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I would like to know whether the following metatheorem on nonabelian $H^2$ has been ever stated and/or proved.

Let $k$ be a perfect field and $k^s$ its fixed separable closure. Let $X^s$ be a variety with additional structure over $k^s$ (I don't want to specify what I mean by additional structure). By a $k$-model of $X^s$ I mean a variety with additional structure $X$ over $k$ together with a $k^s$-isomorphism $$ X\times_k k^s\overset{\sim}{\to} X^s.$$

Metatheorem. Let $k$ be a perfect field and $k^s$ its fixed separable closure. Let $X^s$ be a variety with additional structure over $k^s$. Write $A^s=\mathrm{Aut}(X^s)$, and assume that $A^s$ "is" an algebraic group over $k^s$. Assume that for any $\sigma\in\mathrm{Gal}(k^s/k)$ there exists a $k^s$-isomorphism $$\lambda_\sigma\colon \sigma X^s\to X^s,$$ where $\sigma X^s$ is the variety obtained from $X^s$ by transport of structure. Assume also that $X^s$ admits a $k_1$-model over a finite Galois extension $k_1/k$ contained in $k^s$. Then these data define a $k$-kernel $$\kappa\colon\mathrm{Gal}(k^s/k)\to \mathrm{Out}(A^s)$$ and a cohomology class $\eta\in H^2(k,A^s,\kappa)$. If $\eta$ is not neutral, then $X^s$ has no $k$-model. If $\eta$ is neutral and the variety $X^s$ is quasi-projective, then $X^s$ admits a $k$-model $X$. Moreover, set $A=\mathrm{Aut}(X)$, then there is a canonical bijection between $H^1(k,A)$ and the set of isomorphism classes of $k$-models of $X^s$.

Example of application of the metatheorem: If $k=\mathbb{R}$, $k^s=\mathbb{C}$, $A^s$ is a finite abelian group of odd order, then $H^2(\mathbb{R},A)=1$ and $H^1(\mathbb{R},A)=1$ (because $\mathrm{Gal}(\mathbb{C}/\mathbb{R})$ is of order 2), hence $X^s$ has a unique model over $\mathbb{R}$.

I would be also glad to have references where this metatheorem was proved in special cases.

I know that it was proved in the case when $X^s$ is a principal homogeneous space of $G^s$ dominating $Y^s$, where $Y$ is a given homogeneous space (not necessarily principal) of an algebraic group $G$ defined over $k$, see Springer, Nonabelian $H^2$ in Galois cohomology. In: Algebraic Groups and Discontinuous Subgroups (Proc. Sympos. Pure Math., Boulder, Colo., 1965), pages 164--182. Amer. Math. Soc., Providence, R.I., 1966. Borovoi, Abelianization of the second nonabelian Galois cohomology, Duke Math. J. 72(1), 217--239, 1993. Flicker, Scheiderer, Sujatha, Grothendieck's theorem on nonabelian $H^2$ and local-global principles. J. Amer. Math. Soc. 11(3), 731--750, 1998.

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  • $\begingroup$ Most of the content of the metatheorem is covered in section 12 of mathcs.emory.edu/~brussel/Papers/galoisdescent.pdf. In these terms the H^2 obstruction is the canonical obstruction to the existence of an equivariant object in a groupoid acted upon by a group. $\endgroup$ Jun 29, 2016 at 22:00
  • $\begingroup$ @YonatanHarpaz: Dear Yonatan, Could you please elaborate? Could you please kindly post an answer? The link that you give does not mention $H^2$ at all, and I don't understand what group, groupoid, and action you mean.... $\endgroup$ Jun 30, 2016 at 6:51

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Let me elaborate more on the remark above. Let $k$ be a perfect field. Let $\mathrm{Field}_k$ denote the category of finite extensions of $k$, i.e., the objects of $\mathrm{Field}_k$ are fields $k'$ equipped with an embedding $k \to k'$ such that $k'$ is finite dimensional over $k$. The morphisms are the maps of fields $k' \to k''$ which respect the embedding (here we do not think of all the $k'$ as subfield of a fixed seperable closure). Suppose we have a functor $F:\mathrm{Field}_k \to \mathrm{Grpd}$ to the category of small groupoids. For example, $F$ may be the functor which sends $k'$ to the groupoid whose objects are varieties with a certain structure defined over $k'$ and whose morphisms are structure preserving $k'$-isomorphisms between them. For $k'$ in $\mathrm{Fields}_k$ the inclusion $\iota:k \to k'$ can be considered as a morphism in $\mathrm{Fields}_k$, and we consequently have an associated functor $F(\iota):F(k) \to F(k')$ which we can call the base change functor. If $k'$ is also a Galois extension of $k$ with (finite) Galois group $G$, then the automorphism group of $k'$ in the category $\mathrm{Fields}_k$ is exactly $G$. In particular, $G$ now acts on the groupoid $F(k')$ (via functors). Given an object $X \in F(k')$ let us denote by $X^{\sigma}$ the image of $X$ under the action of $\sigma \in G$ on $F(k')$. Now whenever we have a group $G$ acting on a groupoid $Z$, we have an associated notion of a $G$-equivariant object of $Z$. This is an object $X \in Z$ equipped with a compatible collection of (iso)morphisms $f_{\sigma}: X \to X^{\sigma}$. We may also call this a twisted action of $G$ on $X$. Let us denote by $Z^{hG}$ the groupoid of $G$-equivariant objects in $Z$ (where the notation echoes the fact that we think of $G$-equivariant objects as homotopy fixed points). Now the mere fact that $F$ is a functor implies that if $X$ is an object of $F(k)$ then the object $F(\iota)(X) \in F(k')$ carries a natural twisted action of $G$. We hence obtain a functor $$ T_{k'/k}:F(k) \to F(k')^{hG} .$$ We can now say that $F$ satisfies Galois descent if $T_{k'/k}$ is an equivalence of groupoids for every finite Galois extension $k'/k$.

Now the $H^2$ and $H^1$ business is something that has only to do with computing groupoids of equivariant objects, and has nothing to do with, say, algebraic varieties. Let $Z$ be a groupoid equipped with an action of a group $G$. Let $\pi_0(Z)$ denote the set of isomorphism classes of $Z$, so that we have an induced action of $G$ on $\pi_0(Z)$. If $x \in \pi_0(Z)$ is an isomorphism class fixed by $G$, then we have an induced action of $G$ on the connected component $Z_x \subseteq Z$ corresponding to $x$. Let $X \in Z_x$ be any object and let $A = Aut(X)$ be its automorphism group. Since $Z_x$ is a connected groupoid, the group of connected components of self-homotopy equivalences of $Z_x$ is naturally isomorphic to $Out(A)$. We hence obtain a natural map $G \to Out(A)$, i.e., a pseudo-action of $G$ on $A$. Classical obstruction theory now associates to $X$ an obstruction element $o_X \in H^2(G,A)$, which is neutral if and only if $X$ admits a $G$-equivariant structure (i.e., a twisted action of $G$). The object $o_X$ is the one associated with a certain group extension $$ 1 \to A \to G_X \to G \to 1 $$ where $G_X$ is the group whose elements are pairs $(f,\sigma)$ where $\sigma$ is an element of $G$ and $f:X \to X^{\sigma}$ is a morphism (composition of elements is defined in a natural way). If the obstruction element $o_X$ is neutral then we can choose a section $G \to G_X$. Each such section determines a twisted action of $G$ on $X$. Furthermore, two such twisted actions result in isomorphic $G$-equivariant objects if and only if the two sections are conjugate by an element of $A$. This data is now classified by the cohomology group $H^1(G,A)$, and we obtain a bijection between $H^1(G,A)$ and the set of isomorphism classes of $G$-equivariant objects in the component $Z_x$. This is the way to compute groupoids of $G$-equivariant objects.

Edit:

If $Z,W$ are two groupoids then the functor category ${\rm Fun}(Z,W)$ is a groupoid as well. Two functors $f,g: Z \to W$ are homotopic if they are isomorphic in ${\rm Fun}(Z,W)$, and a functor $f: Z \to W$ is a homotopy equivalence if it has an inverse up to homotopy. For a groupoid $Z$ we have the full subgroupoid ${\rm Equiv}(Z,Z) \subseteq {\rm Fun}(Z,Z)$ spanned by the homotopy equivalences. Then $\pi_0{\rm Equiv}(Z,Z)$ (i.e., the set of isomorphism classes of the groupoid ${\rm Equiv}(Z,Z)$) is naturally a group by composition. This is the "group of connected components of self homotopy equivalences" alluded to in the answer.

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  • $\begingroup$ I'm sure there's a neat homotopical formalization of the last paragraph but i'm struggling to find it. Here's where i'm at: Assume $Z_x= BA$ then a $G$ action on $Z_x$ is the same as a $A$ gerbe over $BG$ so we get a fiber sequence $BA \to BG \to BAut(BA)$ coming from the action of $G$ on $Z_x$ (and the choice of base point $x$). The long exact sequence in homotopy groups looks like: $0 \to Z(A) \to A \to G \to Out(A) \to *$. Here i'm basically stuck... Do you have a suggestion or perhaps a reference for this? $\endgroup$ Sep 7, 2017 at 9:50
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    $\begingroup$ About the second comment, I suppose the case in question corresponds to $F=BG, H = E$, in the slice category over $BG$. In general, calculating mapping spaces in the presence of non-trivial fundamental groups can be quite painful. If fact, even for simply connected spaces $X,Y$, if you want to understand the homotopy groups of a component of $Map(X,Y)$ which is not the component of null homotopic maps then the relevant differentials in the Bousfield Kan spectral sequence will not be linear, and so the problem will not reduce to homological algebra. $\endgroup$ Sep 7, 2017 at 19:56
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    $\begingroup$ $E$ is not the homotopy fiber of the map $BG \to BAut(BA)$, but rather the homotopy fiber product $G \times_{BAut(BA)} BAut_*(BA)$, where $Aut_*(BA) \cong Aut(A)$ is the space of pointed automorphisms of $BA$. We hence obtain a long exact sequence $1 \to Z(A) \to \pi_1(E) \to G \times Aut(A) \to Out(A) \to 1$, which seems to be consistent with the long exact sequence $1 \to A \to \pi_1(E) \to G \to 1$. $\endgroup$ Sep 7, 2017 at 21:46
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    $\begingroup$ and indeed the fiber of $E \to BG$ is $BA$. That's what we want $E$ to be: the family of $BA$'s over $BG$ corresponding to the action of $G$ on $BA$. $\endgroup$ Sep 7, 2017 at 21:58
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    $\begingroup$ @YonatanHarpaz: You write: "Since $Z_x$ is a connected groupoid, the group of connected components of self-homotopy equivalences of $Z_x$ is naturally isomorphic to $Out(A)$." What do you mean by connected components of self-homotopy equivalences? You do not define homotopy equivalences in your answer! $\endgroup$ Nov 24, 2017 at 20:46

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