Let $k$ be a perfect field.

Recall that an algebraic torus $T$ over $k$ is called quasi-split if there exists some finite étale $k$-algebra $A$ such that $$T \cong \mathrm{R}_{A/k} \mathbb{G}_m.$$

A reductive group $G$ over $k$ is called quasi-split if it contains a Borel subgroup $B$ over $k$.

A priori, I see no reason why these two definitions should be related; Indeed any algebraic torus is quasi-split in the second sense (take $B=T$), but generally not in the first sense. However, I have seen it claimed, or at least implicitly used, in proofs that for semisimple $G$, the latter implies the former. Namely:

Let $G$ be a semisimple algebraic group over $k$ with a maximal torus $T$. If $G$ is quasi-split, then is $T$ quasi-split? Does the converse hold?

  • 1
    $\begingroup$ I'm pretty sure that $SU(2, 2)$ over $\mathbf{R}$ is semisimple and quasi-split, but its maximal torus is not. $\endgroup$ – David Loeffler Jun 6 '16 at 8:35
  • 1
    $\begingroup$ If a simply connected semisimple $k$-group $G$ has a Borel subgroup $B$ (defined over $k$), then any maximal torus $T\subset B\subset G$ is quasi-split in your sense. $\endgroup$ – Mikhail Borovoi Jun 6 '16 at 8:35
  • 1
    $\begingroup$ It not true, that any maximal torus in a quasi-split group is quasi-split in your sense (consider a compact torus in the split group $\mathrm{SL}(2,\mathbb{R})$ ). $\endgroup$ – Mikhail Borovoi Jun 6 '16 at 8:39
  • $\begingroup$ Yes, thanks. Shortly after asking my question, I realised it was simple to see that the converse won't hold. $\endgroup$ – Daniel Loughran Jun 6 '16 at 8:40
  • 2
    $\begingroup$ @DavidLoeffler: I can explain the proof in terms of coroots (this is the same as Knop writes). Let $\alpha_1^\vee,\alpha_2^\vee,\alpha_3^\vee$ be the simple coroots of $G=SU(2,2)$ with respect to $T$ and $B$. Since $G$ is simply connected, these simple coroots consitute a basis of the cocharacter group of $T$. Since $B$ is defined over $\mathbb{R}$, the complex conjugation permutes the simple coroots. Thus $T\simeq \mathbb{R}^*\times\mathbb{C}^*$, hence it is quasi-split. $\endgroup$ – Mikhail Borovoi Jun 6 '16 at 10:31

A torus $T$ is quasi-split if its character group is a permutation representation for the Galois group. So a counterexample to your question is: let $G$ be the quasi-split group $SO(n+1,n-1)$, $n\ge2$, $k=\mathbb R$. The weight lattice is $\mathbb Z^n$ with Galois action $$ (x_1,\ldots,x_{n-1},x_n)\mapsto(x_1,\ldots,x_{n-1},-x_n) $$ This is not a permutation module.

On the other hand, the Galois action on $T\subseteq B\subseteq G$ of a quasi-split group is induced by an action on the simple roots. These form a basis if $G$ is of adjoint type. For simply connected groups one can play the same game with fundamental representations. So:

If $G$ is quasi-split, and either simply connected or of adjoint type then the maximal $k$-torus sitting in a $k$-Borel is quasi-split.

I have the suspicion that the two notions of "quasi-splitness" developed independently.

  • $\begingroup$ This is great, thanks. Indeed in my case $G$ is adjoint. Do you know a precise reference in the literature where this result is proved? $\endgroup$ – Daniel Loughran Jun 6 '16 at 8:43
  • 1
    $\begingroup$ This fact for adjoint and simply connected cases was known prior to Harder's paper; e.g., it is Prop. 3.13 Exp. XXIV of SGA3 (expressing the maximal torus in a Borel subgroup as the Weil restriction of ${\rm{GL}}_1$ over the finite etale cover given by the "scheme of Dynkin diagrams"). The setup there may seem forbiddingly general, but the method of proof is exactly the same permutation argument. This fact must have been known to Borel, Tits, Steinberg, et al. prior to SGA3, but at least SGA3 provides an earlier reference. The terminology "quasi-split" for such tori is bad; please avoid it. $\endgroup$ – nfdc23 Jun 6 '16 at 14:52
  • 1
    $\begingroup$ @nfdc23. What is a better term than "quasi-split" for tori? That is an honest question; I do not know what is a good term. Would "permutation torus" be a better term? $\endgroup$ – Jason Starr Jun 7 '16 at 12:37
  • 3
    $\begingroup$ @JasonStarr: "Induced torus". $\endgroup$ – Mikhail Borovoi Jun 7 '16 at 13:59
  • 1
    $\begingroup$ @JasonStarr: Borovoi gave the response I would have given. The motivation for this alternative terminology is that the geometric character lattice of the Weil restriction of a torus through a finite separable extension of fields is the induced representation (for absolute Galois groups) of the geometric character lattice of the given torus. For a hands-on proof of that (natural) identification, see the proof of Theorem 2.4(a) in Chapter II of Oesterle's 1984 Inventiones paper "Nombres de Tamagawa...". $\endgroup$ – nfdc23 Jun 8 '16 at 3:19

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.