# Arithmetic groups and integral points of integral structures

If $$\mathbf{G}$$ is an algebraic group defined over $$\mathbb{Q}$$, a subgroup of $$\mathbf{G}(\mathbb{Q})$$ is arithmetic if it is commensurable to $$\mathbf{G}(\mathbb{Q}) \cap \operatorname{GL}_n(\mathbb{Z})$$ where some representation $$\mathbf{G} < \operatorname{GL}_n$$ has been chosen (and the definition is made so that the choice does not matter).

Fur the purpose of this question let us call a subgroup $$\Gamma$$ of $$\mathbf{G}(\mathbb{Q})$$ strictly arithmetic if there exists a group $$\mathbb{Z}$$-scheme $$\mathbf{G}_\mathbb{Z}$$ with generic fiber $$\mathbf{G}$$ such that $$\Gamma = \mathbf{G}_\mathbb{Z}(\mathbb{Z})$$.

I was recently asked the natural question whether strictly arithmetic is the same as arithmetic. I suspect that the answer is "no". More specifically arithmetic groups can be arbitrarily small (for instance have arbitrarily large covolume in $$\mathbf{G}(\mathbb{R})$$) while I suspect that this is not true of strictly arithmetic groups. But I don't know enough about group schemes to underpin that intuition. So I'm asking here:

Original question: Are there (resp. what are) examples of arithmetic groups that are not strictly arithmetic?

The original question was answered in the comments by David Loeffler using a different obstruction so let me (following YCor's suggestion in the comments) specifically ask:

Additional question: Do there exist "arbitrarily small" strictly arithmetic subgroups, for instance in the sense that the covolume or injectivity radius in $$\mathbf{G}(\mathbb{R})$$ is arbitrarily large?

• Any "strictly arithmetic" subgroup in your sense will be a congruence subgroup. Noncongruence subgroups exist in $SL_2(\mathbf{Z})$, and lots of other groups too. Jul 14 at 8:23
• Since the current main question is answered in comments (non-congruence subgroups are not strictly arithmetic), you might emphasize the covolume question instead (are there strictly arithmetic subgroups of arbitrary large covolume? arbitrary large systole?)?
– YCor
Jul 14 at 10:13
• @DavidLoeffler: It seems difficult to me to define a congruence subgroup of $\textbf{G}(\mathbb{Z})$ without an integral structure. So can I read you assertion as saying "the intersection of two strictly arithmetic subgroups is a congruence subgroup in either"? Do you have a reference? Jul 14 at 11:19
• Congruence subgroup = intersection of $G(\mathbb{Q})$ with an open compact subgroup of $G(\mathbb{A}_f)$. No need for integral structures. Jul 14 at 13:02
• @DavidLoeffler: I seem to remember hearing that congruence subgroups are exactly the strictly arithmetic groups as defined in the question, at least if the group scheme $\mathbf{G}_{\mathbb{Z}}$ is assumed to be smooth and finite type. I have not thought about the proof, but have the feeling that it is not difficult. If anyone cares, I can ask the person who told me this "fact". (The implication from strictly arithmetic to congruence (say with the smoothness condition) is of course easy.)
– naf
Jul 15 at 2:50

Any "strictly arithmetic" subgroup in your sense will, in particular, be a congruence subgroup, i.e. the intersection of $$G(\mathbb{Q})$$ with an open compact subgroup in $$G(\mathbb{A}_f)$$. Since non-congruence subgroups exist in $$SL_2 / \mathbb{Q}$$, and in lots of other groups too, these are examples of arithmetic subgroups which are not strictly arithmetic.
Start with your favourite $$GL_n$$-embedding $$\iota$$, defining some strictly arithmetic $$\Gamma$$. Take some $$g \in G(\mathbb{Q})$$ which isn't in $$\Gamma$$, and consider the embedding into $$GL_{2n}$$ sending $$h$$ to the block-diagonal matrix $$\begin{pmatrix} \iota(h) \\& \iota(g^{-1} h g))\end{pmatrix}$$. This defines a new $$\mathbb{Z}$$-model of $$G$$ whose integral points are $$\Gamma \cap g \Gamma g^{-1}$$. If $$G = SL_2$$, and probably for just about any $$G$$ which isn't abelian, the index of $$\Gamma \cap g \Gamma g^{-1}$$ in $$\Gamma$$ can be made arbitrarily large by a suitable choice of $$\Gamma$$.