# Etale cohomology groups as Galois representations

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?

• 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. Commented Jun 9, 2022 at 13:33

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