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Background: A preorder is a binary relation $\leq$ which is reflexive and transitive. We can write the transitive property as ${\leq}(a,b)\wedge{\leq}(b,c)\to{\leq}(a,c)$. There are additional axioms that give us partial orders etc., so plenty of everyday "order" concepts can be modeled with such a binary relation.

Similarly, a category has a composition of morphisms $\circ$, which has identities and sends $C(a,b)\times C(b,c)\to C(a,c)$. So we can also model an preordered set as a category in which every hom-set has at most one morphism. I haven't really studied this perspective, but I gather that it's useful.

Now, a cyclic ordering is more naturally a ternary relation! Its version of transitivity states that $(a,b,c)\wedge(a,c,d)\to(a,b,d)$. A cyclic ordering is also cyclic: $(a,b,c)\to(b,c,a)$. If you hunt around Wikipedia you can find a few different kinds of cyclic orders with more or less restrictive axioms. For this question, let's understand "cyclic order" broadly, so it might be strict or not, and it might be total or not: whatever is convenient!

Question: Would it be useful to model cyclically ordered sets as categories? How would you do it?

I'm guessing that higher categories might be useful, so that you can replace the ternary relation $(a,b,c)$ with a 2-morphism like $(a\rightarrow b)\Rightarrow(a\rightarrow c)$, and there could be a composition of 2-morphisms like $$\left[(a\rightarrow b)\Rightarrow(a\rightarrow c)\right]\times\left[(a\rightarrow c)\Rightarrow(a\rightarrow d)\right]\mapsto\left[(a\rightarrow b)\Rightarrow(a\rightarrow d)\right].$$ But I'm getting way out of my depth here! I've skimmed over some higher categories on nLab, and I don't see a kind of 2-morphism that does quite what I want. Simplicial 2-morphisms looked promising at first, but they don't seem to compose in the right way. Is that idea a dead end?

Note that I'm not asking about any category of monotone functions between cyclically ordered sets. At least, I don't think that's what I'm asking.

This question was previously asked on Math.SE, but it hasn't prompted an answer. I'm hoping someone here might be able to help, even though it's not a research question for me. I'm just curious. Thanks!

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  • $\begingroup$ I don't see that this would be a particularly useful thing to do. Cyclically ordered sets seem to be much less rich than preorders; one reason it's interesting to regard preorders as categories is that many interesting categorical concepts specialize to interesting concepts involving preorders, whereas I can't think of much to say about cyclic orders. I guess they're transitive $\mathbb{Z}$-sets? $\endgroup$
    – Qiaochu Yuan
    Aug 7, 2013 at 7:17
  • $\begingroup$ (In general a $G$-set can be modeled as a groupoid equipped with a functor to $G$.) $\endgroup$
    – Qiaochu Yuan
    Aug 7, 2013 at 7:17
  • $\begingroup$ @QiauchuYuan Not in general. Only finite cycles, and $\mathbb Z$ itself, are transitive $\mathbb Z$-sets. These cycles can be completely described by a successor relation. On the other hand, more interesting cycles, including dense cycles like the rational numbers $\mathbb Q$ and the circle $S^1$, lack a transitive action of $\mathbb Z$ by monotonic maps. $\endgroup$
    – Chris Culter
    Aug 7, 2013 at 7:29
  • $\begingroup$ As for richness, let's start with the totally ordered case. A cyclic order is basically a linear order that has forgotten where $\infty$ is. So they're about as rich as linear orders. In the partially ordered case, there's a new phenomenon: a partial cyclic order cannot always be extended to a total cyclic order! This situation is more interesting than the situation for partial orders. In the case of cyclic preorders, we should probably expect even stranger behavior. $\endgroup$
    – Chris Culter
    Aug 7, 2013 at 7:39
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    $\begingroup$ @QiaochuYuan No worries, Wikipedia is disappointing. :) Some interesting examples of non-total cyclic orders are products and powers of total cyclic orders, either direct or lexicographic. If the poset of subsets of a set $X$ is $2^X$, then an analogy would be the direct power $3^X$, where $3$ is the smallest nontrivial cycle. A simple example would be $3^2$, in which the elements $(0,2)$, $(1,1)$, and $(2,0)$ are unordered. I think I've read about the possibility of embedding arbitrary partial cyclic orders in sets of the form $3^X$, but I can't remember the conclusion. $\endgroup$
    – Chris Culter
    Aug 8, 2013 at 22:05

1 Answer 1

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It seems like cyclically ordered sets ought to be regarded as cousins of categories rather than as categories themselves. More precisely, a cyclically ordered set $C$ has a nerve $N(C)$ analogous to the nerve of a (higher) category or (higher) groupoid. The nerve is first of all a sequence $N(C)_n$ of sets, namely the set of $n$-tuples $(a_1, ... a_n)$ such that the relation $(a_i, a_j, a_k)$ holds for all $1 \le i < j < k \le n$ (or two of $a_i, a_j, a_k$ are equal). There are various natural maps between the $N(C)_n$ which make $N(C)$ a cyclic set in the sense of Connes.

Cyclic sets lie intermediate between simplicial sets (which is where the nerve construction takes its values for (higher) categories) and symmetric sets (which is where the nerve construction takes its values for (higher) groupoids). They are in particular simplicial sets, so you can think of cyclically ordered sets as modeled by simplicial sets with extra data. This construction is a beefed-up version of the construction of the order complex of a poset (which is the nerve (regarded as an abstract simplicial complex) of the poset (regarded as a category)).

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  • $\begingroup$ Thanks, that sounds like a great approach! I'll chew on this answer a bit to understand it. Feel free to answer the question in Math.SE as well, just for the record. $\endgroup$
    – Chris Culter
    Aug 9, 2013 at 0:28

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