# Finiteness of Galois cohomology

Let $k$ be a field, $X$ a smooth projective variety over $k$, $\overline{X} := X\times_k {k}^{\rm sep}$ for a separable closure ${k}^{\rm sep}$ of $k$, $\ell$ a prime with $\ell\in k^{\times}$.

Are the Galois cohomology groups $$H^i(\text{Gal}({k}^{\rm sep}/k),H^j_{\rm ét}(\overline{X},\mathbf{Z}_{\ell}))$$ finite for $i\ge 1$, $j\ge 0$? Are they torsion?

I would appreciate to get some references.

I have seen stated that $H^1(\text{Gal}({k}^{\rm sep}/k),H^j_{\rm ét}(\overline{X},\mathbf{Z}_{\ell}))$ is finite for $j\ge 0$ and $k$ a finite field, but I had the impression this should be a general fact about cohomology of profinite groups with coefficients in finitely generated $\mathbf{Z}_{\ell}$-modules with appropriate action.

• In Jannsen's Continuous étale cohomology, there is a Hochschild-Serre spectral sequence $H^p(G_k,H^q_\mathrm{et}(\bar{X},\mathbf{Z}_\ell(j)) \Rightarrow H^{p+q}_\mathrm{cont}(X,\mathbf{Z}_\ell(j))$ and for $k$ with finite $G_k$-cohomology for all $\ell$-torsion $G_k$-modules (e.g. $k$ finite or local), $H^p(G_k,H^q_\mathrm{et}(\bar{X},\mathbf{Z}_\ell(j))) = \varprojlim_n H^p(G_k,H^q(\bar{X},\mathbf{Z}/\ell^n(j)))$ and $H^p_\mathrm{cont}(X,\mathbf{Z}_\ell(j)) = \varprojlim_n H^i(X,\mathbf{Z}/\ell^n(j))$. – TKe Feb 5 '18 at 19:04
• Hmm, there is a definition that needs to be clarified, otherwise the two current answers are contradictory (and I believe they are both correct with their own interpretation of the question). What is $H^j(\bar X,\bf Z_l)$? For example what is it when $X$ is a point and $j=0$? Then clearly it is $\bf Z_l$ as a group, but with its discrete topology or with its $l$-adic cohomology. In number theory (since SGA5) we consider $H^j(\bar X,{\bf Z_l})$ as a shorter notation for $\projlim H^j(\bar X,{\bf Z}/l^n{\bf Z})$ – Joël Feb 5 '18 at 20:29
• What do you mean by the Galois cohomology? Are you using the discrete topology on $H^i_{\operatorname{ét}}(\bar X, \mathbb Z_\ell)$, or the $\ell$-adic topology? In the discrete topology, the higher cohomology is trivially torsion (this holds for any discrete module), but in the $\ell$-adic topology it's much more subtle (see e.g. @TKe's comment). – R. van Dobben de Bruyn Feb 5 '18 at 20:43
• @R. van Dobben de Bruyn : Yes, this point ought to be clarified. – Joël Feb 5 '18 at 21:13
• I meant, in my preceding comment, the action is continuous if you use the $l$-adic topology. Otherwise it is not continuous, as shown by Will and ndc's comments under the answer of AG2073951378. – Joël Feb 5 '18 at 22:00

I'm not sure what's going on with this question, but let me drop a few lines to summarize what the official position should be.

First off, I can't think about any situation these groups may occur beyond that of a Hochschild-Serre spectral sequence, so I hope group cohomology here is meant to be continuous group cohomology, and $\mathbf{Z}_{\ell}$-cohomology must then mean continuous étale cohomology.

The given answer (the accepted one) is incorrect.

In the functorial short exact sequences

$$0\to {\lim}^1 H^{i-1}(X_{\rm ét},\mathbf{Z}/(\ell^n))\to H^i\to \lim H^i(X_{\rm ét},\mathbf{Z}/(\ell^n))\to 0$$

where $H^i$ is either $H^i(X_{\rm proét},\underline{\mathbf{Z}}_{\ell})$ or Jannsen's continuous étale cohomology $H^i_{\rm cont}(X,\{\mathbf{Z}/(\ell^n)\})$ (since they agree for any $X$) the ${\lim}^1$ term vanishes as soon as $H^{i-1}(X_{\rm ét},\mathbf{Z}/(\ell^n))$ is finite, by Mittag-Leffler.

In particular, if $X$ is defined over a separably closed field and proper (as in the OP's assumptions), $H^i$ agrees with usual $\ell$-adic cohomology.

Will's example is the typical way to show that geometric $\ell$-adic cohomology is usually not a discrete Galois representation, so any argument trying to infer $H^i(\text{Gal}(k^{\rm sep}/k), H^j)$ is torsion for $i>0$ and $j\ge 0$ from discreteness of $H^j$ is wrong.

Also, it is wrong to say that the same argument as for geometric étale cohomology of abelian sheaves, showing the Galois action is discrete, works for proétale cohomology. One can define the action abstractly and the concrete way as in the accepted answer, but these two actions don't generally agree anymore. It boils down to the fact that evaluation at usual geometric points does not give a conservative family of fiber functors on the proétale topos.

Both questions asked by the OP have negative answer, regardless of what $H^i$ is meant to be, among the possibilities discussed here.

Maybe the OP was meaning to ask something different?

It is not known, I think. For $X$ a point, $k$ a number field, $j=0$, $i=2$, you get the statement that $H^2(k,\mathbb Q_p)=0$ which is Leopoldt's conjecture.

• @user104832 If $H^2(k,\mathbb Q_p)$ is nonzero, it is also nontorsion, so I don't see how this changes much. – Will Sawin Feb 6 '18 at 20:59

I don't think the finiteness part is true, owing to the possibility of torsion in $\mathrm{H}^j(\bar{X},\mathbb{Z}_\ell)$.

For example, take $X$ to be an Enriques surface over $\mathbb{Q}$. Then there is 2-torsion in the Néron–Severi group of $\bar{X}$, so also in $\mathrm{H}^2(\bar{X},\mathbb{Z}_2(1))$, which as a group is isomorphic to $\mathrm{H}^2(\bar{X},\mathbb{Z}_2)$. That means that $\mathrm{H}^1(\mathbb{Q}, \mathrm{H}^2(\bar{X}, \mathbb{Z}_2))$ is infinite.

• That's a right counter-example for the older question, but the OP had changed "finite" into "torsion". – Joël Feb 5 '18 at 21:59
• Oh. From what I see both questions are still there. (But I agree this seems to be the easier part.) – Martin Bright Feb 5 '18 at 22:01
• Ah yes, you're right. – Joël Feb 5 '18 at 23:24