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 DG-functor $F_E : Perf(X) \to Perf(Y)$ 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 Fourier-Mukai transform with kernel $E$. Is it possible to define quasi-functor (i.e right quasi-representable bimodule) between these DG-categories that induces Fourier-Mukai 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 quasi-coherent cohomology. $\endgroup$– Chris BravJan 31, 2012 at 13:57
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$\begingroup$ 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 DG-category $C$ s.t. $C$ is quasi-equivalent 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 PavlovJan 31, 2012 at 14:19
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2$\begingroup$ If you work with all complexes of $\mathcal{O}$-modules having quasi-coherent cohomology sheaves, then isn't $Perf$ identified with the full dg subcategory on complexes that are locally quasi-isomorphic 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 BravJan 31, 2012 at 14:30
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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 NevinsJan 31, 2012 at 19:56
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$\begingroup$ And I second Tom's quibble. $\endgroup$– Chris BravJan 31, 2012 at 22:04
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