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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
$$
Q(H):=\prod_{\alpha\in\Delta^+}(e^{\ …
1
vote
Generating a reductive real Lie group with finitely many maximal real tori
So this is a comment to my question. In general if $G/k$ ($k$ is any field) is a connected linear algebraic $k$-group then one can show that it is generated over $\bar{k}$ by its Cartan subgroups defi …
4
votes
3
answers
4k
views
On the Weyl character formula
So let $G$ be a compact real Lie group. Let $\rho:G\rightarrow GL_n(\mathbb{C})$ be an
irreducible representation of $G$ and let $\chi_{\rho}$ be the character associated to
$\rho$. Let $\Lambda_{\rho …
3
votes
3
answers
465
views
Generating a reductive real Lie group with finitely many maximal real tori
Let $G$ be be a connected real algebraic reductive Lie group. Is it always possible to find finitely many maximal algebraic $\mathbf{R}$-tori $\{T_i\}_{i=1}^n$ such that the group generated by the $T_ …
0
votes
2
answers
641
views
Looking for general approaches to show connectedness of topological groups
Let $G$ be a topological group. One general approach to show that $G$ is connected is the following:
For every subgroup $H\leq G$ (not necessarily closed) we have a projection map:
$$
\pi: G\rightar …
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 …
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 …
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 …
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 …
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 $G$?
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 …