Timeline for Bi-orderability of Baumslag-Solitar group $\langle a,b \mid a^{-1} b^m a = b^n\rangle$ and of $\langle a,b \mid a^{-1} b a^m = b^n\rangle$
Current License: CC BY-SA 4.0
16 events
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| Feb 16, 2020 at 12:53 | history | edited | YCor | CC BY-SA 4.0 |
edited tags, added name in title
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| Feb 16, 2020 at 12:52 | comment | added | YCor | I'd suggest to move Q2 to a separate thread, it's a fairly distinct question (and the groups $D(m,n)$ are far less studied, I'm even not sure that they explicitly appeared anywhere before this question except for a few small values of $m,n$.) | |
| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| May 17, 2015 at 18:59 | vote | accept | Salvo Tringali | ||
| Oct 31, 2013 at 19:52 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
Update on Q1 and correction on Q2
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| Oct 31, 2013 at 14:29 | comment | added | Salvo Tringali | Yes, that's it. But now I wonder if I should have raised Q2 in a separate thread. In any case, thank you much for the fruitful exchange. | |
| Oct 31, 2013 at 14:21 | answer | added | YCor | timeline score: 10 | |
| Oct 31, 2013 at 14:11 | comment | added | YCor | OK, you're right. Thus, to conclude, $BS(m,n)$ is bi-orderable iff $mn>0$ and $\min(|m|,|n|)=1$. | |
| Oct 31, 2013 at 13:44 | comment | added | Salvo Tringali | If $XY \ne YX$, then $X^mY=YX^m$ is impossible for $m \ge 1$. For suppose wlog $XY \prec YX$. Then, $X^m Y \prec X^{m-1} Y X \prec \cdots \prec XYX^{m-1} \prec YX^m$, which is the same argument used in the OP to prove that ${\rm BS}(n,n)$ is not bi-orderable for $n \ge 2$ (it works as well for $|n| \ge 2$). | |
| Oct 31, 2013 at 13:37 | comment | added | YCor | I agree with your first comment but don't understand the second one. | |
| Oct 31, 2013 at 13:34 | comment | added | Salvo Tringali | As for your 2nd comment: $[X^m, Y] = 1$ iff $X^m Y = Y X^m$, and this is impossible if $m \ge 2$ for the same reason expressed in the OP with reference to the orderability of ${\rm BS}(n,n)$ when $n \ge 2$. | |
| Oct 31, 2013 at 13:29 | comment | added | Salvo Tringali | Many thanks for the interest, Yves. Just a minor detail as for your 1st comment: I think it should be $Y = y^q$, so $X^s = Y^s = 1$ isn't true. Nevertheless, $X^s=Y^s$ is enough to conclude: Since $X\ne Y$ (here is where we use that $s =\gcd(m,n) \ge 2$), then either $X\prec Y$, and then $X^s\prec Y^s$, or $Y \prec X$, and then $Y^s\prec X^s$. | |
| Oct 31, 2013 at 13:17 | comment | added | YCor | Also $BS(2,n)$ is not bi-orderable for any $n\ge 2$, because $A(2,n)$ is not bi-orderable. The reason is that in a biorderable group, $[X^2,Y]=1$ implies $[X,Y]=1$. Indeed $X(XYX^{-1}Y^{-1})X^{-1}=YXY^{-1}X^{-1}=(XYX^{-1}Y^{-1})^{-1}$, i.e. $c=XYX^{-1}Y^{-1}$ is conjugate to is inverse, and hence is trivial. This maybe generalizes to showing $[X^m,Y]=1\Rightarrow [X,Y]=1$ for any $m\ge 2$, in which case it would follow that $A(m,n)$ and hence $BS(m,n)$ is never bi-orderable if $m,n\ge 2$. | |
| Oct 31, 2013 at 12:45 | comment | added | YCor | If $m,n\ge 2$ are not coprime then $BS(m,n)$ is not bi-orderable. Indeed it contains the amalgam $A(m,n)=\langle x,y\mid x^m=y^n\rangle$ as a subgroup. Then $A(m,n)$ is not bi-orderable. Indeed write $m=sp,n=sq$ with $s,p,q\ge 2$, $X=x^p$ and $Y=x^q$. Then $X\neq Y$ and $X^s=Y^s=1$. But extraction of $s$-roots is unique in a bi-orderable group. Indeed if $X<Y$, then $YX^{-1}>1$, hence $1<Y^{s-1}(YX^{-1})Y^{-s+1}=Y^sX^{-1}Y^{-s+1}=X^{s-1}Y^{-s+1}$, hence $X^{s-1}>Y^{s-1}$, but also since $X<Y$ we have $X^{s-1}<Y^{s-1}$, contradiction, and similarly contradiction if $Y>X$. | |
| Oct 31, 2013 at 11:11 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
Fixed typos and ruled out some trivial cases
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| Oct 31, 2013 at 10:21 | history | asked | Salvo Tringali | CC BY-SA 3.0 |