Motivic cohomology and cohomology of Milnor K-theory sheaf - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T20:48:09Z http://mathoverflow.net/feeds/question/122072 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/122072/motivic-cohomology-and-cohomology-of-milnor-k-theory-sheaf Motivic cohomology and cohomology of Milnor K-theory sheaf David Loeffler 2013-02-17T15:40:58Z 2013-05-02T14:39:14Z <p>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 $\mathbf{Q}_p$). </p> <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.</p> <blockquote> <p>Is it true that there are isomorphisms $$H^r(X, \mathscr{K}_s^M) = H^{r+s}(X, \mathbf{Z}(s))$$ for all integers $r, s \ge 0$?</p> </blockquote> <p>(Note that the right-hand side is known to be coincide with Bloch's higher Chow group $CH^s(X, s-r)$.) </p> <p>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, $$H^s(\operatorname{Spec} k, \mathbf{Z}(s)) = K_s^M(k).$$ </p> <p>(This is related to my earlier question <a href="http://mathoverflow.net/questions/106021" rel="nofollow">http://mathoverflow.net/questions/106021</a>, 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.)</p> http://mathoverflow.net/questions/122072/motivic-cohomology-and-cohomology-of-milnor-k-theory-sheaf/129410#129410 Answer by Thi for Motivic cohomology and cohomology of Milnor K-theory sheaf Thi 2013-05-02T11:37:35Z 2013-05-02T14:39:14Z <p>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. </p> <p>Ps: Oh I am sorry, I didn't see that $r,s \geq 0$. I think it is still not true. </p>