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This question is regarding the proof strategy presented in the paper, "On The Grayson Spectral Sequence", which its overview is explained in page 1 and 2. It seems a very general approach is presented for proving when two bigraded cohomology theories are the same, here one of the cohomology theories is the motivic cohomology and the other one is the Grayson motivic cohomology. I will give a brief overview of the method and then draw contradiction out of it, which has made me confused and I cannot figure out what am I missing.

Here is the short version of my issue:

The proof outline given in the fist two pages, seems to implies that $\mathbb{Z}(n)$ is quasi-isomorphic to $\mathbb{Z}(n)\oplus \mathbb{Z}(n-1)$, by simply replacing $\mathbb{Z}^{Gr}(n)$ by $\mathbb{Z}(n)\oplus \mathbb{Z}(n-1)$.

longer version:

We are going to work with the motivic cohomology $\mathbb{Z}(n)$ and another weighted cohomology theory $\mathbb{Z}'(n)$ with a natural map $\mathbb{Z}'(n)\rightarrow \mathbb{Z}(n)$, that we intend to show that it is a quasi-isomorphism. If the cohomology theory $\mathbb{Z}'(n)$ satisfies certain properties it can be shown that the quasi-isomorphism can be reduced to the case of fields. We won't care about these properties and only focus on the case of the fields. The idea of proving the quasi-isomorphism for the fields is induction on $n$ (weight). One should check manually that that quasi-isomorphism holds for $n\leq 0$. Let $\Delta^m$ be the $m$-dimensional algebraic simplex over the field $F$. We denote by $\hat{\Delta}^m$ be the semi-localization of the simplex on its vertices. Let $\mathcal{Z}$ be the family of supports consisting of all closed sub-schemes in the simplex that does not intersect the vertices. Because of purity (excision) property of the motivic cohomology we have the long exact sequence of the following form(check the paper for more details of the definition of these notations): $$H^{m+p-1}(\hat{\Delta}^m, \{\hat{\Delta}^m_i\}, \mathbb{Z}(n))\rightarrow H^{m+p}_{\mathcal{Z}}({\Delta}^m, \{{\Delta}^m_i\}, \mathbb{Z}(n))\rightarrow H^{m+p}({\Delta}^m, \{{\Delta}^m_i\}, \mathbb{Z}(n)) \rightarrow H^{m+p}(\hat{\Delta}^m, \{\hat{\Delta}^m_i\}, \mathbb{Z}(n))$$

It been proved in the paper that $H^{m+p}(\hat{\Delta}^m, \{\hat{\Delta}^m_i\}, \mathbb{Z}(n))=0$ for $m+p>n$.

The same sequence can written for $\mathbb{Z}'(n)$. Let's assume that we know that for example $H^{m+p}(\hat{\Delta}^m, \{\hat{\Delta}^m_i\}, \mathbb{Z'}(n))=0$ for $m+p>n+1$. Then for a fixed $p$ and $n$ if $m$ is large enough we have the following isomorphisms:

$$H^{m+p}_{\mathcal{Z}}({\Delta}^m, \{{\Delta}^m_i\}, \mathbb{Z}(n))\cong H^{m+p}({\Delta}^m, \{{\Delta}^m_i\}, \mathbb{Z}(n))\cong H^p(F, \mathbb{Z}(n))$$ The last isomorphism is proved in the paper and follows from the homotopy invariance of motivic cohomology. We will also assume that $\mathbb{Z}'(n)$ is also homotopy invariant. Note that we just needed to assume that $m$ is large enough to get the above isomorphism. Because of the similar assumption on $\mathbb{Z}'(n)$ we get the isomorphism $H^{m+p}_{\mathcal{Z}}({\Delta}^m, \{{\Delta}^m_i\}, \mathbb{Z'}(n))\cong H^p(F, \mathbb{Z'}(n))$. Because of our inductive hypothesis that $\mathbb{Z}$ and $\mathbb{Z}'$ are quasi-isomorphic for weights $\leq n$ and the cohomological purity we have $H^{m+p}_{\mathcal{Z}}({\Delta}^m, \{{\Delta}^m_i\}, \mathbb{Z'}(n)) \cong H^{m+p}_{\mathcal{Z}}({\Delta}^m, \{{\Delta}^m_i\}, \mathbb{Z}(n))$, which implies that $H^p(F,\mathbb{Z}(n))\cong H^p(F, \mathbb{Z}'(n))$, which is the quasi-isomorphism for the weight $n$ and finishes the proof.

As you can see the proof seems to be very general, now I can give a counter-example for this. Assume that $\mathbb{Z}'(n)=\mathbb{Z}(n)\oplus \mathbb{Z}(n-1)$. Then it is easy to check that $\mathbb{Z}'(n)$ is a homotopy invariant sheaf which satisfies cohomological purity and at weights $\leq 0$ it coincides with the motivic cohomology. So the above argument shows that $\mathbb{Z}'(n)$ should be quasi-isomorphic to the motivic cohomology $\mathbb{Z}(n)$ on fields (and consequently on smooth schemes), which is obviously not correct.

So what am I missing about the proof in the paper? I'd really appreciate if anyone point me to the right direction and save me from the confusion.

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  • $\begingroup$ in your proposed contradiction, what is the "natural map Z'(n)-->Z(n)" you are using? is there such a map? $\endgroup$ Jul 19 at 0:58
  • $\begingroup$ @DylanWilson Yes you can take the projection onto the direct summand. $\endgroup$
    – user127776
    Jul 19 at 1:04
  • $\begingroup$ is it clear that choosing this map would be compatible with your use of the induction hypothesis on n? $\endgroup$ Jul 19 at 2:15
  • $\begingroup$ I think it is. What one needs to verify is that such map induces a map on the long exact sequence of the excision, which does. You can think of the cohomology groups of $\mathbb{Z}'$ as direct sum of two motivic cohomology groups (one shifted) that each satisfies the excision. So the projection does induce a map on the long exact sequences. (The long exact sequence used above is induced from excision) $\endgroup$
    – user127776
    Jul 19 at 2:26
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The $Z(n)\oplus Z(n-1)$ should be regarded as an $\mathbb{A}^1$-local (Suslin) complex, i.e., whose cohomologies are $\mathbb{A}^1$-invariant. So if you want to define some morphism between $\mathbb{A}^1$-local objects, you need to go back to the construction of $C^*$ and show that it's an quasi-isomorphism, i.e., their hypercohomologies on fields coincides. So for $Z(n)\oplus Z(n-1)$ you can't argue as simply as before.

On the other hand, the cohomology sheaves of $Z(n)$ are the Nisnevich sheafifications of the presheaves $X\longmapsto H^{p,n}(X,\mathbb{Z})$, so you can't say $Z(n-1)\oplus Z(n)=Z(n)$.

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  • $\begingroup$ In the argument sketched above or the first two pages of the paper if you know that cohomologies of the complexes are $\mathbb{A}^1$-invariant and there is an excision long exact sequence and the vanishing $H^{m+p}(\hat{\Delta}^m, \{\hat{\Delta}^m_i\}, \mathbb{Z}(n))=0$ for large enough $m$, you can conclude the map induces quasi-isomorphism on the fields. All of $C^*$ construction and $K_0^{\oplus}$-presheaf construction has been used to prove the vanishing result and also the excision for the Grayson motivic cohomology. $\endgroup$
    – user127776
    Jul 19 at 5:54
  • $\begingroup$ But if one knows them from start (as for $\mathbb{Z}(n)\oplus \mathbb{Z}(n-1)$) those constructions doesn't seem to be necessary. $\endgroup$
    – user127776
    Jul 19 at 5:54
  • $\begingroup$ I am also skeptical about the Nisnevich sheafification statement. Nisnevish sheafification of cohomology groups gives us a bunch of honest sheaves but $\mathbb{Z}(n)$ is not necessarily an honest sheaf but a complex of sheaves. $\endgroup$
    – user127776
    Jul 19 at 6:02
  • $\begingroup$ @user127776 I mean the cohomology sheaves of the complex $Z(n)$. $\endgroup$ Jul 19 at 6:17
  • $\begingroup$ @user127776 I think the point is the isomorphisms of cohomologies of fields. In the paper the proof reduces to polyrelative cohomologies. I think this part should be carefully read. $\endgroup$ Jul 19 at 6:31

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