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Is every topological (resp. Lie-) group the isometrygroup of a metric space (resp. Riemannian manifold)?

The isometry group of a metric space is a topological group (with the compact open topology). The isometry group of a Riemann Manifold is a Liegroup. (Thm. of Steenrod-Myers) So, is every topological ...
archipelago's user avatar
  • 2,974
17 votes
0 answers
1k views

What groups are Lie groups?

We know how to tell if a topological group is a Lie group: this was famously asked by Hilbert and answered gloriously by Gleason, Montgomery and Zippin in the 50s (a locally compact topological group ...
Mariano Suárez-Álvarez's user avatar
12 votes
4 answers
2k views

Which compact groups have finitely many irreducible representations of each dimension?

If my understanding is correct, this is true of sufficiently nice nonabelian Lie groups (see Ben Webster's answer below), and any finite group. On the other hand, this is false for any infinite ...
Qiaochu Yuan's user avatar
12 votes
2 answers
883 views

Does almost every pair of elements in a compact Lie group generates the connected component?

It is known that almost every pair of elements in a connected compact Lie group (topologically) generates the group. Obviously this isn't true for non-connected groups but Given a compact Lie ...
Jack the Ripper's user avatar
9 votes
1 answer
657 views

Must an inverse limit of simply connected groups be simply connected?

While the fundamental group $\pi_1$ preserves products, it is not true in general that an inverse limit of simply connected topological spaces is simply connected. I would like to know if similar ...
Jeremy Brazas's user avatar
9 votes
2 answers
717 views

Topology on extensions of topological groups

Let $G$ and $H$ be two topological groups and let $\mathcal{E}:0 \to G \to E \to H \to 0$ be an extension of abstract groups. Is there a way to introduce a topology on $E$ such that $\mathcal{E}$ ...
jap's user avatar
  • 125
9 votes
0 answers
367 views

Is every space a classifying space?

Despite a pretty thorough look (I think) I can’t find the answer to the following question: Is every (reasonable?) path connected space weakly equivalent to the classifying space of some topological ...
Josh Lackman's user avatar
  • 1,198
8 votes
1 answer
730 views

Status of Hilbert-Smith conjecture and H-S conjecture for Hölder actions

The Hilbert-Smith conjecture states that If $G$ is a locally compact group which acts effectively on a connected manifold as a topological transformation group then is $G$ a Lie group. It was ...
Zarathustra's user avatar
  • 1,414
8 votes
2 answers
362 views

Is every contractible homogeneous space of a connected Lie group homeomorphic to a Euclidean space?

Problem. Let $G$ be a connected Lie group and $H$ is a closed subgroup of $G$ such that the homogeneous space $G/H$ is contractible. Is $G/H$ homeomorphic to a Euclidean space $\mathbb R^n$ for some $...
Taras Banakh's user avatar
  • 41.9k
8 votes
2 answers
1k views

$p$-adic exponentials for $p$-adic Lie groups

Let $G$ be a $p$-adic Lie group, $\text{Lie}(G)$ its Lie algebra. Is there any reasonable notion of exponential map $\text{exp} : \text{Lie}(G)\to G$?
user avatar
8 votes
0 answers
819 views

Second homotopy group of a topological group

It is well-known that any Lie group $G$ has $\pi_2(G)=0$: see this question. Is the same true for any compact (Hausdorff) topological group? Or even for locally compact ones? Maybe there is a way of ...
Mizar's user avatar
  • 3,146
7 votes
4 answers
2k views

Topological structure of SO(n) as a product

I’m interested in the question for which $n$ the special orthogonal group is homeomorphic to the product $$ \mathrm{SO}(n) \approx S^{n-1} \times \mathrm{SO}(n-1). $$ Allen Hatcher [1, p. 293 f.] ...
Florian Oppermann's user avatar
7 votes
1 answer
490 views

Classification of fibrations $\Bbb S^k\longrightarrow\Bbb S^d\longrightarrow B$

Does there exist a complete classification of all fiber bundles $\Bbb S^k\longrightarrow\Bbb S^d\longrightarrow B$, that is, fibrations of $\smash{\Bbb S^d}$ with each fiber homeomorphic to $\smash{\...
M. Winter's user avatar
  • 13.6k
7 votes
2 answers
2k views

Conditions for a topological group to be a Lie group

In flipping through the Springer lecture notes on Serre's 1964 'Lie Algebras and Lie Groups' lectures at Harvard, I found this pair of suprising results (page 157): Let $G$ be a locally compact group....
Joshua Seaton's user avatar
7 votes
1 answer
342 views

Does a compact Lie group have finitely many conjugacy classes of maximal Abelian Lie subgroups?

Let $G$ be a compact Lie group. An Abelian Lie subgroup $A \leq G$ is a maximal Abelian Lie subgroup if, for any Abelian Lie subgroup $A'$ such that $A \leq A' \leq G$, then $A' = A$. Of course any ...
Dominic Else's user avatar
7 votes
1 answer
270 views

Is $\Gamma(p) := \text{Ker}(SL_2(\mathbb{Z}_p)\rightarrow SL_2(\mathbb{F}_p)$ a "standard" subgroup?

Let $\Gamma(p) := \text{ker}(SL_2(\mathbb{Z}_p)\rightarrow SL_2(\mathbb{Z}_p/p))$. Viewing $SL_2(\mathbb{Z}_p)$ as an analytic group, is there a formal group law $F$ in three variables, defined over $...
stupid_question_bot's user avatar
7 votes
1 answer
205 views

Non-recurrent points of $F(a,b)=(b,ba)$ in a compact metric group $G$

Consider a compact metric group $G$ [A compact topological group $G$ where the topology is generated by an invariant metric]. I am particularly interested in the case where $G$ is the $n$-dimensional ...
Saúl Pilatowsky-Cameo's user avatar
6 votes
2 answers
392 views

Union of conjugates of a closed subgroup of a compact group

Let $G$ be a compact Hausdorff group, $H \leq G$ a closed subgroup of infinite index in $G$. Is it possible that the conjugates of $H$ cover some open neighbourhood of $1$ in $G$ (or the whole of $G$...
Pablo's user avatar
  • 11.3k
6 votes
2 answers
379 views

About Lie group $G$ has this escape property?

Every Lie group $G$ has the following escape property: For every $x \ne e$ in a sufficiently small neighborhood $U$ of the identity $e$ in $G$, there is a integer $n$ such that $x^n$ is not in $U$. ...
free's user avatar
  • 71
6 votes
1 answer
1k views

Classification of compact connected abelian groups

It is known that torsion-free compact abelian groups are exactly the product of the maximal solenoid group $\Sigma_{(2,3,\cdots)}$ (which is the Pontryagin dual of the additive group $\mathbb{Q}$ of ...
Rick Sternbach's user avatar
6 votes
3 answers
1k views

$\pi_1$ Sequence of Topological Groups

Consider a connected topological group $G$ (not necessarily Lie). You have some maps $G\times G\to G$, such as projection to either summand, or multiplication $(g,h)\mapsto gh$. Now let's look at a ...
Chris Gerig's user avatar
  • 17.5k
6 votes
1 answer
235 views

Hausdorff distance in compact Lie groups

Let $G$ be a compact Lie group with a compatible biinvariant metric $d$. The hyperspace $K(G)$ of nonempty compact subsets of $G$ is a compact metric space with the Hausdorff metric, and it is easy to ...
chj's user avatar
  • 157
6 votes
1 answer
287 views

Are nearby crossed homomorphisms from compact Lie groups crossed-conjugate?

Charles Rezk had highlighted in MO:q/123624 that "Nearby homomorphisms from compact Lie groups are conjugate", and in consequence -- further highlighted in Remark 2.2.1 of his Global ...
Urs Schreiber's user avatar
6 votes
1 answer
249 views

Growth function of locally compact groups

Every locally compact second countable group $G$ has a regular left-invariant measure $h$, the Haar measure. On the other hand the Birkhoff–Kakutani Theorem asserts that such groups also admit a ...
Alessandro Carderi's user avatar
6 votes
0 answers
244 views

What can lattices tell us about lattices?

A general group-theoretic lattice is usually defined as something like A discrete subgroup $\Gamma$ of a locally compact group $G$ is a lattice if the quotient $G/\Gamma$ carries a $G$-invariant ...
Mark Schultz-Wu's user avatar
6 votes
0 answers
341 views

When exponential map is 1-1 from vector fields to diffeomorphisms

Let $M$ be a connected and complete Riemannian manifold of positive dimension, $k$ be a positive integer, and let $\mathfrak{X}^k_c$ be the set of class $C^k$-vector fields on $M$ of compact support. ...
ABIM's user avatar
  • 5,405
6 votes
0 answers
2k views

Fourier transforms via Kurzweil-Henstock integral on locally compact commutative groups

Is it possible to define Fourier transforms on locally compact commutative groups using the Kurzweil-Henstock integral instead of the Lebesgue integral?
teil's user avatar
  • 4,351
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
5 votes
1 answer
356 views

Density of matrix coefficients of unitary representations of a locally compact group

Let $G$ be a locally compact group, $C_0(G)$ the $C^*$-algebra of continuous functions on $G$ that vanish at infinity, $C_b(G)$ the $C^*$-algebra of bounded continuous functions on $G$. We know that $...
Rick Sternbach's user avatar
5 votes
0 answers
184 views

Subgroups of $\mathrm{O}_3$ that are the symmetry groups of compact subsets of $\mathbb{R}^3$

Is there a classification theorem for the subgroups of $\mathrm{O}_3$ that are the symmetry groups of compact subsets of $\mathbb{R}^3$? Apparently, there is an almost complete classification in ...
Arshak Aivazian's user avatar
5 votes
0 answers
300 views

Matrix groups with two generators

Given two matrices $A,B\in{\rm{SL}}_2(\Bbb{R})$, is there any criterion guaranteeing that the subgroup they generate is discrete? What if one puts restrictions on $A,B$ e.g. they are both elliptic? ...
KhashF's user avatar
  • 3,599
5 votes
0 answers
129 views

Is there an orbit map without path lifting property?

I am looking for an example of a topological group $G$ acting by homeomorphisms on a metrizable space $X$ such that the orbit map $X\to X/G$ doesn't have the path lifting property, that is, there is a ...
Igor Belegradek's user avatar
5 votes
0 answers
119 views

Characterizing Herz-Schur multipliers using coefficient functions of uniformly bounded representations

Let $G$ be a group and let $c > 1$ be a constant. We denote by $B_c(G)$ the space of all coefficients of the representations of $G$ which are uniformly bounded by $c$; more precisely, a function $f:...
Mahmood Al's user avatar
4 votes
3 answers
2k views

Why is the unipotent radical of a parabolic subgroup unimodular?

Let $G$ be a connected, reductive group defined over a local field $F$. Let $P$ be a parabolic subgroup of $G$ which is defined over $F$, and let $N = \mathscr R_u(P)$ be the unipotent radical of $P$....
D_S's user avatar
  • 6,180
4 votes
1 answer
344 views

Extensions of compact Lie groups

Let $G$, $H$ be two compact Lie groups (possibly disconnected). Two short exact sequences of compact Lie groups $$ 0\rightarrow G\rightarrow M_1 \rightarrow H\rightarrow 0, $$ $$ 0\rightarrow G\...
rori's user avatar
  • 231
4 votes
2 answers
2k views

A question about unitary and anti-unitary matrices

The question is the following: Let $U:\mathbf{C}^n\to \mathbf{C}^n$ be a unitary operator; let $\tilde{U}:\mathbf{C}^n\to\mathbf{C}^n$ be an antiunitary operator. Can one deform $U$ to $\tilde{U}$ ...
jacaboul's user avatar
  • 327
4 votes
1 answer
304 views

Finite covolume of uniform lattice in quotient group

Let $G$ be a locally compact group, let $N \leq G$ be a (proper) closed normal subgroup and let $\Gamma \leq G$ be a uniform lattice, i.e., a discrete subgroup such that $G/\Gamma$ Is compact. Suppose ...
user avatar
4 votes
1 answer
384 views

Invariant integration on principal bundles

Let $G$ be a sufficiently nice topological or Lie group (e.g. compact), and let $H$ be a closed subgroup. This data determines a principal $H$ bundle $G \rightarrow G/H$ defined by the projection $g \...
John von N.'s user avatar
4 votes
0 answers
132 views

Coordinates on quotient manifold $\mathrm{SO}(3)/\Gamma$

$\DeclareMathOperator\SO{SO}$Say I have coordinates for $\SO(3,\mathbb{R})$, e.g., a parametrization by Euler angles. Is there a reasonable way to explicitly prescribe coordinates on the quotient ...
Eric Kubischta's user avatar
4 votes
0 answers
640 views

Closed subgroups of a connected Lie group

Is it true that for any closed subgroup $H$ of a connected Lie group $G$, the group of connected components $\pi_0(H)$ is finite or countable? (inspired by the comment of nfdc23 to this question ).
Mikhail Borovoi's user avatar
4 votes
0 answers
264 views

When can a locally compact group be approximated by discrete subgroups?

This question is about partitioning a (locally) compact group into cells by using discrete subgroups. Let $G$ be a locally compact group. (I am really most interested in the case where $G$ is a ...
Jason Rute's user avatar
  • 6,287
3 votes
1 answer
197 views

Is the exponential map of a locally compact group a local homeomorphism?

We consider a locally compact abelian group $G$. We equip the real vector space $A(G)$ of continuous group homomorphisms $\mathbb{R}\to G$ with the topology of uniform convergence on compact subsets ...
Hans's user avatar
  • 3,031
3 votes
1 answer
454 views

Principal bundles from a fibration of homogeneous spaces

Let $G$ be a compact (Lie) group, and $H \subseteq H'$ two compact (Lie) subgroups. It is clear that we have an obvious surjective map of homogeneous spaces $$ G/H \twoheadrightarrow G/H'. $$ Will it ...
Spyros Olympopolous's user avatar
3 votes
1 answer
201 views

Is each closed subgroups of $\mathbb R^\omega$ isomorphic to a Tychonoff product of locally compact Abelian groups?

It is known that any closed linear subspace of $\mathbb R^\omega$ is topologically isomorphic to $\mathbb R^n$ for some $n\in\omega$. Problem 1. Is each closed subgroup of $\mathbb Z^\omega$ (or ...
Taras Banakh's user avatar
  • 41.9k
3 votes
1 answer
122 views

A converse of Cartan's automatic continuity theorem

Let $G$ be a compact real Lie group. We say that $G$ has property $(*)$ if every abstract automorphism of $G$ is continuous. A theorem of Cartan says that if $G$ has perfect Lie algebra, it has ...
rori's user avatar
  • 35
3 votes
1 answer
380 views

A Comparison between $\pi_{1}$ of cohomology and cohomology of $\pi_{1}$

Let we have a complex of abelian topological or lie groups $$\ldots \to G_{n}\to G_{n+1}\to \ldots$$ such that the image of $G_{n}$ is a closed subgroup of $G_{n+1}$. Then we have a complex of ...
Ali Taghavi's user avatar
3 votes
1 answer
141 views

Measure on orbits of $N$ under conjugation by $H$

Let $G$ be a locally compact topological group with closed subgroups $H, N$ and $H$ normalizing $N$. Then $H$ acts continuously on $N$ by conjugation. If it will help, assume that $N$ is nilpotent, ...
D_S's user avatar
  • 6,180
3 votes
1 answer
142 views

What's the topology on the mapping space $Map_H(G, Y)$ when $G$ is not finite

When $G$ is a finite group and $H$ a closed subgroup of it, the sets of right cosets $H\backslash G$ has the discrete topology on it. Let $Y$ be a $H-$space. We have the $G-$homeomorphism \begin{...
Megan's user avatar
  • 1,040
3 votes
0 answers
58 views

Criteria for density of subgroup of diffeomorphism group

Let $C^{\infty,\star}(\mathbb{R}^d)$ denote the non-commutative topological group of smooth diffeomorphisms from $\mathbb{R}^d$ to itself with $\circ$ as multiplication and let $\emptyset\subset X\...
ABIM's user avatar
  • 5,405
3 votes
0 answers
335 views

Milnor's conjecture on Lie group (co)homology and forgetful functor of extensions

Let $G$ and $H$ be compact Lie groups, Consider $Ext_{Lie}(G,H)$ the set of isomorphism of extensions of Lie groups: $$ 1\rightarrow G\rightarrow M\rightarrow H\rightarrow 1 $$ There exists a ...
Tsemo Aristide's user avatar