# "Haar-like" measure on $S_\omega$

Let $$S_\omega$$ be the collection of bijections $$f:\omega\to \omega$$. Endow $$\omega$$ with the discrete topology and let $$S_\omega$$ be endowed with the subspace topology of $$\omega^\omega$$, where $$\omega^\omega$$ carries the product topology.

EDIT. The following statement of mine from the original post is false:

False statement: $$(S_\omega, \circ)$$ is a locally compact group and so there is a Haar measure on $$S_\omega$$.

But: I would nevertheless like to know whether there is some "Haar-like" measure on $$S_\omega$$. Is there a constructive description of such a measure?

If yes: Let $$M$$ be the set of "finitely bounded permutations of $$\omega$$, that is, $$M=\{\pi \in S_\omega: \exists K\in\omega(\forall n\in \omega(|\pi(n)-n| < K))\}.$$ What is the Haar measure of $$M$$, and of $$S_\omega \setminus M$$?

• This is not a locally compact group. (It has a basis of clopen neighborhoods each being homeomorphic to $S_\omega$ itself, which is not compact.)
– YCor
Feb 23 '21 at 17:44
• Btw your subgroup $M$ is rather the group of "bounded displacement permutations" of $\omega$, and has also been referred to as "wobble group" of $\omega$ (viewed as metric space with the Euclidean topology). Another side historical remark is that this topology on $S_\omega$ was introduced by L. Onofri in 1928 (and rediscovered decades later).
– YCor
Feb 23 '21 at 17:46
• Thanks - will correct. It would be interesting to know nevertheless whether there is a "Haar-like" measure on $S_\omega$ - what's your take on this? Feb 23 '21 at 20:03
• There's no Haar measure on $S_\omega$. I don't know what you mean by "Haar-like".
– YCor
Feb 23 '21 at 21:33

Whenever $$G$$ is a non locally compact Polish group, there does not exist any nonzero $$\sigma$$-finite Borel measure $$\mu$$ on $$G$$ such that all left-translates of $$\mu$$ are absolutely continuous with respect to $$\mu$$. This is due to Weil; there is a nice proof in a 1946 paper of Oxtoby.
• You're missing some assumption, since both the zero measure and the counting measure satisfy this requirement. Maybe you require that some neighborhood of $1_G$ has a positive finite measure?
• An anonymous edit has been proposed to add the hypothesis "nonzero $\sigma$-finite". I have rejected it in case it conflicts with the author's intention, but, if that is the correct hypothesis (and if perhaps the edit was proposed by the original author without being logged in), then perhaps leave a comment saying so. Feb 23 '21 at 22:30
You'll have to restrict your class of observables. What you can do for example is look at a uniformly chosen permutation of $$N$$ elements and only keep track of its cycle decomposition. You'll find that most points belong to cycles of length about $$\alpha N$$ for some $$\alpha \in (0,1)$$. If you then only keep track of cycle lengths, divide their length by $$N$$, and let $$N \to \infty$$, you do get a limiting distribution on the space of unordered partitions of $$[0,1]$$ called the Poisson-Dirichlet distribution. As usual, Terry has a nice write-up on his blog.