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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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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

1 Answer 1

up vote 4 down vote accepted

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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omg, the display is so awful, I don't know how to fix it. I am sorry. –  Thi May 2 '13 at 13:11
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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
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Thanks very much Prof. Arapura –  Thi May 2 '13 at 14:31

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