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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 previous answer contained a major error/misconception, and I apologize for the dealy in correcting it. The answer to the question is "yes" locally in the Zariski topology but "no" globally.

Comparison of Milnor K-cohomology and motivic cohomology via edge maps: Motivic cohomology has Zariski descent and hence there is a descent/hypercohomology/Brown-Gersten-Quillen/coniveau spectral sequence computing the motivic cohomology of a smooth scheme $X$. On the $E_1$-page, the Gersten resolution of the sheaves $\mathcal{H}^q(\mathbb{Z}(n))$ (Zariski sheafification of motivic cohomology) appears. The entry $E_1^{p,q}$ is $\bigoplus_{z\in X^{(p)}}{\rm H}^{q-p}(\kappa(z),\mathbb{Z}(n-p))$ and the differential $d_1:E_1^{p,q}\to E_1^{p+1,q}$ is given by the appropriate residue maps. The $E_2$-spectral sequence has the form $$ E^{p,q}_2={\rm H}^p(X,\mathcal{H}^q(\mathbb{Z}(n)))\Rightarrow {\rm H}^{p+q}(X,\mathbb{Z}(n)) $$ and the differentials go $d_s^{p,q}:E^{p,q}_s\to E^{p+s,q-s+1}_s$.

This spectral sequence can then be used to relate Milnor K-cohomology to motivic cohomology. The relevant thing to note is the identification $$ \tau_{\geq s}\mathbb{Z}(s)_X\to\mathbf{K}^{\rm M}_{s,X}[-s] $$ which can be found e.g. in Section 3.1, Corollary 3.2 of

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

This statement basically says that $\mathcal{H}^q(\mathbb{Z}(n))=0$ for $q>n$ and $\mathcal{H}^n(\mathbb{Z}(n))=\mathbf{K}^{\rm M}_n$, where $\mathcal{H}^q$ denotes the motivic cohomology sheaves. (It also says that the answer is yes locally in the Zariski topology.)

From this vanishing statement, we get edge maps ${\rm H}^p(X,\mathbb{Z}(n))\to {\rm H}^{p-n}(X,\mathcal{H}^n(\mathbb{Z}(n)))$ because in the line $q=n$ there are no incoming differentials and so $E^{p,n}_\infty$ is a subgroup of ${\rm H}^{p-n}(X,\mathcal{H}^n(\mathbb{Z}(n)))$. Since $E^{p,n}_\infty$ is the last subquotient of the filtration of ${\rm H}^p(X,\mathbb{Z}(n))$ this provides the above edge maps. These would be the natural comparison maps (and they are also mentioned in Thi's answer.)

Comparison isomorphisms for $r=s$ and $r=s-1$: To get the Bloch formula and the identification of ${\rm H}^{2n-1}(X,\mathbb{Z}(n))\cong {\rm H}^{n-1}(X,\mathbf{K}^{\rm M}_n)$, we need to show that there can be no outgoing differentials either. In the $E_2$-page, the relevant differentials for Bloch's formula are $d_s^{p,q}:{\rm H}^n(X,\mathcal{H}^n(\mathbb{Z}(n)))\to {\rm H}^{n+s}(X,\mathcal{H}^{n-s+1}(\mathbb{Z}(n)))$. The latter group is trivial, since on the $E_1$-page, the coefficients would be $\mathcal{H}^{1-2s}(\mathbb{Z}(-s))$ and the sheaves $\mathbb{Z}(-s)$ are trivial for $s>0$. The other relevant differentials are $d_s^{p,q}:{\rm H}^{n-1}(X,\mathcal{H}^n(\mathbb{Z}(n)))\to {\rm H}^{n+s-1}(X,\mathcal{H}^{n-s+1}(\mathbb{Z}(n)))$ with the coefficients in the $E_1$-page given by ${\rm H}^{2-2s}(\mathbb{Z}(1-s))$. For $s>1$, this is again trivial. In particular, all the outgoing differentials are trivial, so that $E^{n,n}_\infty={\rm H}^n(X,\mathbf{K}^{\rm M}_n)$ and $E^{n-1,n}_\infty={\rm H}^{n-1}(X,\mathbf{K}^{\rm M}_n)$. This also implies that the only $E^{n+s,n-s}_\infty$ contributing to ${\rm H}^{2n}(X,\mathbb{Z}(n)))$ is ${\rm H}^n(X,\mathbf{K}^{\rm M}_n)$, yielding Bloch's formula. To get ${\rm H}^{n-1}(X,\mathbf{K}^{\rm M}_n)\cong {\rm H}^{2n-1}(X,\mathbb{Z}(n))$ we need to show that ${\rm H}^{n+s-1}(X,\mathcal{H}^{n-s}(\mathbb{Z}(n)))=0$. The relevant coefficients in the $E_1$-page are $\mathcal{H}^{1-2s}(\mathbb{Z}(1-s))$. For $s=1$, we get $\mathcal{H}^{-1}(\mathbb{Z}(0))$ which is trivial, all the other terms are again trivial because $\mathbb{Z}(1-s)=0$ for $s>1$.

Failure of global comparison: The natural comparison isomorphism fails to be an isomorphism in general. The simplest counterexample (pointed out by Jens Hornbostel) is actually $\mathbb{P}^1$. In this case, we have $$ {\rm H}^r(\mathbb{P}^1,\mathbf{K}^{\rm M}_s)\cong \left\{\begin{array}{ll} {\rm K}^{\rm M}_s(F) & r=0\\ {\rm K}^{\rm M}_{s-1}(F) & r=1\end{array}\right. $$ where $F$ denotes the base field. On the other hand, the projective bundle formula for motivic cohomology implies $$ {\rm H}^{r+s}(\mathbb{P}^1,\mathbb{Z}(s))\cong {\rm H}^{r+s}(F,\mathbb{Z}(s))\oplus {\rm H}^{r+s-2}(F,\mathbb{Z}(s-1)). $$ For $r=0$, we have ${\rm H}^0(\mathbb{P}^1,\mathbf{K}^{\rm M}_s)\cong {\rm K}^{\rm M}_s(F)$ and ${\rm H}^s(\mathbb{P}^1,\mathbb{Z}(s))\cong {\rm K}^{\rm M}_s(F)\oplus {\rm H}^{s-2}(F,\mathbb{Z}(s-1))$, so Milnor K-cohomology and motivic cohomology differ whenever ${\rm H}^{s-2}(F,\mathbb{Z}(s-1))\neq 0$. A particular instance where that happens is $s=3$ in which case ${\rm H}^1(F,\mathbb{Z}(2))\cong {\rm K}^{\rm ind}_3(F)$.

In the formulation of the descent spectral sequence, the spectral sequence is contained in the two columns for $p=0,1$ and degenerates at the $E_2$-term. The failure of the global comparison follows since $\mathcal{H}^2(\mathbb{Z}(3))$ has nontrivial first cohomology over $\mathbb{P}^1$, given by ${\rm H}^1(F,\mathbb{Z}(2))$.

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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

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