# Equivalence of rational Voevodsky motives: partial Converse to Conjecture of Orlov

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.

• What is the rational bounded derived category (as opposed to the usual one?)
– dhy
Commented Sep 26, 2017 at 15:37
• 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. Commented Sep 28, 2017 at 7:59

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,
• 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 $n$. Commented Sep 27, 2017 at 22:00