# Weil restriction

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.

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

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 $$G^d$$.

• 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. Jan 8, 2023 at 16:27

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.

• (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)? Aug 25, 2014 at 15:45
• 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. Aug 25, 2014 at 15:49
• 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 $Y$? Aug 25, 2014 at 15:57

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.