Derived equivalent Hilbert schemes of points on K3 surfaces which are not birational
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Abstract
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 nonbirational Hilbert schemes which are derived equivalent.
1 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 algebrogeometric 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 nonbirational 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].
1.1 Various notions of equivalence
Throughout this article, we will use \(\mathrm {D}(X)\) to denote the bounded derived category of coherent sheaves on a smooth complex projective variety X. Moreover, we will say that two smooth complex projective varieties X and Y are Dequivalent if we have an equivalence \(\mathrm {D}(X)\simeq \mathrm {D}(Y)\).
Recall that two varieties X and Y are said to be Kequivalent if there exists a birational correspondence \(X\xleftarrow {\pi _X} Z\xrightarrow {\pi _Y} Y\) with \(\pi _X^*\omega _X\simeq \pi _Y^*\omega _Y\). The Bondal–Orlov conjecture says \(\mathrm {K}\implies \mathrm {D}\). Notice that if the canonical bundles of X and Y are trivial then Kequivalence is the same as birationality.
We will say that two varieties X and Y are Hequivalent if there exists a Hodge isometry \(\mathrm {H}^2(X,{\mathbb Z})\simeq _\mathrm {Hdg}\mathrm {H}^2(Y,{\mathbb Z})\), that is, an isomorphism respecting the Hodge structure and the intersection pairing. For a general hyperkähler, Huybrechts [10, Corollary 4.7] shows that \(\mathrm {K}\implies \mathrm {H}\). However, Namikawa [20] showed that there are Abelian surfaces A for which the associated generalised Kummer fourfolds \(K_2(A)\) and \(K_2({\widehat{A}})\) are Hodgeequivalent but not birational; this was the first major counterexample to the birational Torelli problem for hyperkählers. Also, Verbitsky’s Torelli theorem [27, Theorem 7.19] proves that \(\mathrm {H}\implies \mathrm {K}\) when the hyperkähler is of \(\text {K3}^{[n]}\)type and \(n=p^k+1\) for some prime p and positive integer k. However, if \(n1\) is not a prime power then Markman [16, Lemma 4.11] provides examples of Hilbert schemes which are Hodgeequivalent and yet not birational. Taken together, these results illustrate the relationship between Hequivalence and Kequivalence of hyperkählers is quite delicate. That is, while Huybrechts says that \(\mathrm {K}\implies \mathrm {H}\), the converse only holds when we impose certain extra conditions.
 (i)
as a lattice, we have \({\widetilde{\Lambda }}\simeq U^4\oplus E_8(1)^2\),
 (ii)
the orthogonal complement of \(\mathrm {H}^2(X,{\mathbb Z})\) in \({\widetilde{\Lambda }}\) has rank one and is generated by a primitive vector of square \(2n2\),
 (iii)
if X is a moduli space \(M_S(\varvec{\mathrm {v}})\) of sheaves on a K3 surface S with Mukai vector \(\varvec{\mathrm {v}}\in \mathrm {H}^*(S,{\mathbb Z})\) then there is an isomorphism \({\widetilde{\Lambda }}\simeq \mathrm {H}^*(S,{\mathbb Z})\;;\;\mathrm {H}^2(X,{\mathbb Z})\mapsto \varvec{\mathrm {v}}^\perp \).
2 Examples which are Dequivalent but not Kequivalent
2.1 Degree twelve
We work through a specific example in order to demonstrate how certain Hilbert schemes can be derived equivalent and not birational.
Question 1.1
For which positive integers n, are \(X^{[n]}\) and \(Y^{[n]}\) birational?
Recall that Oguiso [21] has shown that the number of Fourier–Mukai partners of a K3 surface X with \(\mathrm {Pic}(X)={\mathbb Z}[H]\) and \(H^2=2d\) is given by \(2^{\rho (d)1}\), where \(\rho (d)\) is the number of prime factors of d. Thus, to ensure that \(X^{[n]}\) and \(Y^{[n]}\) are not all isomorphic, we must have \(H^2\ge 12\). For simplicity, we choose \(H^2=12\).
Let X be a complex projective K3 surface with \(\mathrm {Pic}(X)={\mathbb Z}[H]\) and \(H^2=12\). Since \(\varvec{\mathrm {w}}=(2,H,3)\) is an isotropic vector, we have another K3 surface \(Y=M_H(\varvec{\mathrm {w}})\). Moreover, since \(\varvec{\mathrm {v}}=(1,H,4)\) is a vector such that \((\varvec{\mathrm {v}},\varvec{\mathrm {w}})=1\) (or \(\gcd (2,3)=1\)), we have a universal family \({\mathcal E}\) and a derived equivalence as above. Now, the proof of [26, Theorem 2.4] shows that \(\mathrm {H}^2(X,{\mathbb Z})\not \simeq _\mathrm {Hdg}\mathrm {H}^2(Y,{\mathbb Z})\) 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!
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 \(Y=M_X(2,H,3)\).
Proposition 1.2
Proof
As discussed above, the derived Torelli theorem of Mukai and Orlov [24] shows that \(\mathrm {D}(X)\simeq \mathrm {D}(Y)\) and hence Ploog’s result [25, Proposition 8] ensures that \(\mathrm {D}(X^{[n]})\simeq \mathrm {D}(Y^{[n]})\) for all \(n\ge 1\). The conditions on when the Hilbert schemes are birational can be found in Theorem 2.2 below. \(\square \)
Remark 1.3
n  1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17 

Birational?  ✗  ✓  ✓  ✓  ✗  ✗  ✗  ✗  ✗  ✓  ✗  ✓  ✗  ✓  ✓  ✗  ✗ 
Remark 1.4
Notice that the Hilbert schemes \(X^{[n]}\) and \(Y^{[n]}\) in Proposition 1.2 are all Mequivalent. Indeed, the Markman–Mukai lattices \({\widetilde{\Lambda }}\) of \(X^{[n]}\) and \(Y^{[n]}\) are given by \(\mathrm {H}^*(X,{\mathbb Z})\) and \(\mathrm {H}^*(Y,{\mathbb Z})\), respectively; these are Hodge isometric by [24]. When \(n=p^k+1\) for some positive integer k, we can use Verbitsky’s Torelli theorem to conclude that the nonbirational examples are also not Hequivalent. For example, when \(n=5,6,8,9,17,...\) the examples in Remark 1.3 are not Hequivalent, but for \(n=7,11,13,16,19,\dots \) Torelli does not hold and so we cannot say whether they are Hequivalent or not. This information should be taken into consideration along with the diagram in (1). Finally, let us point out that all of these pairs of Hilbert schemes are also deformation equivalent to each other, and thus complement the recent articles mentioned in the introduction.
2.2 K3s with many FM partners
A second way to produce examples is by considering K3 surfaces with many Fourier–Mukai partners. However, this produces less “constructible” examples, as the following shows.
Proposition 1.5
Let (X, H) be a K3 surface of degree 2pqr and Picard rank one, where p, q, r are relatively prime integers greater than 1. Then, for all n, there exists a K3 surface Y such that \(\mathrm {D}(X^{[n]})\simeq \mathrm {D}(Y^{[n]})\) 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 nonisomorphic 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 \(n\ge 1\). \(\square \)
3 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 FMpartner \(Y=M_X(\varvec{\mathrm {v}})\), 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 \(\mathrm {Pic}(X)={\mathbb Z}H\).
Proposition 2.1
 (i)
There exist integers p, s, q, t such that \((r,cH,x)=(p^2 s,pq H,q^2 t)\), where \(H^2=2st\) and \(\gcd (p,q)=1\).
 (ii)
If \(M_X(r,cH,x)\) is fine then \((r,cH,x)=(p^2 s,pq H,q^2 t)\) with \(\gcd (ps,qt)=1\). Moreover, in this case \(M_X(p^2 s,pqH,q^2 t) \simeq M_X(s,H,t)\).
 (iii)
\(M_X(s,H,t) \simeq M_X(s',H,t')\) 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 \(c^2 H^2/2=rx\), respectively. Thus, we see that \(p^2 \mid r\). If we set \(r=p^2 s\) and \(c=pq\) then we must have \(q^2H^2/2=sx\) 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 from [7, Corollary 4.6.7] that \(M_X(r,cH,x)\) is a fine moduli space if and only if \(\gcd (r,cH^2,x)=1\). Hence \(\gcd (r,x)=1=\gcd (ps,qt)\).
By [14, §5.2 and Prop. 5.6], the Mukai vector \((p^2 s,pq H,q^2 t)=p^2 s \exp (\tfrac{q}{ps}H)\) corresponds to (k, s) in [14, Equation (21)], where k is any integer. Since this identification is independent of p, q, we get \(M_X(p^2 s,pqH,q^2 t) \simeq M_X(s,H,t)\). The last claim is the content of [9] (see also [14, Prop. 5.6]). \(\square \)
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:
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 \(p^2 s(n1)q^2 t=\pm 1\) and \(Y=M_X(p^2 s,pq H,q^2 t)\). Moreover, \(\{s,t\}\) is uniquely determined by Y.
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 \(\varvec{\mathrm {w}}=\pm (p^2 s,pq H,q^2 t)\in \mathrm {H}^*(X,{\mathbb Z})\) such that \(\langle (1,0,1n),\varvec{\mathrm {w}}\rangle =1\) and \(Y \simeq M_X(\varvec{\mathrm {w}})\). The first condition is equivalent to \(p^2 s(n1)q^2 t=\pm 1\) and, by Proposition 2.1, the pair \(\{s,t\}\) is determined by Y. Notice that the Mukai vector \(\varvec{\mathrm {w}}\) will correspond to the Hilbert–Chow (birational) contraction on \(X^{[n]}\) which has \(Y^{(n)}\) as the base variety. \(\square \)
Remark 2.3
Let us look back at the case analysed in Proposition 1.2.
Example 2.4
If \(H^2=12\) then the only Fourier–Mukai partner of X, other than itself, is given by \(Y:=M_X(2,H,3)\). For \(n=5\), it is easy to see that there is no solution to \(8p^23q^2=\pm 1\) or \(12p^22q^2=\pm 1\). Hence, \(X^{[5]}\) and \(Y^{[5]}\) are not birationally equivalent. For \(n=10\), we have \(\langle (1,0,9),(27,33H,242) \rangle =1\), and so \(X^{[10]}\) and \(Y^{[10]}\) are birational. Similarly, for \(n=12\), we can observe that \(\langle (1,0,11),(3,4H,32) \rangle =1\), hence \(X^{[12]}\) and \(Y^{[12]}\) are birational.
Example 2.5
3.1 Counting birational equivalence classes
An interesting question concerns the number of nonbirational 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 HilbertChow 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.

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.
Definition 2.6
The group we just defined has the following structure:
Lemma 2.7
If \(n>2\), then \(S_{d,n}/\pm 1\) is an infinite cyclic group.
Proof
See [29, Corollary 6.6].\(\square \)
The key use of this group is that its action allows us to determine different presentations of the Mukai vector \((1,0,1n)\) corresponding to the Hilbert scheme of points on X as a sum of two isotropic vectors (which will correspond to the Mukai vectors (1, 0, 0) and \((0,0,1n)\) on a Fourier–Mukai partner of X), as proven in [29, Lemma 6.8]. The number B then depends on a generator of \(S_{d,n}/\pm 1\):
Proposition 2.8
Proof
Example 2.9
Example 2.10
If there are integers p, q satisfying \(dp^2(n1)q^2=\pm 1\), then \(B=N\). In particular, if \(n1=dp^2 \pm 1\), then \(B=N\).
Remark 2.11
Assume that \(\sqrt{d(n1)} \in {\mathbb Z}\). In this case, \(p^2s(n1)q^2 t=\pm 1\) implies \(\gcd (s (n1),t)=1\). Hence \(\sqrt{s (n1)},\sqrt{t} \in {\mathbb Z}\). Then \(p=0\) and \(q^2=t=1\), or \(q=0\) and \(p^2=s=n1=1\). Hence \(M_X(p^2 s,pqH,q^2 t)=X\) in Theorem 2.2.
Summing all of this up, we have the following:
Proposition 2.12
 (1)If \(\sqrt{d(n1)} \not \in {\mathbb Z}\), \(\mathrm {Mov}(X^{[n]})\) is defined by (0, 0, 1) and a primitive \(v_1\), where \(\varvec{\mathrm {v}}_1\) satisfies one of the following.
 (a)
If \(\varvec{\mathrm {v}}_1=(p^2 s,pqH,q^2 t)\) with \(p^2 s (n1)q^2 t=\pm 2\), the primitivity of \(\varvec{\mathrm {v}}_1\) implies \(\gcd (ps,q)=\gcd (p,t)=1\) and \(P(pq \sqrt{d},p^2 s(n1) \mp 1)\) is the generator of \(S_{d,n}/\pm 1\). In this case, \(\varvec{\mathrm {v}}_1\) defines a Li–Gieseker–Uhlenbeck contraction, therefore \(B=N\).
 (b)
If \(\varvec{\mathrm {v}}_1=(r,cH,r(n1))\) with \(c^2 dr^2 (n1)=1\), then the generator of \(S_{d,n}/\pm 1\) is \(P(r,c\sqrt{d})\). In this case, \(\varvec{\mathrm {v}}_1\) defines a Brill–Noether contraction, therefore \(B=N\).
 (c)
If \(\varvec{\mathrm {v}}_1=(p^2 s,pqH,q^2 t)\) with \(p^2 s (n1)q^2 t=\pm 1\), then \(P(p \sqrt{s},q \sqrt{t})\) is the generator of \(S_{d,n}/\pm 1\). In this case, \(\varvec{\mathrm {v}}_1\) defines a Hilbert–Chow contraction, therefore \(B=N\) if \(\{s,t\}=\{1,n\}\) and \(B=N/2\) otherwise.
 (a)
 (2)
Assume that \(d(n1)\) is a perfect square. Then \(\mathrm {Mov}(X^{[n]})\) is defined by (0, 0, 1) and a primitive \(\varvec{\mathrm {v}}_1\). In this case, \(\varvec{\mathrm {v}}_1\) defines a Lagrangian fibration and \(B=N\).
Remark 2.13
Remark 2.14
In Proposition 2.12, the assumption on n is needed as, if \(n=3\), then there may exist a Hilbert–Chow contraction with \(\langle \varvec{\mathrm {v}}_1,(1,0,2) \rangle =\pm 2\).
Footnotes
 1.
The condition that \((\varvec{\mathrm {v}},\varvec{\mathrm {w}})=1\) is equivalent to Mukai’s criterion that \(\varvec{\mathrm {w}}=(r,H,s)\) for some integers r and s satisfying \(\gcd (r,s)=1\) and \(H^2=2rs\).
Notes
Acknowledgements
A very similar result was independently discovered by Shinnosuke Okawa [22]. His result only considers the case when Brill–Noether contractions exist. The first author thanks David Ploog for a very helpful and enjoyable discussion. Special thanks also go to Evgeny Shinder and Joe Karmazyn for pointing out the interesting Example 2.5.
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