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Let us define an abstract "cycle complex" as the following combinatorial object: it is $(V, C)$, where $V$ is a set of $n$ nodes, $C$ is a set of $c$ cyclically ordered subsets of $V$, each of size exactly $s$, and it is assumed that two sets in $C$ intersect on at most one node. Let us define the genus of $(V, C)$ as the minimum genus of an orientable surface in which $(V, C)$ can be embedded. By "embedded," we mean that we can place the nodes $V$ in the surface such that:

  1. Each set $X \in C$ is represented by a connected region of the surface,
  2. If the nodes on the boundary of this region are read in clockwise order, we get exactly the set of nodes in $X$ in their given circular ordering
  3. The intersection of the regions corresponding to $X_1, X_2 \in C$ is the same as $X_1 \cap X_2$. That is, the regions are disjoint if $X_1 \cap X_2 = \emptyset$, or it is exactly their single point of intersection if there is one.

Question 1: Is there a "real name" for this combinatorial object that I can search for to find relevant literature?

Question 2: I am specifically interested in extremal upper bounds on the genus $g$, i.e. results of the form "$g \le f(n, c, s)$" for some function $f$. Is anything like this known?

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1 Answer 1

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Let's first consider the special case that each node appears in exactly two sets in $C$. (Nodes that only appear in one set are irrelevant and can be deleted without changing the possible embeddings, and I will say something about nodes that appear in more than 2 sets later).

Then we can define the "dual" of a cycle complex $(V,C)$ with each set in $C$ of size $s$ to be the $s$-regular graph $G=(V',E')$ where there is a vertex $v_X\in V'$ for each set $X\in C$ and an edge $e_w\in E'$ for each node $w\in V$. Then the cyclic orderings of each set $X$ determine a cyclic ordering of the edges around each vertex $v_X$ and this gives rise to a rotation system. I have not tried to write out a formal proof but I believe your notion of embedding agrees with the notion of embedding there.

So at least in this special case, Question 2 is equivalent to determining the genus of the rotation system. In fact, if you just want a bound in terms of $n,c,s$ then we can just ask about the genus of the graph $G$, or actually the "genus distribution" of the graph. These are well-studied problems and a simple upper bound for the maximum genus of the graph (see e.g. Section 3.4.7 of Gross and Tucker's book "Topological Graph Theory" or Theorem 16.2.3 in Gross and Yellen's book "Graph Theory and Its Applications") is (using the notation from your question):

$$g\leq \frac{\lfloor n-c+1\rfloor}{2}.$$

I'm not sure if $s$-regularity buys us anything, but maybe someone who knows can add a comment / answer.

I think you can reduce general cycle complexes to the above case as follows: For each node $x$ which appears in more than two sets in $C$, do the following. Suppose that $x$ appears in sets $X_1,\dots, X_k$. Then replace $x$ in $X_1,\dots,X_k$ by $k$ new nodes $x_1,\dots,x_k$ (and delete $x$ from $V$). Then create a new set $X_x$ which contains $x_1,\dots,x_k$ with some cyclic ordering $\sigma_x$.

When you're finished, you don't quite have a cycle complex under your definition since the $C'$ are not necessarily of size $s$. However, you can still apply the previous construction to get a graph and a rotation system, and the genus will be the minimum genus of the rotation system under all possible choices of the $\{\sigma_x\}$. I don't immediately see how to find a good bound for this, but it seems possible. In particular I'll leave it to you to figure out if there's a useful analogue of the "simple bound" above.

Note that we had to make choices of the $\sigma_x$ so we can't map general cycle complexes to rotation systems (so they don't give a perfect answer to your Question 1).

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  • $\begingroup$ Thanks for the response! The rotation system dual observation is very nice, I agree that this is exactly what I'm looking for. I'm confused about the point that any embedding of $G$ suffices, though. Isn't there a risk that an arbitrary embedding of $G$ (or the one that realizes $g \le \frac{\lfloor n-c+1 \rfloor}{2}$) wouldn't necessarily respect the prescribed rotation ordering of its edges? $\endgroup$
    – GMB
    Commented Oct 27, 2017 at 16:31
  • $\begingroup$ @GMB What you are saying is true but I'm not sure I understand your confusion. In general, you can associate many rotation systems to a given graph. Then, given a rotation system, you can compute its genus. I thought the point of question 2 was to look for upper bounds on the genus given in terms of a few simple properties of the cycle complex, but perhaps I misunderstood. $\endgroup$
    – j.c.
    Commented Oct 27, 2017 at 17:05
  • $\begingroup$ Sorry if I'm being dense -- my confusion is that when we associate a graph $G$ to a rotation system, I think the genus of the graph will be less than or equal to that of the rotation system, since there is more freedom in the graph embedding. Therefore an upper bound on the genus of the graph won't necessarily imply the same upper bound on the genus of the rotation system. $\endgroup$
    – GMB
    Commented Oct 27, 2017 at 19:11
  • $\begingroup$ @GMB I think I see: I should have made this more clear; the bound I give is for the "maximum genus" of $G$, which is the maximum of the genus over all possible rotation systems that are associated to $G$. Have a look at the discussion in the references I suggested and let me know if what I said still doesn't make sense. $\endgroup$
    – j.c.
    Commented Oct 27, 2017 at 22:58
  • $\begingroup$ Ahh, got it, now I understand. Thank you! $\endgroup$
    – GMB
    Commented Oct 28, 2017 at 2:21

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