The standard probability space $(I, \mathcal B, \lambda)$ consists of the interval $I = [0,1]$, its Borel $\sigma$-algebra $\mathcal B := \mathcal B(I)$ and Lebesgue measure $\lambda$. In applications, most probability spaces of interest are measure isomorphic to the standard probability space.

Let $\Gamma = \operatorname{Aut}(I,\mathcal B, \lambda)$ denote the automorphism group of the standard probability space. Recall that a measurable automorphism is a bijection $f : I \to I$ which pulls back and pushes forward measurable sets. A measure automorphism additionally preserves measure: $\lambda = \lambda \circ f^{-1} = \lambda \circ f$.

Let $\Gamma_{erg}$ be the subgroup of ergodic automorphisms. i.e., $f \in \Gamma_{erg}$ if $f \in \Gamma$ and $$\lambda(A \, \triangle \, f A) = 0 \mathrm{~implies~} \lambda(A) = 0 \mathrm{~or~} \lambda(A) = 1$$ for all Borel sets $A \in \mathcal B(I)$, where $\triangle$ denotes the symmetric difference.

Is there a nice characterization of $\Gamma$ or $\Gamma_{erg}$? These are both very, very large groups, and hard for me to conceive of. What structure do they satisfy? Does every Lie group embed as a subgroup of $\Gamma$ or $\Gamma_{erg}$? Does the quotient $\Gamma / \Gamma_{erg}$ have any meaningful structure?

This is an open-ended question, so I've marked it as big list (and community wiki). I would be happy with a list of some non-trivial subgroups of $\Gamma$ and $\Gamma_{erg}$.

  • 2
    $\begingroup$ If $G$ is a semisimple connected Lie group, then it admits a lattice $\Lambda$ and the action of $G$ on $G/\Lambda$ is faithful, this probably embeds $G$ into your group $\Gamma$ (since $G/\Lambda$ is a standard probability space). $\endgroup$ – YCor Oct 18 '13 at 20:00
  • 2
    $\begingroup$ This paper seems relevant. ams.org/journals/proc/1990-110-02/S0002-9939-1990-1009997-6/… $\endgroup$ – Henrique de Oliveira Oct 18 '13 at 21:17
  • 7
    $\begingroup$ The ergodic transformations do not form a group... $\endgroup$ – Mikael de la Salle Oct 18 '13 at 21:18
  • 1
    $\begingroup$ I don't know what you are looking for. $\Gamma$ has a natural metric structure on it under the Ky-Fan metric which is the metric of convergence in probability. The space $\Gamma$ is a closed sunspace (under this metric) of the space of measurable maps from the standard probability space to [0,1]. The ergodic maps are a closed subspace of $\Gamma$. $\endgroup$ – Jason Rute Oct 18 '13 at 23:14
  • 6
    $\begingroup$ @TomLaGatta: I'm confused; the identity map is not ergodic. $\endgroup$ – Nate Eldredge Oct 19 '13 at 2:58

I can't explain the group theoretic structure of $\Gamma$, but I can explain the topological structure. (Warning, this post is mostly a continuous stream of thoughts. I hope it is well organized and truthful, but you should check the details.)

The space of measurable functions $f\colon(I,\mathcal{B},\lambda) \rightarrow \mathbb{R}$

There are a number of metrics and norms on spaces of measurable functions $f\colon(I,\mathcal{B},\lambda) \rightarrow \mathbb{R}$. Of course there are

  • The $L^p$ norms (on the subspace of $p$-integrable functions).

Also, there are lesser-known metrics which give the topology of convergence in probability (a.k.a. convergence in measure). These two metrics are equivalent:

  • The Ky-Fan metric $$\rho_\textit{Ky-Fan} (f,g) = \inf\left\{\varepsilon > 0 : \lambda \left\{x : |f(x) - g(x)| \geq \varepsilon\right\} \leq \varepsilon\right\}.$$ This definition makes more sense when you consider the definition of convergence in probability.

  • The metric $$\rho (f,g) = \int \min \left\{|f-g|,1\right\} \, d\lambda$$ (If you know the name for this metric, please answer this MO question!) Notice the similarity between this metric and the $L^1$ metric. Also notice, that for indicator functions, this metric becomes the familiar metric $\rho(\mathbf{1}_A,\mathbf{1}_B) = \lambda(A \triangle B)$.

The space of measurable functions $f\colon(I,\mathcal{B},\lambda) \rightarrow I$

For the subspace of functions $f\colon(I,\mathcal{B},\lambda) \rightarrow I$, it is easy to see that this last metric is exactly the same as the $L^1$ metric. Moreover, one can show on this space that all the $L^p$ metrics are equivalent (easy exercise).

The space of measure preserving automorphisms $\textrm{Aut}(I,\mathcal{B},\lambda)$

The measure preserving automorphisms form a subspace of the previous space. It is closed. This is because the push-forward map $f \mapsto \lambda_f$ is continuous in any of the above metrics, where the topology on the codomain is given by the Levy-Prokohorov metric, that is the metric of convergence in distribution.

This space is therefore a complete separable metric space (Polish space) under any of the above metrics. However, the usual candidates for a countable dense set (e.g. polynomials with rational coefficients) don't work. Instead, the following functions form a nice dense set: For each $n$ choose, consider a permutation $\pi$ on $\{0,\ldots,2^n - 1\}$. Then let $f^n_\pi \colon [0,1] \rightarrow [0,1]$ be as follows. Break up $[0,1]$ into $2^n$ equally spaces dyadic intervals and let $f^n_\pi$ rearrange the intervals according to $\pi$.

(Actually, consider the $L^1$ metric on this subset of basic functions. Take two such "basic functions" $f^n_\pi$ and $f^n_\sigma$. (WLOG, they break up $[0,1]$ into the same number of intervals.) Then the distance $\| f -g\|_1$ is $2^{-n}\sum_{i=0}^{2^n} (\pi(i) - \sigma(i))$.

In this way, one can think of this space as a continuum sized extension of the countable group $G = \bigcup_n S_{2^n}$ where we embed $S_{2^n}$ into $S_{2^{n+1}}$. (Although, our metric necessarily breaks the symmetry of $S_{2^n}$.)

This space is not compact. (One can find a sequence of such basic functions which does not have a convergence subsequence.)

The space of ergodic measure preserving automorphisms $\textrm{Aut}_\textrm{Ergodic}(I,\mathcal{B},\lambda)$

This is dense in the previous space. To see this, consider an irrational shift $g_\alpha(x) = x + \alpha \mod 1$. Then compose it with a basic function. It only changes the $L^1$ norm of the basic function slightly, but this composed function is now ergodic. (This takes a little thought.)

  • $\begingroup$ There exist compact spaces with sequences with no converging subsequences (e.g. the sequence $(n)_{n\ge 0}$ in the Stone-Cech compactification of $\mathbf{N}$). Also one usually identifies functions coinciding outside null sets in the automorphism group of a measure space, so we'd expect the topology to be well-defined on the quotient. $\endgroup$ – YCor Oct 19 '13 at 10:05
  • 1
    $\begingroup$ @Jason When you said "this space is not compact", did you mean the dense set you just described? What about the whole space of automorphisms? $\endgroup$ – Henrique de Oliveira Nov 5 '13 at 17:52
  • 2
    $\begingroup$ @HenriquedeOliveira: I meant the space of automorphisms is not compact. There is a sequence of basic functions which does not converge to an automorphism (in the metrics I mentioned). One such sequence is $(f^n_{\pi_n})_n$ where $\pi_{n-1}(i)=2i$ and $\pi_{n-1}(2^n+i)=2i+1$ for $0\leq i \leq 2^{n}$. While there is a limit in the space of measurable functions, namely the map $f(x)=2x \mod 1$, this is not an automorphism. With a little more tweaking we can adjust the sequence so it does not converge to any measurable function. (I'll leave that as an easy exercise.) $\endgroup$ – Jason Rute Nov 5 '13 at 21:23
  • 1
    $\begingroup$ @YvesCornulier: On a metric space, compact and sequentially compact are the same: (en.wikipedia.org/wiki/Sequentially_compact_space). Also, technically the metrics I defined are pseudo-metrics. One needs to work up to a.e. equivalence for them to be metrics. $\endgroup$ – Jason Rute Nov 5 '13 at 21:28
  • 1
    $\begingroup$ @HenriquedeOliveira (1) Nhu's metric is equivalent to mine. (Note, this means his/her metric is equivalent to $L^1$ and $L^2$.) (2) This topology seems "right" because it is the topology which makes the map $A \mapsto f^{-1}(A)$ continuous (where the measurable sets have the topology give by the metric $d(A,B) = \lambda (A \triangle B)$). (3) However, you bring up a good point about compactness. Possibly the weak $L^2$ norm would make the group compact. One would need to check that (a) the space is closed (and hence compact) in weak-$L^2$ and (b) the group is continuous in weak-$L^2$. $\endgroup$ – Jason Rute Nov 15 '13 at 4:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.