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What are the examples of lattices in $\operatorname{SL}_n(\mathbb{Q}_p)$ with $n\geq 3$ or in other semisimple $p$-adic groups of higher rank?

It is known $\operatorname{SO}_n(\mathbb{Z}[1/p])$ is a lattice in $\operatorname{SO}_n(\mathbb{Q}_p)$ and, in general, $G(\mathbb{Z}[1/p_i,1\leq i\leq m])$ is a lattice in $G(\mathbb{R})\times \Pi_{1\leq i \leq m}G(\mathbb{Q}_{p_i})$. How can we exclude the archimedean places and consider a single prime $p$?

It is also known that irreducible lattice in $G$ are (S-)-arithmetic if the rank of $G$ is no less than 2. (See Margulis’ book Discrete Subgroups of Semisimple Lie Groups). But it is always assumed $S$ contains the infinite places.

A similar question: discrete subgroups in p-adic Lie groups?.

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  • $\begingroup$ What is the definition of lattice?—discrete subgroup with finite covolume? In what sense does @KeivanKarai's answer to the question you indicated not answer your question? \\ MathJax note: please use MathJax stars *stars*, not TeX $\it fakery$ $\it fakery$, for italics. I have edited accordingly. $\endgroup$
    – LSpice
    Commented Dec 16, 2022 at 17:46
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    $\begingroup$ One fact is that all lattices in these groups are cocompact. We can exclude the Archimedean place when the real group is compact, i.e. when the arithmetic lattice has real rank zero. $\endgroup$
    – YCor
    Commented Dec 16, 2022 at 17:59
  • $\begingroup$ @LSpice That answer concerns the case that the Archimedean local group is compact and can be excluded. I just want to know a single local group but not any products. $\endgroup$
    – Jun Yang
    Commented Dec 16, 2022 at 18:58
  • $\begingroup$ @YCor Yes, you are absolutely correct. How about the non-compact Archimedean group case? $\endgroup$
    – Jun Yang
    Commented Dec 16, 2022 at 19:00

1 Answer 1

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Here's one example that I like. Consider $\Gamma = \{g \in SL_d\left[\sqrt{-m} / p\right] \mid g^t \cdot g^\sigma= I \}$, where $\sigma$ is the Galois conjugate. Then this is an arithmetic lattice in $SL_d\left(\mathbb{Q}_p\right)$.

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  • $\begingroup$ I don't understand your last assertion. $\endgroup$
    – YCor
    Commented Dec 16, 2022 at 22:31
  • $\begingroup$ Forgot $SL_{d}$... $\endgroup$
    – Asaf
    Commented Dec 17, 2022 at 15:40
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    $\begingroup$ But $\mathrm{SL}_d(\mathbf{Z}[1/pq])$ is not a lattice in $\mathrm{SL}_d(\mathbf{Q}_p)\times\mathrm{SL}_d(\mathbf{Q}_q)$ (it is dense!). It is a lattice in $\mathrm{SL}_d(\mathbf{R})\times\mathrm{SL}_d(\mathbf{Q}_p)\times\mathrm{SL}_d(\mathbf{Q}_q)$. $\endgroup$
    – YCor
    Commented Dec 17, 2022 at 16:48
  • $\begingroup$ @Asaf Thank you for the example in p-adic $SL_d$. Is that arithmetic? Or what is the definition of arithmeticity in p-adic groups without a set S containing any infinite primes? $\endgroup$
    – Jun Yang
    Commented Dec 17, 2022 at 19:51
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    $\begingroup$ @JunYang , yes, it comes from a an arithmetic construction. The full examples starts as a suitable subgroup fo the $\mathbb{Q}$-grp $GL_{2d}(\mathbb{C})$, with $S={\infty, p}$. The thing one needs to observe is that at the infinity place, $G_{\mathbb{R}}\simeq SU_{d}(\mathbb{R})$ is compact, hence the projection of the lattice to the other factor is still a lattice. $\endgroup$
    – Asaf
    Commented Dec 17, 2022 at 20:29

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