Gus Wiseman

## Every Clutter Is a Tree of Blobs NB   CDF   PDF

Given a finite vertex set, one can construct every connected spanning hypergraph by first choosing a spanning hypertree, then choosing a blob on each of its edges.

### Introduction

If is a finite vertex set and is a collection of finite subsets (called edges), none of which is a subset of another, we recursively define the swell of , , to be the collection of all sets that either:

1. belong to
2. are the union of some pair of overlapping sets, both already belonging to

For example, if

then

If we also have , then the set system is called a clutter. This condition means that (except in the case ) each edge contains at least two vertices, and the hypergraph spanned by the edge set is connected. Here the hypergraph spanned by a set of edges is defined to have vertex set and edge set . (There is no agreed-upon definition of hypergraph. For some authors it is any set system; for others it is a simplicial complex; for others it is an antichain of sets.)

Here is a larger clutter.

The number of clutters spanning vertices is given by A048143 (oeis.org/A048143),

This sequence varies as , so the number of digits required roughly doubles with each consecutive term. Our main example is just one of some 56 sextillion members of .

This normalizing function is a universal invariant for the species of labeled clutters, meaning two clutters are isomorphic iff they have the same image.

Here is a list of nonisomorphic representatives for all clutters with up to four vertices, corresponding to unlabeled clutters. This brute-force enumeration may not work for .

### Kernels and Caps in Clutters

A kernel of is a clutter (the restriction of to edges that are subsets of ) for some .

Define .

A set partition is a set of disjoint sets with .

Suppose is a set partition of such that each block is a kernel of (i.e. and ). Since would imply , it follows that the set of unions is itself a clutter, which we call a cap of .

Equivalently, a cap of is a clutter satisfying both:

1. every edge of is a subset of exactly one edge of

To see that this does not establish a partial order of clutters with a vertex set, observe that

is a nontransitive chain of caps. The following is the set of all set partitions of the edge set indices corresponding to each cap of the clutter.

In these plots of clutter partitions, the filled squares correspond to all pairs of a vertex and an edge such that the vertex belongs to the edge; these squares are then shaded according to which block of the partition the edge belongs to.

### Trees and Blobs

The density of a clutter is

where the sum is over all edges .

A clutter with two or more edges is a tree iff . This is equivalent to the usual definition of a spanning hypertree [1].

A clutter is a blob iff no cap of is a tree.

The trees and blobs among the caps and kernels (respectively) of our running example are as follows.

Suppose a clutter decomposes into a cap and corresponding set of kernels . Then

where the sum is over all . In particular, , and iff every is a tree. Using this simple identity, one easily proves the following.

Lemma

Every kernel (with two or more edges) of a tree is a tree.
Every cap (with two or more edges) of a tree is a tree.
The union of a set of trees whose set of unions is a tree, is a tree.

The following is also straightforward.

Proposition

A clutter is a tree iff no kernel of is a blob.

We now come to the main result.

For our running example, the theorem corresponds to the following decomposition into a tree of blobs.

Theorem

Assume is not a blob. Let be the subset-maximal kernels of that are blobs. Then is a set partition of whose set of unions is a tree.

Proof

First we show that any blob (kernel) is contained within a single branch of any tree (cap). Suppose that is a kernel of and is a blob, and that is a cap of that is a tree. Let be the subtree of contributing to the set partition of non-empty intersections for each branch . The set of unions forms a clutter that is obtained from by deleting in turn all vertices not in , a process that weakly decreases density. Let be the set partition comprised of maximal kernels (i.e. connected components) contained in blocks of . Then is a cap of and . Since is a connected clutter, we have , and therefore . But since is a blob, cannot be a tree, hence it must be a maximal cap (viz. , ).

Next we show that . If any two blobs overlap, both blobs must be contained entirely in whatever branch (of any given tree) contains their intersection. This implies that there is another blob containing their union, and hence that the maximal blobs are disjoint. Since every singleton is also a blob, we conclude that is a cap of .

Finally, if any kernel of were a blob, so would be the restriction of to its union, contradicting maximality of . This proves that the set of unions of is a tree.

The following are the decompositions for each nonisomorphic clutter with four vertices.

### Connected Sets of Kernels

Let be the set of all kernels of . If is itself a (connected) clutter with vertex set , then there exists a unique subset-minimal upper bound satisfying both

1. for all

In general, we can only define uniquely for a connected set of kernels, so is not strictly a join operation for the poset of subsets . But if is not connected as a clutter, then letting be its (maximal) connected components, we say that is a connected set of kernels iff is connected as a set of kernels, in which case the join is given by

In practice, the verification of connectedness and the computation of may require several iterations constructing joins of connected components. For example, consider the connected set of kernels.

It has the following sequence of joins of connected components.

One Problem

If is a connected set of kernels, we define its compression to be the number of iterations in the computation of by constructing consecutive joins of connected components. For the previous example, we have . Although it seems unlikely that is a bounded invariant, we do not know how to construct an example with compression greater than .

1. For which positive numbers does there exist a connected set of kernels such that ?
2. Does there exist an infinite chain of connected sets of kernels such that for all ?

Define an invariant by

for all , where the sum is over all clutter partitions . Here denotes the indicator function for a proposition , equal to 1 or 0 depending on whether is true or false, respectively.

Theorem

For any we have , where the sum is over all connected sets of kernels spanning .

Proof

Let be the set of all clutter partitions . What we have essentially shown above is that , regarded as a subposet of the lattice of set partitions ordered by refinement, is a lattice. We have the simple enumerative identity

where the product is over all non-singleton kernels , here regarded as elements of whose only non-singleton block is . Expanding the right-hand side gives

where the sum is over all sets of non-singleton kernels , again regarded as lattice elements. Here is algorithmically the same operation as the connected-join operation on . Expanding and factoring accordingly, this becomes

where the outer sum is over all , the product is over all , and the inner sum is over all connected sets of kernels spanning . For any kernel , define

where the outer sum is over all clutter partitions , and where the product and inner sum are as before. Letting be the set partition of whose only non-singleton block is , we have shown that

Hence our theorized expansion does indeed satisfy the defining identity of .

Note that it is sufficient in the preceding theorem and proof to consider only connected sets of subset-minimal non-singleton kernels, and it is often practical to do so. The hypergraph whose edges are minimal non-singleton kernels is also of some interest. The well-known Möbius function of a hypergraph is defined on the lattice of connected set partitions, and in this context an element of may be called a pseudo-kernel. In comparison, however, our invariant , which is defined on essentially all clutters, seems to be more interesting; we do not know if it has been studied before.

A semi-clutter is any anti-chain of subsets . For each finite set , let be the set of semi-clutters spanning . A species [2] is an endofunctor on the category of finite sets and bijections, so here we have defined a species of semi-clutters. The compound semi-clutter of a decomposition , as defined by Billera [3], is obtained as a disjoint sum of Cartesian products. Interpreted in the language of species theory, this is a certain natural transformation

where denotes the composition operation on species, a generalization of composition of exponential formal power series. Let be the set of (connected) clutters spanning , let be the set containing only the maximal clutter on , and let be the set of clutters having no expression as a compound of a proper decomposition (i.e. is the species of prime clutters). Billeras main theorem (attributed to Shapley) establishes a unique reduced compound representation, which is itself a species of decompositions

From this it is evident that can ultimately be reduced to a nested compound expression using only trivial and prime clutters. Hence the problem of enumerating semi-clutters on a vertex set is reduced to the problem of constructing, for any connected clutter, its maximal proper committees, which is the nontrivial solution of [2] for the enumeration of prime clutters considered. This is a particularly interesting application of formal species.

### References

 [1] R. Bacher, “On the Enumeration of Labelled Hypertrees and of Labelled Bipartite Trees.” arxiv.org/abs/1102.2708. [2] A. Joyal, “Une théorie combinatoire des séries formelles,” Advances in Mathematics, 42(1), 1981 pp. 1–82. doi:10.1016/0001-8708(81)90052-9. [3] L. J. Billera, “On the Composition and Decomposition of Clutters,” Journal of Combinatorial Theory, Series B, 11(3), 1971 pp. 234–245. doi:10.1016/0095-8956(71)90033-5. G. Wiseman, “Every Clutter Is a Tree of Blobs,” The Mathematica Journal, 2017. dx.doi.org/doi:10.3888/tmj.19-7.