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Let $K$ be a perfect field, and let $S$ be a projective $K$-scheme. Denote by $\text{Twist}(S/K)$ the set of twists of $S$ up to $K$-isomorphism. These are (apriori) sheaves $\mathcal{F}$ on the Galois site of $K$, which become isomorphic to $S$ over a finite field extension, i.e. there exists an $L/K$, for which there exists an isomorphism of functors: $$\phi: \mathcal{F}\otimes L\longrightarrow S\otimes L.$$

Now, it is well known that the sheaves $\mathcal{F}\in \text{Twist}(S/K)$ are all representable in the category of projective $K$-schemes, and I am trying to figure out how.

In Serre's book ``Galois Cohomology", Chapter 3, Section 1, Proposition 5 (page 124), Serre says that this representability persists in even greater generality, e.g. when $S$ is quasiprojective, rather than projective. In his proof, he identifies the twist, $\mathcal{F}$, as a quotient of $S\otimes L$ by the (finite) Galois group of $L/K$, $G$, with its twisted action. (One may assume $L/K$ is Galois without loss of generality). I.e. $\mathcal{F} \cong S\otimes L/G$.

My issue is that it is unclear to me how one interprets the functor $\mathcal{F}$ as the sheaf quotient of $S\otimes L/G$.

For clarification, the twisted Galois action is defined as follows: let $\phi$ be the isomorphism above, and let $\sigma\in G$, then $\phi$ defines a cocycle in $Z^1(G, \text{Aut}(S))$ via $\sigma\mapsto {}^{\sigma}\phi\circ\phi^{-1}\in \text{Aut}(S)$. Denote the corresponding cocycle by $\xi$, then the twisted $G$-action is given by the composition of $\xi(\sigma)$ on $1\otimes \sigma$.

Somehow, I think that $\mathcal{F}$ is more naturally associated with the $G$-invariants rather than the $G$-orbits. For example, assume that $\mathcal{F}$ is the trivial twist, so that $\mathcal{F}\cong S$, and let's take $L$ to be a non-trivial Galois extension of $K$ still, then $\mathcal{F}(K) = S(K)$ on the one hand, yet, according to Serre, this set should be the same thing as the $G$-orbits of $S(L)$, however, the $K$-points of $S$ sit in the $L$-points of $S$ as $G$-invariants, not as $G$-orbits.

EDIT: I think my problem is with my definition of the functor $S$, is it the Yoneda functor sending a test object $T$ of the Galois category of $K$ to $\text{Hom}(T,S)$, or the $G(T/\text{Spec}K)$-orbits on the closed points of $S\times_{\text{Spec}K}T$?

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    $\begingroup$ There is an ample invertible sheaf on your projective scheme over $S$, and the tensor product of all pullbacks of this invertible sheaf by the (finitely many) elements of $G$ gives a $G$-invariant ample invertible sheaf. This descends. An algebraic space with an ample invertible sheaf is a scheme. $\endgroup$
    – Jason Starr
    Commented Sep 19, 2023 at 9:33
  • $\begingroup$ My question is basically: descends onto what? How do you interpret the twist as an algebraic space? $\endgroup$
    – kindasorta
    Commented Sep 19, 2023 at 9:55
  • $\begingroup$ There are very robust criteria for representability of a sheaf by an algebraic space due to Mike Artin. However, here I believe this follows directly from the definition of algebraic space. $\endgroup$
    – Jason Starr
    Commented Sep 19, 2023 at 10:02
  • $\begingroup$ I understand that you are trying to imply that the twist may be written as a quotient of the (base extension of the) non-twisted projective scheme by the (twisted action of the) Galois group. My pushback is that this is precisely my question: why is it that the twist could be written in this form? Please consider my last paragraph $\endgroup$
    – kindasorta
    Commented Sep 19, 2023 at 10:15
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    $\begingroup$ An algebraic space over a scheme is a contravariant set-valued function on the small etale site of the scheme that is a sheaf for the etale topology, whose diagonal is represented by schemes, and such that there exist an etale, faithfully flat morphism to the sheaf from the Yoneda sheaf of a scheme. $\endgroup$
    – Jason Starr
    Commented Sep 19, 2023 at 11:51

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