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$\newcommand{\Q}{\mathbb{Q}}\newcommand{\R}{\mathbb{R}}\DeclareMathOperator\PU{PU}$Let $G$ be a simple algebraic group over $\Q$ such that $G(\R) \simeq \prod_i G_i$, with each $G_i$ being the Lie group $\PU(p_i,q_i)$ for some $p_i,q_i$ such that $p_i \geq q_i$ and $p_i + q_i = n$.

If $\Gamma < G(\Q)$ is a subgroup such that the image of $\Gamma$ in $G_i$ is a non-arithmetic lattice (for some $i$), then is there any known restriction on $p_j, q_j$ for $j \neq i$ (other than the condition that $G_j$ should be noncompact for some $j \neq i$)?

If it helps, one can assume that $\Gamma$ is contained in an arithmetic subgroup of $G(\Q)$.

(By Margulis's theorem one knows that $q_i$ must be $1$, so the question is asking for other known restrictions.)

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  • $\begingroup$ What would be the role of other $j$'s? You can take $\Gamma$ to lie inside $G_i$. I also don't understand your "other than the condition...", unless there's a missing assumption on $\Gamma$. $\endgroup$
    – YCor
    Commented Nov 4, 2023 at 9:35
  • $\begingroup$ @YCor: The point is that $G$ is an algebraic group over $\mathbb{Q}$ and $\Gamma < G(\mathbb{Q})$ but the $G_i$ are only assumed to be Lie groups. As for the "other than...", I believe it is known that if all other $G_j$ are compact then $\Gamma$ must be arithmetic. $\endgroup$
    – naf
    Commented Nov 4, 2023 at 14:29
  • $\begingroup$ Ah, thanks for the clarification. $\endgroup$
    – YCor
    Commented Nov 4, 2023 at 16:39
  • $\begingroup$ As @naf would know, all the Deligne Moscow examples of non-arithmetic lattices in $PU(n,1)$ are constructed this way. They are only constructed in PU(2,1) and PU(3,1), where the conditions on $p_j,q_j$ mean that $q_j$ is 1 (if $PU(p_j,q_j)$ is non-compact. $\endgroup$ Commented Nov 9, 2023 at 17:45
  • $\begingroup$ It is easy to show that if all other factors are compact, then $\Gamma$ is arithmetic, only in the case $\Gamma \subset G(\mathbb Z)$. In other cases, this may be true, but may use a theorem that non-arithmetic lattices in $U(n,1)$ have entries in algebraic integers ($(U1,1)$ excluded). $\endgroup$ Commented Nov 9, 2023 at 18:44

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