Timeline for Is Thompson's group definably orderable?
Current License: CC BY-SA 4.0
15 events
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| Aug 11, 2020 at 2:33 | history | edited | James Hyde | CC BY-SA 4.0 |
Sorry for the multiple edits. I now know that while ctrl-enter compiles in overleaf it saves your edit in mathoverflow.
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| Aug 11, 2020 at 2:26 | history | edited | James Hyde | CC BY-SA 4.0 |
added 2014 characters in body
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| Aug 11, 2020 at 2:15 | history | edited | James Hyde | CC BY-SA 4.0 |
added 2014 characters in body
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| Aug 7, 2020 at 10:51 | comment | added | YCor | @shane.orourke OK, I agree. Actually $F$ is similar to $G=\mathrm{PL}(I)$: just defined by restricting to a segment and adding dyadic constraints on slopes and breakpoints, and similarly, $[F,F]$ is simple as well as $[G,G]$. It seems that $F$ and $[F,F]$ are best-known now, but indeed Chehata's work (1952) is anterior to R. Thompson's groups. | |
| Aug 7, 2020 at 10:44 | comment | added | shane.orourke | @YCor But I think the group $G(I)$ (where $I$ is a compact interval) does satisfy your condition, and is thus definably bi-orderable, although the simple subgroup $G(I')$ (where $I'$ is the interior of $I$) does not. | |
| Aug 6, 2020 at 20:59 | comment | added | YCor | @shane.orourke I've actually had a look at Chehata's 1952 paper. His group is the set of piecewise affine increasing homeomorphisms of an ordered field, that are identity near $\pm\infty$. Thus the argument does not apply, since we don't have $a,b$ at disposal. For the same reason, the argument does not apply to the derived subgroup $[F,F]$ (which is simple and consists of elements of $F$ that are identity outside $[1/n,1-1/n]$ for some $n$). | |
| Aug 6, 2020 at 13:28 | vote | accept | YCor | ||
| Aug 6, 2020 at 13:20 | comment | added | YCor | @shane.orourke yes, this works for a wealth of known "rich" subgroups of $\mathrm{Homeo}^+([0,1])$. Actually, with only assumption existence of $a,b$ and density, one already gets a definable partial bi-ordering (which is not trivial, since $a,b>1$). | |
| Aug 6, 2020 at 12:30 | comment | added | shane.orourke | @YCor So if I've followed correctly, this argument implies that Chehata's group $G(I)$ (which contains his example of a simple bi-orderable group) is definably bi-orderable. | |
| Aug 6, 2020 at 8:11 | comment | added | YCor | @VilleSalo just use that if they commute then the support of each one is invariant by the other one. | |
| Aug 6, 2020 at 8:09 | comment | added | YCor | Great! It seems that this definition works under the bare assumption of a subgroup $G$ of $\mathrm{Homeo}^+([0,1])$ with a dense orbit $Gx$ on the open interval $]0,1[$, such that there exists $a,b$ with support $]0,x[$ and $]x,1[$. This defines $S_4$ as the (conjugation-invariant) set of $g\in G$ such that for some $y$ with $yg>y$ and $zg\ge z$ for all $z\le y$. That this defines a (strict) total order requires something on $G$. It seems enough that $G$ acts piecewise analytically. | |
| Aug 6, 2020 at 7:46 | comment | added | Ville Salo | For the non-trivial direction of 1. (I'm sure it's simpler than this though): If $(1/2)g = t$ and $(1/2)h < t$ then $(t)b^h \neq t$. If $(t)b^h < t$ then for small $\epsilon > 0$ we have $(t+\epsilon) a^g b^h = (t+\epsilon) b^h = x < t$ so $(t+\epsilon) b^h a^g = x a^g \neq x$ because $x$ is in the support $[0,t]$ of $a^g$. If $(t)b^h > t$, then let $t'$ be such that $(t')b^h = t$ and we have $(t'+\epsilon) b^h a^g = (t'+\epsilon) b^h$ because $(t'+\epsilon) b^h > t$, while $(t'+\epsilon) a^g b^h = (t'+\epsilon) b^h$ would imply $(t'+\epsilon) a^g = t'+\epsilon$, contradicting $t' < t$. | |
| Aug 6, 2020 at 6:52 | review | Late answers | |||
| Aug 6, 2020 at 8:09 | |||||
| Aug 6, 2020 at 6:34 | review | First posts | |||
| Aug 6, 2020 at 6:50 | |||||
| Aug 6, 2020 at 6:28 | history | answered | James Hyde | CC BY-SA 4.0 |