Galois Action on underlying Topological space of a Group Scheme I have a question arising from the answer of following thread: https://mathoverflow.net/questions/324887/why-is-for-a-group-scheme-of-finite-type-smooth-resp-irreducible-equivale
Let $G/K$ be a group scheme of finite type. Fix the algebraic closure $\bar{K}$ of $K$ and consider the resulting Galois group ${\rm Gal}(\bar K/K)$.
As @Piotr Achinger stated there exist an "canonical" action on $\bar G = G \otimes \bar{K}$ by ${\rm Gal}(\bar K/K)$.
In futher comments in the linked thread there is explained that this action is concretely "the action" of ${\rm Gal}(\bar K/K)$ (considered as profinite group) on the underlying topological space $|\bar G|$.
My question is what action is here meant? Could anybody describe it concretely?
My consideration:
I know that there exist a canonical action of ${\rm Gal}(\bar K/K)$ on $\bar{K}$-valued points $\bar{G}(\bar{K})= Hom(Spec(\bar{K}), \bar{G})$ 
given explicitely by "conjugation" $g^{-1}\circ \phi \circ g$ under abusing of notation: concretely it is described locally on affine subsets $Spec(A \times \bar{K})= U \subset G$ via $g^{-1}\circ \phi \circ (id \otimes g)$ for $g \in {\rm Gal}(\bar K/K)$ and $\phi \in Hom(A \otimes \overline{K}, \overline{K})$. This is ok.
But what is the action of ${\rm Gal}(\bar K/K)$ on the whole $|\bar G|$?
Indeed we can interpret $\bar{G}(\bar{K})$ as closed subset of $|\bar G|$ but can the Galois action as described above on $\bar{G}(\bar{K})$ be extended to $|\bar G|$? 
Maybe by a density argument? I'm not sure.
Remark: This question looks quite similar to this one:Galois Action on Scheme
The essential difference is that the action which I here try to understand seems to base only on topological structure of $G$, namely the of the underlying topological space $|\bar G|$.
 A: $\newcommand{\Spec}{\mathrm{Spec}}$Let $X$ be a scheme over $k$. Note that $X_{\overline{k}}=X\times_{\mathrm{Spec}(k)}\mathrm{Spec}(\overline{k})$. So, to give a morphism $X_{\overline{k}}\to X_{\overline{k}}$ it suffices to give maps of $k$-schemes $X\to X$ and $\mathrm{Spec}(\overline{k})\to\mathrm{Spec}(\overline{k})$. If $\sigma\in\mathrm{Gal}(\overline{k}/k)$ then declare that the action of $\sigma$ on $X$ is trivial and that the action of $\sigma$ on $\mathrm{Spec}(\overline{k})$ is the one induced from the $k$-algebra map $\overline{k}\to\overline{k}$. The induced map $X_{\overline{k}}\to X_{\overline{k}}$ is the map $\sigma$. Since it's a map of schemes, it's also a map of topological space (read the remark at the end of the post to see a different convention).
If $X=\mathrm{Spec}(A)$ then $X_{\overline{k}}=\mathrm{Spec}(A\times_k \overline{k})$ and $\sigma$ is the map of schemes $X_{\overline{k}}\to X_{\overline{k}}$ corresponding to the ring map $A\otimes_k \overline{k}\to A\otimes_k \overline{k}$ given by $a\otimes \alpha\mapsto a\otimes \sigma(\alpha)$. Covering $X$ by affines with $k$-rational gluing data allows you to bootstrap this concrete understanding of the map to the general case.
How does this relate to the action on $X(\overline{k})$? Note that 
$$X(\overline{k})=\mathrm{Hom}_k(\mathrm{Spec}(\overline{k}),X)=\mathrm{Hom}_{\mathrm{Spec}(\overline{k})}(\mathrm{Spec}(\overline{k}),X\times_{\mathrm{Spec}(k)}\mathrm{Spec}(\overline{k}))$$
The action on the first presentation of $X(\overline{k})$ is to take $x:\Spec(\overline{k})\to X$ and then $\sigma(x)$ is defined to be the composition 
$$\mathrm{Spec}(\overline{k})\xrightarrow{\sigma}\Spec(\overline{k})\xrightarrow{x}X$$
where, again, the first map $\sigma$ is that induced by the ring map $\sigma:\overline{k}\to\overline{k}$. What is the action then on the second presentation of $X(\overline{k})$? It can't clearly be the same idea that for $y:\Spec(\overline{k})\to X\times_{\Spec(k)}\Spec(\overline{k})$ we just precompose by $\sigma$ since that won't be a morphism over $\Spec(\overline{k})$. What's true is that $\sigma(y)$ is the composition
$$\Spec(\overline{k})\xrightarrow{\sigma}\Spec(\overline{k})\xrightarrow{y}X\times_{\Spec(k)}\Spec(\overline{k})\xrightarrow{\sigma^{-1}}X\times_{\Spec(k)}\Spec(\overline{k})\qquad (1)$$
To convince ourselves of this, we need to remind ourselves how the identification 
$$\mathrm{Hom}_k(\mathrm{Spec}(\overline{k}),X)=\mathrm{Hom}_{\mathrm{Spec}(\overline{k})}(\mathrm{Spec}(\overline{k}),X\times_{\mathrm{Spec}(k)}\mathrm{Spec}(\overline{k}))$$
works. It takes an $x:\Spec(\overline{k})\to X$ and maps to $y:\Spec(\overline{k})\to X\times_{\Spec(k)}\Spec(\overline{k})$ which, written in coordinates, is the map $(x,\mathrm{id})$. So, we see that $\sigma(y)$ should just be $(x\circ\sigma,1)$. Let's now think about the composition in $(1)$ looks like when written in coordinates. The projection to $X$ is the composition
$$\Spec(\overline{k})\xrightarrow{\sigma}\Spec(\overline{k})\xrightarrow{(x,1)}X\times_{\Spec(k)}\Spec(\overline{k})\xrightarrow{\sigma^{-1}}X\times_{\Spec(k)}\Spec(\overline{k})\xrightarrow{p_1}X$$
Since $p_1\circ\sigma^{-1}$ is just the identity map we see that we get $x\circ\sigma$. What happens we do the second projection? We are looking at the composition
$$\Spec(\overline{k})\xrightarrow{\sigma}\Spec(\overline{k})\xrightarrow{(x,1)}X\times_{\Spec(k)}\Spec(\overline{k})\xrightarrow{\sigma^{-1}}X\times_{\Spec(k)}\Spec(\overline{k})\xrightarrow{p_2}\Spec(\overline{k})$$
which gives us $\sigma^{-1}\circ \sigma=1$. Thus, we see that the composition in $(1)$ is $(x\circ\sigma,1)$ as desired!
Let's do a concrete example. Let's take $X=\mathbb{A}^1_k$. Then, the map $\sigma:X_{\overline{k}}\to X_{\overline{k}}$ is the map corresponding to the ring map $\overline{k}[x]\to \overline{k}[x]$ defined by sending
$$\sum_i a_i x^i\mapsto \sum_i \sigma(a_i)x^i$$
The points of $X_{\overline{k}}$ come in two forms:


*

*The closed ideals $(x-\alpha)$ for $\alpha\in k$.

*The generic point $(0)$.


It's then clear that $\sigma$ sends $(x-\alpha)$ to $(x-\sigma^{-1}(\alpha))$ (recall that the induced map on $\Spec$ is by pullback! See the remark below) and sends the generic point to itself. That's what it looks like topologically. Let's think about what the action of $\sigma$ looks like on $\overline{k}$-points. 
If we take an $\overline{k}$-point corresponding to the map (thinking of $X(\overline{k})$ as $\mathrm{Hom}_k(\Spec(\overline{k}),X)$) 
$$x:k[x]\to \overline{k}:p(x)\mapsto p(\alpha)$$
then $\sigma(x)$, on the level of ring maps, is apply $\sigma$ as a poscomposition:
$$\sigma(x):k[x]\to \overline{k}:p(x)\mapsto \sigma(p(\alpha))=p(\sigma(\alpha)$$
where last commutation was because $p(x)\in k[x]$. Let's now think about the same situation in the second presentation $X(\overline{k})=\mathrm{Hom}_{\overline{k}}(\Spec(\overline{k})),X\times_{\Spec(k)}\Spec(\overline{k}))$. Our point $x$ above now corresponds to ring $\overline{k}$-algebra map
$$\overline{k}[x]\to \overline{k}:q(x)\mapsto q(\alpha)$$
If we just postcompose this ring map with $\sigma$ we get
$$\overline{k}[x]\to \overline{k}:q(x)\mapsto \sigma(q(\alpha))$$
which is NOT the desired map sending $q(x)\mapsto q(\sigma(\alpha))$ since $q$ doesn't have rational coefficients. But, if we apply the map $\sigma_X^{-1}$ (where I'm using the subscript $X$ to not confuse it with $\sigma$ on $\overline{k}$) on $X_{\overline{k}}$ this has the effect of applying $\sigma^{-1}$ to the coefficients of $q(x)$. So then, you can see that 
$$\sigma(\sigma_X^{-1}(q)(\alpha))=q(\sigma(\alpha))$$
yay! 
Hopefully this all makes sense.
NB: Depending on the author one can take the action on $X\times_{\Spec(k)}\Spec(\overline{k})$ to be what I have called $\sigma^{-1}$. This has a couple nice effects:


*

*It's action geometrically is more intuitive (e.g. instead of sending $(x-\alpha)$ to $(x-\sigma^{-1}(\alpha))$ it sends $(x-\alpha)$ to $(x-\sigma(\alpha))$).

*It's a left action (opposed to a right action). 

*In $(1)$ above the inclusion of $\sigma^{-1}$ on $X_{\overline{k}}$ might be jarring, and so if we use this alternative convention we'd just have $\sigma$. I actually like it the way it is because it reminds me of representation theory where if $V$ and $W$ are representations of some $G$ then $\mathrm{Hom}(V,W)$ is a representation of $G$ by $(g\cdot f)(v)=g(f(g^{-1}(v))$. 


It's one downside is that ring theoretically it's less natural since then its action on $a\otimes\alpha\in A\otimes_k\overline{k}$ is $a\otimes\sigma^{-1}(\alpha)$. Either convention is OK--nothing changes theory wise--it's just something you have to pay attention/be consistent about. 
