# Motivic cohomology and cohomology of Milnor K-theory sheaf

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.)

• 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. – Mikhail Bondarko Feb 17 '13 at 20:04

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 $${\rm H}^1(F,\mathbb{Z}(2))$$.

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.

• omg, the display is so awful, I don't know how to fix it. I am sorry. – Thi May 2 '13 at 13:11
• There is a strange bug in the software. Sometimes putting backticks "`" around the math will allow it to display correctly – Donu Arapura May 2 '13 at 14:27