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Let $k$ be an arbitrary field and $X$ be a proper scheme over $k$. Mastumura and Oort proved that the functor $S \longmapsto Aut_{S-sch}(X \times S)$ is representable by a group scheme, locally of finite type, which I denote by $\underline{Aut}(X)$. I am interested in the forms of such a group scheme. More precisely, if $X'$ is a form of $X$ (that is, a (proper) scheme such that $X'_{\overline{k}} \simeq X'_{\overline{k}}$, where $\overline{k}$ is an algebraic closure of $k$ and $X_{\overline{k}}$ stands for the base change $X \times \mathrm{Spec}(\overline{k})$) then by functoriality $\underline{Aut}(X')_{\overline{k}} \simeq \underline{Aut}(X'_{\overline{k}}) \simeq \underline{Aut}(X_{\overline{k}}) \simeq \underline{Aut}(X)_{\overline{k}}$, so that $\underline{Aut}(X')$ is a form of $\underline{Aut}(X)$.

Question 1: Let $G$ be a form of $\underline{Aut}(X)$. Does there exist a form $X'$ of $X$ such that $G \simeq \underline{Aut}(X')$?

Moreover, if $Z$ is a closed subscheme of $X$ then the functor $$S \longmapsto \{g \in \underline{Aut}(X)(S) \mid \text{the automorphism } g \text{ of } X \times S \text{ induces an automorphism of } Z \times S \}$$ is representable by a subgroup scheme of $\underline{Aut}(X)$, which I denote by $\underline{Aut}(X,Z)$ (see Demazure-Gabriel, Théorème II 1.3.6 p165). If $(X',Z')$ is a form of $(X,Z)$ (that is, $Z'$ is a closed subscheme of $X'$ and the isomorphism $X'_{\overline{k}} \simeq X_{\overline{k}}$ induces an isomorphism $Z'_{\overline{k}} \simeq Z_{\overline{k}}$) then as above $\underline{Aut}(X',Z')$ is a form of $\underline{Aut}(X,Z)$.

Question 2: Let $G$ be a form of $\underline{Aut}(X,Z)$. Does there exist a form $(X',Z')$ of $(X,Z)$ such that $G \simeq \underline{Aut}(X',Z')$?

I ask these questions in a quite broad generality. However, in the example I have in mind, $X$ and $Z$ are smooth, projective and geometrically integral, and $\underline{Aut}(X,Z)$ is a smooth connected affine group scheme. I would be happy if the answers were positive with these restrictions.

I suspect that this could be related to cohomology, but I do not want to assume the field $k$ to be perfect so the forms could be trivialised after an inseparable extension.

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That is not true. The basic issue is that for every coherent sheaf $\mathcal{F}$ on $X$ that is "intrinsic", and thus admits a linearization of the automorphism group, every cohomology group of $\mathcal{F}$ gives a linear representation of the automorphism group. So for every form of $X$, not only do you get a form of the automorphism group, you get a collection of linear representations for every intrinstic coherent sheaf $\mathcal{F}$. It can easily happen that some form of the automorphism group does not have "enough" linear representations over $k$ to arise from a form of $X$.

Here is a concrete counterexample. Let $(a,b,c,d)$ be positive integers with $a\neq b$, with $c\neq d$, and with the sums $m=a+b$ and $n=c+d$ distinct. Let $A$ be a copy of $\mathbb{P}^1$ with homogeneous coordinates $[s,t]$. Let $C$ be a copy of $\mathbb{P}^1$ with homogeneous coordinates $[u,v]$. Let $P$ be a copy of $\mathbb{P}^4$ with homogeneous coordinates $[x_0,x_1,x_2,x_3,x_4]$. Let $Q$ be a copy of $\mathbb{P}^4$ with homogeneous coordinates $[y_0,y_1,y_2,y_3,y_4]$. Denote by $f_{a,b}$, resp. $g_{c,d}$, the closed immersions, $$f_{a,b}:A \to P, \ \ [s,t] \mapsto [s^m,s^{m-1}t,s^at^b,st^{m-1},t^m],$$ $$g_{c,d}:C \to Q, \ \ [u,v] \mapsto [u^n,u^{n-1}v,u^cv^d,uv^{n-1},v^n].$$ Let $R$ be a copy of $\mathbb{P}^{24}$, and denote by $h$ the Segre embedding, $$h:P\times Q \to R,\ \ ([x_i],[y_j]) \mapsto [x_iy_j]_{0\leq i,j\leq 4}.$$ Define $Y$ to be the blowing up of $R$ along the image of $P\times Q$, and define $X$ to be the blowing up of $Y$ along the inverse image of $A\times C$.

The automorphism group of $X$ equals the subgroup of the automorphism group of $Y$ that preserves the inverse image of $A\times C$ as a closed subscheme (rather than "pointwise"). The automorphism group of $Y$ is a semidirect product of $\text{Aut}(P)\times \text{Aut}(Q)$ by a cyclic group $\mathfrak{S}_2$ permuting the factors. The automorphism group of $X$ is the subgroup of this group that preserves $A\times C$ as a closed subscheme of $P\times Q$. Because $m$ does not equal $n$, the cyclic group $\mathfrak{S}_2$ does not preserve $A\times C$. Because $a\neq b$, resp. $c\neq d$, the automorphism group of $A$ in $P$, resp. the automorphism group of $C$ in $Q$, is a copy of the multiplicative group $\mathbb{G}_m$. Thus, the automorphism group of $X$ is isomorphic to $\mathbb{G}_m\times \mathbb{G}_m$. There is a linear representation of this automorphism group on the $k$-vector space of global sections on $X$ of the pullback of $\mathcal{O}(1)$ from $R$. This representation is a direct sum of copies of the characters with weights $(i,j)$ for $$i\in\{ 0,1,a,m-1,m\},\ \ j\in \{ 0,1,c,n-1,n\}.$$ Similarly, the representation on global sections of the pullback of $\mathcal{O}(16) \cong \omega_{R/k}^\vee$ has weights $(i,j)$ ranging from $(0,0)$ to $(16m,16n)$.

The automorphism group of the group scheme $\mathbb{G}_m\times \mathbb{G}_m$ is quite large. For instance, for every separable, degree $2$-field extension $L/k$, the restriction $\text{Res}_{L/k}(\mathbb{G}_{m,L})$ is a group $k$-scheme that is a form of $\mathbb{G}_m\times \mathbb{G}_m$. Since $m$ does not equal $n$, there is no realization of this form arising from a form of $X$: if there were, then the linear representation on the global sections of the twist of the pullback of $\omega_{X/k}^\vee$ would have weights $(i,j)$ that were symmetric under permutation of the two factors of $\mathbb{G}_m\times \mathbb{G}_m$.

There should be similar examples obtained from blowing up finite sets in projective space, but this requires explicit computation of the stabilizer group of the finite point set.

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    $\begingroup$ Thank you for this nice example ! I guess that $P$ and $Q$ should be copies of $\mathbf{P}^4$ and $R = \mathbf{P}^{24}$. $\endgroup$ – BrL Mar 5 '18 at 13:01
  • $\begingroup$ @BrL. Yes, that is correct. I will correct the $\mathbb{P}^3$ to $\mathbb{P}^4$ and the $\mathbb{P}^{15}$ to $\mathbb{P}^{24}$. Thank you. $\endgroup$ – Jason Starr Mar 5 '18 at 13:18

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