Questions tagged [topological-groups]

A topological group is a group $G$ together with a topology on the elements of $G$ such that the group operation and group inverse function are both continuous (with respect to the topology).

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43 votes
7 answers
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Why are free groups residually finite?

Why is it that every nontrivial word in a free group (it's easy to reduce to the case of, say, two generators) has a nontrivial image in some finite group? Equivalently, why is the natural map from a ...
Owen Biesel's user avatar
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23 votes
5 answers
4k views

Fundamental groups of topological groups.

Let $G$ be a topological group, and $\pi_1(G,e)$ its fundamental group at the identity. If $G$ is the trivial group then $G \cong \pi_1(G,e)$ as abstract groups. My question is: If $G$ is a non-...
Chris 's user avatar
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24 votes
1 answer
2k views

Which principlal bundles are locally trivial?

If $H$ is a closed subgroup of a topological group $G$, then the orbit map $G\to G/H$ is a principal bundle, yet somewhat surprisingly, it need not be locally trivial. In the wikipedia article on ...
Igor Belegradek's user avatar
20 votes
4 answers
3k views

Find a "natural" group that contains the quotient of the infinite symmetric group by the alternating subgroup

Let $S_\infty$ the group of permutations of $\mathbb{N}$. It can be shown that there is no homomorphism $S_\infty \to \mathbf{Z}/2$ extending the sign on the finite symmetric groups. Is it possible to ...
Martin Brandenburg's user avatar
66 votes
1 answer
6k views

Why can't a nonabelian group be 75% abelian?

This question asks for intuition, not a proof. An earlier question, Measures of non-abelian-ness was thoroughly answered by Arturo Magidin. A paper by Gustafson1 proves that, for a nonabelian group, ...
Joseph O'Rourke's user avatar
21 votes
0 answers
713 views

Closed connected additive subgroups of the Hilbert space

It is a classical result that a closed and connected additive subgroup of $\mathbb{R}^n$ is necessarily a linear subspace. However, this is no longer true in infinite dimension: a very easy example is ...
Pietro Majer's user avatar
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20 votes
2 answers
1k views

Without choice, can every homomorphism from a profinite group to a finite group be continuous?

In ZFC, some homomorphisms from profinite groups to finite groups are discontinuous. For instance, see the examples in this question. However, all three constructions given use consequences of the ...
Will Sawin's user avatar
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15 votes
3 answers
1k views

Countable subgroups of compact groups

What is known about countable subgroups of compact groups? More precisely, what countable groups can be embedded into compact groups (I mean just an injective homomorphism, I don't consider any ...
Konstantin Slutsky's user avatar
14 votes
3 answers
913 views

Examples of locally compact groups that do not admit enough finite dimensional representations

I apologize in advance if this is well-known, but I can't seem to find the answer in the literature. Let me be precise about my question. I am looking for concrete examples of locally compact ...
Rick Sternbach's user avatar
12 votes
0 answers
331 views

Metric completion of an algebraically closed field is algebraically closed?

Let $F$ be a complete metric topological field. Suppose there is a subfield $F_1 \subset F$, algebraically closed and topoolgically dense in $F$. Must $F$ itself be algebraically closed? We can ...
Gerald Edgar's user avatar
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7 votes
1 answer
312 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
6 votes
2 answers
988 views

Bohr compactification as a topological compactification

Let $G$ be a locally compact Hausdorff group. Denote its Bohr compactification by $bG$. Despite group structure, $G$ has several (Hausdorff) compactifications that, in a sense, the smallest one is ...
XIII's user avatar
  • 707
6 votes
1 answer
167 views

Can each non-open analytic subgroup of a Polish abelian group be covered by countably many closed Haar null subsets?

By a result of Laczkovich ('Analytic subgroups of the reals' Proc AMS Vol 126 (1998)), any non-open analytic subgroup of a Polish locally compact group can be covered by countably many closed Haar ...
Taras Banakh's user avatar
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5 votes
2 answers
444 views

Is each locally compact group topology on the permutation group discrete?

Question. Is each locally compact group topology on the permutation group $S_\omega$ discrete? Here $S_\omega$ is the group of all bijections of the countable ordinal $\omega$. A group topology on a ...
Taras Banakh's user avatar
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4 votes
2 answers
1k views

Subgroups of a topological Group such that quotient space is totally disconnected

If $G$ is a topological group and $G_{{e}}$ is the identity component, the it is well known that $G_{{e}}$ is a normal subgroup of $G$ and the quotient group $G/G_{{e}}$ is totally disconnected. What ...
Steve's user avatar
  • 255
3 votes
1 answer
147 views

Totally bounded group topologies on $\Bbb Q$ with trivial intersection of two neighborhoods

Are there totally bounded group topologies $\mathcal S$ and $\mathcal T$ on $\Bbb Q$ such that for some open sets $A\in\mathcal S$ and $B\in \mathcal T$ we have $A\cap B=\{0\}$?
Minimus Heximus's user avatar
3 votes
4 answers
902 views

Is there a compact connected Hausdorff space in which every non-empty $G_\delta$ set has non-empty interior?

Q1. Is there a compact connected Hausdorff space (with at least two points) in which every non-empty $G_\delta$ set has non-empty interior? (Without the requirement for connectedness, every finite $...
Mirko's user avatar
  • 1,345
2 votes
0 answers
306 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 ...
Mikhail Borovoi's user avatar
1 vote
1 answer
226 views

locally closed orbits in metric Hausdorff topology

I learned the following fact from Bruhat and Tits's paper "Homomorphismes “abstraits” de groupes algebriques simples" Section 3.18 that Let $k$ be a local field. Suppose that a $k$-group $H$ acts $k$...
m07kl's user avatar
  • 1,672
49 votes
5 answers
8k views

Fundamental group as topological group

Background Let $(X,x)$ be a pointed topological space. Then the fundamental group $\pi_1(X,x)$ becomes a topological space: Endow the set of maps $S^1 \to X$ with the compact-open topology, endow the ...
Martin Brandenburg's user avatar
32 votes
4 answers
4k views

Compact open topology on $\mathrm{Homeo}(X)$

Let $X$ and $Y$ be topological spaces. Define the compact open topology on the set $\mathrm{M}(X,Y)$ of continuous maps from $X$ to $Y$ via the subbase $[K,O]$ of all maps $f:X\rightarrow Y$ s.t. $f(K)...
Olivier Bégassat's user avatar
32 votes
3 answers
3k views

morphism from a compact group to Z ?

I wonder if it there exists a topological compact group $G$ (by compact, I mean Hausdorff and quasi-compact) and a non-zero group morphism $\phi : G \to \mathbb{Z}$ (without assuming any topological ...
Florent MARTIN's user avatar
20 votes
2 answers
2k views

Is every compact topological ring a profinite ring?

There are a lot of compact (Hausdorff) groups, whereas every compact field is finite. What about rings? Is there a classification theorem for compact rings? If you take a cofiltered limit of finite ...
Gene S. Kopp's user avatar
  • 2,190
19 votes
2 answers
529 views

Ostrowski's Theorem for topological rings?

Ostrowski's theorem classifies all absolute values on a number field $K$. Questions: More generally, can one classify all Hausdorff topologies on $K$ making $K$ into a topological field? In ...
Tim Campion's user avatar
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19 votes
9 answers
6k views

Haar measure on a quotient, References for

I remember reading Weil's "Basic Number Theory" and giving up after a while. Now I find myself thinking of it (thanks to some comments by Ben Linowitz). Right from the very beginning, Weil ...
Anweshi's user avatar
  • 7,262
16 votes
1 answer
1k views

A possible mistake in Walter Rudin, "Fourier analysis on groups"

I have the following lemma 4.2.4 on page 80 in the book (we have locally compact abelian topological groups $G_1, G_2$ and their duals $\Gamma_1, \Gamma_2$): Suppose $E$ is a coset in $\Gamma_2$ ...
Petr Naryshkin's user avatar
15 votes
3 answers
3k views

Are measurable homomorphisms $ (\Bbb{C},+) \to (\Bbb{C},+) $ or $ (\Bbb{C},+) \to (\Bbb{C},*) $ continuous, and do they admit an explicit description?

I am interested in generalizations of the following fact (known as automatic continuity, as pointed out below). I am especially looking for references to papers dating back to 1920’s. I feel that ...
mmm 's user avatar
  • 1,299
14 votes
1 answer
557 views

How flexible is the infinite-dimensional torus?

Let $\mathbb T=\mathbb R/\mathbb Z$ be the circle group and $\mathbb T^\omega$ be the infinite-dimensional torus, considered as an abelian compact topological group. Problem 1. Is it true that for ...
Taras Banakh's user avatar
  • 40.7k
13 votes
2 answers
1k views

Baire Category Theorem for complete uniform spaces

The version of the Baire Category Theorem I have in mind is the statement that a countable intersection of dense open subsets of a complete metric space is dense. The question is: is it likewise ...
Jonathan Gleason's user avatar
13 votes
1 answer
447 views

A generalization of residual finiteness to topological groups

Consider the following generalization of residual finiteness to topological groups. A locally compact Hausdorff group $G$ is called residually compact if for every compact $K \subseteq G$ there is a ...
Jeremias Epperlein's user avatar
12 votes
0 answers
367 views

Does each compact topological group admit a discontinuous homomorphism to a Polish group?

A compact topological group $G$ is called Van der Waerden if each homomorphism $h:G\to K$ to a compact topological group is continuous. By a classical result of Van der Waerden (1933) the groups $SO(...
Taras Banakh's user avatar
  • 40.7k
12 votes
2 answers
874 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
10 votes
1 answer
790 views

Is $k(\!(x,y)\!)$ a topological field?

More generally, let $(R,m)$ be a Noetherian local domain with fraction field $K$. The $m$-adic topology turns $R$ into a topological ring. When $R$ is a discrete valuation ring, this topology extends ...
Laurent Moret-Bailly's user avatar
9 votes
1 answer
288 views

What is known about topological groups of countable spread in ZFC?

A topological space has countable spread if every discrete subspace is at most countable. By Theorem 8.10 in Todorcevic's book "Partition Problems in Topology", PFA implies that each regular space $X$...
Taras Banakh's user avatar
  • 40.7k
9 votes
1 answer
525 views

Existence of infinite groups that are too reluctant to be topological

With ZFC, is there an infinite group $G$ such that there is no non-trivial non-discrete topology on $G$ with the functions $G\times G\to G,~~ (a,b) \mapsto ab$ and $G\to G,~~ a\mapsto a^{-1}$ ...
Minimus Heximus's user avatar
9 votes
1 answer
198 views

Does each discrete solvable group admit an injective homomorphism to a compact topological group?

It is well-known that each abelian group admits an injective homomorphism to some compact topological group (for example to its Bohr compactification). Is the same fact true for solvable groups? ...
Taras Banakh's user avatar
  • 40.7k
9 votes
2 answers
667 views

Potential connected non-Lie subgroup

This painful question is inspired by the question "non-Lie subgroups" . Let $f$ be a discontinuous additive map from $\mathbb{R} \to \mathbb{R}$. Is it possible that the graph of $f$, inside ...
David E Speyer's user avatar
9 votes
1 answer
396 views

Meager subgroups of compact groups

Suppose we have an infinite compact (Hausdorff) group $G$, and a subgroup $H\leq G$ which is meagre. Can $H$ always be covered by a countable family of nowhere dense sets $H_n$ such that $H_n^2$ is ...
tomasz's user avatar
  • 1,184
8 votes
3 answers
2k views

Measures on general topological groups

I am interested in the group algebras of non-locally compact groups. What references can you advise? This is a wide question, so I list more concretely what I would like to see: Here X can be even ...
Yulia Kuznetsova's user avatar
8 votes
2 answers
399 views

Homomorphisms from $\mathbb{R}$ to $\mathrm{Homeo}^+(\mathbb{R})$, or "fractional iterations"

Let $G$ be the group of orientation-preserving homeomorphisms (or, if you prefer, diffeomorphisms) of the real line. Does there exist a natural way to associate, to each function $f \in G$, a ...
Charles Staats's user avatar
8 votes
2 answers
330 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
  • 40.7k
8 votes
2 answers
744 views

Avoiding countable subgroups of general uncountable groups

The following problem is a general form of another problem (motivation is available there). Initially, the problems were posted together, but the first one is solved below, a solution that does not ...
Boaz Tsaban's user avatar
  • 3,102
7 votes
2 answers
1k views

The group of diffeomorphisms with compact support

Let $M$ be a topological/differentiable manifold. Is there any topology on the group of homeomorphisms/diffeomorphisms with compact support, turning it into a (locally-)compact topological group? (My ...
Alex M.'s user avatar
  • 5,197
7 votes
1 answer
661 views

Topological groups in which all subgroups are closed

General question: does there exist a nondiscrete topological group $G$ such that all subgroups of $G$ are closed? Or, does there exist a nondiscrete topological vector space $V$ such that all vector ...
Leonid Positselski's user avatar
6 votes
1 answer
263 views

A property of rapid sequences of natural numbers

$\newcommand{\IR}{\mathbb R}$ $\newcommand{\IT}{\mathbb T}$ $\newcommand{\w}{\omega}$ $\newcommand{\e}{\varepsilon}$ Taras Banakh and me proceed a long quest answering a question of ougao at ...
Alex Ravsky's user avatar
  • 4,092
6 votes
1 answer
766 views

Solid tensor product of pro-discrete space with Laurent series

Consider the category $\operatorname{Solid}_{\mathbf{Z}}$ of solid abelian groups in the sense of Clausen-Scholze. This category is a full subcategory of condensed abelian groups, $\operatorname{Cond}...
user avatar
6 votes
1 answer
419 views

Discrete spectrum and almost periodicity

According to Vershik, an ergodic invertible measure-preserving transformation $T$ on a Lebesgue space $X$ has discrete spectrum if and only if for every bounded measurable function $f\colon X \to \...
Stéphane Laurent's user avatar
6 votes
2 answers
499 views

Hausdorff group topologies on finitely generated groups

Suppose $G$ is a finitely generated Hausdorff topological group. Must $G$ be first countable (or perhaps a sequential space)? What if we restrict to the abelian case? I wonder if this is even true ...
Jeremy Brazas's user avatar
6 votes
1 answer
186 views

Steinhaus number of a group

$\newcommand\Sn{\mathit{Sn}}$A subset $A$ of a group $X$ is called algebraic if $A=\{x\in X: a_0xa_1x\dotsm xa_n=1\}$ for some elements $a_0,a_1,\dotsc,a_n\in X$. Let $\mathcal A_X$ be the family of ...
Taras Banakh's user avatar
  • 40.7k
5 votes
3 answers
1k views

Action of a profinite group

Let $G$ be a finitely generated profinite group, $p$ a prime number. Put $$ V = \prod_{i \in I} \mathbb{Z}_p$$ a (profinite) group equipped with the product topology (for convenience, $I$ may be ...
Pablo's user avatar
  • 11.2k