7
$\begingroup$

Let $X$ be some smooth scheme over $\mathbf C$ equipped with an action of $\mu_n$ (the group of $n$th roots of unity). The étale cohomology groups of X are therefore equipped with an action of $\mu_n$.

Now, let's suppose that the action of $\mu_n$ on $X$ extends to an action of $\mathbf G_m$. Then, the analytic space $X(\mathbf C)$ is equipped with an action of $\mu_n$ which extends to an action og $\mathbf C^\times$. As $\mathbf C^\times$ is connected, one concludes that the action of $\mu_n$ on the singular cohomology of $X(\mathbf C)$ is trivial (the multiplication by each element $\xi\in \mu_n$ is homotopic to the identity).

Going backwards to étale cohomology, using a comparison theorem between étale and singular cohomology, one concludes that the action of $\mu_n$ on the étale cohomology of $X$ is trivial.

Is there a direct proof that the action on the étale cohomology groups is trivial without having to refer to the topological case?

$\endgroup$

2 Answers 2

7
$\begingroup$

Smoothness of $X$ is not needed (neither for the comparison isomorphism nor for the result in question). Let $X$ be any quasi-separated scheme over a separably closed field $k$, equipped with an action by a connected $k$-group scheme $G$ of finite type. Let $n > 0$ be an integer not divisible by the characteristic of $k$ and choose an integer $i \ge 0$. Then we want to show that the action of $G(k)$ on ${\rm{H}}^i(X, \mathbf{Z}/(n))$ is trivial (using etale cohomology here).

[The hypothesis on $n$ is necessary because if $n = p = {\rm{char}}(k)>0$ and $X = {\rm{Spec}}(A)$ is affine then the effect of $G(k)$ on ${\rm{H}}^1(X, \mathbf{Z}/(p)) = A/\wp(A)$ (with $\wp(f) = f^p-f$) is the induced action from the $G(k)$-action on $A = \Gamma(X,O_X)$, and this is generally nontrivial (e.g., $X = G = \mathbf{A}^1_k$ with the translation action corresponding to $c.f(t) = f(t+c)$ on global functions (for $c \in G(k)$).]

By a spectral sequence argument using a covering by quasi-compact $G$-stable open subsets we may reduce to the case when $X$ is quasi-compact (and quasi-separated). It is harmless to make the radiciel extension from $k$ to its algebraic closure ("topological invariance" of etale cohomology), so we may assume that the separably closed $k$ is even algebraically closed. We may then replace $G$ with $G_{\rm{red}}$ so that $G$ is smooth.

Let $f:G \times X \rightarrow G$ be the projection map. The hypothesis on $n$ and smoothness of $G$ allow us to apply the smooth base change theorem to conclude that $\mathscr{F} = {\rm{R}}^if_{\ast}(\mathbf{Z}/(n))$ is the constant sheaf on $G$ attached to ${\rm{H}}^i(X,\mathbf{Z}/(n))$. Consider the action automorphism $$\alpha: G \times X \simeq G \times X$$ defined by $(g,x) \mapsto (g, gx)$. This commutes with $f$, and so induces an automorphism $[\alpha]$ of $\mathscr{F}$ on $G$. The effect on the stalk at $g \in G(k)$ is the $g$-action on ${\rm{H}}^i(X, \mathbf{Z}/(n))$ that we want to be trivial. But $\mathscr{F}$ is a constant sheaf on a connected scheme, so the effect on $\mathscr{F}$ of any automorphism is uniquely determined by the effect on a single stalk. Looking on the stalk at $g=1$ thereby shows that $[\alpha]$ is the identity automorphism. Now pass to the effect of $[\alpha]$ on the stalk at any $g \in G(k)$ to conclude.

$\endgroup$
4
$\begingroup$

Here's an alternative to grghxy's answer. Let me stick to the original assumptions for simplicity, but they can certainly be relaxed. Let $G$ be the image of the $\mu_n$ action on étale cohomology with finite coefficients. In particular, $G$ is finite. Let $g\in G$. Since the action is assumed to extend to a $\mathbb{C}^*$-action, we can solve $g=h^N$ for arbitrary $N>0$. Now choose $N=|G|$.

NB See some comments by grghxy & Yonatan Harpaz below.

$\endgroup$
4
  • $\begingroup$ Since $h$ has even larger order than $g$, and in the end one is only using $N = |G|$ rather than arbitrarily large $N$, why doesn't this "prove" the triviality of the $\mu_p$-action induced by any $\mu_{p^2}$-action (by taking $N=p$)? Please clarify. $\endgroup$
    – grghxy
    Oct 8, 2015 at 15:35
  • $\begingroup$ Good point! If there is a fix, it'll have to wait until after I finishing teaching etc. $\endgroup$ Oct 8, 2015 at 15:45
  • 3
    $\begingroup$ Choosing $N = |G|$ is not enough because the action of $\mu_{nN}$ might involve elements of $Aut(H^n_{et}(X,A))$ which are not in $G$. However, taking $N=|Aut(H^n_{et}(X,A))|$ should be enough (here $A$ is a finite abelian group, and $X$ is sufficiently nice, so this automorphism group is finite). $\endgroup$ Oct 8, 2015 at 20:03
  • $\begingroup$ This fix looks very nice. I accepted the previous answer because it appeared first. Thank you anyway! $\endgroup$
    – Oblomov
    Oct 9, 2015 at 7:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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