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There is a conjecture of Orlov stating that if $X$ and $Y$ are smooth projective complex varieties that are derived equivalent (equivalent bounded derived categories of coherent sheaves), then their rational Voevodsky motives $M(X)_{\mathbb{Q}}$ and $M(Y)_{\mathbb{Q}}$ are equivalent.

My question is whether the equivalence of $M(X)_{\mathbb{Q}}$ and $M(Y)_{\mathbb{Q}}$ implies the partial converse that the rational(!) bounded derived categories of coherent sheaves are equivalent? There is a result of Cisinski and Tabuada that implies that the equivalence of the rational Voevodsky motives implies that the noncommutative rational motives of $X$ and $Y$ are equivalent in Kontsevich's category of rational noncommutative motives.

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    $\begingroup$ What is the rational bounded derived category (as opposed to the usual one?) $\endgroup$
    – dhy
    Commented Sep 26, 2017 at 15:37
  • $\begingroup$ A comment on the relation between Sasha's answer and the theorem of Cisinski and Tabuada: in the example Sasha gave, there are semiorthogonal decompositions of the derived categories. These imply that the noncommutative motives split like the ordinary motives do (which gives the equivalence of noncommutative motives). But semiorthogonal decompositions are far from giving equivalences of derived categories. $\endgroup$
    – Matthias Wendt
    Commented Sep 28, 2017 at 7:59

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Definitely not. Take $X$ to be the blowup of $P^2$ at a point and $Y$ to be $P^1 \times P^1$. Then $$ M(X) = 1 + 2L + L^2 = M(Y), $$ but the derived categories are different, since both varieties are Fano and non-isomorphic,

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  • $\begingroup$ Even the rational bounded derived categories are different? By rational, I mean you rationalize the triangulated category. $\endgroup$
    – user114292
    Commented Sep 27, 2017 at 21:46
  • $\begingroup$ @user114292 In what sense do you rationalize? $\endgroup$
    – Sasha
    Commented Sep 27, 2017 at 21:49
  • $\begingroup$ By this, I mean you take the triangulated category, and you do Verdier quotient so that $Cone(E\xrightarrow{n\cdot id_E} E)\simeq 0$ for every object $E$ and every nonzero integer $n$. $\endgroup$
    – user114292
    Commented Sep 27, 2017 at 22:00
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    $\begingroup$ @user114292 If everything is over a field of zero characteristic this cone is already zero, so your operation changes nothing. $\endgroup$
    – Sasha
    Commented Sep 27, 2017 at 22:02
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    $\begingroup$ And if the characteristic of the field is positive then this operation kills everything.:) $\endgroup$
    – Mikhail Bondarko
    Commented Sep 28, 2017 at 17:02

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