All Questions
Tagged with gr.group-theory gn.general-topology
124 questions
5
votes
0
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135
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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 ...
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 ...
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 ...
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 ?
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 $f(ax)=f(x)a$ ...
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 ...
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 ...
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 ...
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 ...
4
votes
2
answers
2k
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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 ...
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 ...
10
votes
2
answers
2k
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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 ...
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 ...
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 ...
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 ...
10
votes
2
answers
1k
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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 ...
-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?
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?
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 ...
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) ...
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 ...
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 ...
12
votes
4
answers
2k
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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 ...