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Lie Groups are Groups that are additionally smooth manifolds such that the multiplication and the inverse maps are smooth.

1 vote

On the Weyl character formula

So one may prove the second equality of the question (the so-called Weyl integral formula) in the following way: For every $H\in\mathfrak{h}=Lie(T)$ we denote $$ Q(H):=\prod_{\alpha\in\Delta^+}(e^{\ …
Hugo Chapdelaine's user avatar
1 vote

Generating a reductive real Lie group with finitely many maximal real tori

So this is a comment to my question. In general if $G/k$ ($k$ is any field) is a connected linear algebraic $k$-group then one can show that it is generated over $\bar{k}$ by its Cartan subgroups defi …
Hugo Chapdelaine's user avatar
4 votes
3 answers
4k views

On the Weyl character formula

So let $G$ be a compact real Lie group. Let $\rho:G\rightarrow GL_n(\mathbb{C})$ be an irreducible representation of $G$ and let $\chi_{\rho}$ be the character associated to $\rho$. Let $\Lambda_{\rho …
Hugo Chapdelaine's user avatar
3 votes
3 answers
465 views

Generating a reductive real Lie group with finitely many maximal real tori

Let $G$ be be a connected real algebraic reductive Lie group. Is it always possible to find finitely many maximal algebraic $\mathbf{R}$-tori $\{T_i\}_{i=1}^n$ such that the group generated by the $T_ …
Hugo Chapdelaine's user avatar
0 votes
2 answers
641 views

Looking for general approaches to show connectedness of topological groups

Let $G$ be a topological group. One general approach to show that $G$ is connected is the following: For every subgroup $H\leq G$ (not necessarily closed) we have a projection map: $$ \pi: G\rightar …
Hugo Chapdelaine's user avatar
5 votes
3 answers
1k views

On closed totally disconnected subgroups of connected real Lie groups

So the following statement seems to be obvious but I don't see how to prove it: Q: How does one prove that a closed totally disconnected subgroup of a connected real Lie group is discrete? Note that …
Hugo Chapdelaine's user avatar
6 votes
4 answers
871 views

On the determination of a quadratic form from its isotropy group

Let $F:\mathbf{R}^n\rightarrow\mathbf{R}$ be a non-degenerate quadratic forms. Let $$ O(F):=\{g\in GL_n(\mathbf{R}):F(gv)=F(v),\forall v\in \mathbf{R}^n\} $$ be the isotropy group of $F$. Q: So how d …
Hugo Chapdelaine's user avatar
4 votes
0 answers
740 views

The normalizer a maximal compact subgroup of a semi-simple Lie group

Let $G$ be a semi-simple real Lie group such that $|\pi_0(G)|<\infty$ and let $K$ be a maximal compact subgroup of $G$. Q1: How does one prove that $N_G(K)=K$? So I know a nice (and low-tech) proof …
Hugo Chapdelaine's user avatar
4 votes
1 answer
322 views

On the isometry group of a self cartesian product of a Riemannian space

Let $X$ be a complete Riemannian space. Let us denote by $Iso(X)$ the group of isometries of $X$. It is a well-known fact that the group $Iso(X)$, when endowed with the compact-open topology, is a Lie …
Hugo Chapdelaine's user avatar
2 votes
3 answers
538 views

Non-continuous representations of $\operatorname{SL}_2(\mathbf{R})$

How does one construct a non-continuous representation $\rho:\operatorname{SL}_2(\mathbf{R})\rightarrow G$ for some connected (finite dimensional) Lie group $G$?
Hugo Chapdelaine's user avatar
20 votes
4 answers
2k views

homotopy type of connected Lie groups

Is there a simple proof (short and low-tech) of the following fact: (E. Cartan) A connected real Lie group $G$ is diffeomorphic (as a manifold) to $K\times\mathbb{R}^n$ where $K$ is a maximal compac …
Hugo Chapdelaine's user avatar