3
$\begingroup$

In algebraic topology, for any space with finite homology type, the universal coefficient theorem states that for any abelian group $G$, we have $$H^n(X,G)\cong \left( H^n(X,\mathbb{Z})\otimes G\right)\oplus \text{Tor}_1(H^{n+1}(X,\mathbb{Z}),G).$$ My question is whether the analogous statement is true for the pro-étale cohomology, namely if $R$ is a $\mathbb{Z}_\ell$-algebra, do we have $$H^n_{proét}(X,\underline{R})\cong \left(H^n_{proét}(X,\underline{\mathbb{Z}_\ell})\otimes R\right)\oplus \text{Tor}_1(H_{proét}^{n+1}(X,\underline{\mathbb{Z}_\ell}),R)$$ for a sufficiently nice scheme? I'm mostly interested in the case of a smooth, projective scheme over some algebraically closed field (possibly of positive characteristic). Also, would this decomposition respect that Galois action on the cohomology?

$\endgroup$
6
  • $\begingroup$ Note: $\mathbf Z_\ell$ needs to be endowed with a topology for it to give the 'right' cohomology in the pro-étale site. So there should probably be a natural topology on $R$ as well (somehow compatible with the one on $\mathbf Z_\ell$) for this to have any chance of success. $\endgroup$ Jul 16 '20 at 0:58
  • $\begingroup$ Would it be enough to assume that the structure morphism $\mathbb{Z}_\ell \rightarrow R$ is continuous? $\endgroup$ Jul 16 '20 at 1:26
  • $\begingroup$ I do not think it is reasonable to consider any topology on $R$: the point of this formula is to compare the (derived) tensor product with $R$ within the pro-étale topos with the (derived) tensor product externally (i.e. within the ordinary derived category of $\mathbf{Z}_\ell$-modules), where everything is discrete. $\endgroup$ Jul 16 '20 at 7:03
  • $\begingroup$ @Denis-CharlesCisinski I don't understand what you mean. If you take $R = \mathbf Z_\ell$ with the discrete topology (i.e. the true constant sheaf), then you should get something very different than the usual $\mathbf Z_\ell$ cohomology. $\endgroup$ Jul 16 '20 at 13:51
  • $\begingroup$ @R.vanDobbendeBruyn I mean $R$ should be discrete, not $\mathbf{Z}_\ell$. This is what happen when you go from $\mathbf{Z}_\ell$-linear coefficients to $\mathbf{Q}_\ell$-linear ones: $\mathbf{Q}_\ell=\mathbf{Z}_\ell\otimes\mathbf{Q}$ with $\mathbf{Q}$ discrete. My point is that, in the "universal coefficients theorem", there is a tensor/tor with cohomology groups considered as discrete objects, so that $R$ has to be discrete. My answer below explains how to proceed (I underline the continuous $\mathbf{Z}_\ell$ to differientiate from the discrete one). $\endgroup$ Jul 16 '20 at 14:32
5
$\begingroup$

The only condition you need on $X$ for such a formula to hold is that it is coherent (=quasi-compact and quasi-separated).

Let $R$ be a discrete ${\mathbf{Z}_\ell}$-module. We consider the sheaf $\underline{\mathbf{Z}}_\ell$ on the pro-étale site defined as the limit of the constant sheaves $\mathbf{Z}/\ell^i\mathbf{Z}$. This is an algebra on the constant sheaf associated to the discrete ring $\mathbf{Z}_\ell$. Hence we may define $\underline{R}$ as the tensor product of $R$ with $\underline{\mathbf{Z}}_\ell$ (exercise: it is in fact the derived tensor product, since $\underline{\mathbf{Z}}_\ell$ has no $\ell$-torsion stalkwise; this is where pecularities of the pro-étale site have a role to play, in a proof which is otherwise very formal). Let $R\Gamma(X,-)$ denote the derived global sections on the pro-étale topos of $X$. To prove that the canonical map $$R\Gamma(X,\underline{R})\leftarrow R\Gamma(X,\underline{\mathbf{Z}}_\ell)\otimes^L_{\mathbf{Z}_\ell}R$$ is invertible in the derived category of $\mathbf{Z}_\ell$-modules, assuming that $X$ is coherent, we may assume that $R$ is of finite type because $R\Gamma(X,-)$ commutes with filtered colimits (this is precisely what coherence is good for). If $R=S\oplus T$, it is sufficient to prove it for $S$ and $T$ separately. Hence, without loss of generality, we may assume that $R=\mathbf{Z}_\ell$, in which case this is trivial, or that $R=\mathbf{Z}/\ell^i\mathbf{Z}$, in which case this is always true as well (taking the cone of the multiplication by $\ell^i$ commutes with $R\Gamma(X,-)$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.