Let $X$ and $Y$ be two smooth varieties over some field, and let $E$ be a perfect complex on $X \times Y$. It looks like it is not possible to define a DGfunctor $F_E : Perf(X) \to Perf(Y)$ such that induced triangulated functor $$\Phi_E : H^0(Perf(X))=D^b(X) \to H^0(Perf(Y))=D^b(Y)$$ will be a FourierMukai transform with kernel $E$. Is it possible to define quasifunctor (i.e right quasirepresentable bimodule) between these DGcategories that induces FourierMukai transform?
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$\begingroup$ Perhaps you could explain why you think it is not possible. I would say that the answer is more or less obviously 'yes', since pulling, tensoring, and pushing are all naturally defined at the level of dg categories. The only problem is that if you want to do this explicitly, you might have to make some natural identifications between different models for your dg categories corresponding to different kinds of resolutions or different ways to represent a small category, like $Perf$, by embedding it into a big category, like unbounded complex with quasicoherent cohomology. $\endgroup$ – Chris Brav Jan 31 '12 at 13:57

$\begingroup$ For example, for correct direct image we need injective resolution of a given perfect complex and in category $Perf(X)$ we can't consider such resolution. I think you right about different models: instead of $Perf(X)$ I should consider some DGcategory $C$ s.t. $C$ is quasiequivalent to $Perf(X)$ and on $C$ such functor is well defined, but information in such "roof" is equivalent to bimodule and I don't see what $C$ I can take. $\endgroup$ – Sasha Pavlov Jan 31 '12 at 14:19

2$\begingroup$ If you work with all complexes of $\mathcal{O}$modules having quasicoherent cohomology sheaves, then isn't $Perf$ identified with the full dg subcategory on complexes that are locally quasiisomorphic to finite length complexes of vector bundles? Under fairly general hypotheses (certainly under those in your question), you can also intrinsically pick out $Perf$ as the full subcategory of compact objects, those such that $RHom(E,?)$ preserves colimits. $\endgroup$ – Chris Brav Jan 31 '12 at 14:30

1$\begingroup$ Just to point out: you probably want to add "proper" to the hypotheses on $X$. Otherwise direct image indeed won't take compact objects (in this setting, perfect complexes) to compact objects. That quibble aside, I second what Chris says. :) $\endgroup$ – Thomas Nevins Jan 31 '12 at 19:56

$\begingroup$ And I second Tom's quibble. $\endgroup$ – Chris Brav Jan 31 '12 at 22:04