A cyclic order of a group is non-linear if any cut of it is not compatible with the group operation.

A cut of a cyclically ordered set is a linear order $<$ such that

  • $a < b < c \implies [a, b, c]$

for any elements $a$, $b$, $c$ of the set.

A cut of a cyclically ordered group is compatible with the group operation ($+$) iff:

  • $a < b \implies a + x < b + x$ and $x + a < x + b$

for any elements $a$, $b$, $x$ of the group.

I am trying to find properties of a non-linear cyclic order on fields:

  1. Is there an infinite field with a non-linear cyclic order on the additive group?
  2. Is there a field with a non-linear cyclic order on the additive group such that $[0, 1, -1]$ and $[0, x, 1]$ for some element $x$?
  • $\begingroup$ I assume all elements z = -z are non-negative and non-positive. $\endgroup$ – Alex C Mar 20 '17 at 21:36
  • $\begingroup$ The standard compatibility with addition is required. Any compatibility with multiplication is optional. $\endgroup$ – Alex C Mar 21 '17 at 2:13
  • $\begingroup$ If we are not requiring compatibility with multiplication, then there's little to no point in talking about rings or fields; the question should just ask about cyclically ordered abelian groups. $\endgroup$ – Hurkyl Mar 21 '17 at 14:48
  • $\begingroup$ I don't believe cyclically orders are well-known; it would help if you give a definition of the term, and related terms. For example, what is a "nonnegative" element in this context? The term doesn't appear anywhere in Wikipedia's article. $\endgroup$ – Hurkyl Mar 21 '17 at 14:52
  • $\begingroup$ 1. The compatibility with multiplication is not the same as for addition. It is adjusted to fit certain algebraic structures. I don't see any problems adjusting the compatibility for the strict cyclic order. 2. Nonnegative means not negative. The standard definition of a negative element a: $(0, -a, a)$. $\endgroup$ – Alex C Mar 21 '17 at 14:57

By picking any irrational real number $\alpha$, there is an injective map

$$ \mathbb{Q} \to \mathbb{T} : x \mapsto e^{2 \pi i \alpha x} $$

and thus (?) $\mathbb{Q}$ inherits a cyclic ordering from the one on $\mathbb{T}$. Since, in the comments, you say you only care about compatibility with addition, I imagine this serves as an example to both of your questions.

  • $\begingroup$ @AlexC: Typo fixed $\endgroup$ – Hurkyl Mar 21 '17 at 15:16
  • $\begingroup$ Thanks a lot! Where can I read more about this type of fields (the map)? $\endgroup$ – Alex C Mar 21 '17 at 15:23

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