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I want to show that the next groups are polish topological groups, which criteria should I use here?

And also which are locally compact (same question)?

The groups are:

  • The group of permutations on $\mathbb{N}$, $S(\mathbb{N})$;
  • The group of unitary operators on a separable Hilbert space, with the strong operator topology;
  • The group of homeomorphisms on Cantor comb set with the topology of uniform convergence;
  • The group of automorphisms with the usual ordering of $\mathbb{Q}$, with pointwise convergence topology;
  • The group of invertible measure preserving transformations of an atomless standard probability space, $(X,M,\mu)$, with the weak topology that makes the map: $S\rightarrow \mu(S(E))$, $E\in M$ from this group to the real line continuous.

Or if you have some references that deal with these groups and this kind of questions, it will be best.


share|cite|improve this question
What topology do you put on the group of permutations of $\mathbb N$? – Davide Giraudo Feb 1 '12 at 15:07
topology of pointwise convergence. – MathematicalPhysicist Feb 1 '12 at 15:23
We can put the distance $d(f,g)=\sum_{n\in\mathbb N}2^{-n}\min(1,|f(n)-g(n)|)$ and check that $S(\mathbb N)$ is a topological group for this distance, separable since the permutation with finite support are dense, and complete. Maybe I can give the details in an answer; since an answer to all your questions would be long it will probably divide the work between several users. – Davide Giraudo Feb 1 '12 at 15:34
Ok, I appreciate your input. – MathematicalPhysicist Feb 1 '12 at 15:37
up vote 3 down vote accepted

We put on $S(\mathbb N)$ the following distance: if $f,g\in S(\mathbb N)$ then $$d(f,g):=\sum_{n\in\mathbb N}2^{-n}\min(1,|f(n)-g(n)|).$$ It's easy to check that $d$ is indeed a metric. It corresponds to the metric of pointwise convergence, since if $f_k(n)\to f(n)$ for all integer $k$ then taking $\varepsilon$ and $n_0$ such that $\sum_{n\geq n_0}2^{-n}\leq\varepsilon$, we can find $N$ such that if $k\geq N$ then $d(f_k,f)\leq 2\varepsilon$, and conversely if $d(f_k,f)\to 0$ then $f_k(n)\to f(n)$ for all $n$.

We have to show that for this metric, $S(\mathbb N)$ is a topological group, i.e. the maps $(f,g)\mapsto f\circ g$ and $f\mapsto f^{-1}$ are continuous. Since we work in a metric space, it's enough to show the sequential continuity. If $f_k\to f$ and $g_k\to g$, and $n\in\mathbb N$ then $g(n)=g_k(n)$ for $k\geq N(n)$. Let $N_1=N_1(n)$ such that $f_k(g(n))=f(g(n))$ if $k\geq N_1$. Then for $k\geq \max(N,N_1)$ we have $f_k\circ g_k(n)=f_k(g(n))=f\circ g(n)$. We have for $k\geq N_1$: $n=f_k(f^{-1}(n))$ so $f_k^{-1}(n)=f^{-1}(n)$ and $f_k^{-1}\to f^{-1}$ pointwise. $d$ is complete since if $\{f_k\}$ is a Cauchy sequence then $\{f_k(n)\}$ is a Cauchy sequence of integers, which converges to a limit called $f(n)$. We can check that $f$ is indeed a bijection: if $f(n_1)=f(n_2)$, we have for $k\geq N_1$: $f_k(n_1)=f(n_1)$ and for $k\geq N_2$: $f_k(n_2)=f(n_2)$ so for $k\geq\max(N_1,N_2)$, $f_k(n_1)=f_k(n_2)$ and $n_1=n_2$.

$S(\mathbb N)$ is separable since putting $\mathcal D:=\left\{f\in S(\mathbb N,\exists n_0\mid \forall n\geq n_0, f(n)=n\right\}$. Then $\mathcal D$ is countable and dense in $S(\mathbb N)$.

share|cite|improve this answer
Obviously you meant: $lim_{k\rightarrow \infty} f_k(n)=f(n)$. – MathematicalPhysicist Feb 1 '12 at 18:05
So for the second group, I think the suitable metric would be: $d(T,S)=||T-S||$, right? – MathematicalPhysicist Feb 2 '12 at 8:26
Since you required the strong operator topology, yes. – Davide Giraudo Feb 2 '12 at 21:19

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