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I've already asked a similar question in SE, without success, so I've decided to post here a more general version of my question.

Let $f: Y \to X$ be a finite étale morphism of smooth proper varieties over a field $k$ of characteristic 0. Let $G$ be a linear algebraic $k$-group and $G_Y = G \times_k Y$.

Is the Weil restriction $R_{Y/X} (G_Y) \cong G^d \times_k X$, where $d$ is the degree of $f$?

I'm using the definition of Weil restriction as in section 7.6 of Bosch, Lutkebomert, Raynaud's "Néron models" book.

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  • $\begingroup$ What's the role of $Z$ in your question? $\endgroup$ Aug 25, 2014 at 14:54
  • $\begingroup$ Oops, fair point - no role, I will edit it out! $\endgroup$
    – user57469
    Aug 25, 2014 at 14:57

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A nice formula to replace your guess is: $$𝑅_{𝑌/𝑋}(𝐺_𝑌)\simeq \underline{\operatorname{Hom}}_X(Y,G)$$ where $\underline{\operatorname{Hom}}$ means internal Hom; it is representable a posteriori because it verifies the same universal property as $𝑅_{𝑌/𝑋}(𝐺_𝑌)$. (That $G$ is a group is irrelevant here.) This is locally isomorphic to $G^d$ for the étale topology, so you may think of it as a “twisted form” of $G^d$.

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  • $\begingroup$ It would be nice if there was some conomological stuff to control the twistedness of such object. I am an algebraic topologist, so no expert here, but some substitute of simply connectedness on $G$ could be a nice hypothesis to guarantee the triviality and make OP's formula true. $\endgroup$ Jan 8, 2023 at 16:27
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No, that is not true. For instance, let $k$ be $\mathbb{R}$, let $X$ be $\text{Spec}(\mathbb{R})$, let $Y$ be $\text{Spec}(\mathbb{C})$, and let $G$ be the multiplicative group, $\mathbb{G}_m$. Compare the real points of $R_{Y/X}(G_Y)$ with its induced analytic topology to the real points of $\mathbb{G}^2_{m,X}$. The homotopy groups $\pi_0$ and $\pi_1$ distinguish these two real Lie groups immediately.

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  • $\begingroup$ (This is still user57469, I had to reboot my laptop and for some reason I'm not logged in anymore.) Is the statement false if we specialize $k$ to e.g. a number field? If the general statement is still false even in this more restricted setting, is there any well-known case when the statement holds (some guesses: $G$ a torus or maybe $G$ just connected)? $\endgroup$
    – user57473
    Aug 25, 2014 at 15:45
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    $\begingroup$ It is still false, for instance, if $k$ equals $\mathbb{Q}$, if $X$ equals $\text{Spec}(\mathbb{Q})$, if $Y$ equals $\text{Spec}(\mathbb{Q}[i])$, and if $G$ equals $\mathbb{G}_m$. Since you can get the previous counterexample from this one by basechange, there cannot be an isomorphism. $\endgroup$ Aug 25, 2014 at 15:49
  • $\begingroup$ Ok, thanks! One last question: can the statement be made true (at least for some large "classes" of $G$, e.g. tori, etc) by adding further assumptions on $X$ and $Y$? $\endgroup$
    – user57473
    Aug 25, 2014 at 15:57
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This is false whenever $G \ne 1$ is either a split torus or connected semisimple and absolutely simple with $Y \ne X$. By passing to generic fiber over $X$, it suffices to show that if $K'/K$ is a finite separable extension of degree $d > 1$ and $H\ne 1$ is a split $K$-torus or is connected semisimple and absolutely simple then $\operatorname R_{K'/K}(H_{K'})$ is not $K$-isomorphic to $H^d$. (It is also false in plenty of other situations, but is not worth the effort to address even more generally.)

If $H=T$ is a nontrivial split $K$-torus then the inclusion $T \hookrightarrow \operatorname R_{K'/K}(T_{K'})$ is easily seen to be the maximal $K$-split torus, yet the Weil restriction has even bigger dimension.

The semisimple and absolutely simple case is (as always) more interesting. If $\widetilde{H} \rightarrow H$ is the simply connected central cover, then the same holds for $\operatorname R_{K'/K}(\widetilde{H}_{K'}) \rightarrow \operatorname R_{K'/K}(H_{K'})$, so we may assume $H$ is simply connected.
Thus, by 6.21(ii) in Borel and Tits - Groupes Réductifs (IHES 27) (which is ultimately an application of the product structure of root data for split simple connected groups), every connected semisimple $K$-group that is simply connected has the form $\operatorname R_{F/K}(G)$ a unique pair $(F,G)$ up to $K$-isomorphism consisting of a nonzero finite étale $K$-algebra $F$ and a smooth affine $F$-group $G$ such that the fibers of $G$ over the factor fields of $F$ are connected semisimple, simply connected, and absolutely simple.

Since $H^d = \operatorname R_{F/K}(H_F)$ for $F = K^d$ and $H$ that is absolutely simple over $K$, it follows that the only way $H^d$ can be $K$-isomorphic to $\operatorname R_{K'/K}(H_{K'})$ is if $K' \simeq F = K^d$ as $K$-algebras, an absurdity since $d > 1$ and $K'$ is a field.

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