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Let $X$ be a smooth projective variety and write $\mathbf{D}(X)$ for its triangulated category of perfect complexes of quasi-coherent sheaves. Recall that $\mathbf{D}(X)$ determines the Grothendieck group $K_0(X)$. Therefore by the Grothendieck-Riemann-Roch theorem, $\mathbf{D}(X)$ also determines the (direct sum of the) Chow groups $\mathrm{CH}_*(X, \mathbf{Q})$ with rational coefficients.

Are there known counterexamples to this statement for integral coefficients?

(I should note that I consider $\mathbf{D}(X)$ only as a triangulated category, in particular without its monoidal structure.)

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If the canonical bundle is ample or anti-ample then you can reconstruct X from D(X) by Bondal-Orlov, so any counterexample would have to avoid this. –  Dylan Wilson Dec 19 '13 at 16:51
    
Look here: arxiv.org/abs/1301.0707 –  Sasha Dec 19 '13 at 17:02
    
@Sasha: It doesn't look like Adeel wants to use the tensor structure, otherwise the result was known a long time ago by Thomason in more generality that we can reconstruct the whole scheme, and thus the Chow groups. –  Dylan Wilson Dec 19 '13 at 17:03
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D(X) determines the Grothendieck group, but not the Grothendieck ring, so better say D(X) determines the rationalized Chow groups, rather than the Chow ring. –  ya-tayr Dec 19 '13 at 17:09
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@SébastienPalcoux, this is noncommutative geometry in the sense of Kontsevich, when one replaces a variety by the triangulated category of perfect complexes on it. –  Adeel Dec 21 '13 at 8:16

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