Some Remarks about the FM-partners of K3 Surfaces with Picard Numbers 1 and 2

Stellari, Paolo

doi:10.1007/s10711-004-9291-7

Cited by 22 publications

(31 citation statements)

References 12 publications

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“…But the number itself can be arbitrarily large. More precisely, it has been shown in [28,32] that for any N there exist pairwise non-isomorphic K3 surfaces X 1 , . .…”

Section: Counting Twisted Fourier-mukai Partnersmentioning

confidence: 99%

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Equivalences of twisted K3 surfaces

2005

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Abstract. We prove that two derived equivalent twisted K3 surfaces have isomorphic periods. The converse is shown for K3 surfaces with large Picard number. It is also shown that all possible twisted derived equivalences between arbitrary twisted K3 surfaces form a subgroup of the group of all orthogonal transformations of the cohomology of a K3 surface.The passage from twisted derived equivalences to an action on the cohomology is made possible by twisted Chern characters that will be introduced for arbitrary smooth projective varieties.By definition a K3 surface is a compact complex surface X with trivial canonical bundle and vanishing H 1 (X, O X ). As was shown by Kodaira in [23] all K3 surfaces are deformation equivalent. In particular, any K3 surface is diffeomorphic to the four-dimensional manifold M underlying the Fermat quartic in P 3 defined by x 4 0 + x 4 1 + x 4 2 + x 4 3 = 0. Thus, we may think of a K3 surface X as a complex structure I on M . (As it turns out, every complex structure on M does indeed define a K3 surface, see [14].)In the following, we shall fix the orientation on M that is induced by a complex structure and denote by Λ the cohomology H 2 (M, Z) endowed with the intersection pairing. This is an even unimodular lattice of signature (3,19) and hence isomorphic to (−E 8 ) ⊕2 ⊕ U ⊕3 with E 8 the unique even positive definite unimodular lattice of rank eight and U the hyperbolic plane.We shall denote by Λ the lattice given by the full integral cohomology H * (M, Z) (which is concentrated in even degree) endowed with the Mukai pairing ϕ 0 + ϕ 2 + ϕ 4 , ψ 0 + ψ 2 + ψ 4 = ϕ 2 ∧ ψ 2 − ϕ 0 ∧ ψ 4 − ϕ 4 ∧ ψ 0 . In other words, Λ is the direct sum of (H 0 ⊕ H 4 )(M, Z) endowed with the negative intersection pairing and Λ. Hence, Λ ∼ = Λ ⊕ U .An isomorphism between two K3 surfaces X and X ′ given by two complex structures I respectively I ′ on M is a diffeomorphism f ∈ Diff(M ) such that I = f * (I ′ ). Any such diffeomorphism f acts on the cohomology of M and, therefore, induces a lattice automorphism f * : Λ ∼ = Λ.Conversely, one might wonder whether any element g ∈ O(Λ) is of this form. This is essentially true and has been proved by Borcea in [4]. The precise statement is:For any g ∈ O(Λ) there exist two K3 surfaces X = (M, I) and X ′ = (M, I ′ ) and an isomorphism f : X ∼ = X ′ with f * = ±g.(In fact, we can even prescribe the K3 surface X = (M, I), but stated like this the result compares nicely with Theorem 0.1.) The proof of this fact In a next step, we consider a more flexible notion of isomorphisms of K3 surfaces: One says that two K3 surfaces X and X ′ are derived equivalent if there exists a Fourier-Mukai equivalence Φ :is the bounded derived category of the abelian category Coh(X) of coherent sheaves on X. (Usually, derived equivalence is only considered for algebraic K3 surfaces.)Clearly, any isomorphism between X and X ′ given by f ∈ Diff(M ) induces a Fourier-Mukai equivalence Φ := Rf * . By results of Mukai and Orlov one knows how to associate to any Fourier-Mukai equivalence Φ :o...

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“…But the number itself can be arbitrarily large. More precisely, it has been shown in [28,32] that for any N there exist pairwise non-isomorphic K3 surfaces X 1 , . .…”

Section: Counting Twisted Fourier-mukai Partnersmentioning

confidence: 99%

“…Moreover, following [18], this also holds true if ρ ≥ 3 and the determinant of Pic(X) is square free. For the calculation of the number of Fourier-Mukai partners in the untwisted case see [18,32].…”

Section: Counting Twisted Fourier-mukai Partnersmentioning

confidence: 99%

Equivalences of twisted K3 surfaces

2005

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“…Since by degree reasons the class of a polarization is linearly independent to the class of the fiber of an elliptic fibration, the minimal Picard rank of an elliptic K3 surface is equal to 2. Good references about such K3 surfaces are papers of Stellari [31] and van Geemen [32], and [16, Chapter 11]. Lemma Let

X

be an elliptic K3 surface of index

t > 0

and of Picard rank two.…”

Section: Elliptic Surfaces Of Index Fivementioning

confidence: 99%

“…Such elliptic K3 surfaces may have more than one elliptic fibrations (in fact a Picard rank two elliptic K3 has always one or two elliptic fibrations), and by an isomorphism of elliptic K3 surfaces we mean an isomorphism of K3 surfaces, regardless of the elliptic fibration structure. The above Proposition is proved by analyzing lattice theory of the corresponding K3 surfaces, along the lines of [31, 32].…”

Section: Introductionmentioning

confidence: 99%

L‐equivalence for degree five elliptic curves, elliptic fibrations and K3 surfaces

Shinder

Zhang

2020

Bull. London Math. Soc.

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We construct non-trivial L-equivalence between curves of genus one and degree five, and between elliptic surfaces of multisection index five. These results give the first examples of L-equivalence for curves (necessarily over non-algebraically closed fields) and provide a new bit of evidence for the conjectural relationship between L-equivalence and derived equivalence.The proof of the L-equivalence for curves is based on Kuznetsov's Homological Projective Duality for Gr(2, 5), and L-equivalence is extended from genus one curves to elliptic surfaces using the Ogg-Shafarevich theory of twisting for elliptic surfaces.Finally, we apply our results to K3 surfaces and investigate when the two elliptic L-equivalent K3 surfaces we construct are isomorphic, using Neron-Severi lattices, moduli spaces of sheaves and derived equivalence. The most interesting case is that of elliptic K3 surfaces of polarization degree ten and multisection index five, where the resulting L-equivalence is new.

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“…Indeed, [18,Thm. 2.4] shows that there may be many non-isomorphic K3 surfaces dual (in the sense of Mukai) to a K3 surface.…”

Section: Mukai's Duality and The Action Of G 2dmentioning

confidence: 99%