2005
DOI: 10.1007/s00208-005-0662-2
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Equivalences of twisted K3 surfaces

Abstract: 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 … Show more

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Cited by 95 publications
(162 citation statements)
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“…By [HS05,Rem. 7.10], given any K3 surface S and any nontrivial β ∈ Br(S), there is no equivalence between D b (S, β) and D b (S).…”
Section: The Twisted Derived Equivalencementioning
confidence: 97%
“…By [HS05,Rem. 7.10], given any K3 surface S and any nontrivial β ∈ Br(S), there is no equivalence between D b (S, β) and D b (S).…”
Section: The Twisted Derived Equivalencementioning
confidence: 97%
“…satisfies Φ H * E 0 = (−id H 2 ) ⊕ id H 0 ⊕H 4 . As all orientation preserving Hodge isometries do lift to autoequivalences (see [15,21,31] or [17,Ch. 10]), this seemingly weaker form is equivalent to the original Theorem 2.…”
Section: Deformation Of Derived Equivalences Of K3 Surfacesmentioning
confidence: 99%
“…In the twisted case, T (ϕ) ⊥ does not necessarily contain such an hyperbolic plane. In fact, in [13] we give an explicit example of an Hodge isometry T (ϕ) ∼ = T (ϕ ′ ) which cannot be extended simply due to the fact that the Picard groups are not isometric. We furthermore put forward a version of Cȃldȃraru's conjecture that takes into account not only the transcendental lattice T (X, α B ) ∼ = T (ϕ), but the full Hodge structure defined by the generalized Calabi-Yau structure ϕ = σ + B ∧ σ.…”
Section: Generalized Calabi-yau Structures and Derived Categoriesmentioning
confidence: 99%