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As far as I understand, it was Beilinson who proved that the bounded derived category of coherent sheaves $D^b(\mathbb{P}^n)$ is equivalent to the bounded derived category of a certain (non-commutative) ring. My question is: for which (not necessarily projective) varieties or schemes isomorphisms of the form $D^b(X)\cong D^b(R_X)$ are known to exist?

More generally, when there exists a $t$-structure on $D^b(X)$ or on $D(X)$ whose heart is a certain category of $R$-modules? It appears that this question is related to tilting complexes of sheaves, but this does not help me much. What are the standard references for these matters?

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  • $\begingroup$ It may well be that this is only possible when the varieties are cohomologically simple, projective spaces, grassmannians, quadrics. I doubt there is such description in "general type" varieties, but this is just a guess. $\endgroup$
    – Leo Alonso
    Commented Dec 17, 2020 at 19:01
  • $\begingroup$ If you're willing to go to dg-algebras, I think smooth and proper is enough by Bondal and Van Den Bergh arxiv.org/abs/math/0204218 $\endgroup$
    – Aaron Bergman
    Commented Dec 17, 2020 at 19:07
  • $\begingroup$ I am not willing to go to dg algebras since this will not yield a t-structure.:) And I have a trivial family of examples: one can certainly take $X$ to be an affine variety. $\endgroup$
    – Mikhail Bondarko
    Commented Dec 17, 2020 at 19:09
  • $\begingroup$ Well, for what it's worth, I believe all the del Pezzos have tilting objects and I believe Craw has proved it true for a class of toric varieties. I know more has been done, but I'd just end up googling "tilting object" and "tilting sheaf" to get a more up-to-date answer. $\endgroup$
    – Aaron Bergman
    Commented Dec 17, 2020 at 20:13
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    $\begingroup$ If such $R_X$ exists then Hochschild homology of $X$ is concentrated in non-negative homological degrees. In characteristic zero HKR theorem (for smooth $X$) $HH_i(X)=\bigoplus\limits_{p-q=i}H^q(X,\Omega^p)$ then implies that $H^q(X,\Omega^p)=0$ for $q>p$. If $X$ is smooth proper this implies that $H^{p,q}(X)=0$ when $p\neq q$, so the cohomology of such smooth proper variety would have to be Tate. $\endgroup$
    – SashaP
    Commented Dec 17, 2020 at 23:46

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