# Does $G_2(\mathbb{Z})$ depend on the choice of an integral model?

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 $$G_2$$?

• 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).
– YCor
Feb 5, 2021 at 9:04
• 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". Feb 5, 2021 at 22:42

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.

• 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.)
– YCor
Feb 5, 2021 at 11:28
• This is also true, see Remark 5.2 of Conrads notes on nonsplit reductive groups over Z: virtualmath1.stanford.edu/~conrad/papers/redgpZsmf.pdf
– Jef
Feb 5, 2021 at 11:48