Timeline for Symmetries of the standard probability space
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Nov 15, 2013 at 4:43 | comment | added | Jason Rute | @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-$L^2$. | |
| Nov 14, 2013 at 19:55 | comment | added | Henrique de Oliveira | @JasonRute Maybe this is too broad a question to answer in a comment. But in what sense is this topology the "right" one for this group? I wonder if there isn't an alternative topology that would make the group compact, but still be useful. Is your topology the same as the weak topology that Nhu uses? (see the linked paper in the comments to the question) | |
| Nov 6, 2013 at 10:06 | comment | added | YCor | @Jason: I was not only referring to the metrizable topologies you mentioned... anyway the poster of the original question did not clarify which topology he has in mind. | |
| Nov 5, 2013 at 21:28 | comment | added | Jason Rute | @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. | |
| Nov 5, 2013 at 21:23 | comment | added | Jason Rute | @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.) | |
| Nov 5, 2013 at 17:52 | comment | added | Henrique de Oliveira | @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? | |
| Oct 19, 2013 at 10:05 | comment | added | YCor | 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. | |
| S Oct 19, 2013 at 4:21 | history | answered | Jason Rute | CC BY-SA 3.0 | |
| S Oct 19, 2013 at 4:21 | history | made wiki | Post Made Community Wiki by Jason Rute |