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10 votes
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A uniform bound for a "true" non-congruence subgroup

Before stating my question, let me recall the Congruence Subgroup Property/Problem: Given simply connected absolutely and almost simple algebraic group $G$ with fixed realization as a matrix group one ...
Menny's user avatar
  • 638
3 votes
0 answers
602 views

Pointwise stabilizer of an apartment of the Bruhat-Tits building of $\mathrm{SL}_n(\mathbb{Q}_p)$

Denote by $X$ the Bruhat-Tits building of $\mathrm{SL}_n(\mathbb{Q}_p)$. Let $\Sigma$ be the fundamental apartment of $X$. Let $\Gamma=\mathrm{SL}_n(\mathbb{Z}[\frac{1}{p}])$. We can prove that the ...
Luis Jorge's user avatar
3 votes
0 answers
175 views

$\text{PGL}_n(\mathbf{Q}_p)$ and the Congruence Subgroup Property

Suppose $\Gamma$ is a torsion-free lattice of $\text{PGL}_n(\mathbf{Q}_p)$ for $n\geq 3$. Then I know that by the Margulis arithmeticity theorem, $\Gamma$ must be arithmetic. My question is does $\...
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2 votes
0 answers
80 views

Question about lattice with dense projection

Let $H\subset \operatorname{GL}(n,\mathbb{C})$ be a connected, semisimple algebraic group defined over $\mathbb{Q}$. Fix a number field $K$ with $[K:\mathbb{Q}]=3$ that is not totally real. Denote its ...
studiosus's user avatar
  • 305
2 votes
0 answers
111 views

A group generated by all the root subgroups above ideal of $\mathbb Z[\alpha]$ is of finite index in $G(\mathbb Z[\alpha])$

Let $G$ be a simply connected complex Chevalley group and let $T$ be some maximal torus in $G$ with $\dim T\geq 2$. From "A note on generators for arithmetic subgroups of algebraic groups" by ...
Ami's user avatar
  • 332
1 vote
0 answers
151 views

On the existence of non-arithmetic lattices in algebraic groups over $\mathbb{Q}$

$\newcommand{\Q}{\mathbb{Q}}\newcommand{\R}{\mathbb{R}}\DeclareMathOperator\PU{PU}$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 ...
naf's user avatar
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1 vote
0 answers
98 views

$K$-ranks of some algebraic groups in Lubotzky's "Discrete groups, expanding graphs and invariant measures"

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}$Let $G$ be a semisimple algebraic group and $K$ any field. Then the $K$-rank of $G$ is the maximal rank $r$ of a $K$-splitting torus $T \cong (K^...
user267839's user avatar
  • 5,966
1 vote
0 answers
227 views

Why are compact arithmetic surfaces defined through quaternion algebras (usually) only over $\mathbb{Q}$?

As in Chapter 6.2 of "Introduction to arithmetic groups" (by D.W. Morris), compact arithmetic surfaces could be defined through quaternion algebras $\mathbb{H}^{a,b}_F=\big(\frac{a,b}{F}\big)...
Local's user avatar
  • 128
1 vote
0 answers
257 views

Does the standard arithmetic subgroup of a closed $\mathbb{Q}$-algebraic groups have non-trivial $\mathbb{Q}$-characters?

I am trying to understand the Borel-Harish Chandra theorem about arithmetic subgroups being lattices. Suppose $G$ is an algebraic group inside $GL_n(\mathbb{C})$ such that it is definable as a zero ...
Breakfastisready's user avatar
1 vote
0 answers
190 views

Compactifications of group schemes

Let $G$ be a group scheme over a scheme $S$ which is the spectrum of a discrete valuation ring. Let $\eta$ (resp. $s$) be the generic (resp. closed) point. Assume that the generic fiber $G_{\eta}$ is ...
mton's user avatar
  • 41
0 votes
0 answers
121 views

Commensurability of arithmetic, irreducible, nonuniform lattices

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}...
Mishel Skenderi's user avatar