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?

  • 6
    $\begingroup$ 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 at 8:23
  • 1
    $\begingroup$ 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 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 at 11:19
  • 1
    $\begingroup$ 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 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 at 2:50

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 $\Gamma$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.