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.

  • $\begingroup$ I just noticed that this exact argument is present in the answer you linked, so I am now unsure of what kind of answer you were expecting. Should I delete this? $\endgroup$ – Denis Nardin Jun 1 at 10:33
  • $\begingroup$ 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). $\endgroup$ – Mikhail Bondarko Jun 1 at 14:30
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    $\begingroup$ @MikhailBondarko Regarding your last question: not in general, if $D$ is invertible (wrt to the tensor product), then it is dualizable, and so perfect. $\endgroup$ – Denis Nardin Jun 1 at 14:33

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