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I feel experts might be able to answer this question immediately. Let $G$ be a connected $\mathbb Q$-simple and $\mathbb Q$-isotropic algebraic group. Let $S$ be a maximal $\mathbb Q$-split torus of $G$ and
let $T\supset S$ be a maximal torus defined over $\mathbb Q$. Let $T_a$ be the maximal anisotropic subtorus of $T$. Is $T$ a direct product of $S$ and $T_a$?

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  • $\begingroup$ Thanks for the examples of user89334 and nfdc23. A further question is whether it is true that $T(\mathbb Q)=S(\mathbb Q) T_a(\mathbb Q)$. $\endgroup$
    – ronggang
    Commented Apr 17, 2016 at 2:27

3 Answers 3

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The Weil-restriction construction suggested by user89334 is $\mathbf{Q}$-simple but not absolutely simple. To give absolutely simple examples, consider $G = {\rm{SL}}_n(D)$ for a central division algebra $D$ over $\mathbf{Q}$ with dimension $d^2>1$.

A maximal split torus in $G$ is given by the diagonal torus $S$ in the evident subgroup ${\rm{SL}}_n \subset G$. The centralizer $Z_G(S)$ is equal to $$\{(d_1,\dots,d_n) \in \underline{D}^{\times}\,|\,\prod {\rm{Nrd}}(d_j) = 1\}$$ where $\underline{D}^{\times} := {\rm{GL}}_1(D)$, so $\mathscr{D}(Z_G(S))={\rm{SL}}_1(D)^n$. This derived group has center containing $S[d]$, and every maximal torus $T \subset Z_G(S)$ is uniquely an isogenous product $S \cdot T_0$ for a (necessarily anisotropic) maximal torus $T_0 \subset \mathscr{D}(Z_G(S))$. Such a maximal $T_0$ is the maximal anisotropic subtorus of $T$ and must contain the central $S[d]$ in the derived group due to its maximality, so $T$ is never a direct product of $S$ and $T_0$ since $d > 1$. Passing to (central) isogenous quotients of $G$ never makes that central overlap entirely go away (since $S[d]$ is not cyclic), so the phenomenon persists for all members of the (central) isogeny class.

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  • $\begingroup$ What does $S[d]$ mean? $\endgroup$
    – LSpice
    Commented Oct 29, 2018 at 20:31
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No. Here is an easy to understand example to keep in mind.

Consider $\mathbb{C}^*$. It contains both the group $S^1$ and the group $\mathbb{R}^*$, but it is not their direct product (their intersection is $\{-1,1\}$). Both groups are real tori, the compact one is anisotropic and the noncompact is split.

To see this example in a simple group, just consider the diagonal subgroup of $\text{SL}_2(\mathbb{C})$.

Formally, the algebraic group you want too look at here is the extension of scalars of $\text{SL}_2$ from $\mathbb{R}$ to $\mathbb{C}$. Let me call it $\mathbf{G}$. $\mathbf{G}$ happens to be isomorphic as a $\mathbb{C}$-group to $\text{SL}_2\times \text{SL}_2$ which is not simple, but it has a real structure for which it is $\mathbb{R}$-simple. In fact, since you started with an isotropic $\mathbb{Q}$-simple group, the resulting group will be so too. $\mathbf{G}$ will contain the 2-dim maximal $\mathbb{Q}$-torus $T$ which is the extension of scalars of the (1-dim) diagonal subgroup of $\text{SL}_2$. $T$ will contain $T_a\simeq \text{SO}_2$ and $S\simeq G_m$, but will not be their direct product.

Edit: this is an edit made in order to emphasize the $\mathbb{Q}$ structure in the above example and to address the question asked in a comment to the original question.

Let me redo the extension of scalars on $\text{SL}_2$, but this time from $\mathbb{Q}$ to its quadratic extension $\mathbb{Q}[i]$. This gives a $\mathbb{Q}$-structure on $\mathbf{G}$ such that $\mathbf{G}(\mathbb{Q})=\text{SL}_2(\mathbb{Q}[i])$. The torus $T$ becomes $\mathbb{Q}$-isomorphic to the extension of scalars of $\mathbf{G}_m$ from $\mathbb{Q}$ to $\mathbb{Q}[i]$. In particular, $T(\mathbb{Q})\simeq \mathbb{Q}[i]^*$ and under this isomorphism, $S(\mathbb{Q})\simeq \mathbb{Q}^*$ and $T_a(\mathbb{Q})=\{a+ib\mid a,b\in \mathbb{Q},~a^2+b^2=1\}$. Check that the inclusion $S(\mathbb{Q})T_a(\mathbb{Q})<T(\mathbb{Q})$ is not an equation (e.g $1+i$ is in the RHS but not in the LHS).

If you wish for a higher level explanation, note that the difference $T(\mathbb{Q})/S(\mathbb{Q})T_a(\mathbb{Q})$ could be measured by the first Galois cohomology group of the norm torus, a nice discussion of which you could find in p. 73 of the excellent book "Algebraic Groups and Number Theory" by Platonov and Rapinchuk.

Let me conclude that the example I gave above is $\mathbb{Q}$-simple, as asked, but not absolutely simple: as a $\mathbb{C}$-group (in fact, as a $\mathbb{Q}[i]$-group) it is not simple. If you wish for an absolutely simple example you can consult the other excellent answers here, or merely find a $\mathbb{Q}$-embedding of our group $\mathbf{G}$ in $\text{SL}_n$ for some $n$ ($n=4$ will do).

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    $\begingroup$ The statement "The groups happens to be ${\rm{SL}}_2 \times {\rm{SL}}_2$" is potentially confusing: that is as a $\mathbf{C}$-group, not as an $\mathbf{R}$-group (so it may create a mistaken picture for the intended $\mathbf{Q}$-group). $\endgroup$
    – nfdc23
    Commented Apr 16, 2016 at 22:35
  • $\begingroup$ Notation fixed, definitions clarified - Thank you. $\endgroup$
    – Uri Bader
    Commented Apr 17, 2016 at 18:09
  • $\begingroup$ I made further edits in order to emphasize the $\mathbb{Q}$-structure in the above example and to address the question asked in a comment to the original question. $\endgroup$
    – Uri Bader
    Commented Apr 18, 2016 at 6:29
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As the example of nfdc23 shows, the answer is generally no. But maybe it helps to think about this question in a somewhat wider context, where the notions of split and anisotropic tori arise: the study of a connected reductive algebraic group defined over an arbitrary field $k$ (as in the 1965 paper by Borel and Tits). In the structure theory of such groups, it quickly becomes clear that the nature of $k$-anisotropic groups depends heavily on $k$ (and is not understood for many familiar fields). Leaving that aside, Borel and Tits got a lot of unified information about the structure of a $k$-isotropic group. Modulo the knowledge of $k$-anisotropic groups, this leads ultimately to the Tits classification method. Of course, the special case $k=\mathbb{Q}$ is part of this story, but the main ideas are developed for all $k$.

Note especially that the question raised here never gets answered explicitly in the structure theory. Indeed, the maximal $k$-anisotropic subtorus $T_a$ here is mentioned but does not play an important role. The key players include: a (nontrivial!) maximal $k$-split torus $S$ (unique up to $k$-conjugacy), along with its (reductive) centralizer in $G$ (which of course contains $T_a$), a minimal $k$-parabolic subgroup containing $S$, and various data about the associated root systems and Weyl groups.

What the general theory reveals is the existence of an almost-direct product $ T=T_a\, S$: see for example Borel's 8.15 in GTM 126. But toward the end of their respective textbooks, Borel (in his expanded second edition) and Springer (in his later framework of $F$-groups) develop a lot of finer detail about classical groups somewhat in the spirit of the answer by nfdc23.

One extreme, however, is the case of a quasi-split group $G$, in which a minimal $k$-parabolic subgroup is a Borel subgroup (and which is the only type possible for finite or some other special fields).

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  • $\begingroup$ In what sense are quasi-split groups an extreme case? Even here the difficulty of an only almost-direct product can arise. (Of course literally any difficulty that can arise for tori can arise for quasi-split groups, since tori are quasi-split!) I would think that split groups would be an (excessively) extreme case. $\endgroup$
    – LSpice
    Commented Oct 30, 2018 at 21:14

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