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Let $n \in \mathbb{Z}_{\geq 2}$ be arbitrary. Let $r_1$ and $r_2$ be arbitrary elements of $\mathbb{Z}_{\geq 0}$ that satisfy $r_1 + r_2 > 0.$ Let $G := {\rm SL}_n(\mathbb{R})^{r_1} \times {\rm SL}_n(\mathbb{C})^{r_2}.$ Let $K$ be an arbitrary algebraic number field with $r_1$ real embeddings and $2r_2$ complex embeddings. Let $\Gamma$ denote the image of ${\rm SL}_n(\mathcal{O}_K)$ in $G$ under the diagonal embedding. Then $\Gamma$ is an irreducible, arithmetic, and nonuniform lattice in $G.$

Now let $\Lambda$ be any arithmetic, irreducible, and nonuniform lattice in $G.$ Is it then necessarily true that $\Lambda$ and $\Gamma$ are commensurate? If not, then does at least one of the inequalities $[\Gamma : (\Gamma \cap \Lambda) ] < +\infty$ and $[\Lambda : (\Gamma \cap \Lambda) ] < +\infty$ hold?

I am asking this question because I have recently been reading this paper on arXiv. Thanks to this paper, I know that the answer to the commensurability question is affirmative when $n=2.$ I should be very grateful for an answer or a relevant reference for the other cases. Thank you.

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    $\begingroup$ The two "$<\infty$" conditions are equivalent for two lattices in the same locally compact group, so the "If not..." question makes no sense. Also the answer is no (even for $n=2$), because for given $\Gamma$ there exists $g$ such that $\Gamma\cap g\Gamma g^{-1}$ is finite... so maybe you mean "commensurate up to conjugation". Side note: the arithmeticity condition is automatic as soon as $(n,r_1+r_2)\neq (2,1)$, by Margulis. $\endgroup$
    – YCor
    Apr 10, 2020 at 19:08
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    $\begingroup$ Anyway it's far from true (in the sense: up to conjugation) even in $\mathrm{SL}_n(\mathbf{K})$, $\mathbf{K}\in\{\mathbf{R},\mathbf{C}\}$ for $n\ge 3$. For instance if $G$ is a $\mathbf{Q}$-group that is a form of $\mathrm{SL}_3$, is $\mathbf{R}$-split, and of $\mathbf{Q}$-rank 1, then $G(\mathbf{Z})$ is a non-uniform lattice not commensurate up to conjugation to the standard arithmetic lattice. See Witte-Morris' book for detailed information about arithmetic lattices therein. $\endgroup$
    – YCor
    Apr 10, 2020 at 19:16
  • $\begingroup$ Thank you for your answer. Up to conjugation, what makes it true for $n = 2$ but not for $n > 2$? (Is this also in Witte-Morris?) $\endgroup$
    – Mishel Skenderi
    Apr 10, 2020 at 20:13
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    $\begingroup$ It's certainly there. The point (but not a proof) is that non-uniform means that the $\mathbf{Q}$-rank is $>0$, while conjugate to the standard arithmetic lattice essentially means, say for $(r_1,r_2)=(1,0)$, that the $\mathbf{Q}$-rank $r_\mathbf{Q}$ equals the (absolute) rank $r$ (in all cases one needs in addition to have $G$ $\mathbf{R}$-split so that the lattice lies in $\mathrm{SL}_n(\mathbf{R})$). When $n=2$, $r=1$ and there is no room between $0$ and $r$, while one can have $0<r_{\mathbf{Q}}<r$ when $r=n-1\ge 2$. $\endgroup$
    – YCor
    Apr 10, 2020 at 21:07

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