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I was thinking of some example for an Right ordered group ( $RO$-group) which is not an $O-$group (Ordered group) i.e. not left ordered.

I guess looking in matrix groups will be fruitful but how to define an ordering on matrices , say $GL_n{(\mathbb{R})}$. Now $\mathbb{Z}$ has an ordering '$<$', then if I say $A<'B \iff det(A)<det(B)$, <' is not an ordering on group of matrices as $A<'B$ does not imply $ CA <'CB $ when det($C$) = $-1$.

Suggest some ordering on matrices or some other example of some $RO$ group which is not ordered.

Bhaskar Vashishth
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1 Answers1

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Here's an example from Kopytov and Medvedev, Right-Ordered Groups, which they attribute to Smirnov:

Consider the group of 2×2 matrices of the form $[{}^k_0\,{}^a_1]$ where $k$ is a positive rational and $a$ is a rational. The group operations work out to $$ (k_1,a_1)*(k_2,a_2) = (k_1k_2, a_1+k_1a_2) \qquad\quad (k,a)^{-1} = (1/k,-a/k) \qquad\quad e = (1,0)$$ (One might observe that this is a semidirect product $\mathbb Q_{>0}^{\times}\rtimes \mathbb Q^+$).

Choose a fixed irrational number $\varepsilon$ and define $(k,a)>e$ iff $k+a\varepsilon > 1$. One easily sees that for every $g\ne e$, exactly one of $g>e$ and $g^{-1}>e$ holds, so this extends uniquely to a right order $$ (k_1,a_1) > (k_2,a_2) \iff k_1+k_2a_1\varepsilon > k_2+k_1a_2\varepsilon $$ However, this is not a left order. For example, if $\frac13<\varepsilon<\frac12$ then we have $$ (1,1) < (2,0) $$ but $$ (1,1)*(1,1) = (1,2) > (1,1)*(2,0) = (2,1) $$

hmakholm left over Monica
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  • But I think that it should be $\frac13<\varepsilon<\frac{1}{2}$, for this to be valid. Can you double check, my calculations are telling me this. – Bhaskar Vashishth May 28 '15 at 21:59
  • @Bhaskar: Whoops, that's what my notes say, too. Fixed. – hmakholm left over Monica May 28 '15 at 22:00
  • It is nice to know there is a whole book on RO groups. It is costly to purchase for me right now, but some day. – Bhaskar Vashishth May 28 '15 at 22:03
  • Can we define a similar ordering on this group which makes it Left orderable but not right orderable? I was trying but did not get anywhere. – Bhaskar Vashishth May 30 '15 at 04:58
  • @Bhaskar: Yes, once you have a concept of $g>e$ such that (a) $g_1>e$ and $g_2>e$ imply $g_1g_2>e$, and (b) when $g\ne e$, exactly one of $g>e$ and $g^{-1}>e$ is true, then you can extend that uniquely to a right ordering by defining $g_1>g_2$ to mean $g_1g_2^{-1}>e$ (which is what happens above), or to a left ordering by defining $g_1>g_2$ to mean $g_2^{-1}g_1>e$. – hmakholm left over Monica May 30 '15 at 10:30