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Let $X$ be a smooth variety over a field $k$. (Assume $k$ has characteristic 0 if it helps; in fact I'd be happy to assume that $k$ is a finite extension of either $\mathbf{Q}$ or $\mathbf{Q}_p$).

Then there is a sheaf $\mathscr{K}_m^M$ on $X$ (in the Zariski topology), for each $m \ge 1$, which comes from sheafifying the Milnor $K$-theory of function rings of affine opens of $X$. So we can take sheaf cohomology groups $H^*(X, \mathscr{K}_m^M)$. We can also consider Voevodsky's motivic cohomology groups $H^*(X, \mathbf{Z}(i))$, where $\mathbf{Z}(i)$ is Voevodsky's motivic complex.

Is it true that there are isomorphisms $$H^r(X, \mathscr{K}_s^M) = H^{r+s}(X, \mathbf{Z}(s))$$ for all integers $r, s \ge 0$?

(Note that the right-hand side is known to be coincide with Bloch's higher Chow group $CH^s(X, s-r)$.)

Here is why I think this. Both sides are zero if $r > s$ or $r > \dim X$. For $r \le s$, one has a candidate for the isomorphism using Kerz's Gersten complex for Milnor K-theory and a construction due to Landsburg; and for $r = s$ or $r = s-1$ this map is indeed known to be an isomorphism. On the other hand, for $r = 0$ and $X = \operatorname{Spec} k$, this is just the isomorphism of Nesterenko--Suslin--Totaro, $$H^s(\operatorname{Spec} k, \mathbf{Z}(s)) = K_s^M(k).$$

(This is related to my earlier question When do the $\gamma$-filtration and codimension filtration of K-theory agree?, where I asked essentially the same thing for the Quillen $K$-theory sheaf instead of the Milnor one. It was pointed out there that for $r = 0$ and $X = \operatorname{Spec} k$ the motivic cohomology is just the Milnor K-theory, which leads me to wonder if one does get an isomorphism using the Milnor $K$-theory sheaf; for $r = s$ or $r = s-1$ the cohomology of the Milnor and Quillen $K$-theory sheaves agrees.)

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    $\begingroup$ No, there is no such isomorphism in general. The problem is that Milnor K-groups just yield the $s$-th (co)homology of the complex of sheaves $Z(s)$. Probably, there is a comparison morphism (that is an isomorphism when $X$ is local); I have to think about the details here. $\endgroup$ – Mikhail Bondarko Feb 17 '13 at 20:04
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The answer is yes. Philosophically, it should be yes because both sides, cohomology of the Milnor K-sheaf as well as higher Chow groups, provide a way of extending the short exact sequence of classical Chow groups $$ {\rm CH}_\bullet(Z)\to {\rm CH}_\bullet(X)\to {\rm CH}_\bullet(U)\to 0 $$ to a long exact localization sequence. It would be a bit weird if these two would not agree because then higher Chow groups would not tell us everything there is to say about intersection theory.

Away from philosophy to actual argument. The case for $r=0$ (for arbitrary smooth schemes) can be seen rather directly as follows: first, Bloch proved the Gersten conjecture for higher Chow groups (Theorem 10.1 of Bloch's paper "Algebraic cycles and higher K-theory") which shows that $$ {\rm CH}^r(X,r)\cong {\rm H}^0_{\rm Zar}(X,{\scr CH}^r(r)), $$ i.e. the Chow group ${\rm CH}^r(X,r)$ is the $0$-th cohomology of a complex of the form $$ 0\to \bigoplus_{x\in X^{(0)}} {\rm CH}^r(k(x),r) \to \bigoplus_{x\in X^{(1)}} {\rm CH}^{r-1}(k(x),{r-1})\to\cdots\to \bigoplus_{x\in X^{(\dim X)}} {\rm CH}^0(k(x),0)\to 0 $$ Now the Nesterenko-Suslin-Totaro isomorphism allows to replace the individual terms of the complex, and then one has to do a residue computation to show that the residue map in localization for higher Chow groups agrees with the residue in Milnor K-theory. Consequently, one gets isomorphic complexes computing higher Chow groups and Milnor K-cohomology, in the case $r=0$.

I'm not quite sure if it's see easy to see that this also works for the other $0\leq r\leq s$. In any case, an argument has been written up in the paper

  • K. Rülling and S. Saito. Higher Chow groups with modulus and relative Milnor K-theory. arXiv:1504.02669v2.

The relevant statement is proved in Section 3.1. In particular Corollary 3.2 shows that a natural cycle class map provides a quasi-isomorphism of complexes of Zariski sheaves on $X$: $$ \tau_{\geq s}\mathbb{Z}(s)_X\to {\scr K}^{\rm M}_{s,X}[-s]. $$ This implies a positive answer to the question (it's even slightly better - the isomorphisms are all induced from a geometrically constructed chain map, so probably also functorial). A posteriori we know that the cohomology of the Gersten complex for Chow groups ${\rm CH}^r(r)$ gives motivic cohomology in the required range, but that's not quite what is stated in Bloch's paper.

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  • $\begingroup$ I apologize but I don't think this answers the question positively. From this result one can say that there is a distinguished triangle $R\Gamma(X,\tau_{<s}\mathbb{Z}(s))\rightarrow R\Gamma(X,\mathbb{Z}(s))\rightarrow R\Gamma(X,\tau_{\geq s} \mathbb{Z}(s))=R\Gamma(X,\mathscr{K}_s^M[-s])$ but, it's not necessarily true that $R\Gamma(X,\tau_{<s}\mathbb{Z}(s))$ has vanishing cohomology in degrees $>s$. $\endgroup$ – Eoin Apr 9 '18 at 5:00
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The answer is no, even locally. There is canonical map $H^n(X,\mathbb{Z}(m)) \rightarrow H^{n-m}(X,H^m(\mathbb{Z}(m))$, where $H^m(\mathbb{Z}(m))$ denotes the Nisnevich sheaf associated to the presheaf $U \mapsto H^m(U,\mathbb{Z}(m))$, which is your unramified Milnor sheaf $K^M_m$ above (Theorem of Nesterenko-Suslin-Totaro). This map is an isomorphism if $n \geq m + dim(X)$ or $n \geq 2m -2$. In general, it is false. Take simply $H^{1,2}(k,\mathbb{Z}$). This group is non-trivial. Indeed, from the motivic spectral sequence one knows that this group is $K_3(k)_{ind} = Coker (K_3^M(k) \rightarrow K_3^Q(k))$ (the indecomposable class), while the group $H^{-1}(k,K^M_2)$ is obviously trivial.

Ps: Oh I am sorry, I didn't see that $r,s \geq 0$. I think it is still not true.

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  • $\begingroup$ omg, the display is so awful, I don't know how to fix it. I am sorry. $\endgroup$ – Thi May 2 '13 at 13:11
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    $\begingroup$ There is a strange bug in the software. Sometimes putting backticks "`" around the math will allow it to display correctly $\endgroup$ – Donu Arapura May 2 '13 at 14:27
  • $\begingroup$ Do you have a reference for the claim that the map $H^n(X,\mathbb{Z}(m))\rightarrow H^{n-m}(X,H^m(\mathbb{Z}(m)))$ is an isomorphism in degree $n\geq 2m-2$? I would greatly appreciate it! $\endgroup$ – Eoin Apr 9 '18 at 5:02

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