derived categories of twisted sheaves on calabi-yau manifolds

a K3 surface, and choosing a quasi-universal sheaf Ũ we obtain a Fourier-Mukai
transform
Ũ ∨
ϕX→M : ˜ H(X, Q) → ˜ H(M, Q)
(which depends on Ũ ).
There is a natural element α ∈ Br(M) which expresses the obstruction to the
existence <strong>of</strong> a universal sheaf on X × M. The following theorem gives a precise
identification <strong>of</strong> α in terms <strong>of</strong> the initial data X and v.
Theorem 5.3.1. Let u ∈ ˜ H(X, Z) be any vector such that (u, v) = 1 mod n,
U ∨
(where n is the one defined in Theorem 5.1.12) and let w = ϕX→M (u). Denote by
w2 the H2 (M, Q) component <strong>of</strong> w, and let ¯w be the element <strong>of</strong> Br(M) that is the
image <strong>of</strong> w2 under the composite homomorphism
H 2 (M, Q) → H 2 (M, Q/Z) → H 2 (M, Q/Z)/(NS(X) ⊗ Q/Z) = Br(M).
Then ¯w is the obstruction to the existence <strong>of</strong> a universal sheaf on X × M, as
described in Definition 5.1.8.
Remark 5.3.2. Since ˜ H(X, Z) is a unimodular lattice, one can always find a vector
u satisfying (u, v) = 1 mod n, so that one can always compute the obstruction
α = ¯w using this theorem.
Pro<strong>of</strong>. Let’s start <strong>of</strong>f by analyzing the Fourier-Mukai transform when the moduli
problem is in fact fine, but we are using a quasi-universal sheaf instead <strong>of</strong> the
universal sheaf.
So assume that we are in the above setup, with n = 1, and therefore there
exists a universal sheaf E on X × M. If OX(1) is an ample line bundle on X, and
k is large enough, then
E ∨
V = ΦX→M(OX(−k))[2] is a locally free sheaf on M. Indeed, this is a relative version <strong>of</strong> the fact that on a
smooth, projective scheme <strong>of</strong> dimension r, we have
Ext r−i (F (k), ω) = H i (F (k)) = 0
for any coherent sheaf F , i > 0, and large enough k. So let k be large enough to
have V locally free. We are interested in the quasi-universal sheaf U = E ⊗ π∗ MV ,
<strong>of</strong> similitude rk(V ).
It is important to note that we have
E ∨
v(V ) = v(ΦX→M(OX(−k))[2]) E ∨
= v(ΦX→M(OX(−k))) E ∨
= ϕX→M(v(OX(−k))), 74