Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [topological-groups]

A topological group is a group endowed with a topology such that the group operation and inversion are continuous maps. They are useful in various areas of mathematics. Every topological vector space is a topological group. Locally compact groups are important in harmonic analysis.

A topological group is a group endowed with a topology such that both the group operation and inversion are continuous maps. Every group can be understood as a topological group, if we take the discrete topology.

Topological groups are useful in various areas of mathematics. Every topological vector space is a topological group. Locally compact groups are important in harmonic analysis, see e.g. Pontryagin duality.

2270 questions
1
vote
0 answers

Is the following homomorphism continuous?

Let $G$ be a closed, torsion-free and divisible subgroup of locally compact abelian group $X$ such that $nX=G$ for some $n$. For $x\in X$, there exist $g\in G$ such that $nx=ng$. So we can define a homomorphism $f:X\to G$, $f(x)=g$. Now, my…
Aliakbar
  • 3,167
  • 2
  • 17
  • 27
1
vote
1 answer

Is the following descending sequence nonzero?

Let $K_{1}\supseteq K_{2}\supseteq K_{3}\supseteq \cdots$ be a descending sequence of compact subgroups of compact, torsion-free group $G$. Is $\bigcap_{r=1}^\infty n^r K\neq 0$? (for a positive integer $m$, $mK=\{mx;x\in K\}$)
Aliakbar
  • 3,167
  • 2
  • 17
  • 27
1
vote
1 answer

For a metric $d$ on a group $G$, why do $d$ and $d^{-1}$ generate the same topology.

Let $d$ be a metric on a group $G$ and define $d^{-1}$ by $d^{-1}(x,y)=d(x^{-1},y^{-1})$. Why do $d$ and $d^{-1}$ generate the same metric topology on $G$? Let $g \in G$ and $\epsilon >0$. Let $B_d=B_d(g,\epsilon)=\{h \in G : d(g,h) < \epsilon\}$.…
user41728
  • 1,560
1
vote
1 answer

Consequences of differing definitions of topological groups

I'm using this document to learn about the representation theory of compact groups. They define a topological group $G$ to be a group with a topology such that The multiplication and inversion operations are continuous. $\{e\}$ is closed in $G$,…
ABanerjee
  • 175
  • 11
1
vote
1 answer

Is every discrete, torsion and divisible group $\sigma-$compact?

Let $G$ be a discrete, torsion and divisible group. Is $G$, $\sigma-$compact?
Aliakbar
  • 3,167
  • 2
  • 17
  • 27
1
vote
0 answers

Let $H$ be a $\sigma-$compact subgroup of locally compact abelian group $G$ such that $G/H$ is $\sigma-$compact. Is $G$ $\sigma-$compact?

Let $H$ be a $\sigma-$compact, closed subgroup of locally compact abelian group $G$ such that $G/H$ is $\sigma-$compact. Is $G$ $\sigma-$compact?
Aliakbar
  • 3,167
  • 2
  • 17
  • 27
1
vote
1 answer

A question on divisible group

Let $G$ be a divisible, locally compact abelian group and $L_{1}\supseteq L_{2}\supseteq L_{3}\supseteq ...$ be a sequence of compactly generated, open subgroups of $G$. Can we deduce that there exist $n$ such that $L_{i}=L_{n}$ for $i\geq n$?
Aliakbar
  • 3,167
  • 2
  • 17
  • 27
1
vote
1 answer

A question on torsion-free locally compact abelian groups

Let $X$ be a torsion-free, totally disconnected, locally compact abelian group. I want to find a non-zero subgroup of $X$ with the following conditions: 1- compact 2- open 3- $n-$divisible for some $n$. Of course, it is clear that…
Aliakbar
  • 3,167
  • 2
  • 17
  • 27
1
vote
0 answers

Why do we use closed subgroups of topological groups?

I have been working with Topological Groups and have noticed that the subgroups of real interest in the topic are the closed subgroups. I figured this is because closed subgroups inherit most of the nice properties of the groups we work on and the…
YBibbyB
  • 23
1
vote
2 answers

A non discrete, torsion-free and $\sigma-$compact, locally compact abelian group?

We know that every countable, discrete torsion-free group is $\sigma-$compact. Is there a non discrete, torsion-free, $\sigma-$compact, locally compact abelian group?
Aliakbar
  • 3,167
  • 2
  • 17
  • 27
1
vote
0 answers

Is there a compact metrizable non-abelian torsion-free group?

It is quite difficult to google for non-abelian torsion-free groups. I am primarily interested in metrizable (''not too big'') compact groups. However, even if we drop the metrizability condition I do not know the answer.
chj
  • 51
1
vote
1 answer

Closed topological subgroup of $GL(n,\mathbb{R})$

I was looking at the topological group defined by $A=\left \{ \begin{pmatrix} \lambda_1 & & \\ & \ddots & \\ & & \lambda_n \end{pmatrix} : \forall 1 \leq i \leq n, \lambda_j \in \mathbb{R}\backslash \{0\} \right \}$ today, and it's said to be a…
Spida
  • 359
  • 1
  • 10
1
vote
2 answers

Neighborhoods in the product topology.

In the last two lines thread, it is said that $N \times M \subseteq A\times B$ is a neighborhood of $0$ if and only if $N, M$ are neighborhoods of $0$. Here $A, B$ are topological abelian groups. How to prove this result? I searched on the Internet,…
LJR
  • 14,520
1
vote
1 answer

Open morphism maps identity component onto identity component

If $f:G\rightarrow H$ is open morphism of topological groups then $f(G_0)=H_0$. So far I get that since $f$ is continuous that means $f(G_0)$ is connected and since $f$ is homomorphism that means $f(1_G)=1_H$ so we have that $f(G_0)\subset H_0$…
MotionMath
  • 49
1
vote
0 answers

Prove that set of all functions from A to B forms topological group

Hello, I was trying to prove part (b) of problem given above. But I am stuck there. My idea was to use definition of product topology. For example if $U$ is open set in $B^A$ then its projections are open too. On the other hand inverse image in…
MotionMath
  • 49
1 2 3
…
11 12