I am trying to understand how etale cohomology groups are Galois representations. Let $X$ be a scheme over a perfect field $K$ and write $\overline{X}=X \times_{{Spec }K} \textrm{Spec } \overline{K}$ for its base change to $\overline{K}$. For each $\sigma \in \textrm{Gal}(\overline{K}/K)$, we get (by the universal property of fibre products) an induced morphism $\overline{\sigma}: \overline{X} \rightarrow \overline{X}$. By the functorial property of etale cohomology, we get a map $H^i(X, \mathbb Q_\ell) \rightarrow H^i(X, \mathbb Q_\ell)$, thus making $H^i(X, \mathbb Q_\ell)$ into a Galois representation.

My question is: What exactly is the map $\overline{\sigma}$ here and how should one think about it? For instance, if $X$ is a variety defined by a single polynomial equation, what would the map $\overline{\sigma}$ look like?

  • 1
    $\begingroup$ Essentially, if you have a coordinate $x$ then it is mapped to $\sigma (x)$. Similarly the coefficients of the defining equations are mapped according to $\sigma$... but those coefficients are in $K$ so $\sigma$ acts on them trivially. $\endgroup$
    – Zhen Lin
    Jun 9, 2022 at 13:33

1 Answer 1


as already mentioned, if $X$ is defined over the ground field $K$, the Galois group acts trivially on these equations. Maybe its nice to think of a few examples:

  1. Take a finite Galois extension $L/K$ of deg d, say there is a $K$ basis $\{\lambda_1,\dots, \lambda_d\}$ Then define the hypersurface S in $\mathbf{A}_K^d$ by the product equation $$\prod_{\sigma \in Gal(L/K)}(\lambda_1^\sigma x_1+\dots+\lambda_d^\sigma x_d)$$ Geometrically, this is a union $d$ hyperplanes conjugate via the Galois action and you see that $G$ acts on the coefficients of the equations of$S$,i.e. $$\lambda^\sigma := \sigma(\lambda)$$ But: $S$ does not have any $K$ points, because any solution $\alpha \in K$ would imply $\lambda_1^\sigma\alpha_1 +\dots \lambda_n^\sigma\alpha_n=0$ for $\alpha_i \in K$ which is impossible because the $\lambda_i$ form a $K$ basis of $L$.

  2. Take the affine line $\mathbf{A}^1$ over $\mathbf{F}_p$, then $\sigma=\mathrm{Frob}_p \in \mathrm{Gal}_{\mathbf{F}_p}$ acting on e.g.$(t-\alpha) \mapsto (t-\alpha^p)$ for $\alpha \in \bar{\mathbf{F}_p}$. Then this action is trivial if $\alpha \in \mathbf{F}_p$. So for the fixed set under this action $\mathbf{A}^1(\bar{\mathbf{F}}_p){^\sigma}=\mathbf{A}^1(\mathbf{F}_p)$. Note that this Frobenius element is a topological generator of the absolute Galois group as a profinite group so it tells you a lot about it.

  3. This is finally an example of an étale sheaf and a Galois action on it: Namely, fix a field $k$ with algebraic closure $K$ and let $G$ be the absolute Galois group of $k$. Let $\mathcal{F}$ be an étale sheaf (more precisely, a sheaf in the finite étale topology) on $X:=\textrm{Spec}(k)$ and define $\mathcal{F}_K := \varinjlim\mathcal{F}(k')$ where the limit runs over all finite separable extensions $k'$ of $k$. Then $\mathcal{F}_K$ carries a natural $G$ action such that for each section $f \in \mathcal{F}_K$ the subgroup $\{\sigma \in G: f^\sigma = f\}$ is of finite index so this $G$ action is called continuous inducing an equivalence of categories $$ \{\text{sheaves on} X_{fét}\} \rightarrow \{\text{sets with cont. G-action}\} $$ coming from sending $\mathcal{F} \mapsto \mathcal{F}_K$ . To see how this action works, just observe that on each $\mathcal{F}(k')$, $\sigma$ acts after restriction to $k'$ by pullback $\mathcal{F}(k') \mapsto \sigma|_{k'}^*\mathcal{F}(k')$. Note that the subgroup $\textrm{Gal}(K|k')$ of $G$ acts trivially on $\mathcal{F}_k'$ having finite index #$\textrm{Gal}(k'|k)$. This example is from Olssons book on algebraic spaces p.66


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