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

Question

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 $G_2$?

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.