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
3
votes
3 answers

Does every locally compact group $G$ have a nontrivial homomorphism into $\mathbb{R}$?

Does every locally compact group (second countable and Hausdorff) topological group $G$ that is not compact have a nontrivial continuous homomorphism into $\mathbb{R}$? Obviously for compact groups it is not possible since continuous functions send…
nullUser
  • 27,877
3
votes
0 answers

Haar measure on $G \rtimes_\phi H$

Let $G, H$ be locally compact, $\sigma$-compact metric groups equipped with left Haar measures $m_G, m_H$ respectively. Let $\Phi: G \times H \to G$ be continuous such that $\phi: H \to Aut(G), \, h \mapsto \Phi( \cdot, h)$ is a group homomorphism.…
user302234
  • 612
3
votes
1 answer

Is there a "nice" discontinuous, bijective homomorphism $f: (\mathbb{R},+) \to (\mathbb{R},+)$?

Consider $(\mathbb{R},+)$ as a topological group. Using the axiom of choice, we can construct a $\mathbb{Q}$-basis for $\mathbb{R}$ and using this basis, we can define a discontinuous, bijective homomorphism from $(\mathbb{R},+)$ to itself. Is it…
Huy
  • 6,674
3
votes
1 answer

An interesting question about topological group

A subset of a topological space $X$ is called nowhere dense in $X$ if the interior of its closure is empty. A subset of a topological space $X$ is called the first category (or meagre) in $X$ if it is a union of countably many nowhere dense…
David Chan
  • 1,960
2
votes
2 answers

Zero Dim Topological group

I have this assertion which looks rather easy (or as always I am missing something): We have $G$ topological group which is zero dimensional, i.e it admits a basis for a topology which consists of clopen sets, then every open nbhd that contains the…
MathematicalPhysicist
  • 4,191
2
votes
0 answers

A $T_0$ topological groups is $T_{3.5}$ (and consequently $T_3$)

I don't understand: why is this so? I've just seen the proof that a $T_0$ topological group is $T_1$, but don't know how to show that it's $T_{3.5}$. BTW, the fact that $x\overline{V}=\overline{xV}$, one inclusion is obvious $\overline{xV} \subset…
MathematicalPhysicist
  • 4,191
2
votes
1 answer

question on uniform structure

It should be a triviality, I believe. The topology induced by a uniform structure $\mathcal{U}$ with $\cap \mathcal{U} =\Delta$, where $\Delta$ is the diagonal, is Hausdorff. Now I think that if I define the product topology $X \times X$ to be such…
MathematicalPhysicist
  • 4,191
2
votes
0 answers

a measurable function on a LCA group coincide with an mulitplicative character almost everywhere

Let $G$ be an LCA group. We say $\tilde \chi$ is a multiplicative character if it is a continuous function : $\tilde \chi : G \to S^1 $ where $ S^1 : = \{ z \in \mathbb C : |z| =1 \}$ is the unit circle, which is also a homomorphism, i.e., $…
user112564
  • 3,552
  • 1
  • 23
  • 51
2
votes
1 answer

In what topological abelian groups does convergence to zero imply summability?

Let $\; \langle G\hspace{-0.02 in},\hspace{-0.04 in}+,\hspace{-0.04 in}\mathcal{T}\hspace{.03 in}\rangle \;$ be a $\hspace{.02 in}\big(\hspace{-0.03 in}$$\text{T}_{\hspace{-0.02 in}0}$$\hspace{-0.03 in}\big)$ topological abelian group, and let $0$…
user57159
2
votes
1 answer

Exactness of completion of topological abelian groups

Let $0\to G^{\prime}\to G\to G^{\prime\prime}\to 0$ be an exact sequence of abelian groups. Suppose $G$ is a topological group and then give topologies on $G^{\prime}$ and $G^{\prime\prime}$ which are induced by the topology on $G$. If $H$ is…
user119882
  • 1,442
2
votes
1 answer

Convex subsets of a group

Assume that $(G,+)$ is an Abelian topological group (maybe locally compact, if necessary) and assume that $V$ is an open connected neighbourhood of zero. Does there exist an open "convex" neighborhood of zer0 $W$ such that $W \subset V$? A subset…
Richard
  • 4,432
2
votes
1 answer

Double Coset Closed

Let $G$ be a locally compact group and $H$ a closed subgroup. Under what conditions can we say that the double cosets $H\cdot x \cdot H$ are closed? Is this always true? I am interested mainly in the case when $H$ is discrete.
Carlos De la Mora
  • 77
  • 6
2
votes
0 answers

Compact subset of compactly generated group

Let $G$ be a locally compact topological group, that is also Hausdorff and second countable. Let $S$ be a compact subset that generates $G$ as a group, which contains the identity and is closed under taking inverses. Let $T$ be any compact subset…
Josh
  • 21
2
votes
1 answer

In how many ways is $\mathbb R$ a topological group?

This is an easier version of a more general question I proposed, which hasn't received much attention. How many binary operations can we assign to $\mathbb R$ which make it into a group, where group multiplication and inversion are continuous with…
Daron
  • 10,300
2
votes
2 answers

A lemma on a compact subset of $G/H$, where $G$ is locally compact and $H$ is a closed subgroup of $G$

I have this lemma from "Kazhdan’s Property (T)" book (page 343 in the link). Here, $G$ is locally compact and $H$ is a closed subgroup of $G$. Can't really understand why $K$ is compact. The union is obviously compact, but $p^{-1}(Q)$ not…
N17Math
  • 201
1 2 3
…
11 12