4
$\begingroup$

Reading the proof of lemma 4.4.4 in Huybrechts and Lehn Geometry of Moduli Spaces of sheaves I come across an isomorphism relating a fibre of an invertible sheaf and its cohomology, and I really don't see where it comes from. Here is the setup, I hope it is clear.

Let $Q := Quot(\mathcal{H},P)$ be a Quot scheme, where $\mathcal{H}$ is a coherent sheaf on a Noetherian scheme $X$ over an alg. closed field $k$. Let $\mathcal{H} \rightarrow \mathcal{U}$ be the universal family of sheaves parametrized by $Q$. Let $\mathcal{L}_m := \text{det}(\pi_{Q*}(\mathcal{U} \otimes \pi_{X}^*\mathcal{O}_X(m)))$ be an invertible sheaf on $Q$, where $\pi_Q : Q \times X \rightarrow Q$ is a projection and $\pi_X$ similar. Let $q : \mathcal{H} \rightarrow \mathcal{F} \in Q$ a coherent quotient sheaf over $X$ i.e. a $k$-point of $Q$. Then they use the following isomorphism: $\mathcal{L}_m(q) = \bigotimes_i \text{det}(H^i(X,\mathcal{F}(m))^{\otimes (-1)^i})$.

An explanation or a reference would be of great help. Thanks.

$\endgroup$
  • $\begingroup$ I don't have time right now to write more, but to prove it, take a resolution of $H$ by sheaves with no higher cohomology. $\endgroup$ – user149792 Apr 15 '16 at 23:33
  • $\begingroup$ @KevinDong I haven't been able to do this. Could you elaborate? Thanks in any case. $\endgroup$ – baltazar Jun 14 '16 at 15:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.