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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
8 votes
0 answers
201 views

Monodromy groups that are profinitely dense in Sp(2g,Z)

$\DeclareMathOperator\Sp{Sp}$Assume $g\geq 2$. It is known that there exist finitely generated subgroups of $\Sp(2g,\mathbb{Z})$ of infinite index that surject onto all finite quotients of $\Sp(2g,\...
Gabriele Mondello's user avatar
6 votes
0 answers
202 views

Index of subgroups of $\mathrm{Sp}(4,\mathbb{Z})$ conjugate in $\mathrm{GL}(4,\mathbb{Q})$

$\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\SL{SL}\DeclareMathOperator\GL{GL}$Assume that we have two subgroups $G_1,G_2$ of $\Sp(4,\mathbb{Z})$ that are conjugate in $\GL(4,\mathbb{Q})$ $\big($...
T.Ch.'s user avatar
  • 141
6 votes
0 answers
365 views

Cohomology of $\operatorname{SL}_n(\mathbb Z)$ with coefficients

Let $K$ be a number field, which can be $\mathbb Q(\zeta)$ with $\zeta$ a root of unity if that helps, and let $\operatorname{SL}_n(\mathbb{Z})$ act on $(K^\times)^n:=K^\times\times\cdots\times K^\...
Mariano Suárez-Álvarez's user avatar
6 votes
0 answers
164 views

Is a presentation of the hyperbolic orthogonal group of rank 2 over the integers known?

The hyperbolic orthogonal group $O_{g,g}(\mathbb{Z})$ often appears in the study of high-dimensional manifolds, see e.g. work of Kreck or Galatius and Randal-Williams. Let $H$ denote the lattice $\...
skupers's user avatar
  • 8,167
5 votes
0 answers
150 views

Lattices of minimal covolume in $\operatorname{SL}_2(\mathbb{R}) \times \operatorname{SL}_2(\mathbb{R})$

What are the (uniform/non-uniform) irreducible lattices of minimal (or even small) covolume in $\operatorname{SL}_2(\mathbb{R}) \times \operatorname{SL}_2(\mathbb{R})$? Context: Such a lattice will ...
Stefan Witzel's user avatar
5 votes
0 answers
267 views

Generating congruence subgroups of SL_n over totally imaginary number rings

Fix some $n \geq 3$. Let $k$ be an algebraic number field with ring of integers $\mathcal{O}$ and let $\alpha$ be an ideal of $\mathcal{O}$. Define $\text{SL}_n(\mathcal{O},\alpha)$ to be the ...
Frank's user avatar
  • 51
4 votes
0 answers
238 views

What is known about the cohomology of the U-duality group?

$\newcommand{\Es}{E_{7(7)}}\newcommand{\Z}{\mathbb Z}$Let $\Es$ denote the split form of $E_7$, which is a real Lie group. It can be characterized as the subgroup of $\mathrm{Sp}_{56}(\mathbb R)$ ...
Arun Debray's user avatar
  • 6,881
4 votes
0 answers
149 views

Generators of special linear group of a projective module over a number ring

Let $\mathcal{O}$ be the ring of integers in an algebraic number field $K$ and let $M$ be a rank-$n$ projective $\mathcal{O}$-module. By definition, this means that $M \otimes K \cong K^n$, so the ...
Julia's user avatar
  • 41
3 votes
0 answers
103 views

Colimits in cohomology of profinite arithmetic groups

Let $G\subset \operatorname{GL}_n$ be a linear algebraic group over $\mathbb{Q}$ and let $\Gamma\subset G\cap \operatorname{GL}_n(\mathbb{Z})$ be an arithmeric subgroup without torsion. Using a result ...
Jrodri26's user avatar
  • 133
3 votes
0 answers
110 views

Is this a lattice?

Let $R$ be a locally compact ring (commutative with unit) and let $D\subset R$ be a discrete cocompact subring (cocompact means the additive group $R/D$ is compact). Let $G$ be a semisimple linear ...
user avatar
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
205 views

Two basic questions on congruence subgroups

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\GL{GL}$I have two questions related to congruence subgroups. Let $$\Gamma=\Gamma_0(N)=\Big\{\begin{pmatrix} a & b \\ c & d \end{pmatrix} \...
Andrew's user avatar
  • 1,019
2 votes
0 answers
467 views

How does $SL_2(\mathbb Z[\sqrt 2])$ embed into $SL(2,\mathbb R)\times SL(2,\mathbb R)$, and what are the centralizers in $Mat(4\times 4,\mathbb Z)$?

I am trying to learn arithmetic groups, from a dynamical point of view. These questions (maybe silly) come to my mind, but I do not know the answer. How does $SL_2(\mathbb Z[\sqrt 2])$ embed into $SL(...
Changguang's user avatar
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
  • 10.5k
1 vote
0 answers
154 views

Definition of arithmetic subgroups of Lie groups

In Maclachlan-Reid we can read Let $G$ be a connected semisimple Lie group with trivial centre and no compact factor. Let $\Gamma\subset G$ be a discrete subgroup of finite covolume. Then $\Gamma$ is ...
Jacques's user avatar
  • 563
1 vote
0 answers
179 views

Matrix factorizations over $GL_2$ of a real quadratic ring of integers

tl;dr: The groups $GL_2(K)$, or $SL_2(K)$, where $K = \mathbb{C,R}$ admits several factorizations (the polar decomposition, the KAN decomposition, the Schur triangular form, etc). Those ...
Adrián González Pérez's user avatar
1 vote
0 answers
69 views

On equality of two quotients of a congruence subgroup

Related question: Non-torsion part of the abelianisation of congruence subgroups Let $A = \mathbb{F}_q[T]$ be the ring of polynomials with coefficients in a finite field, with $N$ a nonconstant ...
Liam Baker's user avatar
0 votes
0 answers
267 views

Definition of reducible lattice

I am reading Raghunathan's book on discrete subgroups of Lie groups. In particular I am stuck on Corollary 5.19 which gives several equivalent conditions for a lattice in a semisimple Lie group to be ...
user551642's user avatar