# Which complexes of coherent sheaves are dual to perfect ones?

Let $$X$$ be a Noetherian scheme that is not Gorenstein but possesses a dualizing complex $$D$$ of coherent sheaves. Then (if I understand these matters and the answer to the question Characterization of schemes whose dualizing complex is perfect correctly) the bounded derived category $$D^b_{coh}(X)$$ of coherent sheaves on $$X$$ is self-dual, but this duality $$D_X$$ does not send (all) perfect complexes into perfect ones (recall that an object of $$D^b_{coh}(X)$$ is a perfect complex if it is locally quasi-isomorphic to a bounded complex of free sheaves).

My question is: did anybody study the image of the triangulated subcategory of perfect complexes $$D^{perf}(X)\subset D^b_{coh}(X)$$ under the coherent duality $$D_X$$? Is there any name for this subcategory $$D_X(D^{perf}(X))\subset D^b_{coh}(X)$$? Is it true that $$D_X(D^{perf}(X))\cong D^{perf}(X)\otimes D$$? Does it follow that $$D_X$$ induces an equivalence $$D^{perf}(X)^{op}\cong D^{perf}(X)$$?

Since perfect complexes are dualizable, for every perfect complex $$P$$ and any complex $$Q$$ we have $$\mathrm{hom}(P,Q)\cong \mathrm{hom}(P,1)\otimes Q\,.$$ Moreover $$\mathrm{hom}(P,1)$$ is perfect too (since for qcqs schemes perfect=dualizable). In particular, by taking $$Q=\omega$$ the dualizing sheaf this shows that the image of the duality $$D(-):=\mathrm{hom}(-,\omega)$$ is exactly $$\mathrm{Perf}(X)\otimes\omega$$.
I don't think you can imply any more than this (in particular it's unclear to me why you would expect this to give an equivalence $$\mathrm{Perf}(X)^{op}\cong \mathrm{Perf}(X)$$ different from the standard duality.
• Dear Denis, no, you shouldn't delete this. I am not sure that I understand these matters well; so you answer is quite helpful for me; thank you! Moreover, I do not understand whether $\hat D$ gives a tensor inverse to $D$ in this more general case as well (cf. Greg Stivenson's answer). Jun 1, 2019 at 14:30
• @MikhailBondarko Regarding your last question: not in general, if $D$ is invertible (wrt to the tensor product), then it is dualizable, and so perfect. Jun 1, 2019 at 14:33