Skip to main content

All Questions

Filter by
Sorted by
Tagged with
7 votes
1 answer
467 views

Finite index subgroup of $\mathrm{GL}_n(\Bbb C)$ and Chevalley groups

I'm trying to show that if $G$ is a Chevalley group, then every finite index subgroup of $G(\Bbb Z)$ is Zariski dense in $G(\Bbb C)$. ($G(\Bbb Z)$ is the Chevalley group over $\Bbb Z$ and similarly ...
Ami's user avatar
  • 332
6 votes
1 answer
214 views

Are the integer matrices in SO(3,2) "boundedly generated"?

Let $G$ be the subgroup of integer matrices in $\mathrm{SO}(3,2)$. (The invertible linear maps from a $5$ dimensional real vector space to itself which leave invariant a nondegenerate symmetric ...
Pablo's user avatar
  • 11.3k
5 votes
1 answer
599 views

If $G$ is absolutely simple simply connected, why is G(F_v) quasisimple for almost every valuation v?

Let $G$ be an absolutely simple simply connected and connected algebraic group defined over a global field $k$ with ring of integers $\mathcal{O}$. Fix an embedding of $G$ into $GL_n$. Given $v$ a non-...
JGR's user avatar
  • 141
3 votes
2 answers
1k views

Examples of groups for which Margulis superrigidity theorem applies

I am not an expert at all in the subject of Lie groups, lattices, arithmetic groups and rigidity. But, lately I am interested in Margulis superrigidity theorem, which in most versions can be stated as ...
Luis Jorge'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
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