Twisted Fourier-Mukai number of a K3 surface

We give a counting formula for the twisted Fourier-Mukai partners of a projective K3 surface. As an application, we describe all twisted Fourier-Mukai partners of a projective K3 surface of Picard number 1.


Introduction
Fourier-Mukai (FM) partners of a projective K3 surface S have been studied since Mukai's work [8]. The basic results due to Mukai and Orlov ([10]) are as follows : (1) If a 2-dimensional moduli space M of stable sheaves on S is non-empty and compact, then M is a K3 surface. When M is fine, a universal sheaf induces an equivalence D b (M ) ≃ D b (S).
(2) Every FM partner of S is isomorphic to a certain 2-dimensional fine moduli space of stable sheaves on S.
(3) A projective K3 surface S ′ is a FM partner of S if and only if their Mukai lattices are Hodge isometric.
Using Mukai-Orlov's theorem (especially (3)), Hosono-Lian-Oguiso-Yau ( [5]) derived a counting formula for the FM partners of S. On the other hand, it has been recognized that studying twisted FM partners as well as usual (i.e, untwisted) FM partners of S is important when analyzing the derived category D b (S) of S. Recall that a twisted FM partner of S is a twisted K3 surface (S ′ , α ′ ) such that there is an equivalence D b (S ′ , α ′ ) ≃ D b (S). Cȃldȃraru ([1], [2]), Huybrechts-Stellari ( [6], [7]), and Yoshioka ( [13]) generalized Mukai and Orlov's results to twisted situation. When S is untwisted, their results can be stated as follows : (1) When a 2-dimensional coarse moduli space M of stable sheaves on S is a K3 surface, a twisted universal sheaf induces an equivalence D b (M, α) ≃ D b (S), where α is the obstruction to the existence of a universal sheaf.
In fact, we shall derive more simple formulas for #FM d (S) for several classes of K3 surfaces : Jacobian K3 surfaces, in particular K3 surfaces with rk(N S(S)) ≥ 13, K3 surfaces with 2-elementary N S(S), and K3 surfaces with rk(N S(S)) = 1.
As a by-product of the proof of Theorem 1.1, we obtain an upper bound for the twisted FM number of a twisted K3 surface (Proposition 4.3). The sharpness of the estimate is related with a Cȃldȃraru's problem stated in [1] (Remark 3.7). Note that if a twisted K3 surface (S ′ , α ′ ) has an untwisted FM partner, e.g, if rk(N S(S ′ )) ≥ 12 ( [6]), then Theorem 1.1 gives a counting formula for the twisted FM number of (S ′ , α ′ ).
As an application of Theorem 1.1, we shall describe all twisted FM partners of a K3 surface of Picard number 1 as follows.  (5). Let M σ,k be the moduli space of Gieseker stable sheaves on S with Mukai vector v σ,k , endowed with the obstruction α σ,k ∈ Br(M σ,k ) to the existence of a universal sheaf. ( (2) Assume that d 2 = n and d ≥ 3. Choose a set {j} ⊂ (Z/dZ) × of representatives of (Z/dZ) × /{±id}. Then Theorem 1.2 is a twisted generalization of a result of [4]. This paper is organized as follows. In Sect.2.1, we prepare some lattice theory following [9]. In Sect.2.2, we recall from [1] and [6] several facts about twisted K3 surfaces. In Sect.3.1, we describe a quotient set of FM d (S) in terms of lattice theory. In Sect.3.2 and 3.3, we describe the fibers of the quotient map in terms of lattice theory. In Sect.4, we derive Theorem 1.1 and its corollaries. In Sect.5, we prove Theorem 1.2.
Notation. By an even lattice, we mean a free Z-module L equipped with a non-degenerate symmetric bilinear form (, ) : L × L → Z satisfying (x, x) ∈ 2Z for all x ∈ L. The group of isometries of L is denoted by O(L). For two lattices L and M , Emb(L, M ) is the set of the primitive embeddings of L into M . An element l ∈ L is said to be isotropic if (l, l) = 0. The hyperbolic plane U is the even indefinite unimodular lattice Ze + Zf, (e, e) = (f, f ) = 0, (e, f ) = 1.
By a K3 surface, we mean a projective K3 surface over C. The Neron-Severi (resp. transcendental) lattice of a K3 surface S is denoted by N S(S) (resp. T (S)). Set N S(S) : Acknowledgement. The author would like to thank Professor Shinobu Hosono for conversations that inspired this work. His thanks also go to Professors Keiji Oguiso and Ken-Ichi Yoshikawa for valuable comments. The idea of constructing moduli spaces in Section 5 is due to Professor Oguiso.

Even lattices
For an even lattice L, we can associate a finite Abelian group D L := L ∨ /L equipped with a quadratic form q L : , whose kernel is denoted by O(L) 0 . We often write just (D L , q) or D L (resp. r) instead of (D L , q L ) (resp. r L ). Set Proposition 2.1 ( [9]). Let M be a primitive sublattice of an even unimodular lattice L with the orthogonal complement M ⊥ . Then (1) There is an isometry λ : Two even lattices L and L ′ are said to be isogenus if L ⊗ Z p ≃ L ′ ⊗ Z p for every prime number p and sign(L) = sign(L ′ ). By [9], these are equivalent to the conditions that (D L , q) ≃ (D L ′ , q) and sign(L) = sign(L ′ ). The set of the isometry classes of the lattices isogenus to L is denoted by G(L). For a subgroup Γ ⊂ O(L), let Γ + be the subgroup of Γ consisting of orientation-preserving isometries in Γ. The group Γ + is of index at most 2 in Γ. We define the subsets G 1 (L), G 2 (L) ⊂ G(L) by

Twisted K3 surfaces
Let (S, α) be a twisted K3 surface, where α is an element of the Brauer group Br(S). Via the exponential sequence there is a canonical isomorphism Br(S) ≃ Hom(T (S), Q/Z). We identify a twisting α ∈ Br(S) with a surjective homomorphism α : T (S) → Z/ord(α)Z, whose kernel is denoted by T (S, α). So the identity of Br(S) is denoted by 0 and the inverse of α ∈ Br(S) is denoted by −α.
where O Hodge (T (S)) is the group of the Hodge isometries of T (S), and λ : We need the following corollary of Huybrechts-Stellari's theorem. [6], [13], [7]). Let S be a K3 surface and let (S ′ , α ′ ) be a twisted K3 surface. Then there exists an equivalence Proof. By the theorem of Canonaco-Stellari ( [3]), every equivalence between derived categories of twisted K3 surfaces is of Fourier-Mukai type. Thus only "if" part requires an explanation. Assume that there is a Hodge isometry , we obtain a Hodge isometry g : T (S) ≃ T (S ′ , B ′ ). By Proposition 2.1, N S(S) and N S(S ′ , B ′ ) are isogenus. Because N S(S) admits an embedding of the hyperbolic plane U , it follows from Proposition 2.2 that g extends to a Hodge isometry Φ : When α ′ = 0, Proposition 2.4 is Mukai-Orlov's theorem ( [8], [10]).
We remark that for a twisted K3 surface (S, α) with a B-field lift B ∈ H 2 (S, Q), the period of S is determined by the natural Hodge isometry In Section 5, we shall calculate twistings in terms of discriminant group.
More precisely, the divisibility of (0, 0, 1) ∈ N S(S, B) is equal to d, but we do not need this fact in the following.
3 Lattice-theoretic descriptions 3.1 Isotropic elements of the discriminant form Let (S, α) be a twisted K3 surface. We write T := T (S, α), which is an even lattice of sign(T ) = (2, 20 − ρ(S)) equipped with a period. For a twisted FM partner (S 1 , α 1 ) ∈ FM d (S, α) there exists a Hodge isometry g 1 : T (S 1 , α 1 ) ≃ T by Huybrechts-Stellari's theorem ( [6]). Then g 1 and α 1 induce an isomorphism Proof. The ambiguity of the choice of a Hodge isometry g 1 : We prove the injectivity of µ. For two partners Proof. It suffices to show the surjectivity. Let S ′ be an untwisted FM partner of (S, α). Since there exists a Hodge isometry T (S ′ ) ≃ T , we may assume from the first that (S, α) itself is untwisted. Take an isotropic element x ∈ I d (D T ) = I d (D T (S) ). Via the isometry λ : (D T (S) , q) ≃ (D g N S(S) , −q), we obtain an isotropic element λ(x) ∈ I d (D g N S(S) ). Set which are even overlattices of N S(S), T (S) respectively. Via the isometry we obtain an embedding M x ⊕ T x ֒→ Λ K3 with both M x and T x embedded primitively. Since N S(S) ⊂ M x , the lattice M x admits an embedding ϕ of the hyperbolic plane U . Then the lattice Λ ϕ := ϕ(U ) ⊥ ∩ Λ K3 is isometric to the K3 lattice Λ K3 and has the period induced from T x . Now the surjectivity of the period map ( [12], [14]) assures the existences of a K3 surface S ϕ and a Hodge isometry Φ : Pulling back the homomorphism by Φ| T (Sϕ) , we obtain a twisted K3 surface (S ϕ , α ϕ ).
Since there exists a Hodge isometry T (S ϕ , α ϕ ) ≃ Ker(α x ) = T (S), it follows from Proposition 2.4 that (S ϕ , α ϕ ) ∈ FM d (S). By the construction, we have In general, there is a condition on the image of µ. Let B ∈ H 2 (S, Q) be a Bfield lift of α ∈ Br(S). There is a natural isometry λ : One can verify that the set J d (D T ) is independent of the choice of a lift B.
We make the following identification tacitly in Section 4. For x ∈ I d (D T ) such that [x] = µ([(S 1 , α 1 )]), set T x := x, T . For a Hodge isometry g 1 : Here the Hodge isometryḡ 1 : T (S 1 ) ≃ T x is induced from g 1 , and the isometrȳ λ :
Proof. We may assume that (S, α) itself is untwisted. To prove the surjectivity, we shall construct a twisted K3 surface (S ϕ , α ϕ ) from an arbitrary embedding ϕ : U ֒→ N S(S 1 ). The lattice Λ ϕ := ϕ(U ) ⊥ ∩ H(S 1 , Z) is isometric to the K3 lattice Λ K3 and possesses the period induced from T (S 1 ). On the other hand, the lattice M ϕ := ϕ(U ) ⊥ ∩ N S(S 1 ) has the orientation induced from ϕ and the orientation of N S(S 1 ). That is, we can choose a connected component M + ϕ of the open set {v ∈ M ϕ ⊗ R, (v, v) > 0} so that for each vector v ∈ M + ϕ the oriented positive-definite two-plane Rϕ(e + f ) ⊕ Rv ⊂ N S(S 1 ) ⊗ R is of positive orientation. By the surjectivity of the period map, there exist a K3 surface S ϕ and a Hodge isometry Φ : Pulling back α 1 by Φ| T (Sϕ) , we obtain a twisted K3 surface (S ϕ , α ϕ ).
Remark 3.7. In his thesis [1], Cȃldȃraru proposed the following problem: ). Let (S 1 , α 1 ) be a twisted K3 surface. For each untwisted FM partner S 2 ∈ FM(S 1 ) and each Hodge isometry g : From the construction of the twisted K3 surface (S ϕ , α ϕ ) in the proof of Proposition 3.6, it is immediately verified that the map ν in Lemma 3.5 is bijective if and only if the answer to Cȃldȃraru's question is positive for the twisted K3 surface (S 1 , α 1 ) and each S 2 , g.
(ii) ⇒ (iii) : For an anti-symplectic automorphism f , Then we can extend the isometry −γ ⊕ −id T (Sϕ) to the antisymplectic Hodge isometry Φ : H 2 (S ϕ , Z) ≃ H 2 (S ϕ , Z) which preserves the positive cone. Composing Φ with an element of the Weyl group W (S ϕ ), we may assume that Φ is effective. Let (S, α) be a twisted K3 surface with a B-field lift B ∈ H 2 (S, Q). For an isotropic element x ∈ I d (D T (S,α) ) we define the lattice T x by T x = x, T (S, α) and define the homomorphism α x : T x ։ Z/dZ by

The counting formula
From Section 3 and Proposition 4.1, we deduce the formula for #FM d (S).
Theorem 4.2. For a K3 surface S the following formula holds.
Here x runs over the set O Hodge (T (S))\I d (D T (S) ), and the lattices M , M ′ run over the sets G 1 ( λ(x), N S(S) ), G 2 ( λ(x), N S(S) ) respectively.
We also obtain the following inequality.  By [9], N S(S, B) admits an embedding of U ⊕ U if rk(N S(S)) ≥ 13.
Here ϕ is the Euler function and τ (m) is the number of the prime factors of m.
Proof. We have D N S(S) = H 2n ≃ Z/2nZ and O Hodge (T (S)) = {±id} (cf. [5]). Since it follows that #FM d (S) = 0 unless d 2 |n. Assume that d 2 |n. Then we have For d ≥ 3 and k ∈ (Z/dZ) × , we have ZH, k H d = Z H d and it follows from Theorem 4.2 that We have N S(M σ,k ) = ZH σ,k andkH σ,k = [(r σ , 0, −s σk 2 )]. Since the Hodge isometry (6) is orientation-preserving, H σ,k is the ample generator of N S(M σ,k ). For the transcendental lattice we have Here L means the primitive hull of the lattice L in H(S, Z).
(2) Assume d 2 = n and d ≥ 3. Choose a set {j} ⊂ (Z/dZ) × of representatives of (Z/dZ) × /{±id}. then Proof. That (M σ,k , α σ,k ) ∈ FM d (S) is a direct consequence of Cȃldȃraru's theorem ( [2]). By Proposition 5.1, it suffices to show that the twisted K3 surfaces in the right hand sides are not isomorphic to each other.