Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [polish-groups]

Polish group are topological groups which are complete and separable, i.e., have a countable dense subset and are completely metrizable. Polish groups are homeomorphic to separable complete metric spaces.

Use this tag when asking a question regarding these groups.

A Polish group is a topological group $G$ that is also a Polish space, i.e., homeomorphic to a separable complete metric space.

A remarkable fact about Polish groups is that Baire-measurable (i.e., the preimage of any open set has the property of Baire) homomorphisms between them are automatically continuous. The group of homeomorphisms of the Hilbert cube $[0,1]^N$ is a universal Polish group, in the sense that every Polish group is isomorphic to a closed subgroup of it.

Standard Examples of Polish Groups include

  • All finite-dimensional Lie groups with a countable number of components are Polish groups.
  • The unitary group of a separable Hilbert space (with the strong topology) is a Polish group.
  • The group of homeomorphisms of a compact metric space is a Polish group.
  • The product of a countable number of Polish groups is a Polish group.
  • The group of isometries of a separable complete metric space is a Polish group.

Source: https://en.wikipedia.org/wiki/Polish_space#Polish_groups

16 questions
3
votes
1 answer

If $\tau$ and $\tau'$ define the same borel sets, then $\tau=\tau'$

I have a group $G$ with two topologies $\tau, \tau'$ on $G$ that makes it a Polish group (a completely metrizable and separable topological group). I need to show that if $\mathcal{B}(\tau)=\mathcal{B}(\tau') $ , that is they define the same borel…
Marcelo
  • 1,462
3
votes
0 answers

Group of isometries of a Polish metric space acting on the space – the orbit of a point

Assume that $(X,d)$ is a separable complete metric space a denote by $G$ the group of all isometries of $(X,d)$. It is well-known that $G$ equipped with the pointwise convergence topology is a Polish group. Moreover, this group naturally acts on $X$…
Jenda358
  • 406
3
votes
1 answer

Equivalence relation induced by a group action is an analytic set

We say that $X$ is a standard Borel space iff it is a Polish space equipped with the Borel $\sigma$-algebra. Similarly, a standard Borel group is a Polish group s.t. multiplication and inversion are both Borel maps. These concepts are very common in…
LBJFS
  • 1,385
2
votes
1 answer

Small index property

The following is an excerpt of Truss's 1989 paper Infinite Permutation Groups II.: Subgroups of Small Index. (The transcript will appear later.) Here, $2^\omega$ the Cantor space, automorphisms mean autohomeomorphisms, and I take "fr" to mean "the…
Pteromys
  • 7,462
2
votes
1 answer

Polish group actions: if an orbit is non-meager in itself, it is a Baire space?

Assume that $G$ is a Polish group continuously acting on a Polish space $X$. Let $x \in X$ be a point such that $G \cdot x$, the orbit of $x$, is non-meager in its relative topology. I would like to know why this implies that $G \cdot x$ is a Baire…
jenda358
  • 533
2
votes
1 answer

Katetov-maps with same support

Let (X,d) be a metric space. f: X $\rightarrow \mathbb{R}$ ist called "Katetov map" iff $\forall x, y, \in X : |f(x)-f(y)| \leq d(x,y) \leq f(x) + f(y)$. The set of all Katetov-maps on X is denoted by E(X). If f $\in$ E(X) and S $\subset$ X are such…
Antigone
  • 115
2
votes
0 answers

Are compact groups acting on Polish spaces essentially Polish?

Suppose $G$ is a compact Hausdorff (not necessarily Polish) group, acting continuously on a Polish (not necessarily compact) space $X$ (i.e. the map $G\times X\to X$ is continuous). Is it true that there is a compact Polish group $H$ acting on $X$…
tomasz
  • 37,896
2
votes
0 answers

Radon-Nikodym derivative of absolutely continuous part as a measurable function in a product space?

My question is similar to Radon-Nikodym derivative as a measurable function in a product space but I am interested in the case where no quasi-invariance is assumed. Suppose $G$ is a Polish group and $P$ is a probability measure on $G$, and $P_x(A)…
nullUser
  • 28,703
1
vote
1 answer

$\{g\in G: gU=V\}$ in a Polish group

Let $G$ be a Polish group, that is, a topological group which is separable and metrizable by a complete metric. Let $U,V\subset G$ be open, and consider the set $$ \{ g\in G: gU=V\}.$$ I'm trying to show that this set is open. From standard results…
Reveillark
  • 13,699
0
votes
1 answer

Polish group whose closed normal subgroups are countable

I asked myself this question recently, and I can't find the answer : Let $G$ be a Polish group such that for every closed normal subgroup $H$ of $G$, $H$ is countable. Is $G$ necessary equal to $\mathbb{R}$ or $\mathbb{S_1}$ ? It is obviously the…
Paul Tristant
  • 324
0
votes
0 answers

Action on Effros Borel space is Borel

Let $G$ be a Polish group, consider the standard Borel space $F(G)$ of closed subsets of $G$ whose borel structure is generated by the sets of the form $\{F\in F(G)\mid F\cap U \neq \emptyset\}$ for $U$ open in $G$. It is stated in Gao's Invariant…
robbert
  • 102
0
votes
0 answers

Is a Polish group with a countable left-invariant open basis a closed subgroup of $S_\infty$?

I've been looking at Theorem 2.4.1 from Su Gao's "Invariant Descriptive Set Theory" which deals with the characterization of closed subgroup of $S_\infty$. In particular I'm having difficulties with the (iv) $\Rightarrow$ (i) implication, which…
Luca Marchiori
  • 43
0
votes
1 answer

Are all topological group structures on the Baire Space of the form $G^{\mathbb N}$ for $G$ a countable discrete group?

This question was accidentally trivial: For any countably infinite discrete group $G$ we have that $G^{\mathbb N}$ is a topological group structure on the Baire Space, as pointed in Qiaochu Yuan's answer. What about the coverse? Are all topological…
Carla only proves trivial prop
  • 752
  • 3
  • 11
0
votes
1 answer

Is there a topological group structure on the Baire Space?

The sum of two irrationals might not be irrational so we can't use that; Also $\mathbb N$ is not a group so there's no obvious way to define a group structure in it seen as $\mathbb N^{\mathbb N}$ either. This answer gives some necessary conditions…
Carla only proves trivial prop
  • 752
  • 3
  • 11
0
votes
1 answer

Approximating Open Sets With Compact Sets in a $\sigma$-compact Polish Space

Suppose that $X$ is a $\sigma$-compact Polish Space. Suppose $\mu$ is a Borel measure on $X$ that is finite on compact sets. How can we approximate any open set with compact sets so as to prove the inner regularity of $\mu$ on open sets?
user263626
1
2