# Is it true that $H^{2r} ( X , \, \mathbb{Q}_{ \ell } (r) ) \simeq H^{2r} ( \overline{X} , \, \mathbb{Q}_{ \ell } (r) )^G$?

Let $$k$$ be a field and let $$X$$ be a smooth projective variety over $$k$$ of dimension $$d$$. We denote by $$\overline{X} = X \times_k \overline{k} \$$ the base change of $$X$$ to the algebraic closure $$\overline{k}$$. Then he Galois group $$G = \mathrm{Gal} ( \overline{k} / k )$$ acts on $$\overline{X}$$ via the second factor.

Question. Is it true that the $$\ell$$ - adic étale cohomology vector space $$H^{2k} ( X , \, \mathbb{Q}_{ \ell } (r) )$$ verifies $$H^{2r} ( X , \, \mathbb{Q}_{ \ell } (r) ) \simeq H^{2r} ( \overline{X} , \mathbb{Q}_{ \ell } (r) )^G \, ?$$ If so, how can we prove it?

• I don't think so. The two are related by the Hochschild-Serre spectral sequence, but I am not qualified to elaborate on exactly how all that works. I think it might fail even for $X=Spec(k)$. Apr 13, 2021 at 17:39
• @Wojowu In his paper "Continuous étale cohomology", Jannsen says that this Hochschild-Serre spectral sequence for ordinary étale cohomology does not exist in general. It exists for continuous étale cohomology though (but I would be unable to give an example). Apr 13, 2021 at 18:15
• @FrançoisBrunault If the mod $\ell^n$ cohomology groups are finite then continuous and usual etale cohomology agree, so it's OK. (Luckily, they are finite in my examples.) Apr 13, 2021 at 18:17
• @FrançoisBrunault Thanks for the remark. As I mentioned I am not well acquainted with the theory. Apr 13, 2021 at 18:19

This is false for a general field $$k$$. It is true for some special fields, like finite fields.

Counterexample: Take $$k = \mathbb C((t))$$, $$E$$ an elliptic curve over $$\mathbb C$$ base-changed to $$\mathbb C((t))$$, $$r=1$$. Because we're over $$\mathbb C$$, the twists don't matter and can be ignored.

$$H^2(\overline{E}, \mathbb Q_\ell) =\mathbb Q_\ell$$ and so its Galois-invariants are one-dimensional. However, we'll show the left side is three-dimensional.

The Galois group of $$\mathbb C((t))$$ is $$\hat{\mathbb Z}$$, so the Hochschild-Serre spectral sequence (which exists because the torsion etale cohomology groups in this setting are finite, so usual etale cohomology agrees with continuous etale cohomology, which has Hochschild-Serre) $$H^p ( \mathbb C((t)), H^q( \overline{E}, \mathbb Q_\ell )) \to H^{p+q} ( E, \mathbb Q_\ell)$$ is nontrivial only for $$p=0,1$$, $$q= 0,1,2$$. Because the Galois action on the cohomology groups is trivial, we will have $$\dim H^{0,0} = \dim H^{1,0}=1, \dim H^{0,1}= \dim H^{1,1}=2, \dim H^{0,2} = \dim H^{1,2} =2$$. There are no possible nonvanishing differentials, and so $$\dim H^2 =2 + 1 = 3$$.

Over finite fields, we again have finite torsion cohomology groups and thus a Hochschild-Serre sequence. We again have cohomology only for $$p=0,1$$, but now the Tate twist does play a role, and the Galois action on cohomology is nontrivial.

$$H^0 ( \mathbb F_q, V)$$ is only nontrivial when the eigenvalues of Frobenius acting on $$V$$ include $$1$$. Since $$H^q (\overline{X}, \mathbb Q_\ell(r))$$ is pure of weight $$q-2r$$, this can happen only for $$q=2r$$. Again there are no possible nontrivial differentials, but now the only group that contributions to $$H^{2r} (X, \mathbb Q_\ell(r))$$ is $$p=0, q=2r$$, which indeed is $$H^{2r}(\overline{X}, \mathbb Q_\ell(r))^{\operatorname{Gal}(\mathbb F_q)}$$.

For number fields $$K$$, it fails for $$r=1$$, $$X$$ a point. This may be what gdchtf meant.

In this case, the right side vanishes and the left side is the inverse limit of $$H^2 ( K , \mathbb Z/\ell^n (1))$$ as $$n$$ goes to $$\infty$$.

By Brauer theory, say for $$\ell$$ odd, this torsion cohomology group is the set of finitely supported $$\mathbb Z/\ell^n$$-valued functions on the set of finite places of $$K$$ which sum to $$0$$.

The inverse limit would be $$\mathbb Z_\ell$$-valued functions on the set of finite places whose sum converges to $$0$$ in the $$\ell$$-adic topology. (For the sum to converge at all, there must be only finitely many terms nonzero mod $$\ell^n$$ for each $$n$$.)

Inverting $$\ell$$, we get $$\mathbb Q_\ell$$-valued functions on the set of finite places whose sum converges to $$0$$.

This vector space is infinite-dimensional and does not vanish.

• What about for number fields ? Thank you. :) Apr 13, 2021 at 18:21
• @YoYo For number fields, it is also false - see edit. Apr 13, 2021 at 18:27
• What is the crucial point here? Is that essentially that taking homotopy invariants does not reduce to invariants of cohomology groups?
– Z. M
Apr 13, 2021 at 18:31
• @Z.M. Certainly, but the additional restriction that we only look at cohomology in degree $2r$ of $\mathbb Q_\ell(r)$ makes it harder to find an example where they differ - impossible, in the finite field case. Apr 13, 2021 at 18:32

No. Take $$k=\mathbb{Q}$$, $$X=\mathrm{Spec}\:k$$, $$r=0$$.

• I get $\mathbb Q_\ell = \mathbb Q_\ell^{ \operatorname{Gal}(\mathbb Q)}$, which is correct, what's the problem? Apr 13, 2021 at 17:55