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Let $X$ be a regular scheme and consider Grothendieck's $\gamma$-filtration $F^nK(X)$ on $K(X)$. For the graded pieces, one has $Gr^0K(X) = CH^0(X)$ and $Gr^1K(X) = \mathrm{Pic}(X) = CH^1(X)$. Does this continue to hold, i.e., do we have $Gr^pK(X) = CH^p(X)$?

I found that for $X/k$ smooth quasi-projective, $CH^q(X,p) \otimes \mathbf{Q} = K_p(X)^{(q)} \otimes \mathbf{Q}$, so this holds after rationalising.

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    $\begingroup$ Note that I've rewritten the second paragraph of my answer, which originally contained too strong a statement. $\endgroup$ Commented Jun 10, 2013 at 19:53

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The map between the graded $K$-theory ring on the one hand and the Chow ring on the other is defined via Chern classes and requires denominators. I know of no good reason to expect an integral isomorphism (or even a map), but I'm not aware of an explicit counterexample (though I'm vaguely aware that the experts think the place to look for that counterexample is over a field with large etale cohomological dimension).

On the other hand, if you replace the $\gamma$-filtration with the filtration by codimension of support, then you do get $Gr^p(X)$ as a quotient (with torsion kernel) of $Ch^p(X)=H^p(X,K_p)$ (over the integers) provided $X$ is both regular and of finite type over a field --- though it would follow from Gersten's conjecture that this holds for all regular $X$.

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  • $\begingroup$ Thank you. By the way, one has $F^1\gamma K(X)=F^1_{top}K(X)$ if there is an ample sheaf. $\endgroup$
    – user19475
    Commented Jun 10, 2013 at 16:07
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This does not hold in general.

An explicit counterexample is given in: Karpenko, Nikita A. Codimension 2 cycles on Severi-Brauer varieties. K-Theory 13 (1998), no. 4, 305–330. (Reviewer: Jean-Pierre Tignol) 16K20 (14C15 19E15). (Available on the author's webpage https://sites.ualberta.ca/~karpenko/publ/ch2.pdf).

Fix a field $k$. Let $p$ be an odd prime. Let $A$ be a central division $k$-algebra of degree $p^2$, exponent $p$, and assume $A$ decomposes into a product $D_1\otimes D_2$ of two smaller algebras (both necessarily of degree $p$). Let $X$ be the Severi-Brauer variety associated to $A$. By Proposition 4.7 of the article above, the quotient $\text{Gr}^2K(X)$ contains torsion but, by Proposition 5.3 of the article above, the group $\text{CH}^2(X)$ is torsion free.

Similarly, one can show if $A$ is a division algebra of index $8$ and exponent $2$ decomposing into a product of smaller algebras then the same conclusion holds. A more general statement is true for the prime $2$ but it depends on the indices of the tensor powers of $A$.

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    $\begingroup$ What is $X$ here? In general, if $X$ is smooth, you have a natural surjection $CH^r(X)\to Gr^r K(X)$ and the chern class map $c_r$ in the opposite direction. The composite is (by Grothendieck-Riemann-Roch without denominators), multiplication by $(r-1)!$ and so, if $X$ is smooth, this map is an isomorphism for $r=2$ ($(2-1)!=1$). This does fail for higher codimension. $\endgroup$
    – Mohan
    Commented Dec 28, 2017 at 0:56
  • $\begingroup$ @Mohan Let me just mention that this is not true. (There is no surjection from the Chow ring to the associated graded for the gamma filtration. There is however a natural surjection to the associated graded for the topological filtration but this doesn't appear in this question.) $\endgroup$
    – Eoin
    Commented Jun 19, 2019 at 20:00

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