Tverberg-Type Theorems with Altered Intersection Patterns (Nerves)

Abstract

Tverberg’s theorem says that a set with sufficiently many points in ${\mathbb {R}}^d$ can always be partitioned into m parts so that the $(m-1)$-simplex is the (nerve) intersection pattern of the convex hulls of the parts. The main results of our paper demonstrate that Tverberg’s theorem is just a special case of a more general situation, where other simplicial complexes must always arise as nerve complexes, as soon as the number of points is large enough. We prove that, given a set with sufficiently many points, all trees and all cycles can also be induced by at least one partition of the point set. We also discuss how some simplicial complexes can never be achieved this way, even for arbitrarily large sets of points.

Appendix A: Proofs of Auxiliary Lemmas

We include proofs of some supplementary lemmas mentioned in the introduction.

Proof of Lemma 1.3

Suppose by contradiction that ${{\,\mathrm{Tv}\,}}(K,d)< 2n$. Let $S \subset {\mathbb {R}}^d$ be a set of points in convex position with $|S|={{\,\mathrm{Tv}\,}}(K,d)$. By the pigeonhole principle, if we partition S into n disjoint subsets, there must be at least one subset that is a singleton $\{{\varvec{x}}\}$. Since K is connected, the node corresponding to the singleton $\{{\varvec{x}}\}$ is connected, by an edge, to at least one other node, implying that $\{{\varvec{x}}\}$ is in the convex hull of another subset. However, this is a contradiction as the points are in convex position. $\square$

Proof of Lemma 1.6

To show that $\mathcal {N}^1(\mathcal {P}) = \mathcal {N}^1(\sigma ({\mathcal {P})})$ it suffices to show that ${{\,\mathrm{conv}\,}}(P_{i_1})\cap {{\,\mathrm{conv}\,}}(P_{i_2}) \ne \emptyset$ if and only if ${{\,\mathrm{conv}\,}}(\sigma (P_{i_1}))\cap {{\,\mathrm{conv}\,}}(\sigma (P_{i_2})) \ne \emptyset$ for all $i_1, i_2 \in [n]$. Suppose ${{\,\mathrm{conv}\,}}(P_{i_1})\cap {{\,\mathrm{conv}\,}}(P_{i_2}) \ne \emptyset$. Then they contain respectively $P_{i_1}'$ and $P_{i_2}'$, which are an inclusion minimal Radon partition of $S_1$. Since $\sigma$ is an order-preserving bijection, $\sigma$ is an isomorphism between oriented matroids (see, for instance, [21]) determined by $S_1$ and $S_2$. The minimal Radon partitions in $S_1$ correspond to the circuits of the oriented matroids and therefore are preserved under $\sigma$. Thus ${{\,\mathrm{conv}\,}}(\sigma (P_{i_1}'))\cap {{\,\mathrm{conv}\,}}(\sigma (P_{i_2}')) \ne \emptyset$. The reverse implication is shown by the reasoning applied to $\sigma ^{-1}$. $\square$

As we mentioned in the introduction, the graph K in Fig. 12 is 2-partition induced (in particular, by the partitioned point set in Fig. 13), but not 2-Tverberg, as implied by the following lemma:

Lemma A.1

Suppose S is any set of points in convex position in ${\mathbb {R}}^2$. Then the graph K in Fig. 12 is not partition induced on S.

Fig. 12

Graph K

Fig. 13

Partitioned point set with nerve K

Proof

We note that since K is a triangle free graph, it suffices to show that it is not the intersection graph of any partition of points in convex position. We argue by contradiction. Suppose that there is a set of points in convex position partitioned so that the graph above is their intersection graph. By Lemma 1.6 we may assume the points are arranged on the boundary of a disc $\mathcal {D}$. Call the convex hull of the points corresponding to each node i by region i. In the rest of the proof of Lemma A.1, we will rely on the following.

Claim

Consider the independent set of nodes $\{A,B,C\}$ in Fig. 12. Up to exchanging their labels (note that the graph is symmetric about A, B, C), there are two possible arrangements of the regions A, B, and C, pictured in Fig. 5.

Proof of the claim

The region $\mathcal {M}-B$ has two connected components. If regions A and C lie in different connected components of $\mathcal {M}-B$, then regions A, B,  and C must be arranged as in Fig. 5. Otherwise, A and C lie in the same connected component, say $\mathcal {N}$, of $\mathcal {M}-B$. If we walk clockwise around the boundary of $\mathcal {N}$, we can only alternate twice between being in regions A and C, reducing to the two possibilities shown. $\square$

By the claim, we see that A, B, and C must be arranged (up to symmetry) as in one of the two cases pictured above. If they are arranged as in Fig. 5, note that regions E and F both intersect regions A, B, and C. In that case it is easy to see that regions E and F must intersect, which is a contradiction.

If the regions are arranged as in Fig. 5, consider those regions D, F, G, and H. Note that the region D intersects A, B, C. Also, region F is disjoint from all the regions B through H, while is intersecting A. Similarly, region G is disjoint from all the regions A through H except B. Also region H is disjoint from all the regions A through H except C. Considering the two cases: F, G, H lie in the same connected component of $\mathcal {M} - D$, or F, G, H lie in different connected components of $\mathcal {M}-D$, it is easy to see that, in both cases, F, G, and H must be arranged as A, B, and C are in Fig. 5. Then I, J are disjoint but both intersect F, G, and H, which is a contradiction by the argument above. Thus K cannot be the nerve of a set of points in convex position. $\square$