I am trying to understand constructions of exceptional groups of type $G_2$ (over rings). In this post, by a model (of type $G_2$) I mean an affine smooth group scheme over $\mathbb{Z}$ such that the fibres are connected simple algebraic groups of type $G_2$.

In Gross' paper Groups over Z (see Page 272) a model $\mathbb{G}$ is given using Coxeter's integral octonions, and it is mentioned there that $\mathbb{G}(\mathbb{Z})$ is isomorphic to $G_2(\mathbb{F}_2)$ as abstract groups. Models are not unique: For example, this $\mathbb{G}$ is a non-split model, and there is also a split one as given in Appendix B of Conrad's paper Non-split reductive groups over Z.

Question: Do we have $\mathbb{G}'(\mathbb{Z})\cong G_2(\mathbb{F}_2)$ for every model $\mathbb{G}'$ of type $G_2$?

  • $\begingroup$ What do you mean by "for any"? do you mean "for every" or "for some" (but "for some" seems to be the claim 2 lines earlier)? If the model $\mathbb{G}'$ is $\mathbf{R}$-split, then $\mathbb{G}'(\mathbf{Z})$ is infinite (being a lattice in a noncompact Lie group, by Harish-Chandra). $\endgroup$
    – YCor
    Feb 5, 2021 at 9:04
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    $\begingroup$ I deleted the '~' ties, which do nothing in MathJax; and, since, as @YCor observes, the reading of "for any" as "for some" is unsupported by the rest of the post, I changed it to "for every". $\endgroup$
    – LSpice
    Feb 5, 2021 at 22:42

1 Answer 1


The short answer is no, because a Chevalley group has infinitely many $\mathbb{Z}$-points (even Zariski dense by the Borel density theorem).

For the long answer, let me first completely describe all $\mathbb{Q}$-models and $\mathbb{Z}$-models of groups of type $G_2$. Let $G_0/\mathbb{Q}$ be the split reductive group of type $G_2$. (It is both simply connected and adjoint since the $G_2$ Cartan matrix has determinant $1$.) This group has a reductive $\mathbb{Z}$-model $\underline{G}_0$ which is unique up to isomorphism by the theory of Chevalley groups.

Now the $\mathbb{Q}$-forms of $G_0$ are classified by $\mathrm{H}^1(\mathbb{Q},G_0)$, bearing in mind that $G_0$ is adjoint and has no outer automorphisms. By the work of many people, the restriction map $\mathrm{H}^1(\mathbb{Q},G_0) \rightarrow \mathrm{H}^1(\mathbb{R},G_0)$ is an isomorphism (for references see Theorems 5.12.24 and 5.12.31 of Poonen's rational points book). By table 1.3 in the Gross' paper you mention, $\mathrm{H}^1(\mathbb{R},G_0)$ has size $2$, where the nontrivial element is represented by the compact form.

Let's call the corresponding $\mathbb{Q}$-form $G$. It can be constructed as the automorphism group of (a $\mathbb{Q}$-form of) the octonions (as done in the Gross' paper) and its real points $G(\mathbb{R})$ are compact. Gross constructs a $\mathbb{Z}$-model $\underline{G}$ of $G$, again using octonions. It has the property that $\underline{G}(\mathbb{Z})\simeq \underline{G}_0(\mathbb{F}_2)$ as abstract groups. He shows using the mass formula that $\underline{G}$ is the only $\mathbb{Z}$-model of $G$.

Conclusion Up to isomorphim, there exist exactly two semisimple groups of type $G_2$ over $\mathbb{Q}$. Both admit unique models over $\mathbb{Z}$.

This is a very special property of $G_2$! Usually a semisimple group has infinitely many $\mathbb{Q}$-forms, most of them will not have integral models, and if such integral models exist they might not be unique. (Considering $\mathrm{SL}_2$ should already give you an idea.)

Back to your question: you are asking whether we have an isomorphism $\underline{G}_0(\mathbb{Z}) \simeq \underline{G}_0(\mathbb{F}_2)$. But the left hand side is not even finite: by a theorem of Borel and Harish-Chandra, $\underline{G}_0(\mathbb{Z})$ is a lattice in $G_0(\mathbb{R})$. By the Borel density theorem, this implies that $\underline{G}_0(\mathbb{Z})$ is Zariski dense in $G_0$, so in particular $\underline{G}_0(\mathbb{Z})$ is infinite.

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    $\begingroup$ I didn't know that $G_2$ has a single non-split $\mathbf{Q}$-form, this sounds quite unexpected to me. Is the same true with octonions? (I.e., does the 8-dimensional complex non-associative algebra obtained as complexification of octonions have exactly 2 $\mathbf{Q}$-forms up to isomorphism? This might be straightforward if the automorphism group is the same, but I'm not 100% sure.) $\endgroup$
    – YCor
    Feb 5, 2021 at 11:28
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    $\begingroup$ This is also true, see Remark 5.2 of Conrads notes on nonsplit reductive groups over Z: virtualmath1.stanford.edu/~conrad/papers/redgpZsmf.pdf $\endgroup$
    – Jef
    Feb 5, 2021 at 11:48

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