Dgdrxrt

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 $barK$ of $K$ and consider the resulting Galois group $rm Gal(bar K/K)$.

As @Piotr Achinger stated there exist an "canonical" action on $bar G = G otimes barK$ 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 $barK$-valued points $barG(barK)= Hom(Spec(barK), barG)$

given explicitely by "conjugation" $g^-1circ phi circ g$ under abusing of notation: concretely it is described locally on affine subsets $Spec(A times barK)= U subset G$ via $g^-1circ phi circ (id otimes g)$ for $g in rm Gal(bar K/K)$ and $phi in Hom(A otimes overlineK, overlineK)$. This is ok.

But what is the action of $rm Gal(bar K/K)$ on the whole $|bar G|$?

Indeed we can interpret $barG(barK)$ as closed subset of $|bar G|$ but can the Galois action as described above on $barG(barK)$ 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 $|bar G|$.

$newcommandSpecmathrmSpec$Let $X$ be a scheme over $k$. Note that $X_overlinek=Xtimes_mathrmSpec(k)mathrmSpec(overlinek)$. So, to give a morphism $X_overlinekto X_overlinek$ it suffices to give maps of $k$-schemes $Xto X$ and $mathrmSpec(overlinek)tomathrmSpec(overlinek)$. If $sigmainmathrmGal(overlinek/k)$ then declare that the action of $sigma$ on $X$ is trivial and that the action of $sigma$ on $mathrmSpec(overlinek)$ is the one induced from the $k$-algebra map $overlinektooverlinek$. The induced map $X_overlinekto X_overlinek$ 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 $X=mathrmSpec(A)$ then $X_overlinek=mathrmSpec(Atimes_k overlinek)$ and $sigma$ is the map of schemes $X_overlinekto X_overlinek$ corresponding to the ring map $Aotimes_k overlinekto Aotimes_k overlinek$ given by $aotimes alphamapsto aotimes sigma(alpha)$. 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 $X(overlinek)$? Note that
$$X(overlinek)=mathrmHom_k(mathrmSpec(overlinek),X)=mathrmHom_mathrmSpec(overlinek)(mathrmSpec(overlinek),Xtimes_mathrmSpec(k)mathrmSpec(overlinek))$$

The action on the first presentation of $X(overlinek)$ is to take $x:Spec(overlinek)to X$ and then $sigma(x)$ is defined to be the composition

$$mathrmSpec(overlinek)xrightarrowsigmaSpec(overlinek)xrightarrowxX$$

where, again, the first map $sigma$ is that induced by the ring map $sigma:overlinektooverlinek$. What is the action then on the second presentation of $X(overlinek)$? It can't clearly be the same idea that for $y:Spec(overlinek)to Xtimes_Spec(k)Spec(overlinek)$ we just precompose by $sigma$ since that won't be a morphism over $Spec(overlinek)$. What's true is that $sigma(y)$ is the composition

$$Spec(overlinek)xrightarrowsigmaSpec(overlinek)xrightarrowyXtimes_Spec(k)Spec(overlinek)xrightarrowsigma^-1Xtimes_Spec(k)Spec(overlinek)qquad (1)$$

To convince ourselves of this, we need to remind ourselves how the identification

$$mathrmHom_k(mathrmSpec(overlinek),X)=mathrmHom_mathrmSpec(overlinek)(mathrmSpec(overlinek),Xtimes_mathrmSpec(k)mathrmSpec(overlinek))$$

works. It takes an $x:Spec(overlinek)to X$ and maps to $y:Spec(overlinek)to Xtimes_Spec(k)Spec(overlinek)$ which, written in coordinates, is the map $(x,mathrmid)$. So, we see that $sigma(y)$ should just be $(xcircsigma,1)$. Let's now think about the composition in $(1)$ looks like when written in coordinates. The projection to $X$ is the composition

$$Spec(overlinek)xrightarrowsigmaSpec(overlinek)xrightarrow(x,1)Xtimes_Spec(k)Spec(overlinek)xrightarrowsigma^-1Xtimes_Spec(k)Spec(overlinek)xrightarrowp_1X$$

Since $p_1circsigma^-1$ is just the identity map we see that we get $xcircsigma$. What happens we do the second projection? We are looking at the composition

$$Spec(overlinek)xrightarrowsigmaSpec(overlinek)xrightarrow(x,1)Xtimes_Spec(k)Spec(overlinek)xrightarrowsigma^-1Xtimes_Spec(k)Spec(overlinek)xrightarrowp_2Spec(overlinek)$$

which gives us $sigma^-1circ sigma=1$. Thus, we see that the composition in $(1)$ is $(xcircsigma,1)$ as desired!

Let's do a concrete example. Let's take $X=mathbbA^1_k$. Then, the map $sigma:X_overlinekto X_overlinek$ is the map corresponding to the ring map $overlinek[x]to overlinek[x]$ defined by sending

$$sum_i a_i x^imapsto sum_i sigma(a_i)x^i$$

The points of $X_overlinek$ come in two forms:

• The closed ideals $(x-alpha)$ for $alphain k$.


• 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 $overlinek$-points.

If we take an $overlinek$-point corresponding to the map (thinking of $X(overlinek)$ as $mathrmHom_k(Spec(overlinek),X)$)

$$x:k[x]to overlinek:p(x)mapsto p(alpha)$$

then $sigma(x)$, on the level of ring maps, is apply $sigma$ as a poscomposition:

$$sigma(x):k[x]to overlinek:p(x)mapsto sigma(p(alpha))=p(sigma(alpha)$$

where last commutation was because $p(x)in k[x]$. Let's now think about the same situation in the second presentation $X(overlinek)=mathrmHom_overlinek(Spec(overlinek)),Xtimes_Spec(k)Spec(overlinek))$. Our point $x$ above now corresponds to ring $overlinek$-algebra map

$$overlinek[x]to overlinek:q(x)mapsto q(alpha)$$

If we just postcompose this ring map with $sigma$ we get

$$overlinek[x]to overlinek:q(x)mapsto sigma(q(alpha))$$

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 $overlinek$) on $X_overlinek$ 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 $Xtimes_Spec(k)Spec(overlinek)$ 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 $X_overlinek$ 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 $mathrmHom(V,W)$ is a representation of $G$ by $(gcdot f)(v)=g(f(g^-1(v))$.

It's one downside is that ring theoretically it's less natural since then its action on $aotimesalphain Aotimes_koverlinek$ is $aotimessigma^-1(alpha)$. Either convention is OK--nothing changes theory wise--it's just something you have to pay attention/be consistent about.