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Lie Groups are Groups that are additionally smooth manifolds such that the multiplication and the inverse maps are smooth.
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 …
16
votes
Non-isomorphic complex Lie groups with the same exceptional Lie algebra for $\mathfrak{g_2,f...
I prefer to use the language of algebraic groups.
All algebraic groups and Lie algebras are defined over $\Bbb C$.
1. Let ${\mathfrak g}$ be a semisimple Lie algebra.
Consider the automorphism group $ …
16
votes
Accepted
In a compact lie group, can two closed connected subgroups generate a non-closed subgroup?
The abstract subgroup generated by $H$ and $K$ is closed.
We may assume that $G$ is connected.
The groups $G$, $H$, $K$ are the groups of real points of real algebraic groups $\mathbf{G}$, $\mathbf{H …
14
votes
Is there a formula for the Frobenius-Schur indicator of a rep of a Lie group?
Let us write the Frobenius-Schur indicator of the representation $V$ with highest weight $\lambda$
as $\mathrm{FS}(\lambda)$.
Let $R$ denote the root system, and let $\Pi=\{\alpha_1,\dots,\alpha_l\}$ …
11
votes
Does $SU(N)$ have pseudo-real representation?
Let $G$ be a compact (anisotropic) real algebraic group.
Let $\rho\colon G\to {\rm GL}(n,{\mathbb{C}})$ be a complex linear (polynomial) representation of $G$.
Following OP, we say that $\rho$ is pse …
9
votes
Accepted
About the conjugation of semi-simple subgroups
The answer is YES. It suffices to assume that $H_1$ and $H_2$ are conjugate over $\mathbb{C}$ or, what is the same, that they are conjugate over $\overline{\mathbb{Q}}$.
Theorem 1. Let $G$ be a co …
9
votes
Real Lie groups versus real linear algebraic groups: differences in connexity and fundamenta...
In the book Lie groups and algebraic groups by Onishchik and Vinberg, Theorem 3 in Section 5.2.1 on page 240 says: Let $S$ be a real structure on a simply connected complex semisimple Lie group $G$. …
8
votes
2
answers
464
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. …
8
votes
When is the normalizer of the maximal torus maximal?
$
\newcommand{\g}{{\mathfrak g}}
\newcommand{\h}{{\mathfrak h}}
\newcommand{\t}{{\mathfrak t}}
\newcommand{\C}{{\mathbb C}}
\newcommand{\Ad}{{\rm Ad}}
\newcommand{\ad}{{\rm ad}}
$Theorem. Let $G$ be …
7
votes
Accepted
Conjugacy classes of involutions in compact simple Lie group
For the case of a connected semisimple compact Lie group $G$, see this preprint, Section 3, where, following ideas of Kac and Vinberg, we describe set of conjugacy classes of $n$-th roots of a given …
7
votes
different Shimura data with common underlying group?
The question is essentially about ${\mathbf{R}}$-groups, so we shall assume that $G$ is defined over $\mathbf{R}$.
It is not true that for a given connected reductive ${\mathbf{R}}$-group $G$, there …
7
votes
2
answers
666
views
Élie Cartan's paper "Les groupes réels simples, finis et continus" of 1914
Question 1.
Does Élie Cartan's paper
Les groupes réels simples, finis et continus,
Ann. Sci. École Norm. Sup. (3) 31 (1914), 263–355
contain a classification of $\Bbb C$-linear involutions of simple …
7
votes
2
answers
715
views
What does a homogeneous space of a linear algebraic group know about the group?
Let $X=G/H$, where $G$ is a connected linear algebraic group over the field $\mathbf{C}$ of complex numbers
and $H\subset G$ is an algebraic subgroup.
In general, we can write the algebraic variety $X …
7
votes
Accepted
Compact dual of a noncompact Lie group
(0) The group ${\rm Aut}(\mathfrak{g}_0)$ does not have to be connected (even over $\mathbb C$), take $\mathfrak{g}_0=\mathfrak{su}(2,2)$ as a counter-example. So let $G={\rm Inn}(\mathfrak g_0)$, tha …
7
votes
2
answers
228
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 $\si …