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derived categories of twisted sheaves on calabi-yau manifolds

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All we need to do now is show that we can find an F as above for which (v(F ), v) =<br />

1 mod n. Indeed, we can then take u = v(F ) (using the fact the we have the<br />

freedom <str<strong>on</strong>g>of</str<strong>on</strong>g> choosing u as l<strong>on</strong>g as (u, v) = 1 mod n) and get<br />

U ∨<br />

ϕX1→M1 (u)2<br />

U ∨<br />

= ϕX1→M1 (v(F ))2<br />

= (v(F ), v)<br />

rk(V1) c1(V1) + integral part<br />

= (u, v)<br />

rk(V1) c1(V1) + integral part,<br />

which is what we want.<br />

Since we assume that n = 1 at t = 1 we can find a line bundle F <strong>on</strong> X1 with<br />

c1(F ) = d such that gcd(v0, v2.d, v4) = 1 (where v = (v0, v2, v4)). Since n divides<br />

v0 and v4, we must have gcd(v2.d, n) = 1. By possibly replacing F by some tensor<br />

power <str<strong>on</strong>g>of</str<strong>on</strong>g> itself, we can assume (v2.d) = 1 mod n. Then we have<br />

v(F ).v = v2.d − v4 − v0 − v0(d.d) = 1 mod n,<br />

as desired. Twisting F by OX1(−k) for k large enough, we get the F we were<br />

looking for, thus finishing the pro<str<strong>on</strong>g>of</str<strong>on</strong>g>. (Note that twisting by O(−k) does not change<br />

(v(F ), v) mod n, because (c1(OX1(1)).v2) must be divisible by n, since it equals<br />

(c1(OX0(1)).v2).)<br />

5.4 The Map <strong>on</strong> Brauer Groups<br />

We have seen in the previous secti<strong>on</strong> that we can identify the obstructi<strong>on</strong> α solely<br />

in terms <str<strong>on</strong>g>of</str<strong>on</strong>g> the Fourier-Mukai transform. In this secti<strong>on</strong> we prove that there is in<br />

fact a map between the Brauer groups <str<strong>on</strong>g>of</str<strong>on</strong>g> X and <str<strong>on</strong>g>of</str<strong>on</strong>g> M, whose kernel is generated<br />

by α.<br />

Lemma 5.4.1. Let M be a K3 surface, and let TM be the transcendental lattice<br />

<str<strong>on</strong>g>of</str<strong>on</strong>g> M, i.e. the orthog<strong>on</strong>al lattice to NS(M) inside H 2 (M, Z) (endowed with the<br />

intersecti<strong>on</strong> pairing). Then there is a natural isomorphism<br />

where T ∨<br />

M<br />

is the dual lattice to TM.<br />

Br(M) = T ∨<br />

M ⊗Z Q/Z,<br />

Pro<str<strong>on</strong>g>of</str<strong>on</strong>g>. C<strong>on</strong>sider the exact sequence <str<strong>on</strong>g>of</str<strong>on</strong>g> Propositi<strong>on</strong> 1.1.3,<br />

0 → NS(M) ⊗ Q/Z → H 2 (M, Q/Z) → Br(M) → 0,<br />

(note that for a K3 M we have NS(M) = Pic(M)); using the universal coefficient<br />

theorem, we also know that<br />

H 2 (M, Q/Z) = H 2 (M, Z) ⊗ Q/Z,<br />

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