All Questions
Tagged with lie-groups topological-groups
79 questions
3
votes
0
answers
64
views
Metrically homogeneous spaces as inverse limits
Let $(X,d)$ be a locally compact, separable, connected and $\sigma$-compact metric space such that the group of isometries $G$ acts transitively on $X$. The question is the following:
Is $X$ ...
2
votes
2
answers
423
views
Lie (and topological) group extensions of $\mathbb{R}^2$ by $\mathbb{R}$
What are all the non-split Lie (and topological) group extensions $0 \to \mathbb{R} \to G \to \mathbb{R}^2 \to 0$? Here, $\mathbb{R}$ and $\mathbb{R}^2$ are regarded as Lie (and topological) groups ...
2
votes
1
answer
282
views
Does every locally compact group G contain a maximal open subgroup P which is a pro-Lie group?
EDIT 1: All topological groups in this question are assumed to be second countable. In particular, this forces every group to be metrizable and every Lie group to have at most countably many ...
2
votes
1
answer
101
views
Compact Lie groups as quotients of torsion-free compact metrizable groups
The question:
(1) Is every compact Lie group $G$ isomorphic (as a topological group) to some quotient $H/N$ where $H$ is a torsion-free compact metrizable group?
Or equivalently:
(2) Is every compact ...
2
votes
1
answer
216
views
How to prove that Chevalley groups over $\mathbb R$ have no compact factors
I am trying to see why the Chevalley groups (not limited to the adjoint group) over $\mathbb R$ are without compact factors in order to use the Borel density theorem.
I've been told in another thread ...
2
votes
1
answer
182
views
Gelfand-Naimark and Peter-Weyl for the unitary group
Consider the compact Lie groups $U(l)$ (the unitary group) and $U(1) \times SU(l)$ for some natural number $l$. Both the groups have the same Lie algebra $\frak{gl}_l$. Which means that they both have ...
2
votes
1
answer
82
views
Structure of extensions arising in Lie approximation of connected groups
My imperfect understanding is that, by the work of various authors (Gleason, Yamabe, Montgomery, Zippin ...), the following result is known:
Let $G$ be a connected, locally compact, Hausdorff group, ...
2
votes
1
answer
263
views
Profinite extension of a Lie group
Let $H,G,K$ be three topological groups, we say that $G$ is an extension of $K$ by $H$ if the following short sequence
$$0\rightarrow H\rightarrow G\rightarrow K\rightarrow 0$$
is exact. (If $H$ is a ...
2
votes
1
answer
571
views
On homeomorphic compact connected topological groups
I wish to thank Professor Claudio Gorodski for his very helpful
answers to my question on the webcite:
If compact connected Lie groups are homeomorphic as topological space, are they isomorphic as Lie ...
2
votes
0
answers
85
views
Coherent states on compact abelian state spaces and complexification
First, to establish notation, let $T^*(M)$ denote the cotangent bundle of a manifold $M$. Let $\widehat{(-)}:= \hom_{\sf LCAbGrp}(-,\mathbb{T}):{\sf LCAbGrp}^{\sf op}\simeq {\sf LCAbGrp}$ denote the ...
2
votes
0
answers
86
views
Homomorphisms from circle to $GL(k,\mathbb{R})$ [duplicate]
Example 3 at the website tricki proves that every measurable homomorphism of groups from the circle to the non-zero complex numbers is continuous. Is there any analogous (true) statement for ...
2
votes
0
answers
91
views
Fixed point set with non-empty interior
Let $G$ be an infinite compact separable Hausdorff metric group, and $H\subset G$ a closed subgroup, such that the left $G$-action on $G/H$ is effective (i.e., $H$ doesn't contain a non-trivial closed ...
2
votes
0
answers
80
views
Are the roots of an infinitely divisible probability infinitely divisible themselves?
Let $\mu$ be an infinitely divisible probability on a topological group $G$. If $\nu ^{* n} = \mu$ for some $n$, is $\nu$ an infinitely divisible probability too?
A sufficient criterion would be to ...
2
votes
0
answers
321
views
Surjective homomorphisms of non-connected Lie groups
Let $\psi\colon B\to C$
be a homomorphism of real Lie groups, where the group $C$ is connected.
Let $B^0$ denote the identity component of $B$, and we set $\pi_0(B)=B/B^0$, then $\pi_0(B)$ is a ...
2
votes
0
answers
82
views
Uniquely divisible neighborhoods of identity in topological groups
Let $G$ be a (finite dimensional real) Lie group, and let $A\subset G$ be an open neighborhood of identity. If $A=\operatorname{Exp}(\mathcal{A})$ is the injective range of the exponential map from a ...
2
votes
0
answers
62
views
Codistal subgroups of locally compact groups
Let $G$ be a topological group and let $H$ be a closed subgroup of $G$. Say $H$ is codistal in $G$ if the translation action of $G$ on the coset space $G/H$ is distal (meaning that no non-diagonal ...
1
vote
2
answers
481
views
Cross section for closed Lie subgroup in a Lie group
Let $G$ be a Lie group and $H$ a closed Lie subgroup. Is there an explicit way to construct a local cross section of $H$ in $G$ so that $\pi: G\to G/H$ is a fiber bundle?
1
vote
1
answer
111
views
Is every compact quasisimple group a Lie group?
Let $ G $ be a compact topological group which is quasisimple in the sense that
$$
[G,G]=G
$$
and
$$
G/Z(G)
$$
is simple as an abstract group. Must $ G $ be a Lie group?
This is a follow-up question ...
1
vote
1
answer
338
views
distance between unitary and anti-unitary matrices
This question is related to the previous post, "A question about unitary and anti-unitary matrices". Following the suggestion of Lspice, I am posting it as a separate question, as it might ...
1
vote
0
answers
43
views
Continuous surjection from $X(D_n)$ onto $\operatorname{Homeo}_0(D_n)$
Let $n>1$ and let $\mathfrak{X}(D_n)$ denote the set of continuous vector fields on the closed disc $D_n\subseteq \mathbb{R}^n$. Let $\operatorname{Homeo}_0(D_n)$ be the set of homeomorphism of ...
1
vote
0
answers
164
views
Continuous vs $L^2$ homomorphisms from circle to non-zero complex numbers
Let $T:S^1\to C^\ast$ be a group theoretic homomorphism from the circle to the non-zero complex numbers.
Presumably it is true that if $T$ is $L^2$, then it is continuous. Is there a simple proof, or ...
1
vote
0
answers
140
views
Describing compact Lie groups in purely topological terms
Compact Lie groups are a very special type of compact group, namely those which admit a differentiable structure. Is it possible to describe compact Lie groups in purely topological terms, that is, ...
1
vote
0
answers
121
views
A section over an orbit space
Let $G$ be a compact second countable Hausdorff group, and let $X=G/H$ be a homogeneous space with $H\subset G$ a closed subgroup. Let further $K\subset G$ be another closed subgroup.
Questions:
...
1
vote
0
answers
128
views
The group of polynomial homeomorphism of the plane
Let $G$ be the set of all homeomorphisms $f$ of $\mathbb{R}^2$ such that
both $f$ and $f^{-1}$ are polynomial maps.
We equip $G$ with the compact open topology and the obvious group ...
0
votes
1
answer
434
views
Reference request: Any connected Lie group has a countable base for its topology
I am looking for a reference for the assertion in the title. This assertion is proved in a comment of user nfdc23 to this question. Has any proof of this assertion been published?
0
votes
0
answers
98
views
An application of the Gleason-Montgomery-Zippin Theorem
In the book How groups grow by Avinoam Mann, the author cites the following theorem attributed to Gleason-Montgomery-Zippin.
Theorem 6.4 (Gleason–Montgomery–Zippin: solution of Hilbert’s Fifth ...
0
votes
1
answer
83
views
Is the union of 1-dimensional pro-tori in a finite dimensional pro-torus dense?
Is the union of 1-dimensional compact connected abelian subgroups in a finite dimensional compact connected abelian group dense?
0
votes
0
answers
267
views
Definition of reducible lattice
I am reading Raghunathan's book on discrete subgroups of Lie groups.
In particular I am stuck on Corollary 5.19 which gives several equivalent conditions for a lattice in a semisimple Lie group to be ...
-3
votes
4
answers
5k
views
What is the situation with Hilbert's Fifth Problem?
The common knowledge in this regard seems to be that Hilbert's Fifth Problem was completely solved by Gleason, Montgomery, and Zippin. However, such wisdom was contested by Peter Olver:
Olver, Peter ...