Tverberg’s theorem with constraints

(of Theorem 3) Set $N:=(d+1)(q-1)$, and let $q>2$ be of the form $p^{r}$ for some prime number $p$. As in the proof of topological Tverberg theorem in the version of [9], we consider the space ${\sf K}:=(\sigma^{N})^{*q}_{\Delta(2)}$ as configuration space. It models all possible partitions of the vertex set into $q$ blocks: A maximal simplex of ${\sf K}$ encodes a (Tverberg) partition as shown in Figure 5, and it can be represented as a hyperedge using one point from each row of ${\sf K}$.

Remember that $\|{\sf K}\|$ is $N-1$-connected. In the original proof of Theorem 1, the assumption that there is no Tverberg partition for $f$ leads to the existence of a $(\mathbb{Z}_{p})^{r}$-map $f^{q}:\|{\sf K}\|\rightarrow(\mathbb{R}^{d})^{*q}_{\Delta}$. However, there is not such a map due to Volovikov’s Lemma 9. Hence a Tverberg partition exists for $f$.

In the following, we construct for each graph a good subcomplex ${\sf L}$ of ${\sf K}$ such that: i) ${\sf L}$ is invariant under the $(\mathbb{Z}_{p})^{r}$-action, and ii) ${\rm conn}({\sf L})\geq N-1$. Here good means that ${\sf L}$ does not contain any of Tverberg partitions using an edge of our graph. As in the subsequent paragraph, the assumption that there is no Tverberg partition leads to a $(\mathbb{Z}_{p})^{r}$-map $f^{q}:\|{\sf L}\|\rightarrow(\mathbb{R}^{d})^{*q}_{\Delta}$. Finally Volovikov’s Lemma 9 implies a contradiction, and so that there is a Tverberg partition not using any edge of our graph. Hence, our graph is a constraint graph.

Our construction of good subcomplexes is based in its simplest case – for $K_{2}$ – on the following observation:

If two points $i$ and $j$ end up in the same partition set, then the maximal face representing this partition uses one of the vertical edges between the corresponding rows $i$ and $j$ in ${\sf K}$.

To prove the $K_{2}$ case, we have to come up with a subcomplex ${\sf L}$ that does not contain maximal simplices using vertical edges between rows $i$ and $j$. Let ${\sf L}$ be the join of the chessboard complex $\Delta_{2,q}$ on rows $i$ and $j$, and the remaining rows. Figure 6 shows this construction of ${\sf L}$ for $q=3$ and $d=2$. The chessboard complex $\Delta_{2,q}$ does not contain any vertical edges. Moreover, ${\sf L}$ is $(\mathbb{Z}_{p})^{r}$-invariant as only the orbit of the vertical edges is missing. For the connectivity of ${\sf L}$ see the next paragraph.

i) Construction of ${\sf L}$ for complete graphs $K_{l}$: Let $i_{1},i_{2},\ldots,i_{l}$ be the corresponding rows of ${\sf K}$. ${\sf L}$ must not contain any maximal faces with vertical edges between any two of these rows. The chessboard complex on these rows is such a candidate. Let ${\sf L}$ be the join of the chessboard complex $\Delta_{l,q}$ on the corresponding $l$ rows, and the remaining rows:

The subcomplex ${\sf L}$ is closed under the $(\mathbb{Z}_{p})^{r}$-action. Using Theorem 8 on the connectivity of the chessboard complex, and inequality (1) on the connectivity of the join, we obtain:

In the last step, we use that $\Delta_{l,q}$ is $(l-2)$-connected for $2l<q+2$.

ii) Construction of ${\sf L}$ for complete bipartite graphs $K_{1,l}$: We first construct an $(\mathbb{Z}_{p})^{r}$-invariant subcomplex $C_{l,q}$ on the corresponding $l+1$ rows. For this, let $i$ be the row that corresponds to the vertex of degree $l$, and $j_{1},j_{2},\ldots j_{l}$ be the corresponding rows to the $l$ vertices of degree $1$. Let $C_{l,q}$ be the maximal induced subcomplex of ${\sf K}$ on the rows $i,j_{1},j_{2},\ldots,j_{l}$ that does not contain any vertical edges starting at a vertex of row $i$. Then $C_{l,q}$ is the union of $q$ many complexes $L_{1},L_{2},\ldots,L_{q}$, which are all of the form of ${\rm cone}([q-1]^{*l})$. Here the apex of $L_{m}$ is the $m$th vertex of row $i$ for every $m=1,2,\ldots,q$. In Figure 7, the maximal faces of the complex $L_{3}$ are shown for $q=4$, and $l=2$.
Let ${\sf L}$ be the join of the complex $C_{l,q}$ and the remaining rows of ${\sf K}$:

Now ${\sf L}$ is good and $(\mathbb{Z}_{p})^{r}$-invariant by construction. Let’s assume

for $1<l<q-1$. The connectivity of ${\sf L}$ is then shown as above:

We prove assumption (3) in Lemma 12 below.

iii) Construction of ${\sf L}$ for paths $P_{l}$ on $l+1$ vertices: We construct recursively a good subcomplex ${\sf L}$ on $l+1$ rows such that ${\rm conn}({\sf L})\geq l-1$. The case $l=1$ is covered in the proof of i) so that we can choose ${\sf L}$ to be the complex $D_{2,q}:=\Delta_{2,q}$. For $l>1$, choose ${\sf L}$ to be the complex $D_{l,q}$ which is obtained from $D_{l-1,q}$ in the following way: Order the corresponding rows $i_{1},i_{2},\ldots,i_{l+1}$ in the order they occur on the path. Take $D_{l-1,q}$ on the first $l$ rows. A maximal face $F$ of $D_{l-1,q}$ uses a point in the last row $i_{l}$ in column $j$, for some $j\in[q]$. We want $D_{l,q}$ to be good so that we cannot choose any vertical edges between row $i_{l}$ and $i_{l+1}$. Let $D_{l,q}$ be defined through its maximal faces: All faces of the form $F\uplus\{k\}$ for $k\not=j$. Let $D_{l,q}^{k}$ be the subcomplex of all faces $D_{l,q}$ ending with $k$. Then $D_{l,q}=\bigcup_{k=1}^{q}D_{l,q}^{k}$. In Figure 8 the recursive definition of the complex $D_{l,5}^{2}$ is shown.

The complex is $(\mathbb{Z}_{p})^{r}$-invariant, and the connectivity of $D_{l,q}$

is shown in Lemma 13 below using the decomposition $\bigcup_{k=1}^{q}D_{l,q}^{k}$.

iv) Construction of ${\sf L}$ for cycles $C_{l}$ on $l$ vertices: Choose ${\sf L}$ to be the complex $E_{l,q}$ obtained from $D_{l-1,q}$ on $l$ rows by removing all maximal simplices that use a vertical edge between first and last row. The following result on the connectivity of $E_{l,q}$ is shown in Lemma 14 below:

v) Construction of ${\sf L}$ for disjoint unions of constraint graphs: For every graph component construct a complex on the corresponding rows as above. Let ${\sf L}$ be the join of these subcomplexes, and of the remaining rows. Then ${\sf L}$ is a good $(\mathbb{Z}_{p})^{r}$-invariant subcomplex by the similar arguments as above. The connectivity of ${\sf L}$ follows analogously from inequality (1) on the connectivity of the join. ∎

In our proof, we use the decomposition of $C_{l,q}$ into subcomplexes $L_{1},L_{2},\ldots$ $L_{q}$ from above.

The nerve $\cal N$ of the family $L_{1},L_{2},\ldots,L_{q}$ is a simplicial complex on the vertex set $[q]$. The intersection of $t$ many $L_{m_{1}},L_{m_{2}},\ldots,L_{m_{t}}$ is $[q-t]^{*l}$ for $t>1$ so that the nerve $\cal N$ is the boundary of the $(q-1)$-simplex. Hence $\cal N$ is $(q-3)$-connected.

Let’s look at the connectivity of the non-empty intersections $\bigcap_{j=1}^{t}L_{m_{j}}$. For $t=1$, every $L_{m}$ is contractible as it is a cone. For $1<t<q-1$, the space $[q-t]^{*l}$ is $(l-2)$-connected, and for $t=q-1$ the intersection is non-empty, hence its connectivity is $-1$. All non-empty intersections $\bigcap_{j=1}^{t}L_{m_{j}}$ are thus $(l-t)$-connected. The $(l-1)$-connectivity of $C_{l,q}$ immediately follows from the nerve theorem using $q>2$, and $l<q-1$. ∎

In our proof, we use the decomposition of $D_{l,q}$ into subcomplexes$D^{1}_{l,q},D_{l,q}^{2},\ldots,D_{l,q}^{q}$ from above. We prove the following connectivity result by an induction on $l\geq 1$:

Let $l=1$, then $D_{1,q}=\bigcup_{j\in[q]}D_{1,q}^{j}$ is the chessboard complex $\Delta_{2,q}$ which is $0$-connected for $q>2$. The union of complexes $D_{1,q}^{i}$ is a union of contractible cones which is $0$-connected. For $l\geq 2$, look at the intersection of $t>1$ many complexes $D_{l,q}^{i}$. Let $T\subset[q]$ be the corresponding index set of size $1<t<q-1$, and $\bar{T}$ its complement in $[q]$. Then their intersections are

The nerve $\cal N$ of the family $D_{l,q}^{1},D_{l,q}^{2},\ldots,D_{l,q}^{q}$ is a simplicial complex on the vertex set $[q]$. The nerve is the $(q-1)$-simplex, which is contractible.

For $l=2$, let’s apply the nerve theorem. For this, we have to check that the non-empty intersection of any $t\geq 1$ complexes is $(2-t)$-connected. Every $D_{2,q}^{j}$ is $1$-connected as it is a cone. The intersection of $t=2$ many complexes is $0$-connected for $q>3$ by equation (5). Note that this is false for $q=3$. The intersection of $t=3$ many complexes is non-empty.

For $l=3$, we have to show that the non-empty intersection of any $t$ complexes is $(3-t)$-connected. Every $D_{3,q}^{j}$ is $2$-connected as it is a cone. The intersection of $t<q-1$ many complexes is $1$-connected by equation (5). The intersection of $t=q-1$ many complexes is a union of two cones due to equation (6). The intersection of these two cones is:

which is non-empty. Using the nerve theorem, we obtain for their union:

The intersection of $t=q\geq 4$ many complexes is non-empty by equation (7).

Let now $l>3$, we apply again the nerve theorem to obtain inequality (4). It remains to check that the non-empty intersection of any $t$ complexes is $(l-t)$-connected. The complex $D_{l,q}^{j}$ is $(l-1)$-connected as it is a cone for every $j\in[q]$. The intersection of any $1<t<q-1$ complexes is $(l-2)$-connected by equation (5) and by assumption. The intersection of $t=q-1$ many complexes is a union of two cones due to equation (6). The intersection of these two cones is:

which is $(l-4)$-connected by assumption. Using the nerve theorem, we obtain for their union:

The intersection of $q$ many complexes is $(l-3)$-connected by equation (7) and by assumption. ∎

The proof is similar to the proof of Lemma 13. The case $l=3$ has already been settled in the proof of case i) of Theorem 3. The cases $l=4,5$ are analogous for $q\geq 5$, but need some tedious calculations. Observe that the inductive argument in the proof of Lemma 13 also works for $E_{l,q}$, which was obtained from $D_{l-1,q}$ by removing some maximal faces.

Let’s describe the differences to the proof of Lemma 13. We consider the decomposition $E^{1}_{l,q},E^{2}_{l,q},\ldots,E^{q}_{l,q}$ of $E_{l,q}$. Here $E^{i}_{l,q}$ is the complex that is obtained from $D^{i}_{l-1,q}$ by removing all maximal faces that contain the $i$th vertex of the first row. In Figure 9 the complex $E^{1}_{l,5}$ is shown: Any face of $D^{1}_{l-1,q}$ containing one of the broken edges is removed.

The intersection of this family is non-empty, in fact:

Thus its nerve is a simplex. Using the nerve theorem it remains to show that the intersection of $t\geq 1$ complexes is $(l-2-t+1)$-connected. For $t=1$, the complex $E^{i}_{l,q}$ is a cone. For $t=q$, this follows from equation (8). For $1<t<q$, this follows as in the proof of Lemma 13 from the equations:

where $\tilde{D}^{i,S}_{l,q}$ is the following subcomplex of $D^{i}_{l,q}$ for $S\subset[q]$: Delete all faces that contain a vertex in $S$ of the first row. In other words $\tilde{D}^{i,\{i\}}_{l,q}=E^{i}_{l+1,q}$, see also Figure 10 for equation (10). There any face containing a broken edge is deleted from $D^{i}_{l,q}$.

Using again the nerve theorem, one then shows the necessary connectivity results for equations (9) and (10). This can be done for $q\geq 5$, inductively on $l\geq 5$:

and for $T\subset[q]$, $1<|T|<q-1$:

∎

(of Theorem 5) Let $X$ be a set of $(d+1)(q-1)+1$ points in $\mathbb{R}^{d}$, and $p_{1}$ is a Tverberg point which is not of type I. The Tverberg point $p_{1}$ is the intersection point of $\bigcap_{i=1}^{k}{\rm conv}(F_{i}^{1})$, where $k\in\{2,3,\ldots,d\}$. Choose an edge $e_{1}$ in some $F_{i}$, and apply Theorem 3 with constraint graph $G_{1}=\{e_{1}\}$. Then there is a Tverberg partition that does not use the edge $e_{1}$ so that there has to be second Tverberg point $p_{2}$. Now add another edge $e_{2}$ from the corresponding $F_{i}^{2}$ to the constraint graph $G_{1}$, and apply again Theorem 3 with constraint graph $G_{2}=\{e_{1},e_{2}\}$. Hence there is another Tverberg point $p_{3}$ and so on. This procedure depends on the choices of the edges, and whether $G_{i}$ is still a constraint graph.
Figure 11 shows an example for $d=2$ and $q=3$: A set of seven points in $\mathbb{R}^{2}$. There are exactly four Tverberg points – highlighted by small circles – in this example. A constraint graph – drawn in broken lines – can remove only three among them.

Constraint graphs for $q$ are also constraint graphs for the subsequent prime power $q^{\prime}$ so that our constant $c_{d,q}$ is weakly increasing in $q$. The constant $c_{d,q}$ also depends on $d$ as the simplex $\sigma^{(d+1)(q-1)}$ grows in $d$.

It remains to prove $c_{2,3}>3$. For this, suppose we have three Tverberg partitions of type II for the set $\{a,b,c,d,e,f,g\}$ of seven points in $\mathbb{R}^{2}$.

If some edge, e. g. $\{a,b\}$, belongs to two partitions, we could find an edge in the third partition disjoint with $\{a,b\}$. The union of these two edges is a constraint graph.

If no edge belongs to two partitions, we have up to permutation the Tverberg partitions $\{a,b,c\},\{d,e\},\{f,g\}$ and $\{a,d,f\},\{b,e\},\{c,g\}$and the third partition could be either $\{a,e,g\},\{b,d\},\{c,f\}$ or $\{b,d,g\},\{a,e\},\{c,f\}$. In the former case the constraint graph $\{b,c\},\{d,f\},\{e,g\}$ contains an edge from every partition, and shows that there has to be a fourth Tverberg partition. In the later case, the same is true for the graph $\{b,c\},\{a,f\},\{d,g\}$. ∎