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We say that a group $(A, \cdot)$ is bi-orderable if there exists a total order $\preceq$ on $A$ such that $xz \prec yz$ and $zx \prec zy$ for all $x,y,z \in A$ with $x \prec y$.

Let $m,n$ be non-zero integers, and let ${\rm BS}(m,n)$ denote the Baumslag-Solitar group $\langle a, b \mid a^{-1} b^m a = b^n\rangle$. I was happily surprised to learn from Yves Cornulier (see here) that ${\rm BS}(1,n)$ is bi-orderable for $n \ge 1$, which naturally leads to the following:

Q1. Is there a ``nice'' characterization of those pairs $(m,n)$ such that ${\rm BS}(m,n)$ is bi-orderable?

E.g., ${\rm BS}(m,n)$ is not bi-orderable if $mn < 0$ (wlog, say that $1 \prec b$. Then, $b^{-1} \prec 1$, with the result that $ 1 \prec a^{-1} b^m a$ and $b^n \prec 1$ for $m > 0$, and dually $a^{-1} b^m a \prec 1$ and $1 \prec b^n$ otherwise) or $m = n \ge 2$ (wlog, say that $ab \prec ba$. Then, $a^n b \prec a^{n-1} b a \prec \cdots \prec aba^{n-1} \prec ba^n$). But what about the other cases?

Update on Q1. The question has been completely answered: ${\rm BS}(m,n)$ is bi-orderable if and only if $mn > 0$ and $\min(|m|,|n|) = 1$. For details, see Yves' answer below.

On a related note, let ${\rm D}(m,n)$ be the two-generator one-relator group $\langle a, b\mid a^{-1} b a^m = b^n\rangle$. This is sort of a variant of ${\rm BS}(m,n)$, and, while different in many ways, I'd like to know if both have a similar behavior as for orderability (by the way, does ${\rm D}(m,n)$ have a conventional name, so that I can look for references by myself?). In particular:

Q2. Is there a ``nice'' characterization of those pairs $(m,n)$ such that ${\rm D}(m,n)$ is bi-orderable?

In fact, ${\rm BS}(1,n) = {\rm D}(1,n)$, so the above (and, especially, Yves' argument) proves that ${\rm D}(1,n)$ is bi-orderable if and only if $n \ge 1$. But what about the other cases? E.g., is ${\rm D}(n,n)$ bi-orderable for $n \ge 2$?

Here is a question similar to Q1 (with total orders replaced by partial ones).

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    $\begingroup$ 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 $Y>X$. $\endgroup$
    – YCor
    Commented Oct 31, 2013 at 12:45
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    $\begingroup$ 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 $m,n\ge 2$. $\endgroup$
    – YCor
    Commented Oct 31, 2013 at 13:17
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    $\begingroup$ 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 $Y^s\prec X^s$. $\endgroup$
    – Salvo Tringali
    Commented Oct 31, 2013 at 13:29
  • $\begingroup$ 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 $n \ge 2$. $\endgroup$
    – Salvo Tringali
    Commented Oct 31, 2013 at 13:34
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    $\begingroup$ 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 $|n| \ge 2$). $\endgroup$
    – Salvo Tringali
    Commented Oct 31, 2013 at 13:44

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(After the discussion in the comments.) $BS(m,n)$ is bi-orderable iff $mn>0$ and $\min(|m|,|n|)=1$.

Since it was already mentioned that $mn<0$ implies $BS(m,n)$ not bi-orderable and that $BS(1,n)$ is bi-orderable for all $n\ge 2$, all remains is to check that $2\le m\le n$ implies $BS(m,n)=\langle t,x\mid tx^mt^{-1}=x^n\rangle$ is not bi-orderable.

Define $y=txt^{-1}$. Then $y^m=x^n$. So $x$ commutes with $y^m$. So (*) $x$ commutes with $y$. But this is not the case.

(*) in a biorderable group, as mentioned by Salvo, $[X,Y^m]=1$, $m\ge 1$ implies $[X,Y]=1$, because otherwise if $XY<YX$, $X^mY<X^{m-1}YX<\dots<YX^m$ and similarly $YX<XY$ implies $YX^m<X^mY$.

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