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5 votes
2 answers
504 views

A finiteness property for semi-simple algebraic groups

Let $G$ be a semi-simple algebraic group over a field $K$, I am considering a question about whether there exists a finite set of semi-simple $K$-subgroups, say $H_1,...,H_r$, such that for any semi-...
Golden Wave 's user avatar
4 votes
1 answer
615 views

About the conjugation of semi-simple subgroups

Let $G$ be a semi-simple algebraic group over $\mathbb{Q}$, I would like to find an integer $d>0$ only depending on $G$ with the following property. For any two semi-simple $\mathbb{Q}$-subgroups $...
Golden Wave 's user avatar
6 votes
1 answer
658 views

Anti-holomorphic involutions of a complex linear algebraic group

Let $G$ be a connected linear algebraic group over the field of complex number ${\Bbb C}$. Let $G({\Bbb C})$ denote the complex Lie group of ${\Bbb C}$-points of $G$. Let $\sigma$ be an anti-...
Mikhail Borovoi's user avatar
24 votes
3 answers
2k views

Spin group as an automorphism group

Consider the real algebraic group $SO(p,q)$, this is the automorphism group of the vector space $\mathbb{R}^n$ of dimension $n=p+q$ over $\mathbb{R}$, endowed with the diagonal quadratic form with $p$...
Mikhail Borovoi's user avatar
11 votes
2 answers
935 views

Non-isomorphic complex Lie groups with the same exceptional Lie algebra for $\mathfrak{g_2,f_4,e_6,e_7,e_8}$?

An exceptional complex Lie algebra is a simple Lie algebra whose Dynkin diagram is of exceptional (nonclassical) type. There are exactly five such Lie algebras: $\mathfrak{g}_{2}$, ${\mathfrak {f}}_{4}...
annie marie cœur's user avatar
10 votes
1 answer
719 views

what is the intersection of all congruence subgroups of the profinite completion of SL(2,Z)?

Let $\widehat{SL(2,\mathbb{Z})}$ be the profinite completion of $SL(2,\mathbb{Z})$. Let $\Gamma(N)$ denote the typical principal congruence subgroup of $SL(2,\mathbb{Z})$ (ie, all matrices congruent ...
Will Chen's user avatar
  • 10.7k
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
5 votes
0 answers
298 views

What are the matrix coefficients associated with the irreducible representations of compact real linear algebraic groups?

What are the matrix coefficients associated with the irreducible representations of a compact real linear algebraic group $G$? Peter-Weyl tells us that $L^2(G)$ is the (closure of) $\bigoplus_\pi A_{\...
Andrew NC's user avatar
  • 2,071
3 votes
1 answer
269 views

A more precise description of conjugation of semi-simple subgroups

Let $G$ be a semi-simple algebraic group over $\mathbb{Q}$, I would like to find an integer $d>0$ only depending on $G$ with the following property. For any two semi-simple $\mathbb{Q}$-subgroups $...
Golden Wave 's user avatar
2 votes
2 answers
757 views

Abstract Commensurator Group of $\mathbb{Z}^n$ $Comm(\mathbb{Z}^n)\cong GL(n,\mathbb{Q})$?

Hello! In a paper I read that $\mathrm{Comm}(\mathbb{Z}^n)\cong \mathrm{GL}(n,\mathbb{Q})$. Why is that true? How can I find an isomorphism of this groups? I know that $\mathrm{Aut}(\mathbb{Z}^n)\...
Peter's user avatar
  • 33
2 votes
0 answers
290 views

Automorphisms group of complex and real simple Lie algebras

$\DeclareMathOperator{\Inn}{\operatorname{Inn}}\DeclareMathOperator{\Aut}{\operatorname{Aut}}\DeclareMathOperator{\Out}{\operatorname{Out}}\DeclareMathOperator{\g}{\mathfrak{g}}$According to Wikipedia,...
annie marie cœur's user avatar
1 vote
0 answers
370 views

Characterizing the big Bruhat cell of the universal Chevalley groups over $\mathbb C$

Is there a simple characterization of the big Bruhat cell of the universal (simply-connected) Chevalley groups over $\mathbb C$? For example, it is known that the Borel subgroup of $\mathrm{SL}_n(\...
Ami's user avatar
  • 332