Hyperbolic groups with finitely presented subgroups not of Type $F_3$

Abstract

We build on the constructions in Brady (J Lond Math Soc 60(2):461–480, 1999) and Lodha (A hyperbolic group with a finitely presented subgroup that is not of type FP3, London Mathematical Society Lecture Note Series, Cambridge University Press, Cambridge, pp 67–81, 2017) to give infinite families of hyperbolic groups, each having a finitely presented subgroup that is not of type $F_3$. By calculating the Euler characteristic of the hyperbolic groups constructed, we prove that infinitely many of them are pairwise non isomorphic. We further show that the first of these constructions cannot be generalised to dimensions higher than 3.

Minimal sizeable graphs—by Giles Gardam

In this appendix we determine minimal sizeable graphs: we construct a sizeable graph (as defined in Definition 4.1) on 24 vertices (Fig. 6), and prove that there exists no such graph on 23 or fewer vertices (Theorem A.5). We also determine the smallest sizeable graphs in certain classes: where the 4 subgraphs are all paths (as in [12]), all cycles, and all cycles defined ‘arithmetically’ as in [14].

Notation

For our sizeable graphs, we will denote the two bipartitions of the vertex sets as $A = A_0 \sqcup A_1$ and $B = B_0 \sqcup B_1$, rather than using $+$ and − as indices. The 4 induced subgraphs ${\varGamma }(A_s, B_t)$ will be referred to as the defining subgraphs of ${\varGamma }$.

First, let us sketch why constructing sizeable graphs is a delicate matter. On the large scale, a sizeable graph “looks like” a 4-cycle: think of shrinking each $A^s$, $B^t$ to a single vertex, and draw an edge to represent an induced subgraph that is connected. This runs contrary to containing no 4-cycle, which makes constructing such graphs difficult.

Suppose that each $A^s$ and $B^t$ has n edges. Connectivity of the defining subgraphs requires that they have average degree approximately 2 (so average degree 4 in the whole graph), since the sum of degrees over the 2n vertices must be at least $2(2n-1)$.

Proposition A.1

Let ${\varGamma }$ be a random bipartite graph on $A \sqcup B$ with $\left|A\right| = \left|B\right| = 2n$, constructed by including any edge from A to B independently with probability $\frac{2}{n}$. Then the expected number of subgraphs of ${\varGamma }$ isomorphic to the 4-cycle is $(1-\frac{1}{2n})^2 64$.

This underlines the difficulty of constructing such graphs.

Proof

The number of possible 4-cycles is ${\left( {\begin{array}{c}2n\\ 2\end{array}}\right) }^2$, and the probability of any given 4-cycle occurring is $\left( \frac{2}{n}\right) ^4$. $\square$

Definition A.2

We call a graph arithmetic if it can be constructed in the following manner. The vertices are divided into 4 sets $A_0, A_1, B_0, B_1$ of equal size n, identified with the group $\mathbb {Z} / n$. The edges are determined by 8 numbers: $h_{s,t}$ and $k_{s,t}$ for $s, t \in \{0, 1\}$. The vertex $a_s^i \in A^s$ is joined by an edge to $b_t^{i+h_{s,t}}$ and $b_t^{i+k_{s,t}}$ in $B^t$.

Lodha constructed arithmetic sizeable graphs for $n = 11$, that is, on 44 vertices. The graph in Fig. 5b is an example for $n = 9$, with the values

$$\begin{aligned} h_{0,0}&= 0&h_{0,1}&= 0&h_{1,0}&= 0&h_{1,1}&= -1 \\ k_{0,0}&= -1&k_{0,1}&= 2&k_{1,0}&= 2&k_{1,1}&= -2 \end{aligned}$$

(the difference in sign is due to orientation of the edges from $A_0$ and towards $A_1$, used for convenience in the proof of Theorem A.3).

Theorem A.3

The smallest number of vertices for an arithmetic sizeable graph is 36.

Proof

The example in Fig. 5b shows that 36 is possible; we now show that less than 36 is not. Consider an arithmetic graph. For any vertex $a_0 \in A_0$, there are 8 paths of length 2 from $a_0$ to a vertex $a_1 \in A_1$: such a path can go via $B_0$ or $B_1$, and there are two choices of a neighbour b of $a_0$ in each corresponding $B^t$, and b then has 2 neighbours in $A_1$. If any of these paths had the same endpoint, this would give a 4-cycle. Thus $A_1$ must have at least 8 vertices, so the whole graph must have at least 32.

Suppose for the sake of contradiction that we had an arithmetic sizeable graph on 32 vertices, that is, for $n = 8$. After a change of coordinates, we can assume that $h_{0,0} = h_{1,0} = h_{0,1} = 0$. To be precise, we relabel the vertex $b_0^i$ with $b_0^{i-h_{0,0}}$, $b_1^i$ with $b_1^{i-h_{0,1}}$, and $a_1$ with $a_1^{i-h_{0,0}+h_{1,0}}$ and adjust the values of $h_{s,t}$ and $k_{s,t}$ accordingly afterwards. For notational convenience, as indicated in Fig. 5a we let $a = k_{0,0}$, $b = -{k_{1,0}}$, $c = k_{0,1}$, $d = -h_{1,1}$ and $e = -k_{1,1}$.

Fig. 5

Arithmetic graphs

The 8 paths of length 2 from $a_0^0$ to $A_1$ end at the vertices $a_1^i$ for $i = 0, a, b, a+b, d, c+d, e, c+e$, and thus these are a permutation of $0, \dots , 7 \pmod {8}$. Summing these 8 numbers, we have

$$\begin{aligned} 2(a + b + c + d + e) \equiv 28 \equiv 4 \pmod {8} \end{aligned}$$

However, for the 4 defining subgraphs to be connected (and not to split as 2 or more disjoint cycles) we need a, b and c to be odd, and d and e to be of opposite parity, so without loss of generality assume d is even and e is odd. Now the 4 endpoints $a_1^i$ for $i = a$, b, $c + d$ and e must be a permutation of 1, 3, 5, 7, so that

$$\begin{aligned} a + b + c + d + e \equiv 1 + 3 + 5 + 7 \equiv 0 \pmod {8}. \end{aligned}$$

This is a contradiction. $\square$

For the lower bound of 32, all we used was the assumption that in each of the 4 defining subgraphs, every vertex has degree 2. For instance, this is true if the subgraphs are (not necessarily arithmetically defined) cycles. By exhaustive computational search, we have determined there is no sizeable graph of 32 vertices of this type either, so a minimal sizeable graph with all defining subgraphs cycles is our arithmetic example on 36 vertices.

An immediate observation is that we can assume the defining subgraphs of any sizeable graph are trees: the definition requires only that they be connected, but whenever one is not a tree, we could remove an edge from a cycle and obtain a sizeable graph on the same number of vertices (since we clearly will not have introduced a 4-cycle, nor affected connectivity of subgraphs) but with one less edge. So it is natural to ask how small we can make a sizeable graph where the 4 defining subgraphs are not cycles, but rather paths. In fact, intuition suggests that one could well construct sizeable graphs of the absolute minimum number of vertices in this way, because one anticipates that the branching of any tree that is not a path creates central, highly connected vertices that would serve to create 4-cycles.

However, with the 4 defining subgraphs as paths we cannot achieve the minimum of 24, or even very close. Most of this gap is explained by the fact that having the total number of edges in each subgraph more evenly distributed over the degrees of the vertices (which have degree 2 except for in 2 cases) means that every vertex in B joins sufficiently many pairs in $A_0 \times A_1$ that unless there are 28 vertices, creating a 4-cycle is unavoidable by the pigeonhole principle. We now prove this.

Proposition A.4

Any sizeable graph with all 4 defining subgraphs a path has at least 28 vertices.

Proof

Let the $n = \left|B_0\right|$ vertices in $B_0$ be labelled $1, 2, \dots n$ and let $d_i$ denote the degree of vertex i in the subgraph ${\varGamma }(A_0, B_0)$ and let $e_i$ be its degree in ${\varGamma }(A_1, B_0)$. Since the defining subgraphs are paths, every $d_i$ and every $e_i$ is either 1 or 2. The sum of degrees over one side of a bipartite graph is the number of edges, so $\sum _{i=1}^n d_i = \left|A_0\right| + \left|B_0\right| - 1$ and it follows that the number of $d_i$ that are equal to 1 is $\left|B_0\right| + 1 - \left|A_0\right|$ (which is either 0, 1, or 2, noting that in a bipartite path graph the two sides have size differing by at most 1). Similarly, precisely $\left|B_0\right| + 1 - \left|A_1\right|$ of the $e_i$ are equal to 1.

The number of pairs in $A_0 \times A_1$ that are joined by a vertex in $B_0$ is $\sum _{i=1}^n d_i e_i$. By the rearrangement inequality, this is bounded below by the value for when we never have $d_i = e_i = 1$, that is, when there are $\left|B_0\right| + 1 - \left|A_0\right|$ pairs $(d_i, e_i) = (2, 1)$, $\left|B_0\right| + 1 - \left|A_1\right|$ pairs $(d_i, e_i) = (1, 2)$, and the remaining pairs are (2, 2) (if such an arrangement is not possible due to there being too many 1’s, the following bound still holds). Thus we have

$$\begin{aligned} \sum _{i=1}^n d_i e_i \ge 4n - 2 ( 2\left|B_0\right| + 2 - \left|A_0\right| - \left|A_1\right|) = 2 (\left|A_0\right| + \left|A_1\right| - 2 ) \end{aligned}$$

We have the same lower bound on the number of pairs in $A_0 \times A_1$ joined by a vertex in $B_1$, so since these pairs must be distinct, we have

$$\begin{aligned} \left|A_0\right| \left|A_1\right| \ge 4 (\left|A_0\right| + \left|A_1\right| - 2) \end{aligned}$$

which gives $(\left|A_0\right| - 4) (\left|A_1\right| - 4) \ge 8$. This is possible (for natural numbers) only if $\left|A_0\right| + \left|A_1\right| \ge 14$. We similarly conclude $\left|B_0\right| + \left|B_1\right| \ge 14$ so the graph has at least 28 vertices in total. $\square$

Remark 3

By exhaustive computational search, we know that actually the minimum size we can attain is 31 vertices.

Theorem A.5

The minimal number of vertices of a sizeable graph is 24.

Fig. 6

A sizeable graph on 24 vertices, depicted as a subgraph of the incidence graph for the projective plane of order 3 (Fig. 7), shown together with the 4 induced subgraphs

With no topological assumptions on the subgraphs (such as being cycles or paths) the best lower bound we can get with the techniques used to prove Proposition A.4 is 22, which is close to the actual minimum of 24.

Proposition A.6

Every sizeable graph has at least 22 vertices.

Proof

As in the proof of the above proposition, let the degrees of vertices of $B_0$ be $d_i$ and $e_i$. Connectivity requires that $d_i, e_i \ge 1$ and thus $(d_i - 1)(e_i - 1) \ge 0$, so $d_i e_i \ge d_i + e_i - 1$. Thus $\sum _{i=1}^n d_i e_i \ge \sum _{i=1}^n d_i + \sum _{i=1}^n e_i - \left|B_0\right| = (\left|A_0\right| + \left|B_0\right| - 1) + (\left|A_1\right| + \left|B_0\right| - 1) - \left|B_0\right| = \left|A_0\right| + \left|A_1\right| + \left|B_0\right| - 2$.

We have a similar bound for $B_1$, and thus considering the total number of pairs in $A_0 \times A_1$ at distance 2 in the graph metric, we see that

$$\begin{aligned} \left|A_0\right| \left|A_1\right| \ge \left|A_0\right| + \left|A_1\right| - 4 + (\left|A_0\right| + \left|A_1\right| + \left|B_0\right| + \left|B_1\right|) \end{aligned}$$

We can assume without loss of generality that $\left|B_0\right| + \left|B_1\right| \ge \left|A_0\right| + \left|A_1\right|$, so after substituting into the above inequality and factorizing we have

$$\begin{aligned} (\left|A_0\right| - 3) (\left|A_1\right| - 3) \ge 5 \end{aligned}$$

Since $\left|A_0\right|, \left|A_1\right|$ are positive integers this implies that $\left|A_0\right| + \left|A_1\right| \ge 11$, so the graph has at least 22 vertices.

While we can use such a pleasantly simple argument to get a lower bound of 22, to improve the lower bound to 24 we now apply results in extremal graph theory, concerning the Zarankiewicz problem. To do this, we weaken the requirements on our graphs somewhat: we forget the bipartitions of A into $A_0$ and $A_1$ and of B into $B_0$ and $B_1$, and thus do not ask for connectivity of these subgraphs, but rather ask simply that there be at least as many edges as connectivity would require, which is $2N-4$ for a sizeable graph on N vertices.

Definition A.7

([9, Definition 1.2]) A bipartite graph $G = (A, B; E)$ is called $K_{s,t}$-free if it does not contain s vertices in A and t vertices in B that span a subgraph isomorphic to the complete bipartite graph $K_{s,t}$. The maximum number of edges that a $K_{s,t}$-free bipartite graph of size (m, n) may have is the Zarankiewicz number $Z_{s,t}(m,n)$.

Even for $(s,t) = (2,2)$ – the case we are interested in, corresponding to having no 4-cycles – not all Zarankiewicz numbers are known exactly, but they are known for the range of values relevant to us [9, Table 1]. This smallest value of $m+n$ for which $Z_{2,2}(m, n) \ge 2(m+n) - 4$ holds is 23: $Z_{2,2}(11,12) = 42$ (and is the only possibility with $m+n = 23$ up to swapping m and n), whereas for $m+n = 22$ we have

(m, n)	(7, 15)	(8, 14)	(9, 13)	(10, 12)	(11, 11)
$Z_{2,2}(m,n)$	33	35	37	39	39

Corollary A.8

A sizeable graph has at least 23 vertices.

A sizeable graph on 23 vertices lies somehow just beyond the cusp of what is possible. This makes the situation incredibly constrained; we suppose that we were to have a sizeable graph on 23 vertices and progressively determine more and more of its structure, until we zero in on a specific contradiction. Up to symmetry, there is at each step of the argument only one way for the graph to develop.

Fig. 7

The projective plane of order 3 with 13 points and 13 lines. Each pair of points defines a line, and each pair of lines intersects in one point

The incidence graph of the projective plane of order 3, Fig. 7, is a natural candidate for a small sizeable graph. The projective planes are incidence geometries that are optimal in various senses. Of particular importance for us is that they are bipartite and have no 4-cycles: a 4-cycle would correspond to a pair of lines that intersected in two distinct points. The projective plane of order 3 then presents itself since its points and lines all have $3+1=4$ incidences, so that the total number of edges is larger than the minimum needed for connectivity of the 4 induced subgraphs. Exhaustive computation revealed that the incidence graph of the projective plane of order 3 cannot give a sizeable graph: no bipartition of A and of B will give 4 connected subgraphs. However, our minimal sizeable graph is a subgraph of this incidence graph; given a sizeable graph on n vertices one cannot necessarily extend it to a sizeable graph on $n+1$ vertices.

We reproduce here the relevant part of Corollary 3.19 of [9]. This is a stability result, saying that graphs close to having the desired combinatorial prorties of the incidence graph of the projective plane can be embedded in it.

Corollary A.9

Let $c \in \mathbb {N}$. Then

$$\begin{aligned} Z_{2,2} (n^2 + c, n^2 + n) \le n^2(n+1) + cn \end{aligned}$$

Moreover, if $c \le n+1$, then graphs reaching the bound can be embedded into a projective plane of order n.

Now we are equipped to prove that no sizeable graph has fewer than 24 vertices, giving the main theorem of this appendix.

Fig. 8

The stages of the proof by contradiction that there is no sizeable graph on 23 vertices

Proof of Theorem A.5

By Corollary A.8 and the example of Fig. 6, it only remains to rule out a sizeable graph on 23 vertices. For $n=3$ and $c=2$, Corollary A.9 tells us that any sizeable graph on 23 vertices, which must achieve the bound on $Z_{2,2}(11,12)$ (by tabulated values), is embedded in the incidence graph of the projective plane of order 3. This incidence geometry has 13 points and 13 lines. Since moving between the incidence geometry and the incidence graph it defines can be a source of confusion, we emphasize the following: points and lines refer to the incidence geometry, whereas vertices and edges refer to the corresponding graph. By duality (there is an automorphism of the incidence geometry that interchanges points and lines), we can assume that it is formed by removing 2 points and 1 line. Every point is incident to 4 lines, and vice versa, so the total number of incidences is 52. After removing two points we will have removed 8 incidences, leaving 44. Thus the line we remove must in fact be the unique line on which the two points lie: this would mean its removal only affects the 2 remaining points on the line, leaving 42 incidences, whereas the removal of any other line would destroy at least 3 incidences. Since the group of automorphisms of the projective plane is 2-transitive on the set of points (since $GL_3 \mathbb {F}_3$ is 2-transitive on lines in $\mathbb {F}_3^3$, being transitive on linearly independent tuples), we have only one possibility for the bipartite graph on A, the points, and B, the lines. For concreteness, we remove the points labelled 9 and 11 in our standard figure of the projective plane of order 3, Fig. 7, and the line they define, as shown in Fig. 8a. The incidence structure left is an affine plane on the 9 points labelled from 0 to 8, together with two points at infinity, labelled by 10 and 12.

Now all that remains is to show that we cannot give A and B the requisite bipartite structures with 4 connected subgraphs. First we show that there is only one possibility for the bipartition of the points. For the induced subgraphs to be connected, we of course need every line to have degree at least 1 in the two subgraphs in which it occurs; this means precisely that each line must contain at least one point both from $A_0$ and from $A_1$. This has a very strong consequence, as a small counting argument now reveals. Since there can only be one line incident to a given pair of points, the total number of triples $(a_0, a_1, b)$ where points $a_0 \in A_0$ and $a_1 \in A_1$ both lie on the line b is at most $\left|A_0\right| \cdot \left|A_1\right|$. Counting over the lines first, we see that a line containing k points creates at least $k-1$ such tuples, with equality if and only if all but 1 point on the line are in the same $A_i$. There are 6 lines containing 4 points and 6 lines containing 3, so this gives at least $6 \times (4-1) + 6 \times (3-1) = 30$ such triples. On the other hand, since $\left|A\right| = \left|A_0\right| + \left|A_1\right| = 11$, the maximum value possible for $\left|A_0\right| \cdot \left|A_1\right|$ is 30. Thus without loss of generality, $\left|A_0\right| = 5$ and $\left|A_1\right| = 6$, and moreover no line can contain 2 points of $A_0$ and 2 points of $A_1$.

Consider now the points “at infinity” labelled 10 and 12. A line passing through them contains either 0 or 2 points in the same side of the bipartition of points. Summing over the three lines, we see that of the 9 points (labels 0 to 8) on the affine plane, an even number is contained in the same side of the partition. Thus the two points at infinity must be in the same side as each other, and this must be $A_1$, of size 6, since it thus contains an even number of points. Thus the three lines incident to one of these points at infinity will comprise 1 line that is entirely $A_0$ inside the affine plane, and 2 lines that have only 1 point in $A_0$. The projective plane with the two points at infinity added admits symmetries permuting the 3 lines incident to any point at infinity, so without loss of generality, we can suppose that the lines $\{0, 1, 2, 10\}$ and $\{0, 3, 6, 12\}$ are the two relevant lines which, in the affine plane, are entirely in $A_0$. That is, up to symmetry, there is only one possible way of forming the bipartition $A = A_0 \sqcup A_1$, which is indicated in Fig. 8b, with $A_0$ in white and $A_1$ in black.

We now consider the two induced subgraphs that involve $A_1$. Any line that only contains one point from $A_1$ can be removed, since it will correspond to a leaf vertex in the corresponding bipartite graph, and its removal will not affect connectivity. This leaves us with Fig. 8c. The points 10 and 12 at infinity are now of degree 2 in the bipartite graph, so they must have degree 1 in two subgraphs. So they will be leaves, and we can safely remove them, retaining the condition that each of them forces the two lines incident to it to be in different sides of $B = B_0 \sqcup B_1$. The incidence geometry now has 4 points, with each pair of points defining a line (it is in fact the affine plane of order 2). There are two pairs of non-intersecting lines which must go in different sides of the partition (the condition we retained when removing the points at infinity); this forces the same of the third pair, since otherwise one side has only 2 lines and the corresponding graph has 6 vertices but only 4 edges. Thus, up to symmetry, we have a bipartition of the lines as indicated in Fig. 8d. The graph corresponding to the dashed lines is not connected, so we have reached the desired contradiction. $\square$