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I am trying to understand how Weil restriction affects the structure theory of a reductive group over the two fields involved. As a toy example, I looked at the following:

Consider $SL_2$ as an algebraic group over a quadratic extension $F / \mathbb{Q}$. This group splits over $F$ and its Dynkin diagram is of type $A_1$ so it behaves as nicely as one can wish for.

Let $G$ be $\operatorname{Res}_{F/\mathbb{Q}} \ SL_2$, the restriction of scalars to $\mathbb{Q}$. By the general theory $G$ is at least quasi-split, since its anisotropic kernel is the restriction of the one in $SL_2$ over $F$ which is trivial. Further its Dynkin diagram is a disjoint union of copies of $A_1$ indexed by the Galois group $\operatorname{Gal}(F / \mathbb{Q})$, so something like $A_1 \sqcup A_1$. This should also be true for the set of positive roots of $G$ with respect to some maximal $\mathbb{Q}$-split torus.

  1. What can be said about the structure theory of $G$ aside from the above? For example since $G$ is quasi split it should have a Borel subgroup $B = TU$ defined over $\mathbb{Q}$, $T$ being a maximal torus over $\mathbb{Q}$ and $U$ a maximal unipotent subgroup. Is it true, that $T$ is the Weil restriction of the diagonal Torus in $SL_2$ and if yes, is something like that true in general? What can be said about $U$?

  2. Is there some description of the Lie Algebra of the Weil restriction? I especially want to understand the root Eigenspaces $\mathfrak{g}_\alpha$ for positive roots $\alpha$ which might not be one dimensional in case $G$ is not split. Again, is my example too basic to see this effect here?

Thank you.

Torsten Schoeneberg
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Chris
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    I'm not too firm in the group setting, but as for Lie algebras, I am firm that restriction here just translates to good old restriction of scalars (i.e. base field of the LA as vector space). So here indeed, if you look at e.g. $\mathfrak{sl}_2(F)$ as a LA over $\mathbb Q$, then there is a one-dimensional split CSA (torus) $\pmatrix{c&0\0&-c} (c \in \mathbb Q)$ with two relative roots (restrictions of the standard roots of $\mathfrak{sl}_2$), and each has a two(-$\mathbb Q$)-dimensional root space, e.g. $\pmatrix{0&f\0&0}, f \in F$. – Torsten Schoeneberg Mar 18 '21 at 02:35

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  1. A good reference for some of these issues is Borel's article ''Automorphic $L$-functions,'' section I.5 and the references therein. For example, if $K/k$ is a finite extension of fields, and $\mathrm{G}= \mathrm{Res}_{K/k}(\mathrm{G}')$ is a restriction of scalars, then $\mathrm{G}$ is quasi-split over $k$ if and only if $\mathrm{G}'$ is quasi-split (as an $K$-group). More precisely, the map $$ B\longmapsto \mathrm{Res}_{K/k}(B) $$ is a bijection between $K$-rational Borel subgroups of $\mathrm{G}'$ and ${k}$-rational Borels of $\mathrm{G}$, with similar statements for other parabolics, unipotent radicals, etc. It is also true that if $T$ is a maximal torus of $\mathrm{G}'$, then $ \mathrm{Res}_{K/k}(T)$ is a maximal torus of $\mathrm{G}$.

  2. As for the Lie algebra, as in @Torsten Schoeneberg's answer, one has $$ \mathrm{Lie}_k(\mathbb{G})=\mathrm{Lie}_K(\mathbb{G}') $$ with the right hand side viewed as a $k$-vector space via the extension $K/k$. I'm not sure precisely what the question about root spaces is; if you clarify, perhaps I can say more.

Something rather nice occurs in the special case that we fix a reductive $k$-group $\mathrm{G}''$ such that we may realize $\mathrm{G}'=\mathrm{G}''_K$ as the base-change of $\mathrm{G}''$ from $k$ to $K$ (for example, if $\mathbb{G}$ is split over $K$). In this case, $\mathrm{G}= \mathrm{Res}_{K/k}(\mathrm{G}_K'')$ is equipped with an action of $\mathrm{Gal}(K/F)$ such that $$ \mathrm{G}^{\mathrm{Gal}(K/k)}=\mathrm{G}''. $$ On the Lie algebra, if $\mathfrak{g}''=\mathrm{Lie}(\mathrm{G}'')$ and $\mathfrak{g}=\mathrm{Lie}(\mathrm{G})$, then $$ \mathfrak{g}= \mathfrak{g}''\otimes_k K. $$ For example, if $K/k = \mathbb{C}/\mathbb{R}$, then $$ \mathfrak{g}= \mathfrak{g}''\otimes_\mathbb{R}\mathbb{C} = \mathfrak{g}''\oplus \mathfrak{g}''\sqrt{-1}. $$

SLeslie
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  • That very last direct sum is just a direct sum of vector spaces though, right? There is a decomposition of (extension followed by restriction) [and also, (restriction followed by extension)] to the $[K:k]$-fold direct product of the original under certain good conditions though, cf. https://math.stackexchange.com/q/3884795/96384. – Torsten Schoeneberg May 19 '21 at 17:26
  • As vector spaces, yes. I was thinking of it as a graded Lie algebra, though. – SLeslie May 20 '21 at 22:16