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5 votes
0 answers
135 views

Possible homogeneity of infinite dimensional Sierpinski carpet analogues?

Start with the Hilbert cube $H=I^\omega$, thinking of its coordinates as written in ternary expansion. Construct subsets $S_n$ by removing points from $H$ if for any $m$, at least $n$ of the ...
David Feldman's user avatar
11 votes
2 answers
2k views

Two Definitions of "Character" of topological groups

When I first met the concept of "characters" of topological groups in Pontryagin's book "Topological groups", it was defined as follows: Let $G$ be a topological group. A ...
Hiro's user avatar
  • 945
12 votes
1 answer
1k views

(Closures of sets of) operations in topological groups.

Let $G$ be a topological group. For each $n \in \mathbb{Z}$, consider the continuous functions $f_{n} \colon G \to G : x \mapsto x^{n}$, and set $F := \{f_{n} \mid n \in \mathbb{Z}\}$. Is there a ...
Niemi's user avatar
  • 1,498
3 votes
1 answer
401 views

Action on a compact group

If $G$ is an infinite compact group, how many orbits can $G$ have under the group action of its continuous automorphisms ?
Drike's user avatar
  • 1,555
0 votes
1 answer
296 views

homeomorphism of topological group

Let G be a topological group and a be an element of order 2 in G. Further suppose the element a does not belong to center of G. Then is it true that only homeomorphism f of G such that $f(ax)=f(x)a$ ...
jasp's user avatar
  • 1
3 votes
2 answers
483 views

When does a LCA group not contain a (closed) infinite cyclic subgroup?

If $G$ is an LCA (locally compact abelian) group, is there any 'nice' sufficient (or preferably necessary and sufficient) criteria for when $G$ does not contain a closed (and hence discrete in the ...
Iian Smythe's user avatar
  • 3,115
6 votes
2 answers
492 views

Distinct, non-homeomorphic, profinite topologies on a given abstract group ?

Just a silly little question which arose in connection with infinite Galois groups and their Krull topology:- can a given abstract group be endowed with distinct, non-homeomorphic, profinite ...
Stephan F. Kroneck's user avatar
13 votes
1 answer
736 views

Idempotent measures on the free binary system?

Let $(S,*)$ be the free (non associative) binary system on one generator (so $S$ is just the set of terms in $*$ and $1$). There is an extension of $*$ to the space $P(S)$ of finitely additive ...
Justin Moore's user avatar
  • 3,547
18 votes
7 answers
2k views

Superfluous definitions

It is well known that the axioms of a ring R with unity 1 imply that the underlying group must be commutative. For if a and b are elements of R, and writing + for the group operation then applying ...
7 votes
2 answers
594 views

Computational cost of converting between 3-manifold presentations

Given a 3-manifold presented as a triangulation, a Heegaard splitting, or a Dehn surgery, what is the computational cost of converting to the other two presentations? I would like Heegaard splittings ...
Gorjan Alagic's user avatar
4 votes
2 answers
2k views

Irreducible unitary representations of locally compact groups

Let $G$ be a locally compact group and let $\mu$ be a left Haar measure. We know that $\mu$ is unique up to a scalar in $\mathbf{R}_{>0}$. I don't know so much about unitary representations of ...
Hugo Chapdelaine's user avatar
5 votes
0 answers
501 views

Profinite topologies

We can define two topologies on a group $G$ by considering all normal subgroups of finite index (resp. of index a finite power of $p$ - where $p$ is a prime) as basis of $1\in G$. My questions: Under ...
Alves's user avatar
  • 51
10 votes
2 answers
2k views

pro-discrete = locally compact and open normal subgroups have trivial intersection?

EDIT: After talking to some experts on the subject, I have concluded that a) the answer is not obvious or well-known for locally compact groups in general, b) the answer should be 'no' and I have some ...
Colin Reid's user avatar
  • 4,728
11 votes
9 answers
1k views

Proving the impossibility of an embedding of categories

A number of topological invariants take the form of functors $\mathscr{T}\to\mathscr{G}$, where $\mathscr{T}$ is the category of all topological spaces and continuous functions, and $\mathscr{G}$ is ...
Daniel Miller's user avatar
11 votes
2 answers
843 views

covers of $Z^\infty$

Is it possible to cover $Z^\infty$ (the infinite direct sum of $Z$'s with the $l_1$-metric) by a finite set of collections of subsets $U^0,...,U^n$ such that each collection $U^i$ consists of ...
user avatar
3 votes
1 answer
405 views

Topological simplicity and dense subgroups

Let $G$ be a (topologically) simple Hausdorff topological group. Let $H$ be a dense subgroup of $G$. Now throw away the topology. What restrictions are known on the structure of $H$ as an abstract ...
Colin Reid's user avatar
  • 4,728
10 votes
2 answers
1k views

Is there a way to see a topological group as the "Cayley graph" of its "infinitesimal generators"?

At the time of writing, the most recent blog post over at What's new by Terrence Tao is Cayley graphs and the geometry of groups, and that (excellent, as with most of Tao's writing) post most ...
Theo Johnson-Freyd's user avatar
-2 votes
1 answer
476 views

Countable open subgroup

In a Hausdorff topological group, how can I show that every infinite topological group has a countable open subgroup?
user7188's user avatar
9 votes
3 answers
951 views

Is there a non-trivial topological group structure of $\mathbb{Z}$?

More specificaly, is there a haussdorf non-discrete topology on $\mathbb{Z}$ that makes it a topological group with the usual addition operation?
Cristos A. Ruiz's user avatar
9 votes
2 answers
4k views

Quotient of a Hausdorff topological group by a closed subgroup

Sorry if this question is below the level of this site: I've read that the quotient of a Hausdorff topological group by a closed subgroup is again Hausdorff. I've thought about it but can't seem to ...
Dyke Acland's user avatar
  • 1,479
28 votes
8 answers
4k views

Is there a compact group of countably infinite cardinality?

Apologies for the very simple question, but I can't seem to find a reference one way or the other, and it's been bugging me for a while now. Is there a compact (Hausdorff, or even T1) (topological) ...
Harrison Brown's user avatar
2 votes
1 answer
265 views

Hausdorff Derived Series

There is a short section in the book Locally Compact Groups by Markus Stroppel (Chapter B7) on the notion of a "Hausdorff Solvable Group", which he defines as a topological group with a descending ...
QPeng's user avatar
  • 33
21 votes
5 answers
1k views

Explanation for E_8's torsion

To study the topology of Lie groups, you can decompose them into the simple compact ones, plus some additional steps, such as taking the cover if necessary. After that, the structure of $SO(n)$'s is ...
Ilya Nikokoshev's user avatar
12 votes
4 answers
2k views

Elements of infinite order in a profinite group

Say G is a profinite group with elements of arbitrarily large order. Do elements of infinite order exist (A) if we assume G is abelian? (B) in general? A start for (A): we can ask the same question ...
Andrew Critch's user avatar

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