If a linear order is a way of arranging the elements of a set ‘on a line’ (with a direction), then a *cyclic order* is a way of arranging them ‘on a circle’ (with a chosen direction, say clockwise or counterclockwise).

Unlike a linear order (and most other things called ‘order’) which are defined as a sort of binary relation, a cyclic order is defined as a sort of *ternary* relation, since on a circle we can’t say “$x$ comes before (or after) $y$” only “$x$, $y$, and $z$ come in clockwise (or counterclockwise) order.”

A **cyclic relation** on a set $S$ is a ternary relation $R$ on $S$ such that, if $R(x,y,z)$ for any elements $x, y, z$ of $S$, then $R(y,z,x)$ (and hence also $R(z,x,y)$).

Then a **cyclic order** on $S$ is a cyclic relation $R$ that satisfies ternary versions of the properties of a linear order:

- Cyclic irreflexivity: If $R(x,y,z)$, then $y \ne z$.
- Cyclic asymmetry: If $R(x,y,z)$ and $R(x,z,w)$, then $y \ne w$.
- Cyclic transitivity: If $R(x,y,z)$ and $R(x,z,w)$, then $R(x,y,w)$.
- Cyclic comparison: If $R(x,y,z)$, then for any $w$, $R(x,y,w)$ or $R(x,w,z)$ or $R(w,y,z)$.
- Cyclic connectedness: If $x \ne y$, $x \ne z$, and $y \ne z$, then $R(x,y,z)$ or $R(x,z,y)$.

Actually, none of these order axioms (except for comparison) is really complete as stated; any cyclic permutation of the axiom should also be assumed. However, these permutations all follow automatically for a cyclic relation.

Note that for a constructive version, the set $S$ needs to be already equipped with a (tight) apartness relation for the connectedness axiom to make sense; unlike with a linear order, we can't recover the apartness relation from the cyclic order. For a similar reason, it's difficult to state antisymmetry correctly for a nonstrict (like a total order) version of a cyclic order.

As with a linear order, not all of these axioms are needed. In particular, using excluded middle, it's enough if a cyclic relation is transitive and satisfies

- Cyclic trichotomy: For all $x, y, z$, ($x = y$ or $x = z$ or $y = z$) xor $R(x,y,z)$ xor $R(x,z,y)$.

An injective function $f:S\to T$ between cyclically ordered sets is **strictly monotone** if it preserves the ternary relation $R$, i.e. if $R_S(x,y,z)$ implies $R_T(f(x),f(y),f(z))$. This is the cyclic counterpart of a strictly monotone function between linear orders (one which preserves the relation $\lt$). As in that case, irreflexivity then implies that $f$ is injective as long as $S$ has at least three distinct elements; but in general, this should be stated explicitly (and interpreted in the strongest sense for constructive mathematics).

We say that $f$ is merely **monotone** if it satisfies (either of) the following *equivalent* conditions:

- $f$ reflects the ternary relation $R$, i.e. if $R_T(f(x),f(y),f(z))$ then $R_S(x,y,z)$.
- $f$ preserves the ternary relation $R$ for elements with pairwise distinct images, i.e. if $R_S(x,y,z)$ holds and $f(x)$, $f(y)$, and $f(z)$ are pairwise distinct, then $R_T(f(x),f(y),f(z))$.

This is the cyclic counterpart of a monotone function between linear orders (one which reflects the relation $\lt$).

- One would like the cycle category $\Lambda$ to be the category of finite nonempty cyclically ordered sets and monotone functions, just as the simplex category $\Delta$ is the category of finite nonempty linearly ordered sets and monotone functions. However, this seems to fail for the $0$-cycle, which has a nontrivial loop and is therefore not terminal (unlike the cyclically ordered singleton, which is terminal).

- a round chord diagram is a cyclic order equipped with a pairing of all its elements

- Tim Campion, Jinhe Ye?,
*Non-trivial higher homotopy of first-order theories*(arXiv:2306.12011)

Last revised on June 22, 2023 at 05:59:24. See the history of this page for a list of all contributions to it.