Question

Because I have heard the phrase "totally ordered abelian group", I imagine there should be non-abelian ones. By this I mean a group with a total ordering (not to be confused with a well-ordering) which is "bi-translation invariant": a < b should imply cad < cbd.

Does anyone know any examples?

Totally ordered abelian groups are easy to come up with: any direct product of subgroups of the reals, with the lexicographic ordering, will do. Knowing some non-abelian ones would help reveal what aspects of totally ordered abelian groups really depend on them being abelian...

Edit: Via Andy Putman's answer below, I found this great summary of results about ordered and bi-ordered groups (i.e. groups with bi-translation invariant orderings) on Dale Rolfsen's site:

Lecture notes on Ordered Groups and Topology

He shows numerous examples of non-abelian bi-orderable groups, including a bi-ordering (bi-translation invariant ordering) on the free group with two generators. As well, he mentions, due to Rhemtulla, that a left-orderable group is abelian iff every left-ordering is a bi-ordering, which I think really highlights the relationship between ordering and abelianity.

Answer 1

This concept is usually called biorderability (there is also left- and right-orderability). There are many examples, such as free groups and surface groups. Most spectacularly, the pure braid groups are biorderable, while the full braid groups are left orderable but not biorderable. The left ordering on the braid groups is usually attributed to Dehornoy, though it was discovered even earlier by Thurston (but not published).

Dale Rolfsen has several nice surveys of material related to this on his webpage here. In particular, there is the complete text of a nice book called "Why are braids orderable?" that he wrote with Patrick Dehornoy, Ivan Dynnikov, and Bert Wiest. I believe that a new and much expanded edition of this book was just published.

EDIT 1 : I just found the website for the much-expanded version of Rolfsen et al's book here.

EDIT 2 : Thurston's construction of a left-ordering on the braid groups (which, of course, uses hyperbolic geometry) is very beautiful. It is explained very nicely in the first few pages of the paper "Orderings of mapping class groups after Thurston" by Short and Wiest, which is available on the arXiv here. The intro sections of this paper also contain a brief but enlightening account of the general theory of group orderings.

Also, I have not read it, but there is a book entitled "Orderable Groups" by Rehmtulla and Mura. However, it is from 1977 and will thus omit a lot of recent work.

Answer 2

What you want is to define a positive subset of the group that satisfies trichotomy (every group element is either positive, negative, or the identity), that is closed under the group law, and that is invariant under conjugation.

I think that the Heisenberg group is an example. This is the group of matrices M = [[1,a,c],[0,1,b],[0,0,1]]. Say, integer matrices. Then we can say that M is positive if a > 0, or if a = 0 and b > 0, or if a = b = 0 and c > 0. I think that this works?

Answer 3

Thurston's construction of a left-order on braid groups uses hyperbolic geometry, but it doesn't have to. For the sake of argument, let S be a closed, oriented surface of non-positive Euler characteristic. The universal cover of S is homeomorphic to the plane. An essential embedded loop in S lifts to a properly embedded line in the plane. The mapping class group of S permutes the set of isotopy classes of essential embedded loops in S, and a certain extension of this group acts on the universal cover, permuting the set of embedded lines covering properly embedded curves on the surface. Any collection of proper embedded rays in the plane whose pairwise intersections are compact inherits a natural circular order from the topology of the plane; it is this circular order that gives rise to circular orders on (certain extensions of) mapping class groups, and left orders on braid groups.