector bundle $E$ of rank $r$ and degree $d$ on a curve $C$ of genus $g$, one can associate to $E$ in a natural way several other vector bundles. For example, one can take wedge powers of $E$. If $E$ is generated by global sections, the kernel of the evaluation map of sections is again a vector bundle. Also, new vector bundles can be produced by taking elementary transformations centered at a fixed point. Under suitable conditions on degree and rank, these constructions can be carried out globally. While all this processes seem quite elementary, very little is known about the resulting maps. The purpose of this paper is to fill in this gap.


Introduction
Given a vector bundle E of rank r and degree d on a curve C of genus g, one can associate to E in a natural way several other vector bundles. For example, one can take wedge powers of E. If E is generated by global sections, the kernel of the evaluation map of sections is again a vector bundle. Also, new vector bundles can be produced by taking elementary transformations centered at a fixed point. Under suitable conditions on degree and rank, these constructions can be carried out globally.While all this processes seem quite elementary, very little is known about the resulting maps. The purpose of this paper is to fill in this gap. We shall start by considering the kernel of the evaluation map of sections. If E is generated by global sections, consider M E the kernel of the evaluation map of sections of E. Then M E can be defined from the exact sequence ( * ) 0 → M E → H 0 (E) ⊗ O C → E → 0. Our main result is 1.1. Theorem Let d ≥ (2r − 1)g + 1. Then the map between the moduli spaces of vector bundles of rank r and degree d and rank d − rg and degree d given by E → M E is generically injective For the following result we assume that we are in characteristic zero. It is known then that the wedge powers of a semistable vector bundle are again semistable. Therefore, there is a rational map between the moduli space U (r, d) of vector bundles of rank r and degree d and the moduli space of vector bundles of rank r i and degree r i i d r . We show 1.2. Theorem Let C be a projective non-singular curve defined over a field of characteristic zero. Let i be an integer with 0 < i < r. Then the map between the moduli space U (r, d) and the moduli space U r i , r i i d r given by E → ∧ i E has finite fibers. The same statement is true when replacing wedge powers by symmetric powers.
Finally consider elementary transformations. Starting with a vector bundle, one can produce another vector bundle of degree one more (or one less) that essentially differs from the original one only at one point. This construction can be done in many different ways and in general one can get an r-dimensional family of such transforms starting with a single vector bundle. We show here that if we start with a family of vector bundles and do all possible such transforms, the dimension of the vector bundles so obtained never goes down.
We then give an application in finding lower bounds on the dimension of the base locus of the theta divisor in the moduli space of vector bundles (see Theorem 5.2 and Corollary 5.3). For many values of the rank, this improves known results of Arcara ( [A]), Popa ([P]) and Schneider ([S]).

Injectivity of the map E → M E
If E is a generic vector bundle of rank r and degree at least rg + 1, then it is generated by global sections. The same is true for every semistable vector bundle if the degree d satisfies d > r(2g − 1). Whenever E is generated by global sections, one can define the vector bundle M E as in (*). Our goal in this section is to prove that the assignment E → M E is generically injective when it is globally defined (namely d ≥ 2rg + r. What we show in fact is a little bit more, namely that M (ME ) * = E * . This in turn is equivalent to h 0 ((M E ) * ) = h 0 (E). We shall reduce this statement to the surjectivity of a map between spaces of sections of vector bundles and prove the latter by induction on the rank.

Lemma
Let C be a generic curve. Let L be a line bundle on C such that the complete linear system L has no fixed points. Then, the map C → |L| is composed with an involution if and only if L =L 2 withL a line bundle corresponding to a g 1 g+2 2 on C.
Proof. Assume that the map associated to the line bundle L is composed with an involution. As the curve C is generic, it is not a covering of any curve of genus at least one. Therefore the map corresponding to the linear system |L| factors through where C ′ is a rational curve and the first map is given by a g 1 k . Because of the genericity of C, this implies that the corresponding Brill-Noether number ρ = g − 2(g − k + 1) ≥ 0 is positive. Equivalently, k ≥ g+2 2 . Moreover, L = rg 1 k with h 0 (L) = r + 1.
) and this kernel is zero by the injectivity of the Petri map for the generic curve. Hence, Here the last inequality comes from the fact that r ≥ 2. Hence, all inequalities must be equalities and then r = 2, k = g+2 2 as stated.

Proposition
Let L be a line bundle generated by global sections on a generic curve. Assume L =L 2 whereL is a line bundle corresponding to a g 1 g+2 2 on C. Then, the map is onto. The result is also true on any curve if L is generic.
Proof. From 2.1, the map corresponding to the complete linear system |L| is not composed with an involution and by assumption it has no fixed points. The same is true on any curve if L is taken to be generic. By where h = h 1 (L) = h 0 (K ⊗ L −1 ). Choose P 1 , . . . , P k−2 generic points on C. Then, Moreover L(−P 1 − · · · − P k−2 ) has no fixed points and is therefore generated by global sections. Consider the mapψ : By the base point free pencil trick, the kernel ofψ is H 0 (K ⊗ L −1 (P 1 + · · · + P k−2 )) and by our choice this space has dimension h. Therefore, the image of the above map has dimension 2g − h. Now, the P i have been chosen to impose independent conditions on the linear system |L|. Hence, they impose independent conditions to the image of the map ψ. Then, The next result is not needed in the sequel. For the sake of completeness, we show that the the restriction in the previous Proposition is not arbitrary.

Lemma
Assume that L =L 2 whereL is a line bundle corresponding to a g 1 g+2 2 on C. Then the morphism ψ : is not onto.
Proof. Let s 0 , s 1 be a basis of section of |L|. Then s 2 0 , s 0 s 1 , s 2 1 are sections of |L|. From the base-point-free pencil trick, the map and as in the proof of 2.1, the latter is zero. Hence, the image has dimension 2g. As h 0 (K ⊗ L) = 2g + 1, the map is not onto. In order to prove the claim, it suffices to show now that the image of the map ψ is the same as the image of the mapψ. This is equivalent to proving that the elements of the form s 0 s 1 t, t ∈ H 0 (K) are in the image of ψ.
We show first that the map is onto. In fact, the kernel of this map is zero because this is a Petri map for a generic curve. Therefore the image has dimension 2h 0 (K ⊗L −1 ) = 2(h 0 (L) + g − 1 − deg(L)) = 2g as stated. Then, given t ∈ H 0 (K), we can write Then, s 0 s 1 t = (as 1 )s 2 0 + (bs 0 )s 2 1 as needed.
2.4. Theorem. Let C be a generic curve. Let L be a line bundle on C generated by global sections. Assume L =L 2 whereL is a line bundle corresponding to a g 1 g+2 2 on C. Consider the vector bundle M L defined by the exact sequence The same is true for every curve if L is taken to be generic.
Proof. Dualizing the sequence above and taking global sections, one obtains Therefore, the statement of the proposition is equivalent to the injectivity of the map This in turn is dual of the map ψ : which has been proved to be onto in 2.2.

Proposition
Let E be a generic vector bundle of rank r and degree d ≥ (2r − 1)g + 1. Then, the map is onto.
Proof. By Lange's Theorem [RT], a generic vector bundle can be written as an extension of generic stable bundles 0 → E 1 → E → E 2 → 0 so long as the slopes satisfy We prove the result by induction on the rank r. The case r = 1 has been proved in 2.2. Assume now that r ≥ 2 and write E as an extension where L is a line bundle of degree g+1 and E 2 is a vector bundle of degree at least (2r−2)g ≥ (2(r−1)−1)g+1. Then both L and E 2 are generated by global sections and by induction assumption the corresponding maps ψ L , ψ E2 are onto. Consider the exact diagram The zeros on the right are obtained because h 1 (L) = 0 as L is generic of degree g + 1 and also h 1 (K ⊗ L) = 0 As the left and right vertical maps are onto, so is the middle one.
2.6. Theorem. Let C be a generic curve. Let E be a generic vector bundle on C of degree at least (2r − 1)g + 1. Consider the vector bundle M E defined by the exact sequence Proof. The proof is now identical to the one in the line bundle case, replacing 2.2 by 2.5.
The proof of Theorem 1.1 immediately follows from the above as E = M * (ME ) * .

Elementary transformations
Given a vector bundle E a point P and a linear map of the fiber E P of E at P to the base field C, one has an exact sequence 0 → E ′ → E → C P → 0. Then, E ′ is said to be obtained from E by a (direct) elementary transformation.
Dualizing the above sequence, one obtains So, E * is obtained from E ′ * by a direct elementary transformation and E is said to be obtained from E ′ by an inverse transformation.
3.1. Lemma Let C be a curve, A a subvariety of U (r, d). Let B be the set of bundles obtained from bundles in A by doing an elementary transformation. Then dim B ≥ dim A.
Proof. Given an E ∈ A, one obtains elements in B by considering extensions of the form where P ∈ C is a point and k P is a one dimensional sky-scraper sheaf with support on P . Equivalently, one has an exact sequence Given E, one can choose a point P in C and a map from the fiber of E * at P to k P up to multiplication with a constant. This then determines E ′ . Assume that dim B < dim A. As every element in A gives rise to an r-dimensional family of elements in B, this would imply that every element in B comes from a family of elements in A that is at least r + 1-dimensional. From the first exact sequence a fixed E ′ in B comes from a family of dimension at most r in A. Hence, this is impossible.

Symmetric and wedge maps
In this section, we need to work over a field of characteristic zero, as otherwise the wedge and symmetric powers of semistable bundles are not necessarily semistable. We prove Theorem 1.2: Proof. The proof that follows was suggested to the authors by S.Ramanan.

The map
Pic is a finite map. Hence it suffices to prove the result for moduli spaces of vector bundles with fixed determinant. The Picard group of the moduli spaces of vector bundles of given rank and determinant is Z generated by an ample divisor θ. Consider the map The pull-back of θ ( r i ) is a multiple kθ r of the generator θ r of the Picard group of U (r, L) and therefore is ample. Hence, it intersects any positive dimensional subvariety. As it cannot intersect the fibers of the map above, it follows that the fibers are finite.
The proof for symmetric powers is carried out in the same way.

A lower bound for the base locus of the theta divisor in moduli spaces of vector bundles
Denote by U (r, L) the moduli space of vector bundles of rank r and fixed determinant L of degree r(g − 1). The Picard group of the moduli space is generated by the theta divisor that can be described as Consider the linear system in U (r, L) associated to the theta divisor. Its base locus consists of those vector bundles E such that h 0 (E ⊗ L) > 0 for every L of degree zero. It was shown first by Raynaud (see [R] ) that this base locus is non-empty. Later Popa showed in [P] that it is in fact of positive dimension. Work of Schneider and Arcara generalize Popa's results. In all these examples, the bound on the dimension depends on the rank (usually quite large compared with the genus). We give here a lower bound for the dimension of the base locus which, for some values of the rank, improves known bound.
We need some preliminary results on properties of M E that are analogous to known properties for M L when L is a line bundle.

Proposition
Let E be a vector bundle of rank r and degree d. Let M E be the dual of the kernel of the evaluation map of E. If j ≤ d − rg − 1 and P 1 , . . . , P j are generic points on C, there exists an exact sequence Proof. Choose generic points P 1 , . . . , P j and consider generic one-dimensional quotients k P1 , . . . , k Pj of E P1 , . . . , E Pj respectively. Let E 1 be the elementary transform of E associated to these quotients, namely E 1 is defined by the exact sequence Let V be the kernel of the map of vector spaces H 0 (E) → k P1 ⊕ ... ⊕ k Pj and W = H 0 (E)/V . Let F * be the kernel of the map V ⊗ O C → E 1 . We get then a commutative diagram Here A denotes the cokernel of the map F * → M E . Then, from the lower exact row, A = O(−P 1 ) ⊕ · · · ⊕ O(−P j ). Dualizing the left exact sequence, the result follows.

Theorem
Let r be an integer r ≥ 2. Let β = rg(2rg − 1) and L a line bundle of degree β(g − 1). The theta linear system in the moduli space of vector bundles of rank β and fixed determinant L has a base locus of dimension at least (r 2 − 1)(g − 1).
Proof. Let L 1 be a line bundle of degree 3g. Let E be a vector bundle of rank r and determinant L 1 . Note that E moves in a family of dimension r 2 (g − 1) + 1 − g = (r 2 − 1)(g − 1). Then M E has rank 2rg and det((M E ) * ) = det(E) = L 1 . Then From our choices, the slope of Let P 1 , . . . , P g−2 , Q 1 , Q 2 be generic points on C. From 5.1 and the condition on d, O(Q 1 ) ⊕ O(Q 2 ) is a subsheaf of (M E ) * . Hence O(Q 1 + Q 2 ) is a subsheaf of ∧ 2 M E . Given a line bundle T of degree g − 4, there exist points P 1 , · · · , P g−2 , Q 1 , Q 2 in C such that T = O(P 1 + . . . Let L 2 be a line bundle such that L rg(2rg−1) 2 L 2rg−1 1 = L. This implies that L 2 has degree g − 4 and is determined up to a finite number of choices.
Using 1.2, 1.1, the dimension of the set of vector bundles E and the dimension of the set of vector bundles ∧ 2 (M E ) * is the same, so the claim follows.

Corollary
Let j, β be defined as in the previous statement. If α > β, then the moduli space of vector bundles of rank α and given determinant L of degree α(g − 1) has a base locus of dimension at least ((r 2 − 1) + (α − β) 2 )(g − 1) + 1 Proof. Let E 1 be a vector bundle of rank α − β and slope g − 1. Let E 2 be a vector bundle of rank β and determinant L ⊗ det(E 1 ) −1 in the base locus of the theta divisor. Then, E = E 1 ⊕ E 2 is in the base locus of the theta divisor as h 0 (E ⊗ L 0 ) ≥ h 0 (E 2 ⊗ L 0 ) > 0 for every line bundle of degree zero. Hence, our claim follows from the previous theorem.