Derived equivalent Hilbert schemes of points on K3 surfaces which are not birational

We provide a criterion for when Hilbert schemes of points on K3 surfaces are birational. In particular, this allows us to generate a plethora of examples of non-birational Hilbert schemes which are derived equivalent.


Introduction
The Bondal-Orlov conjecture [6] provides a fundamental bridge between birational geometry and derived categories. It claims that if two varieties with trivial canonical bundle are birational then their bounded derived categories of coherent sheaves are equivalent. Whilst this conjecture is of paramount importance to the algebro-geometric community, it is examples where the converse fails that we are most interested in. The most famous example of this kind is Mukai's derived equivalence [18] between an Abelian variety and its dual. Calabi-Yau examples have been the focus of a recent flurry of articles: [2][3][4][11][12][13]15,23], but there were no such examples in the hyperkähler setting until very recently. Indeed, the first examples of derived equivalent non-birational hyperkählers were exhibited in [1,Theorem B] as certain moduli spaces of torsion sheaves on K3 surfaces. This article complements this discovery with further examples coming from Hilbert schemes of points and, in some sense, completes the investigation initiated by Ploog [25].
In the special case of Hilbert schemes of points on K3 surfaces, the implication K ⇒ D was established in [25,Proposition 10]. Halpern-Leistner [8] has announced a generalisation of this result to moduli spaces of sheaves on K3 surfaces.
1 Examples which are D-equivalent but not K-equivalent

Degree twelve
We work through a specific example in order to demonstrate how certain Hilbert schemes can be derived equivalent and not birational.
Let X be a complex projective K3 surface with Pic(X ) = Z[H ] and w ∈ H * alg (X , Z) a primitive vector with w 2 = 0. Then Mukai [19] shows that the moduli space Y = M H (w) of Gieseker H -stable sheaves is a K3 surface. Moreover, the derived Torelli theorem of Mukai and Orlov [24] shows that if there exists a vector v ∈ H * alg (X , Z) with (v, w) = 1 1 then there is a universal family E on X × Y which induces a derived equivalence: By [25,Proposition 8] H , 4) is a vector such that (v, w) = 1 (or gcd(2, 3) = 1), we have a universal family E and a derived equivalence as above. Now, the proof of [26,Theorem 2.4] shows that H 2 (X , Z) Hdg H 2 (Y , Z) and so the K3 surfaces X and Y are not birational. That is, when n = 1 the answer to our Question 1.1 above is no. However, when n = 2, 3, 4 the answer to Question 1.1 is yes! To see this, we use [28, Lemma 7.2] (with d 0 , d 1 , l = 1, r 0 = 2 and k = 3) which shows that the cohomological Fourier-Mukai transform acts as follows: where H is an ample divisor class on Y . In particular, since F H E is a Hodge isometry and X [2] and hence a birational map F E : X [2] Y [2] . In this case, [28,Theorem 7.6] shows that F E is actually an isomorphism! See [28,Example 7.2] for details.
Similarly, when n = 3 we have X [3] M X (1, 0, −2) M X (1, −H , 4), where the second isomorphism is given twisting by O X (H ). Thus, we see that is a birational map X [3] Y [3] . For n = 4, X [4] M X (1, −H , 3), and it is enough, by [5,Corollary 1.3], to find an equivalence : H , 3). If we set D X := Hom( , ω X ) [2] to be the dualising functor and T O X to be the spherical twist around O X then For n = 5, we first note that X [5] M X (1, −H , 2) is birational to M X (2, H , 1) and then observe that the second moduli space has a Li-Gieseker-Uhlenbeck contraction. Indeed, the Hodge isometry T H O X [1] sends (1, −H , 2) to (2, H , 1) and the Mukai vector w = (0, 0, −1) is an isotropic class which pairs with (2, H , 1) to give 2. Now, if X [5] and Y [5] are birational then we have an induced map between their second integral cohomology groups. Moreover, since a birational map preserves the movable cone (cf. [17,Section 6]), this map can either send the exceptional divisor of the Hilbert-Chow (HC) contraction to itself or to the exceptional divisor of the Li-Gieseker-Uhlenbeck (LGU) contraction. In particular, if HC were mapped to LGU then we would contradict the fact that a birational map between Hilbert schemes necessarily sends primitive classes to primitive classes, whereas if HC were mapped to HC then the orthogonal complements must be Hodge-isometric as well, i.e. H 2 (X , Z) Hdg H 2 (Y , Z), and hence the underlying K3s would be isomorphic which they are not. Thus, by contradiction, X [5] and Y [5] cannot be birational.
The key thing about the previous argument is that the movable cone of X [5] has two different boundary walls. For a Picard rank one K3 surface X , the movable cone of the Hilbert scheme X [n] has two boundary walls. At least one of these boundaries is a Hilbert-Chow wall, and so we are essentially looking to see if the other wall is a different type: Brill-Noether (BN), Li-Gieseker-Uhlenbeck (LGU), or Lagrangian fibration (LF). If it is then a similar argument to case of n = 5 above shows that X [n] and Y [n] cannot be birational, where Y = M X (2, H , 3).
) for all n ≥ 1. Moreover, the Hilbert schemes X [n] and Y [n] are birational if and only if there is a solution to either of the Pell's equations: In particular, when n = 5, 6, 7, 8, 9, 11, . . ., these Hilbert schemes are not birational.
Proof As discussed above, the derived Torelli theorem of Mukai and Orlov [24] shows that D(X ) D(Y ) and hence Ploog's result [25,Proposition 8]   ) and X [n] is not birational to Y [n] .
Proof From the condition on the degree of the polarisation of X , it follows from [21] that X has four non-isomorphic Fourier-Mukai partners: X , Y , Z , W . Since X has Picard rank one, we see that for every n > 1, the movable cone of X [n] has exactly two extremal rays. Suppose, for a contradiction, that three of the four Hilbert schemes X [n] , Y [n] , Z [n] and W [n] are all birational. Then these three birational maps induce maps between the movable cones of these Hilbert schemes, sending extremal rays to extremal rays. In particular, it follows from the arguments given in Sect. 1.1 that two of these Hilbert schemes must have Hodge isometric second integral cohomology groups with an isometry preserving the class of the exceptional divisor. Hence, the second integral cohomology groups of the underlying K3s must be Hodge isometric as well, meaning that two of the Fourier-Mukai partners are isomorphic, which they are not. Thus, by contradiction, two of the Hilbert schemes cannot be birational for any n ≥ 1.

Criterion for birationality of Hilbert schemes
We give a criterion for when Hilbert schemes of points on certain K3 surfaces are birational. More specifically, given a K3 surface X and a FM-partner Y = M X (v), we provide a criterion which determines precisely when X [n] is birational to Y [n] . First of all, let us start by recalling some properties of moduli spaces on K3 surfaces of Picard rank one: let (X , H ) be a polarised K3 surface such that Pic(X ) = ZH .

Moreover, in this case M X ( p 2 s, pq H , q 2 t) M X (s, H , t). (iii) M X (s, H , t) M X (s , H , t ) if and only if {s, t} = {s , t }.
Proof We set p := gcd(r , c). Since v is primitive and isotropic, we have gcd( p, x) = 1 and c 2 H 2 /2 = r x, respectively. Thus, we see that p 2 | r . If we set r = p 2 s and c = pq then we must have q 2 H 2 /2 = sx and gcd( p 2 s, pq) = p. This implies that we have gcd(q, s) = 1, and hence x = q 2 t and H 2 /2 = st.
Recall  (21)], where k is any integer. Since this identification is independent of p, q, we get M X ( p 2 s, pq H , q 2 t) M X (s, H , t). The last claim is the content of [9] (see also [14,Prop. 5.6]).
The previous proposition, together with Verbitsky's global Torelli theorem [27], Markman's computation of the monodromy group [16], and Bayer and Macrì's results about the ample cone of moduli spaces [5], gives the following: , pq H , q 2 t). Moreover, {s, t} is uniquely determined by Y .

Theorem 2.2 Let X and Y be two derived equivalent K3 surfaces of Picard rank one. Then, X [n] is birationally equivalent to Y [n] if and only if p
Proof By the global Torelli theorem for irreducible symplectic manifolds, X [n] and Y [n] are birational if and only if they are Hodge isometric through a monodromy operator. By Markman's computation of the monodromy groups (see [5,Corollary 1.3]), this means that such an isometry extends to the Mukai lattice associated to the two K3s X and Y . Thus, X [n] is birationally equivalent to Y [n] if and only if there is a primitive isotropic Mukai vector w = ±( p 2 s, pq H , q 2 t) ∈ H * (X , Z) such that (1, 0, 1 − n), w = 1 and Y M X (w). The first condition is equivalent to p 2 s(n − 1) − q 2 t = ±1 and, by Proposition 2.1, the pair {s, t} is determined by Y . Notice that the Mukai vector w will correspond to the Hilbert-Chow (birational) contraction on X [n] which has Y (n) as the base variety. ( p 2 s, pq H , q 2 t) and H , s) by Proposition 2.1(iii), we actually have two Pell's equations governing the birationality. That is, if we want to check whether two Hilbert schemes are birational then we need to find a solution to either:

Remark 2.3 Notice that because the pair {s, t} is determined by M X
Let us look back at the case analysed in Proposition 1.2.

Example 2.5
If H 2 = 130 = 2 · 5 · 13 then we have two Fourier-Mukai partners and the non-trivial partner is given by Y = M X (5, H , 13). By Theorem 2.2, we see that X [n+1] is birational to Y [n+1] if and only if there is a solution to either: Reducing these equations modulo 5 and 13, respectively, we see that there are no solutions to 13q 2 = ±1 mod 5 or 5q 2 = ±1 mod 13. This shows that there are no solutions to the original equations for any n. In other words, whilst these Hilbert schemes are always derived equivalent, they are, in fact, never birational!

Counting birational equivalence classes
An interesting question concerns the number of non-birational derived equivalent Hilbert schemes that we can produce starting from the set of Fourier-Mukai partners of X . As we analysed in the previous sections, the two numbers are strictly linked: for any X as above, the Hilbert scheme X [n] has precisely two boundaries of the movable cone and they only depend on the algebraic part of its Hodge structure, hence the Hilbert scheme Y [n] on any Fourier-Mukai partner Y of X has the same geometry of rays making up the movable cone, see [5,Prop. 13.1]. If X [n] and Y [n] are birational for two different Mukai partners X and Y , then one ray from each cone has to correspond to the Hilbert-Chow contractions. Therefore, if N is the number of Fourier-Mukai partners of X , the number B of birational equivalence classes of Hilbert schemes of points on these partners is either N or N /2. Indeed, the former occurs when X [n] is not birational to any other Y [n] , and the latter occurs when there is a single Y [n] birational to X [n] which represents the second Hilbert-Chow wall. Note that as soon as X [n] is birational to Y [n] for a single Fourier-Mukai partner Y of X then the same happens for all Fourier-Mukai partners of X .
When N = B, we have one of the following: • There is a Hilbert-Chow wall and a different divisorial contraction on X [n] .
• X [n] has a Lagrangian fibration.
• There are two Hilbert-Chow walls in the movable cone of X [n] which are exchanged by a birational map.
To state the result properly, we need to introduce a few more notations and results contained in [29]. In loc. cit., the results are stated for Abelian surfaces but they still hold for K3 surfaces mutatis mutandis. We assume that √ (n − 1)d / ∈ Z, where d = H 2 /2 is half the degree of the K3 surface as before and n > 2.

Definition 2.6
For (x, y) ∈ R 2 , set We also set The group we just defined has the following structure: .
is birationally equivalent to X j [n] if and only if X i = M X j (a 2 s, abH , b 2 t). In particular, we have using the same arguments as presented at the start of Sect. 2.1.

Example 2.9
If n = 2 then S d,n (Z/2Z) ⊕2 ⊕ Z and the torsion subgroup is For a generator P(a √ s, b √ t) of a cyclic subgroup, we have a similar claim to (2).
Summing all of this up, we have the following: Proposition 2.12 Let X be a K3 surface of degree 2d with Picard rank one and let n > 3 be an integer.  (c) If v 1 = ( p 2 s, pq H , q 2 t) with p 2 s(n − 1) − q 2 t = ±1, then P( p √ s, q √ t) is the generator of S d,n / ± 1. In this case, v 1 defines a Hilbert-Chow contraction, therefore B = N if {s, t} = {1, n} and B = N /2 otherwise.
(2) Assume that d(n − 1) is a perfect square. Then Mov(X [n] ) is defined by (0, 0, 1) and a primitive v 1 . In this case, v 1 defines a Lagrangian fibration and B = N .