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.



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


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.)


