4
$\begingroup$

Let $f:X\to B$ and $g:Y\to B$ be smooth morphisms of complex projective varieties. Assume that for every closed point $b\in B$, the fibres $X_b=X\times \kappa(b)$ and $Y_b$ are derived equivalent. Assume furthermore that the Fourier-Mukai transforms are induced by complexes of sheaves $F_b \in D^b(X_b \times Y_b)$ that glue to a sheaf $F\in D^b(X \times_B Y)$. Does it follow that the total spaces $X$ and $Y$ are derived equivalent?

Edit: As pointed out by pro, the answer to the original question (without the italic part) is negative. An explicit example is provided by taking $P^1\times P^1$ and any Hirzebruch surface $F_n$. Both are $P^1$ bundles over $P^1$, so the fibres are even isomorphic, but since $P^1\times P^1$ has an antiample canonical bundle both varieties could only be derived equivalent if they were isomorphic, which they are not.

$\endgroup$
  • 2
    $\begingroup$ I can't imagine this could ever be true, there's no gluing data. Take X and Y to be the total spaces of projectivizations of vector bundles on B. I would bet you could already find a counterexample there (perhaps with total spaces being Fano so you can use Bondal-Orlov to reduce to the case of vector bundles with non-isomorphic projectivizations). $\endgroup$ – pro Dec 22 '15 at 0:19
2
$\begingroup$

With the added assumptions the answer is yes. In other words, one can check equivalences by looking at fibres.

This is Proposition 2.15 of http://arxiv.org/pdf/math/0610319.pdf (it might be that when everything is smooth this was already known).

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.