Let $\pi: X \rightarrow B$ be a proper flat morphism of varieties and $0 \in B$ be a closed point such that on $B \setminus 0$ the morphism $\pi$ is trivial, isomorphic to $(B \setminus 0) \times F$. Do there exist examples where the fibre $X_0$ is derived equivalent to $F$ but not isomorphic? What if I ask that the derived equivalence be induced by $\pi$?
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$\begingroup$ I believe this already happens for the family $F\times (B\setminus\{0\}) = (\mathbb{P}^1\times \mathbb{P}^1)\times (B\setminus\{0\})$ specializing to a Hirzebruch surface $\Sigma_2$, i.e., the blowing up of the vertex of a singular quadric cone in $\mathbb{P}^3$. $\endgroup$Jason Starr– Jason Starr2018-02-14 16:46:41 +00:00Commented Feb 14, 2018 at 16:46
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4$\begingroup$ @JasonStarr I am afraid this cannot happen because $\mathbb{P}^1 \times \mathbb{P}^1$ is Fano, so by Bondal-Orlov a variety derived equivalent to it would also have to be isomorphic. $\endgroup$Piotr Achinger– Piotr Achinger2018-02-14 17:11:43 +00:00Commented Feb 14, 2018 at 17:11
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$\begingroup$ @PiotrAchinger I was thinking the same. I have to say that the situation I am thinking about is a little more general, I just want to work with the derived categories, not the varieties. I don't know how this affects the question! $\endgroup$Lawrence Jack Barrott– Lawrence Jack Barrott2018-02-14 17:12:05 +00:00Commented Feb 14, 2018 at 17:12
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$\begingroup$ @PiotrAchinger. You are correct! The two schemes are deformation equivalent, and they have isomorphic Chow groups and K-groups, but they are not derived equivalent. $\endgroup$Jason Starr– Jason Starr2018-02-14 17:26:59 +00:00Commented Feb 14, 2018 at 17:26
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$\begingroup$ @JasonStarr I think there may be a misunderstanding then, I want an example where the central fibre is derived equivalent to a general fibre (but not isomorphic). $\endgroup$Lawrence Jack Barrott– Lawrence Jack Barrott2018-02-14 17:33:26 +00:00Commented Feb 14, 2018 at 17:33
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I do not think this is possible. What I write below is not a proof, rather this is the reason why I think so.
Assume $\pi \colon X \to B$ is such a family. Let us consider it as a deformation family of the central fiber $F_0$. Note that $F_0$ is smooth if the general fiber is smooth, since smoothness can be detected by the derived category.
Let $\kappa \in H^1(F_0,T_{F_0})$ be the tangent vector for the deformation. Note that there is a natural morphism $$ H^1(F_0,T_{F_0}) \to HH^2(D^b(F_0)) \cong \bigoplus_{p+q = 2} H^q(F_0,\Lambda^p T_{F_0}) $$ to the second Hochschild cohomology group, which is the tangent space to deformations of $D^b(F_0)$. This map is injective, so if $\kappa \ne 0$, the deformation of the category in the family is non-trivial as well.
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$\begingroup$ This seems like a good way to think about it, there is a similar question in Hartshorne's book on deformation theory if I remember. $\endgroup$Lawrence Jack Barrott– Lawrence Jack Barrott2018-02-14 19:11:45 +00:00Commented Feb 14, 2018 at 19:11
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$\begingroup$ Thinking about this I feel I need to ask, is the Hochschild cohomology defined without the tensor product structure? Otherwise how are we comparing deformations of the central fibre to the general fibre? $\endgroup$Lawrence Jack Barrott– Lawrence Jack Barrott2018-02-16 12:56:30 +00:00Commented Feb 16, 2018 at 12:56
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$\begingroup$ @LawrenceJackBarrott: Yes, the definition of Hochschild cohomology does not require tensor structure. Morally, it is equal to $Ext$-groups from the identity functor to itself; to define it rigorously one only needs an enhancement (for instance a DG-enhancement) for the category. $\endgroup$Sasha– Sasha2018-02-16 16:15:25 +00:00Commented Feb 16, 2018 at 16:15
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$\begingroup$ I was playing about with it earlier today and indeed found of course you don't need the tensor structure. How strange, that answers so many of my questions! $\endgroup$Lawrence Jack Barrott– Lawrence Jack Barrott2018-02-16 16:27:01 +00:00Commented Feb 16, 2018 at 16:27