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?

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    $\begingroup$ 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. $\endgroup$ Jul 14, 2021 at 8:23
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    $\begingroup$ 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?)? $\endgroup$
    – YCor
    Jul 14, 2021 at 10:13
  • $\begingroup$ @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? $\endgroup$ Jul 14, 2021 at 11:19
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    $\begingroup$ Congruence subgroup = intersection of $G(\mathbb{Q})$ with an open compact subgroup of $G(\mathbb{A}_f)$. No need for integral structures. $\endgroup$ Jul 14, 2021 at 13:02
  • $\begingroup$ @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.) $\endgroup$
    – naf
    Jul 15, 2021 at 2:50

1 Answer 1


First question (do non-strictly-arithmetic subgroups exist?):

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.

Second question (can strictly arithmetic subgroups be small?):

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 $\Gamma$.


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