Is the action of $\textrm{Gal}(\overline{k}/k)$ on $G \times_k \overline{k}$ a group homomorphism? [closed]

Question

I don't know how to state my question precisely in the language of group schemes. You have an algebraic group $G$, defined over a perfect field $k$, and the Galois group $\Gamma = \textrm{Gal}(\overline{k}/k)$ acts on $G$. The $k$-rational points $G(k)$ consist of those $x \in G$ which are fixed by $\Gamma$.

For fixed $\sigma \in \Gamma$, $x \mapsto x^{\sigma}$ is a bijection of $G$, but not an isomorphism of varieties.

My questions:

1 . Is this mapping $x \mapsto x^{\sigma}$ in general an abstract group isomorphism?

2 . How do you state these results in the language of group schemes?

An example, consider $G = U(2,1)$ as a group over $\mathbb{R}$. As a set, $G = \textrm{GL}_3(\mathbb{C}$). If $\sigma \in \Gamma = \textrm{Gal}(\mathbb{C}/\mathbb{R})$ is complex conjugation, then $\Gamma$ acts on $G$ as

$$x^{\sigma} = w \, ^t \bar{x}^{-1}w$$

where $w = \begin{pmatrix} & & 1 \\ & -1 & \\ 1 & & \end{pmatrix}$ and $\bar{x}$ denotes entrywise complex conjugation. This is a group homomorphism.

Answer 1

A group scheme over a scheme $S$ is a scheme $G$ such that the functor $\text{Hom}_S(*,G)$ from $\textbf{Sch}/S\rightarrow\textbf{Sets}$ factors through $\textbf{Groups}$. This means, that for any $S$-scheme $T$, the set $\text{Hom}_S(T,G)$ (called the set of $T$-valued points of $G$) has the structure of a group, and for any morphism of $S$-schemes $f : T'\rightarrow T$, the induced map $$f^* : \text{Hom}_S(T,G)\rightarrow\text{Hom}_S(T',G)$$ is a homomorphism of groups.

In your situation, if you set $S = \text{Spec }k$, $T = T' = \text{Spec }\bar{k}$, and $f = \sigma$ (or more precisely, $f$ is the morphism $\text{Spec }\bar{k}\rightarrow\text{Spec }\bar{k}$ induced by $\sigma$, which we also call $\sigma$) then by noticing that the map $x\mapsto x^\sigma$ can be realized in the above situation as the map $$\sigma^* : \text{Hom}_k(\text{Spec }\bar{k},G)\rightarrow \text{Hom}_k(\text{Spec }\bar{k},G)$$ it follows that $x\mapsto x^\sigma$ is a group homomorphism (in fact isomorphism).

Note that the fact that $f^*$ is a group homomorphism on Hom sets can be deduced purely formally from the fact that the group operation and inversion can be given in terms of morphisms of schemes/varieties over $S$.

Caveat: The fact that everything above is done in the category of schemes over $S$ (or $k$) is critical. For example, if you set $S = \text{Spec }\bar{k}$, then the map $\sigma:\text{Spec }\bar{k}\rightarrow\text{Spec }\bar{k}$ is still a morphism of schemes, but is not a morphism of schemes over $\bar{k}$, and hence does not induce a map $$\sigma^* : \text{Hom}_{\bar{k}}(\text{Spec }\bar{k},G\times_k\bar{k})\rightarrow \text{Hom}_{\bar{k}}(\text{Spec }\bar{k},G\times_k\bar{k})$$

Answer 2

Here is the story for a general (group) scheme: let $k$ be a field (no need for perfectness) and let $k_{s}$ be a field extension (in practice, the separable closure of $k$). Let $X$ be a scheme defined over $\text{Spec }k$, and let $\alpha \in \text{Aut }(k_s/k)$. Then you have a commutative square (in the category of schemes) $$\require{AMScd} \begin{CD} X\times_{\text{Spec }k} \text{Spec }k_s @>{f_{\alpha}}>> X\times_{\text{Spec }k} \text{Spec }k_s\\ @VVV @VVV \\ \text{Spec }k_s @>{\text{Spec }\alpha}>> \text{Spec }k_s \end{CD}$$ The isomorphism $f_{\alpha}$ is found by universal property. Now, since all horizontal maps are isomorphisms, this readily gives you an isomorphism $X(k_s)\to X(k_s)$. Finally, if $X$ is an algebraic group defined over $k$ (i.e. a group object in the category of $k$-schemes), since base change behaves naturally with product, you will indeed get that the isomorphism $X(k_s)\to X(k_s)$ respects the group law!

To go back to your example, let me clear up some apparent confusion: in the beginning of the question, you use $G$ for what I denoted above $X\times_{\text{Spec }k} \text{Spec }k_s$ (which is standard notation in algebraic group theory). Sticking with this notation, $G$ is $\text{GL}_3$ as an algebraic group over $\mathbf{C}$ (not merely as a set, and not merely at the level of rational points). What it means to consider $U(2,1)$ as an algebraic group over $\mathbf{R}$ is that there exists a variety (and even an algebraic group) $X$ defined over $\mathbf{R}$ such that $X(\mathbf{R}) = \lbrace x\in M_3(\mathbf{C})\vert \text{det }(x)\neq 0 \text{ and } x=w^{t}\bar{x}^{-1}w \rbrace $. This guy $X$ is given to you by descent theory, but in this case, it is also straightforward to find explicitly the equations defining $X$. An important feature (which descent theory gives by construction) is that $X\times_{\text{Spec }\mathbf{R}} \text{Spec }\mathbf{C}$ is (isomorphic to ) $\text{GL}_3$ (as an algebraic group over $\mathbf{C}$).

It is also interesting to check that the isomorphism $X(k_s)\to X(k_s)$ that I described at the beginning is indeed the one you are describing on $\text{GL}_3(\mathbf{C})$. This might be a bit of work if you first give the equations of $X$ over $\mathbf{R}$, then give explicitly the isomorphism $X\times_{\text{Spec }\mathbf{R}} \text{Spec }\mathbf{C} \cong \text{GL}_3$ (and you might not get what you want depending on the identifications you made). But if instead you use descent theory (and in this case, this is just Galois descent), this is essentially a tautology.