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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
20 votes
2 answers
1k views

Solving equations in SO(3) : an open problem by Jan Mycielski

I am interested in a problem closely related to a problem stated by Jan Mycielski in his paper Can One Solve Equations in Group? (The American Mathematical Monthly, 1977, http://www.jstor.org/stable/...
Alexander Belov's user avatar
15 votes
2 answers
613 views

Existence of a regular semisimple element over $\mathbb{F}_{q}$

This is probably old, a Chevalley level of old, but I'm not at all an expert in this field so I need help. Let $G$ be a simply connected (almost) simple linear algebraic group defined over $K=\mathbb{...
D. Dona's user avatar
  • 455
11 votes
2 answers
934 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
11 votes
2 answers
1k views

Finite subgroups of $PGL(3,K)$

It is well-known that finite subgroups of $PGL_2(\mathbb{C})$ are cyclic groups, dihedral groups, A4, S4 and A5 and each of these groups occurs exactly once (up to conjugacy). These facts are ...
Tim's user avatar
  • 125
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
10 votes
2 answers
1k views

Is there a way to see a topological group as the "Cayley graph" of its "infinitesimal generators"?

At the time of writing, the most recent blog post over at What's new by Terrence Tao is Cayley graphs and the geometry of groups, and that (excellent, as with most of Tao's writing) post most ...
Theo Johnson-Freyd's user avatar
9 votes
2 answers
865 views

Multiplication in Peter-Weyl theorem

$\DeclareMathOperator\SL{SL}$It is known that the coordinate algebra $\mathcal O(\SL_n(\mathbb C))$ decomposes as direct sum of $V \otimes V^*$ for $V$ finite-dimensional irreducible representations ...
Yellow Pig's user avatar
  • 2,964
8 votes
1 answer
534 views

Representation theory of $\mathrm{GL}_n(\mathbb{Z})$

I want to understand the (complex) representation theory of $\mathrm{GL}_n(\mathbb{Z})$, the general linear group of the integers. I have gone through several representation theory texts but all of ...
Kenji's user avatar
  • 81
8 votes
1 answer
337 views

What is the subgroup of $\mathrm{SL}(n,\mathbb{C})$ which preserves the discriminant?

$\DeclareMathOperator{\SL}{\operatorname{SL}}$Let $\mathcal{P}_{n-1}$ be the space of complex polynomials in one variable, say $z$, of degree at most $n-1$. As a complex vector space, it is clearly $n$...
Malkoun's user avatar
  • 5,215
8 votes
2 answers
482 views

Parabolics and simple roots for a special unitary group: reference request

I am looking for a reference where the relative root system, the relative system of simple roots, and parabolic $\Bbb R$-subgroups for the real algebraic group ${\rm SU}(p,q)$ are explicitly computed. ...
Mikhail Borovoi's user avatar
7 votes
1 answer
442 views

Does every cocompact lattice admit a homomorphism (with infinite image) into a compact Lie group?

Let $\Gamma$ be a cocompact arithmetic lattice in a semisimple algebraic group. Does it admit a homomorphism $\Gamma \to K$ with infinite image into a compact real Lie group $K$?
Jarek Kędra's user avatar
  • 1,782
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
7 votes
2 answers
231 views

Non-semisimple symmetric subgroups of simply connected simple algebraic groups

Let $G$ be a simply connected simple algebraic group over the field of complex numbers $\mathbb C$. Let $H$ be a symmetric subgroup of $G$. This means that there exists an automorphism of order 2 $\...
Mikhail Borovoi's user avatar
6 votes
1 answer
352 views

All surjections onto trivial irrep split equivalent to being reductive

$\DeclareMathOperator\Hom{Hom}$Let $ G $ be linear algebraic group over a field $ k $. Is it true that every short exact sequence of algebraic $ G $-representations $$ 0 \to W \to V \to k \to 0 $$ ...
Ian Gershon Teixeira's user avatar
6 votes
1 answer
390 views

Can these two irreducible $GL_n \mathbb Z$-representations be isomorphic?

Fix $n\in \mathbb N$ and a partition $\lambda$ with at most $n-1$ parts (of length at most $n-1$). Let $V$ be the irreducible $GL_n \mathbb R$-representation with highest weight $\lambda$ and $D$ the ...
Peter Patzt's user avatar
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
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
6 votes
1 answer
255 views

Which Lie groups are a central extension of an algebraic group?

Suppose $G$ is a connected real Lie group. The quotient $G/Z(G)$ is the image of the adjoint representation, so a linear group. Is it known for which groups this quotient is Lie isomorphic to an ...
Luis's user avatar
  • 161
6 votes
0 answers
200 views

Why should Serre's conjecture on congruence subgroup property hold?

There seem to be several related properties of an algebraic group, exhibiting the dichotomy between rank 1 and rank $\ge2$. Whether a lattice in the group satisfies the congruence subgroup property, ...
GTA's user avatar
  • 1,024
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
5 votes
1 answer
420 views

Analogue of the special orthogonal group for singular quadratic forms

The special orthogonal group $SO(n)$ is the subgroup of the special linear group $SL(n)$ of $n\times n$ matrices with determinant one that preserve a non-degenerate symmetric bilinear form. If such a ...
user avatar
5 votes
1 answer
729 views

The normalizer of block diagonal matrices

Let $G=\mathrm U_n$ or $\mathrm{GL}_n(\mathbf C)$ and $H$ the subgroup of block diagonal matrices respecting a partition $n=n_1+\dots+n_k$. Is the normalizer $N=N_G(H)$ computed anywhere in the ...
Francois Ziegler's user avatar
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
5 votes
0 answers
123 views

Conjugacy classes of plane k-jet group

Define $G(n, k)$ as a subgroup of $\rm{Aut}(\Bbb C[[x_1, \dots, x_n]]/\mathfrak m^{k+1})$ with identity linear part (so, group of $k$-jets of selfmaps of $\Bbb C^n$). I'm interested in the map from $G(...
Denis T's user avatar
  • 4,600
5 votes
0 answers
140 views

Intermediate subgroups between $H(\mathbb{Z})$ and $H(\mathbb{Z}[\sqrt{2}])$, for anisotropic form of $SL_2$

Let $H$ be an anisotropic $\mathbb{Q}$-form of $SL_2$, with $H(\mathbb{R}) = SL_2(\mathbb{R})$. Are there any results or conjecture regarding the existence, or non-existence of intermediate subgroups ...
JadeSnail's user avatar
  • 474
4 votes
2 answers
578 views

Proper compact connected subgroup of $Spin(n)$

What are the proper compact connected subgroups of $Spin(n)$ of maximal rank where $Spin(n)$ is the spin group, that is, the universal cover of the special orthogonal group $SO(n)$? In fact, I am ...
berl13's user avatar
  • 471
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
4 votes
1 answer
1k views

How to think about the simple reflection $s_0$ in the affine Weyl group?

Let $G$ be a simply connected algebraic group over $\mathbb{C}$, $W$ be the Weyl group for $G$ and $W_{aff}$ be the affine Weyl group for the loop group $G(\mathbb{C}((t)))$, $\Phi$ be the coweight ...
Qiao's user avatar
  • 1,719
4 votes
2 answers
1k views

Center of the algebraic group $G_{\mathbb{R}}$ for a centerless $G$

This must be an easy question but I don't have a good argument for it and have not found a counterexample: Let $G$ be a connected semisimple algebraic group over $\mathbb{Q}$ such that the center of $...
Jack's user avatar
  • 637
4 votes
2 answers
181 views

The orbits of an algebraic action of a semidirect product of a unipotent group and a compact group are closed?

We consider real algebraic groups and real algebraic varieties. It is known that the orbits of an algebraic action of a unipotent algebraic group $U$ on an affine variety are closed. The orbits of an ...
Vladimir47 's user avatar
4 votes
1 answer
436 views

Universal covering groups of simple linear algebraic group schemes

Let $R$ be a Dedekind domain with fraction field $K$, and let $G$ be a smooth affine group scheme over $S = \text{Spec }R$ whose geometric fibers are connected and simple linear algebraic groups (i.e.,...
stupid_question_bot's user avatar
4 votes
1 answer
119 views

Reference for nonquasi-split groups of type $E_6$ and $E_7$ over local fields

The semisimple groups over a local field have been classified by Tits, cf. [1] "Classification of algebraic semisimple groups" in Boulder and [2] "Reductive groups over local fields" in Corvallis. In ...
Arkandias's user avatar
  • 991
4 votes
2 answers
1k views

Dimension of Unipotent Radicals

A parabolic subgroup of a linear algebraic group $G$ defined over a field $k$ is a subgroup $P\subseteq G$, closed in the Zariski topology, for which the quotient space $G/P$ is a projective algebraic ...
Stanley Chang's user avatar
4 votes
1 answer
171 views

Which Lie groups admit finite generation by a set of Lie algebra elements? And what are some known choices of generators which realize this?

Consider a (finite-dimensional) real connected Lie group $G$ with Lie algebra $\frak{g}$. Take a generating set $\mathcal{G} = \{ X_1, \cdots X_n \} $ of $\frak{g}$, i.e. such that any element of $\...
Another User's user avatar
4 votes
1 answer
701 views

Centralizers of semisimple subgroups

$\DeclareMathOperator\GL{GL}$If $G$ is a simple Lie group, and $\rho: G \to \GL(V)$ is a representation, then by Schur's lemma, the group of automorphisms of $\rho$ is a reductive subgroup of $\GL(V)$....
unknownymous's user avatar
4 votes
1 answer
350 views

A generalization of Siegel property

In reduction theory of arithmetic groups, one has the following finiteness property. Proposition 1 (Siegel property). Let $G$ be a reductive group over $\mathbb{Q}$ and let $\Gamma\subset G(\mathbb{...
Golden Wave 's user avatar
4 votes
1 answer
633 views

Homomorphisms from binary polyhedral group to compact Lie groups

Are homomorphisms from binary polyhedral groups to (simple and simply connected) compact Lie groups classified? For cyclic groups, the result is well known (see e.g. Kac's "Infinite dimensional Lie ...
Yuji Tachikawa's user avatar
4 votes
1 answer
1k views

Reductive Lie Groups and Complexification

Let $G$ be a complex Lie group (not necessarily connected) with reductive Lie algebra $\frak{g}$. (We may assume that $G$ has finitely many connected components and is linear-algebraic.) Of course, $G$...
Peter Crooks's user avatar
  • 4,920
4 votes
0 answers
93 views

Homomorphism from a product of spin groups to a bigger spin group

In the paper "Essential dimension of spinor and clifford groups" by Chernousov and Merkurjev, it says that there is a natural homomorphism $\operatorname{Spin}(n)\times \operatorname{Spin}(m)...
YJ Kim's user avatar
  • 321
4 votes
0 answers
180 views

Zariski density for certain subsemigroups

$\DeclareMathOperator\GL{GL}$Suppose that we have a Zariski-dense subgroup $\Gamma$ of $\GL(d,\mathbb{R})$. Let $\delta$ be the abscissa of convergence of the series $$ \sum_{x \in \Gamma} e^{-s \log\|...
Zestylemonzi's user avatar
4 votes
0 answers
68 views

The weak restriction of the Jacquet module

Let $P= MN$ be a parabolic subgroup of a reductive p-adic group $G$, and $(\pi, V)$ is an irreducible, admissible representation of $G$. The Jacquet module is the representation $(\pi_N, V_N)$, where $...
IMED's user avatar
  • 41
3 votes
1 answer
276 views

For $G$ an adjoint Chevalley group, are all of $G(\mathbb Z)$'s finite-index subgroups congruence subgroups?

Let $G$ be an adjoint Chevalley group. Are all of $G(\mathbb Z)$'s finite-index subgroups congruence subgroups? I read a theorem that states: When $G$ is the universal Chevalley group and it's not of ...
Ami's user avatar
  • 332
3 votes
2 answers
555 views

Counterexamples to Margulis Normal subgroup theorem in rank 1

Margulis' normal subgroup theorem states that any normal subgroup of a higher rank irreducible lattice is either finite or of finite index. What are the known counter-examples in rank $1$ ? I am ...
Xavier Roulleau's user avatar
3 votes
2 answers
2k views

Is there an almost-direct product decomposition for disconnected reductive algebraic groups?

$\textbf{Some definitions:}$ Let $G$ be an algebraic group (for me that is the complex points of an affine algebraic group). We say $G$ is reductive if its unipotent radical (maximal connected normal ...
Maxime's user avatar
  • 397
3 votes
1 answer
2k views

center of the centralizer of semisimple element

Let $G$ be an adjoint group over an algebraically closed field $k$ and $s\in G$ a semisimple element. Let $H=C_{G}(s)^{0}$ the neutral component of the centralizer of $s$. Do we have that the center ...
prochet's user avatar
  • 3,472
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
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
3 votes
0 answers
237 views

Centralizers and algebraic groups

Suppose $G$ is a linear algebraic group - I am also happy to assume $G$ is a simple algebraic group over an algebraically closed field of characteristic zero, but the question won't require this. The ...
James Freitag's user avatar
3 votes
0 answers
71 views

Conjugacy classes in reductive group under adjoint action of parabolic subgroup

Given a reductive group $G$ over a finite field and a parabolic subgroup $P$ , I wonder what are the orbits in $G$ under the adjoint action of $P$. This should be standard, but I can only find results ...
Simon Lentner's user avatar