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I'm encountering a lot of problems when dealing with the root of unity sheaf $\mu_\infty := \mathrm{colim}_n\mu_n$.

Let $X$ be a smooth geometrically integral variety over a number field $k$. Although we have the canonical inclusion $\mu_\infty \subset \mathbb{G}_m$, the cohomology groups with coefficients in the latter sheaf are better understood. For example, in Galois cohomology for our given $k$ we know that $$H^1(k,\mathbb{G}_m) = 0,\,\,H^2(k,\mathbb{G}_m) = \mathrm{Br}(k), \,\, H^3(k, \mathbb{G}_m) = 0.$$

For etale cohomology on $X$, we have $$H^0(X,\mathbb{G}_m) = k[X]^*, \,\, H^1(X,\mathbb{G}_m) = \mathrm{Pic}(X), \,\, H^2(X,\mathbb{G}_m) = \mathrm{Br}(X).$$

Now consider the spectral sequence $$H^p(k,H^q(\bar{X},\mu_\infty)) \implies H^{p+q}(X,\mu_\infty).$$ One obtains a long exact sequence of low degree terms: $$0 \rightarrow H^1(k,H^0(\bar{X},\mu_\infty)) \rightarrow H^1(X,\mu_\infty) \rightarrow H^0(k,H^1(\bar{X},\mu_\infty)) \rightarrow H^2(k,H^0(\bar{X},\mu_\infty))$$

$$\rightarrow \mathrm{Ker}[H^2(X,\mu_\infty) \rightarrow H^0(k,H^2(\bar{X},\mu_\infty))] \rightarrow H^1(k,H^1(\bar{X},\mu_\infty)) \rightarrow H^3(k,H^0(\bar{X},\mu_\infty)).$$

Question 2. Does $H^2(\bar{X},\mu_\infty)$ have trivial Galois action? If so, is it true for all $H^i(\bar{X}, \mu_\infty)$?

This question came about because one of the papers I'm reading defined some term to be the kernel of $H^2(X,\mu_\infty) \rightarrow H^2(\bar{X},\mu_\infty)$, so I have a feeling it came from this spectral sequence.

There are many other questions I can think of but I would like to end off with this:

Question 3. Is there some connection between $H^1(X,\mu_\infty)$ and $\mathrm{Pic}(X)$?

EDIT. I've removed Question 1 due to some confusion.

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    $\begingroup$ You are asking a lot of questions on this topic, and to be honest I am confused by them (for example your question 1 is a defining property for étale cohomology!). What is your background? $\endgroup$ Jul 24, 2021 at 6:22
  • $\begingroup$ @DenisNardin My reference is Etale Cohomology by Milne. This is my first year dealing with this topic, and I'm working without the completeness property of a variety, which is assumed in most of the book. Furthermore, I couldn't find any text dealing with the constant sheaf $\mu_\infty$, hence I asked these questions. I'm sorry if these three questions are too basic for the site, I can take them down. $\endgroup$ Jul 24, 2021 at 6:35
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    $\begingroup$ The sheaf $\mu_\infty$ is not a constant sheaf. And for every sheaf $F$ it's always true that $H^0(X,F)=F(X)$. This is by definition of cohomology as derived functor of the global sections. I don't think you should take down the questions, but you seem to have some fundamental confusion and it would be good to address that rather than trying to chase down minor lemmas $\endgroup$ Jul 24, 2021 at 6:42
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    $\begingroup$ What do you mean with $\mu_\infty$? Because in my head the two sides of the equality live in different categories, and so it makes no sense to ask whether they're isomorphic.. I think you are being led astray by some common abuse of notation that is normally harmless but is causing confusion since you just started learning this. $\endgroup$ Jul 24, 2021 at 6:47
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    $\begingroup$ The problem here is that you're using $\mu_\infty$ to denote two different things: both the étale sheaf sending a scheme $X$ to the group of roots of unity in $\mathcal{O}(X)$ and the subgroup of $\overline{k}$ consisting of roots of unity (so in particular this is what we might write $\mu_\infty(\overline{k})$ in the first notation). $\endgroup$ Jul 24, 2021 at 9:55

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Question 2. Does $H^2(\bar{X},\mu_\infty)$ have trivial Galois action? If so, is it true for all $H^i(\bar{X}, \mu_\infty)$?

The answers are "often not" and "almost never".

The exact sequence $1 \to \mu_{\infty} \to \mathbb G_m \to \mathbb G_m \otimes \mathbb Q \to 1$ you mention gives a map on cohomology $\operatorname{Pic}(\overline{X} ) \to \operatorname{Pic} (\overline{X} )\otimes \mathbb Q \to H^2 (\overline{X} , \mu_{\infty} )$.

Now $\operatorname{Pic} (\overline{X} )$ will map to a finitely-generated abelian group, the Neron-Severi group, with big divisible kernel, the Picard variety. When we tensor with $\mathbb Q$, we lose the torsion part of the finitely generated abelian group, and the divisible kernel will be canceled by the map from $\operatorname{Pic} (\overline{X} )$, but the rank part is preserved.

Let's say this finitely generated abelian group is $\mathbb Z^r$ times something torsion. We'll get $\mathbb Q^r$ appearing as a quotient of $\operatorname{Pic} (\overline{X} )\otimes \mathbb Q$ which modulo the original $\mathbb Z^r$ gives a $\mathbb Q^r / \mathbb Z^r$ inside $H^2 (\overline{X} , \mu_{\infty} )$.

The Galois action on this is trivial if and only if the Galois action on the rank part of the original Neron-Severi group is trivial. That will be true for some varieties and not others. The simplest example where it fails is a quadric hypersurface $x^2-y^2 + z^2 - d w^2$ where $d \in k$ is not a perfect square. Then Picard will be $\mathbb Z^2$ with Galois group elements swapping the two copies, and $H^2( \overline{X}, \mu_{\infty})$ will be $(\mathbb Q/\mathbb Z)^2$ with a similar swap.

For general $H^i (\overline{X}, \mu_{\infty})$, note that $\mathbb Q_\ell / \mathbb Z_\ell (1)$ is a summand of $\mu_{\infty}$ so $H^i ( \overline{X} , \mathbb Q_\ell / \mathbb Z_\ell (1 ))$ is a summand of $H^i (\overline{X}, \mu_{\infty})$.

We have a short exact sequence $0\to \mathbb Z_\ell(1) \to \mathbb Q_\ell(1) \to \mathbb Q_\ell/\mathbb Z_\ell (1)\to 0$ giving a long exact sequence

$$ H^i ( \overline{X} , \mathbb Z_\ell (1) ) \to H^i(\overline{X}, \mathbb Q_\ell(1)) \to H^i ( \overline{X} , \mathbb Q_\ell / \mathbb Z_\ell (1 ))$$

Here $H^i ( \overline{X} , \mathbb Z_\ell (1) )$ will look like $\mathbb Z_\ell^n$ plus some torsion, which we can see will force $H^i(\overline{X}, \mathbb Q_\ell(1))$ to look like $\mathbb Q_\ell^n$ and $H^i ( \overline{X} , \mathbb Q_\ell / \mathbb Z_\ell (1 ))$ to look like $(\mathbb Q_\ell/\mathbb Z_\ell)^n$ plus some torsion coming form $H^{i+1} ( \overline{X} , \mathbb Z_\ell (1) )$.

Thus, if the Galois action on $H^i(X, \mathbb Q_\ell(1))$ is nontrivial, the Galois action on $H^i ( \overline{X} , \mathbb Q_\ell / \mathbb Z_\ell (1 ))$ is nontrivial.

By the Weil conjectures, all the eigenvaluse of a Frobenius element (at a prime $q$ at which $X$ has good reduction) in the Galois group of $k$ on $H^i (\overline{X}, \mathbb Q_\ell)$ have size $q^{i/2}$, so all the eigenvalues on $H^i (\overline{X}, \mathbb Q_\ell(1))$ have size $q^{i/2-1}$. So the eigenvalues are never equal to $1$ unless $i=2$.

In other words, in every other degree, the Galois action is always nontrivial unless the cohomology group vanishes.

(The Weil conjectures are, unsurprisngly, more advanced than what you're learning right now, but keeping what they tell you in mind can be helpful for intuition.)

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  • $\begingroup$ You left me with plenty to think about, but thank you so much for your effort in going into detail with all these stuff. In the case of a curve, we have the isomorphism $$(\mathrm{Pic}(\bar{X}) \otimes \mathbb{Q})/\mathrm{Im}(\mathrm{Pic}(\bar{X})) \cong H^2(\bar{X},\mu_\infty).$$ Since $\mathrm{Pic}(\bar{X})$ maps to $\mathbb{Z}$, would it be safe to say that here we have $H^2(\bar{X},\mu_\infty) \cong \mathbb{Q}/\mathbb{Z}$, which is a trivial Galois module? $\endgroup$ Jul 24, 2021 at 17:19
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    $\begingroup$ @KelvinLian Yes, in the case of a curve, the Galois action is trivial. $\endgroup$
    – Will Sawin
    Jul 24, 2021 at 17:40
  • $\begingroup$ Regarding the line "$\mathrm{Pic}(\bar{X})$ will map into a finitely generated abelian group, the Neron-Severi group", I am not sure about one thing. We know that the image of $\mathrm{Pic}(\bar{X})$ in its rationalization is its maximal free quotient. Does this imply that this quotient is simply $\mathbb{Z}$ for a curve $\bar{X}$? In other words, for any curve $X$, the maximal free quotient of its Picard group is $\mathbb{Z}$? I don't think this is true, I must have been confused somewhere. $\endgroup$ Jul 26, 2021 at 5:33
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    $\begingroup$ @KelvinLian There certainly is a map to $\mathbb Z$, given by degree. I claim that any class in the kernel of this map is divisible. This follows from the fact that classes in the kernel of this map form an abelian variety, and the multiplication-by-$n$ map from the abelian variety to itself is surjective. $\endgroup$
    – Will Sawin
    Jul 26, 2021 at 12:30
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    $\begingroup$ @KelvinLian Well, $\operatorname{Pic}(\overline{X} ) \otimes \mathbb Q$ admits a degree map into $\mathbb Q$, and the kernel is $\operatorname{Pic}^0(\overline{X}) \otimes \mathbb Q = \operatorname{Pic}^0(\overline{X})$, and the image of $\operatorname{Pic}(\overline{X} ) $ contains $\operatorname{Pic}^0(\overline{X})$ and contains elements of every integral degree and thus contains every element whose degree is an integer. $\endgroup$
    – Will Sawin
    Jul 26, 2021 at 14:13

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