Maximum $\mathcal H$-free subgraphs

TL;DR

This paper investigates the maximum size of hypergraphs free of certain substructures, establishing bounds and exploring conditions under which these bounds remain finite as the hypergraph size grows.

Contribution

The paper develops bounds for the function $f(m, \\mathcal{H}_m)$ and explores the conditions for boundedness across sequences of hypergraph families.

Findings

Derived bounds for $f(m, \\mathcal{H}_m)$ in various cases

Identified conditions for boundedness of $f(m, \\mathcal{H}_m)$ as $m$ increases

Showed that a complete characterization of bounded sequences is unlikely

Abstract

Given a family of hypergraphs $\mathcal H$, let $f(m,\mathcal H)$ denote the largest size of an $\mathcal H$-free subgraph that one is guaranteed to find in every hypergraph with $m$ edges. This function was first introduced by Erd\H{o}s and Koml\'{o}s in 1969 in the context of union-free families, and various other special cases have been extensively studied since then. In an attempt to develop a general theory for these questions, we consider the following basic issue: which sequences of hypergraph families $\{\mathcal H_m\}$ have bounded $f(m,\mathcal H_m)$ as $m\to\infty$? A variety of bounds for $f(m,\mathcal H_m)$ are obtained which answer this question in some cases. Obtaining a complete description of sequences $\{\mathcal H_m\}$ for which $f(m,\mathcal H_m)$ is bounded seems hopeless.

Equations

$$f(m,\mathcal{H}):=\underset{|F|=m}{\min }\text{mbox}ex(F,\mathcal{H}).$$

$$f(m,\mathcal{H})=\lfloor \sqrt{4m+1}\rfloor -1.$$

$$\text{mbox}Forwhichsequenceoffamilies\{{\mathcal{H}}_m{\}}_{m=1}^{\infty }\text{mbox}isf(m,{\mathcal{H}}_m)\text{mbox}bounded(asm\to \infty \text{mbox})?$$

$$\text{mbox}Forwhichsequenceofk\text{mbox}-edgehypergraphs\{H_m{\}}_{m=1}^{\infty }\text{mbox}isf(m,H_m)\text{mbox}bounded(asm\to \infty \text{mbox})?$$

$$b_{\ell }:=\left|\bigcap _{i\in I}{A}_i\setminus \bigcup _{i\in [k]\setminus I}{A}_i\right|.$$

$$\alpha (\vec{\mathbf{b}})(m):=\underset{1\le i\le k-2}{\min }\left(\frac{b_i(m)}{m{b}_{i+1}(m)}\right).$$

$${\left(\frac{1}{2\left(\alpha (\vec{\mathbf{b}})+\frac{1}{m}\right)\left(\genfrac{}{}{0ex}{}{b_{k-1}+b_k}{b_k}\right)}\right)}^{\frac{1}{k}}\le f(m,H(\vec{\mathbf{b}}))\le \frac{k(k-1)}{\alpha (\vec{\mathbf{b}})}+k-1.$$

$$f(m,H(\vec{\bf b}))\geq\left\{\begin{array}[]{lc}m^{\frac{1}{k(b_{k}+1)}}\left(\frac{b_{k-1}}{4(b_{k-2}+2b_{k-1})}\right)^{\frac{1}{k}},&k\geq 4,\\ m^{\frac{1}{b_{3}+2}}\left(\frac{b_{2}}{4(b_{1}+2b_{2})}\right)^{\frac{b_{3}+1}{b_{3}+2}},&k=3.\end{array}\right.$$

$${\left(\frac{1}{2\left(\frac{b_1}{m{b}_2}+\frac{1}{m}\right)\left(\genfrac{}{}{0ex}{}{b_2+b_3}{b_3}\right)}\right)}^{\frac{1}{3}}\le f(m,H(\vec{\mathbf{b}}))\le \frac{6m{b}_2}{b_1}+2.$$

$$\text{mbox}{twr}_{\ell -1}(c^{\prime }{t}^2)<r_{\ell }(t,t)<\text{mbox}{twr}_{\ell }(ct).$$

$$\begin{array}[]{l}\mbox{Let }F\mbox{ be any }\ell\mbox{-graph on }n\mbox{ vertices. Then there is an absolute constant }c_{\ell}\mbox{ such that there is}\\ \mbox{a subgraph }F^{\prime}\subset F\mbox{ with }|V(F^{\prime})|\geq c_{\ell}\cdot\log_{(\ell)}(n)\mbox{, which is either a clique or an independent}\\ \mbox{set. Here }\log_{(\ell)}\mbox{ denotes iterated logarithms.}\end{array}$$

$$\min \left\{\frac{m}{2^{n-\ell +1}},c_{r_{\ell }}{\left(\frac{m}{2^{n-\ell +1}}\right)}^{\frac{1}{r_{\ell }}}\right\}\ge c_{\mathcal{H}}^{\prime }\cdot m^{\frac{1}{r_{\ell }}}.$$

$$|F_{\ell }|\ge C_{\mathcal{H}}\cdot m^{\frac{1}{r_1\dots r_{\ell }}}.$$

$$F_{i_{1}}\cap\ldots\cap F_{i_{\ell}}=\left\{\begin{array}[]{lc}\varnothing,&\ell>L,\\ \neq\varnothing,&\ell\leq L.\end{array}\right.$$

$$\left\{\{B_1,\dots ,B_{\ell }\}:|B_1\cap \cdots \cap B_{\ell }|=a\right\}.$$

$$b_i=a_i-\left(\genfrac{}{}{0ex}{}{k-i}{1}\right)a_{i+1}+\left(\genfrac{}{}{0ex}{}{k-i}{2}\right)a_{i+2}-\cdots +(-1{)}^{k-i}(\genfrac{}{}{0ex}{}{k-i}{k-i})a_k.$$

$$\sum _{j=i}^{k-1}(-1{)}^{j-i}(\genfrac{}{}{0ex}{}{m-k+j-i-1}{j-i})b_j\ge 0.$$

$$Y_{j}^{\ell}=\left\{\begin{array}[]{cc}\{v_{j}^{S}:S\in{[m]\choose\ell}\},&\ \ell<k\\ \{w_{j}\},&\ \ell=k\end{array}\right\}$$

$$Y=\bigcup _{\ell =1}^k\bigcup _{j=1}^{d_{\ell }}{Y}_j^{\ell }.$$

$$\left\{\begin{array}{rl}{A}_1& =\left\{v_1^1;v_1^{12},v_2^{12},v_1^{13},v_2^{13},v_1^{14},v_2^{14};w_1,w_2,w_3\right\}\\ A_2& =\left\{v_1^2;v_1^{12},v_2^{12},v_1^{23},v_2^{23},v_1^{24},v_2^{24};w_1,w_2,w_3\right\}\\ A_3& =\left\{v_1^3;v_1^{13},v_2^{13},v_1^{23},v_2^{23},v_1^{34},v_2^{34};w_1,w_2,w_3\right\}\\ A_4& =\left\{v_1^4;v_1^{14},v_2^{14},v_1^{24},v_2^{24},v_1^{34},v_2^{34};w_1,w_2,w_3\right\}\end{array}\right\}.$$

$$a_i=d_i+\left(\genfrac{}{}{0ex}{}{m-i}{1}\right)d_{i+1}+\cdots +\left(\genfrac{}{}{0ex}{}{m-i}{k-1-i}\right)d_{k-1}+d_k.$$

$$|A_{r_1}\cap \cdots \cap A_{r_i}|=\sum _{G^{\prime }\supseteq G}|U_{G^{\prime }}|=\sum _{G^{\prime }\supseteq G}\left|\bigcap _{X\in G^{\prime }}X\setminus \bigcup _{X\notin G^{\prime }}X\right|.$$

$$|U_{G^{\prime}}|=\left\{\begin{array}[]{cc}d_{|G^{\prime}|},&i\leq|G^{\prime}|<k,\\ 0,&k\leq|G^{\prime}|<m,\\ d_{k},&|G^{\prime}|=m.\end{array}\right.$$

$$a_i=d_i+\left(\genfrac{}{}{0ex}{}{m-i}{1}\right)d_{i+1}+\cdots +\left(\genfrac{}{}{0ex}{}{m-i}{k-1-i}\right)d_{k-1}+d_k$$

$$d_i:=\sum _{j=i}^{k-1}(-1{)}^{j-i}(\genfrac{}{}{0ex}{}{m-k+j-i-1}{j-i})b_j,$$

$$A_{k,m}=(a_{ij}^{(m)}{)}_{1\le i,j\le k},\text{mbox}and{B}_{k,m}=(b_{ij}^{(m)}{)}_{1\le i,j\le k},$$

$$\vec{g}=B_{k,k}\vec{a}.$$

$$\vec{a}=D\vec{d}.$$

$$\vec{d}=W^{\prime }\vec{b}.$$

$$\vec{g}=B_{k,k}D{W}^{\prime }\cdot \vec{b}.$$

Full text

††footnotetext: **MSC Subject Classification: **05D05, 05C35, 05C65

1 Introduction

A hypergraph $H$ on vertex set $V(H)$ is a subset of $2^{V(H)}$. $H$ is an $\ell$-uniform hypergraph, or simply, an $\ell$-graph, if $H\subseteq\binom{V(H)}{\ell}$. All hypergraphs in this paper have finitely many vertices (and edges). Given a family of hypergraphs $\mathcal{H}$, a hypergraph $F$ is said to be $\mathcal{H}$-free if $F$ contains no copy of any member of $\mathcal{H}$ as a (not necessarily induced) subgraph. Given a hypergraph $F$ and a family $\mathcal{H}$, let $\mbox{ex}(F,\mathcal{H})$ be the maximum size of an $\mathcal{H}$-free subgraph of $F$. Define

[TABLE]

Note that $f(m,\mathcal{H})\geq c$ means that every $F$ with $m$ edges contains an $\mathcal{H}$-free subgraph $F^{\prime}\subseteq F$ with $|F^{\prime}|=c$. When the family $\mathcal{H}$ consists of a single hypergraph $H$, we abuse notation and write $f(m,H)$ instead of $f(m,\{H\})$.

This function was introduced by Erdős and Komlós in 1969 [1], who considered the case when $\mathcal{H}$ is the (infinite) family of hypergraphs $A,B,C$ with $A\cup B=C$. The problem was further studied by Kleitman [2], and later by Erdős and Shelah [3], and finally settled by Fox, Lee and Sudakov [4] who proved that

[TABLE]

Erdős and Shelah also considered the case when $\mathcal{H}$ is the family of hypergraphs $A_{1},A_{2},A_{3},A_{4}$ with $A_{1}\cup A_{2}=A_{3}$ and $A_{1}\cap A_{3}=A_{4}$. They called this family $B_{2}$, proved that $f(m,B_{2})\leq(3/2)m^{2/3}$ and conjectured that this bound is asymptotically tight. This conjecture was settled by Barát, Füredi, Kantor, Kim and Patkós in 2012 [5], who also considered more general problems (see [4] for further work).

The same problem has been studied in the special case when $\mathcal{H}$ is a family of graphs. Let $f_{2}(m,\mathcal{H})$ denote the maximum size of an $\mathcal{H}$-free subgraph that every graph with $m$ edges is guaranteed to contain. These investigations began with a question of Erdős and Bollobás [6] in 1966 about $f_{2}(m,C_{4})$, followed up by a conjecture of Erdős in [7]. Consequently the problem of determining $f_{2}(m,H)$ for various graphs has received considerable attention in the recent years [8, 9, 10]. The authors of [9, 10] also considered the problem in the case of $\ell$-graphs.

In the hope of obtaining a general theory for these problems, we investigate the following basic question:

[TABLE]

Question (1) is too general to solve completely, so we focus on special cases. In subsection 2.1 we state our results for constant $\{\mathcal{H}_{m}\}_{m=1}^{\infty}$, and in subsection 2.2 we consider non-constant $\{\mathcal{H}_{m}\}_{m=1}^{\infty}$.

2 Our Results

2.1 Constant Sequences

Suppose $\{\mathcal{H}_{m}\}_{m=1}^{\infty}$ is a sequence such that $\mathcal{H}_{m}=\mathcal{H}$ for every $m$. First, we note that if $\mathcal{H}$ consists of finitely many members, then the answer to Question (1) is given by the following characterization. A $q$-sunflower is a hypergraph $\{A_{1},\ldots,A_{q}\}$ such that $A_{i}\cap A_{j}=\bigcap_{s=1}^{q}A_{s}$ for all $i\neq j$. This common intersection is referred to as the core of the sunflower.

Theorem 2.1**.**

Fix a family of hypergraphs $\mathcal{H}$ with finitely many members. If $\mathcal{H}$ contains a $q$-sunflower with sets of equal size, then $f(m,\mathcal{H})\leq q-1$. Otherwise, $f(m,\mathcal{H})\to\infty$ as $m\to\infty$.

Next, in the same spirit as the properties of being union-free and having no $B_{2}$, if the (infinite) family $\mathcal{H}$ specifies the intersection type of $k$ sets (ie whether they are empty or not), then a characterization can be obtained in the form of Theorem 2.3. Before stating the theorem, we first define what we call an even hypergraph and an $\ell$-uneven hypergraph. A $k$-edge hypergraph is a hypergraph with $k$ edges.

Definition 2.2** (Even and $\ell$-uneven hypergraphs).**

A $k$-edge hypergraph $H=\{A_{1},\ldots,A_{k}\}$ is said to be even if for every $1\leq\ell\leq k$ and for every $I,J\in\binom{[k]}{\ell}$, $\bigcap_{i\in I}A_{i}=\varnothing\iff\bigcap_{j\in J}A_{j}=\varnothing$. It is said to be $\ell$-uneven if there exist $I,J\in\binom{[k]}{\ell}$ such that $\bigcap_{i\in I}A_{i}\neq\varnothing$ but $\bigcap_{j\in J}A_{j}=\varnothing$.

Theorem 2.3**.**

Let $1\leq\ell<k$. Let $\mathcal{H}$ be the (infinite) family of all $\ell$-uneven $k$-edge hypergraphs. Then, $f(m,\mathcal{H})\to\infty$ as $m\to\infty$. Conversely, if $\mathcal{H}$ is the family of all even $k$-edge hypergraphs, we have $f(m,\mathcal{H})=k-1$.

2.2 Non-constant Sequences

As a first step towards understanding the general problem in (1), we focus on the case when for every $m\geq 1$, $\mathcal{H}_{m}=\{H_{m}\}$ for a single hypergraph $H_{m}$, and further assume that all these hypergraphs $H_{m}$ have the same number of edges. Thus we ask the following question:

[TABLE]

We are unable to answer question (2) completely, even for $k=3$. Our main results provide several necessary, or sufficient conditions that partially answer (2). Before presenting them, we introduce the following crucial definition:

Definition 2.4** (Equal Intersection Property).**

For $k\geq 2$, Let $\mathbf{EIP}_{k}$ denote the set of all $k$-edge hypergraphs $H=\{A_{1},\ldots,A_{k}\}$ such that for every $1\leq\ell\leq k$ and $I,J\in\binom{[k]}{\ell}$, we have $\left|\bigcap_{i\in I}A_{i}\right|=\left|\bigcap_{j\in J}A_{j}\right|$.

Every $H=\{A_{1},\ldots,A_{k}\}\in\mathbf{EIP}_{k}$ can be encoded by $k$ parameters $(b_{1},\ldots,b_{k})$, corresponding to the $k$ distinct sizes appearing in the Venn diagram of $H$. More precisely, for $1\leq\ell\leq k$, and for all $I\in\binom{[k]}{\ell}$,

[TABLE]

By inclusion-exclusion, $b_{1},\ldots,b_{k}$ are well-defined for hypergraphs in $\mathbf{EIP}_{k}$. We denote $H\in\mathbf{EIP}_{k}$ with parameters $b_{1},\ldots,b_{k}\geq 0$ by $H(\vec{b})$, where $\vec{b}=(b_{1},\ldots,b_{k})$. We shall see later (Lemma 4.1) that every sequence of $k$-edge hypergraphs $\{H_{m}\}$ such that $f(m,H_{m})$ is bounded, can only have finitely many members not in $\mathbf{EIP}_{k}$. For sequences $\{H_{m}\}_{m=1}^{\infty}$ such that $H_{m}\in\mathbf{EIP}_{k}$ for every $m\geq 1$, we obtain a sequence of length $k$ vectors $\{\vec{b}(m)\}_{m=1}^{\infty}$, where $\vec{b}(m)=(b_{1}(m),\ldots,b_{k}(m))$. We use boldface and write $\vec{\bf b}$ for the sequence $\{\vec{b}(m)\}_{m=1}^{\infty}$. Since every two edges form a $2$-sunflower, this discussion together with the construction from Theorem 2.1 gives us

Proposition 2.5**.**

Let $H_{m}$ be a 2-edge hypergraph for each $m\geq 1$. Then $f(m,H_{m})$ is bounded as $m\rightarrow\infty$ if and only if $H_{m}\in\mathbf{EIP}_{2}$ for all but finitely many $m$.

We may therefore assume in what follows that $k\geq 3$.

Definition 2.6** ($\alpha(\vec{\bf b})$).**

For every sequence of length $k$ vectors $\vec{\bf b}=\{\vec{b}(m)\}_{m=1}^{\infty}$ and $m\geq 1$, let

[TABLE]

Now we state our main results. To simplify notation we will often write $b_{i}$ instead of $b_{i}(m)$ and $\alpha(\vec{\bf b})$ instead of $\alpha(\vec{\bf b})(m)$.

Theorem 2.7**.**

Let $k\geq 3$. Suppose the sequence of length $k$ vectors $\vec{\bf b}$ satisfies $b_{1},\ldots,b_{k-2}>0$, $b_{k-1},b_{k}\geq 0$ for every $m$. Then, for $m\geq 6$,

[TABLE]

Theorem 2.7 implies that when $\binom{b_{k-1}+b_{k}}{b_{k}}$ is bounded from above, $f(m,H(\vec{\bf b}))$ is bounded from above if and only if the sequence $\alpha(\vec{\bf b})$ is bounded away from zero.

We also have the following additional lower bound on $f(m,H(\vec{\bf b}))$:

Theorem 2.8**.**

Fix $k\geq 3$. Let $\vec{\bf b}=\{\vec{b}(m)\}_{m=1}^{\infty}$ be such that $b_{k}(m)=b_{k}$ for every $m$. Then, for $m\geq 6$,

[TABLE]

We now focus on $k=3$. In this case $\alpha(\vec{\bf b})=b_{1}/mb_{2}$ and Theorem 2.7 reduces to

[TABLE]

When $b_{3}=0$, (3) implies that $f(m,H_{3}(b_{1},b_{2},0))$ is bounded if and only if $b_{1}=\Omega(mb_{2})$. We now turn to $b_{3}=1$ which already seems to be a very interesting special case that is related to an open question in extremal graph theory (see Problem 7.3 in Section 7). Here (3) and Theorem 2.8 yield the following.

Corollary 2.9**.**

Let $m\rightarrow\infty$. Then $f(m,H_{3}(b_{1},b_{2},1))$ is bounded when $b_{1}=\Omega(mb_{2})$ and it is unbounded when either $b_{1}+b_{2}=o(m)$ or $b_{1}=o(\sqrt{m}\,b_{2})$.

Corollary 2.9 can be summarized in Figure 2. The light region corresponds to a bounded $f(m,H(\vec{\bf b}))$, and the dark region corresponds to unbounded $f(m,H(\vec{\bf b}))$. White regions correspond to areas where we do not know if $f(m,H(\vec{\bf b}))$ is bounded or not.

We are able to refine our results slightly via the following result.

Theorem 2.10**.**

For every odd prime power $q$ we have $f(q^{2}+1,H(q^{2}-q-1,q,1))=2.$

As a consequence, we obtain the following:

Corollary 2.11**.**

When $b_{1}\geq b_{2}^{\,2}-b_{2}-1$ and $b_{2}^{\,2}\geq m$, we have $f(m,H_{3}(b_{1},b_{2},1))=2$.

Corollary 2.11 yields the following improvement on Figure 2. Note that we are using the parabola $b_{1}=b_{2}^{\,2}$ as an asymptotic approximation of Corollary 2.11. Therefore on this parabola, $f(m,H_{3})=2$. We also know, by virtue of Theorem 7.2, that in the white region to the right of the parabola and between the two lines, we have $f(m,H_{3}(b_{1},b_{2},1))>2$.

3 Proofs of Theorems 2.1 and 2.3

In this section, we prove Theorems 2.1 and 2.3, which answer question (1) for constant sequences. We use the following well-known facts about sunflowers and diagonal hypergraph Ramsey numbers.

Recall that a $q$-sunflower is a hypergraph $\{A_{1},\ldots,A_{q}\}$ such that $A_{i}\cap A_{j}=\bigcap_{s=1}^{q}A_{s}$. The celebrated Erdős-Rado sunflower Lemma [11] states the following.

Lemma 3.1** (Erdős-Rado).**

Let $H$ be an $r$-graph with $|H|=r!(\alpha-1)^{r}$. Then, $H$ contains an $\alpha$-sunflower.

Next, recall that the hypergraph Ramsey number $r_{\ell}(s,t)$ is the minimum $N$ such that any $\ell$-graph on $N$ vertices, admits a clique of size $s$ or an independent set of size $t$. The following is a well-known theorem of Erdős, Hajnal and Rado [12]:

Theorem 3.2**.**

There are absolute constants $c(\ell),c^{\prime}(\ell)$ such that

[TABLE]

Here the tower function $\mbox{twr}_{k}(x)$ is defined by $\mbox{twr}_{0}(x)=1$ and $\mbox{twr}_{i+1}(x)=2^{\mbox{twr}_{i}(x)}$.

The right side of this theorem can be rewritten as follows:

[TABLE]

Now we are prepared to prove Theorems 2.1 and 2.3. Recall that a hypergraph is uniform if all its edges have the same size, otherwise it is non-uniform.

Proof of Theorem 2.1.

Fix a family of hypergraphs $\mathcal{H}$ with $n$ members, $\mathcal{H}=\{H_{1},\ldots,H_{n}\}$. Let $H_{i}\in\mathcal{H}$ be an $r$-uniform $q$-sunflower with core $W$. For every $m\geq q$, let $F$ be an $r$-uniform $m$-sunflower with core $W$. Then every subset of $F$ of size $q$ is isomorphic to $H_{i}$, thus proving $f(m,\mathcal{H})\leq q-1$.

Now, suppose $\mathcal{H}$ contains $\ell$ many uniform hypergraphs labelled $H_{1},\ldots,H_{\ell}$, and $(n-\ell)$ many non-uniform hypergraphs labelled $H_{\ell+1},\ldots,H_{n}$. For $1\leq i\leq\ell$, let $r_{i}$ be the uniformity of $H_{i}$. Given any hypergraph $F$ with $m$ edges, we find a large $\mathcal{H}$-free subgraph as follows. First, since $H_{n}$ is non-uniform, it contains a set of size $a$ and a set of size $b\neq a$. Clearly, at least half of the edges of $F$ have size $\neq a$, or at least half of them have size $\neq b$. Take the appropriate subgraph $F_{1}\subset F$ of size $\geq\frac{m}{2}$. By successively halving the sizes, we obtain a chain of hypergraphs $F_{n-\ell}\subset F_{n-\ell-1}\subset\cdots\subset F_{1}\subset F$ such that $F_{n-\ell}$ is $\{H_{\ell+1},\ldots,H_{n}\}$-free, and $|F_{n-\ell}|\geq\frac{m}{2^{n-\ell}}$.

We now deal with the uniform part of $\mathcal{H}$. Notice that by Lemma 3.1, any $r$-graph $G$ with $|G|=m$ contains an $\alpha$-sunflower, as long as $m>r!\alpha^{r}$. Taking $\alpha=\left\lfloor c_{r}m^{1/r}\right\rfloor$ where $c_{r}=\left((2r)!\right)^{-1/r}$, satisfies the required condition. So, every $r$-graph $G$ of size $m$ contains a sunflower of size $\left\lfloor c_{r}m^{1/r}\right\rfloor$.

Since $H_{\ell}$ is $r_{\ell}$-uniform, we note that either $F_{n-\ell}$ contains a subgraph of size $\frac{1}{2}|F_{n-\ell}|$ which has no sets of size $r_{\ell}$ (and hence is $H_{\ell}$-free), or there is a subgraph of size $\frac{1}{2}|F_{n-\ell}|$ which is $r_{\ell}$-uniform. In the second case, using Lemma 3.1 on this subgraph, we obtain an $H_{\ell}$-free subgraph of $F_{n-\ell}$ of size at least

[TABLE]

We iterate the same argument $\ell-1$ more times, to finally obtain a constant $C_{\mathcal{H}}$ and a subgraph $F_{\ell}\subset F$ such that $F_{\ell}$ is $\mathcal{H}$-free, and

[TABLE]

∎

Proof of Theorem 2.3.

Let $F=\{F_{1},\ldots,F_{m}\}$ have size $m$. Suppose $\mathcal{H}$ is $\ell$-uneven for some $1\leq\ell<k$. Then, there are distinct subsets $I,J\in\binom{[k]}{\ell}$, such that $\bigcap_{i\in I}A_{i}=\varnothing$ and $\bigcap_{j\in J}A_{j}\neq\varnothing$ for every $H=\{A_{1},\ldots,A_{k}\}\in\mathcal{H}$. Then, we construct an $\ell$-graph $G$ with vertex set $F$, and hyperedges $\left\{\{F_{1},\ldots,F_{\ell}\}:\ F_{1}\cap\cdots\cap F_{\ell}=\varnothing\right\}$. By (4), there is a a constant $c_{\ell}$ and a subset $F^{\prime}\subseteq F$ of size $\geq c_{\ell}\cdot\log_{(\ell)}(m)$, such that $F^{\prime}$ is either a clique or an independent set in $G$. In either case, $F^{\prime}$ is $\mathcal{H}$-free.

On the other hand, if $\mathcal{H}$ is such that that for every $1\leq\ell\leq k$ and $I,J\in\binom{[k]}{\ell}$, $\bigcap_{i\in I}A_{i}\neq\varnothing\iff\bigcap_{j\in J}A_{j}\neq\varnothing$, then there exists $L$ such that for every $1\leq\ell\leq L$, the $\ell$-fold intersections $A_{i_{1}}\cap\cdots\cap A_{i_{\ell}}$ are nonempty, whereas for $L<\ell\leq k$, the $\ell$-fold intersections $A_{i_{1}}\cap\cdots\cap A_{i_{\ell}}$ are all empty for $i_{1},\ldots,i_{\ell}\in[k]$.

For every $m\geq k$, we construct a hypergraph $F=\{F_{1},\ldots,F_{m}\}$ in the following manner. Consider the bipartite graph $B=\left([m],\binom{[m]}{L}\right)$ where $x\in[m]$ is adjacent to $y\in\binom{[m]}{L}$ iff $x\in y$. Let $F_{i}$ be the set of neighbors in $B$ of the vertex $i\in[m]$. Notice that for any $i_{1},\ldots,i_{\ell}$,

[TABLE]

This construction therefore shows that $f(m,\mathcal{H})=k-1$. ∎

4 Proof of Theorem 2.7

In this section, we prove Theorem 2.7. We begin with some preliminary exploration of the family $\mathbf{EIP}_{k}$.

First, we make the crucial observation regarding question (2) that every sequence of $k$-edge hypergraphs $\{H_{m}\}$ such that $f(m,H_{m})$ is bounded, can only have finitely many members not in $\mathbf{EIP}_{k}$. This follows immediately from Lemma 4.1. Furthermore, for any $H(\vec{b})\in\mathbf{EIP}_{k}$, one can explicitly determine the relation between the intersection sizes and the parameters $b_{1},\ldots,b_{k}$ by inclusion-exclusion. We state this relation in Lemma 4.2.

Lemma 4.1**.**

Suppose $H=\{A_{1},\ldots,A_{k}\}$ satisfies the following for some $1\leq\ell\leq k$: there are two sets of indices $I,J\in\binom{[k]}{\ell}$ such that $|\bigcap_{i\in I}A_{i}|=a$ and $|\bigcap_{j\in J}A_{j}|=b$ with $a\neq b$. Then there is a constant $c_{\ell}$ such that $f(m,H)\geq c_{\ell}\cdot\log_{(\ell)}(m)$.

Proof.

Let $F$ be any hypergraph with $m$ edges. Construct an $\ell$-graph $G$ with $F$ as its vertex set, and hyperedges

[TABLE]

By (4), there exists a subset $F^{\prime}\subseteq F$ of size $c_{\ell}\cdot\log_{(\ell)}(m)$ which is either a clique or an independent set in $G$. In either case, $H$ cannot be contained in $F^{\prime}$. ∎

Lemma 4.1 implies that if there are infinitely many $m$ such that $H_{m}\not\in\mathbf{EIP}_{k}$, then for each such non-EIP hypergraphs we have $f(m,H_{m})\geq c^{\prime}\cdot\log^{(k)}(m)$, where $c^{\prime}$ is the absolute constant $c^{\prime}=\min\{c_{1},\ldots,c_{k}\}$. This is an infinite subsequence of $\{H_{m}\}$. Therefore, if $f(m,H_{m})$ is bounded, then by looking at the tail of $\{H_{m}\}$, we may assume WLOG that $H_{m}\in\mathbf{EIP}_{k}$ for every $m\geq 1$.

Recall that hypergraphs $H\in\mathbf{EIP}_{k}$ are characterized by the length $k$-vector $\vec{b}$, and for every sequence of hypergraphs $\{H_{m}\}_{m=1}^{\infty}$, we have a corresponding sequence of length $k$ vectors $\vec{\bf b}$.

We now state the relation between the intersection sizes and the parameters $b_{1},\ldots,b_{k}$ for $H(\vec{b})\in\mathbf{EIP}_{k}$.

Lemma 4.2**.**

Let $H(\vec{b})\in\mathbf{EIP}_{k}$, and $a_{i}=|A_{1}\cap\cdots\cap A_{i}|$, for each $1\leq i\leq k$. Then,

[TABLE]

$\square$

Before proving Theorem 2.7, we prove an auxiliary upper bound in Lemma 4.3, which provides a better upper bound on $f(m,H(\vec{\bf b}))$ with tighter constraints on $\vec{\bf b}$.

Lemma 4.3**.**

Suppose $\vec{\bf b}=(b_{1},\ldots,b_{k})$ is such that $b_{i}\geq 0$, and for every $1\leq i\leq k-1$,

[TABLE]

Then $f(m,H(b_{1},\ldots,b_{k}))=k-1$.

Proof of Lemma 4.3. Let $\vec{\bf b}$ satisfy the restrictions given in (6). Note that we need to construct a hypergraph sequence $\{F_{m}\}_{m=1}^{\infty}$, such that every $k$-edge subgraph of $F_{m}$ is isomorphic to $H(\vec{b})$. To achieve this, we define the following general construction:

Construction $F_{m}^{d_{1},\ldots,d_{k}}$:

Given $d_{1},\ldots,d_{k}\geq 0$ and $m\geq k$, let $B=([m],Y)$ be the bipartite graph with parts $[m]$ and $Y$, where $Y$ is defined as follows. For $1\leq\ell\leq k$ and $1\leq j\leq d_{\ell}$, let

[TABLE]

and suppose that $v_{j}^{S}\neq v_{j^{\prime}}^{S^{\prime}}$ for every $(j,S)\neq(j^{\prime},S^{\prime})$. Then

[TABLE]

For $x\in[m]$ and $v_{j}^{S}\in Y$, let $(x,v_{j}^{S})\in E(B)$ iff $x\in S$, and let $(x,w_{j})\in E(B)$ for every $x\in[m]$ and $w_{j}\in Y$. Then, define $F_{m}^{d_{1},\ldots,d_{k}}=\{A_{1},\ldots,A_{m}\}$, where $A_{i}=N_{B}(i)\subset Y$ for $i=1,\ldots,m$. $\blacksquare$

For example, the construction $F_{4}^{1,2,3}$ is given by:

[TABLE]

Informally, $A_{i}$ consists of one vertex $v_{1}^{i}$ corresponding to $\{i\}$, two vertices $v_{1}^{ij}$ and $v_{2}^{ij}$ corresponding to two-element subsets $\{i,j\}$, and three vertices $w_{1},w_{2},w_{3}$ that are in the common intersection of all the $A_{i}$’s, $1\leq i\leq 4$.

We observe the following property of the intersection sizes of the edges of $F_{m}^{d_{1},\ldots,d_{k}}$.

Claim 4.4**.**

$F_{m}^{d_{1},\ldots,d_{k}}\in\mathbf{EIP}_{k}$. Furthermore, for $1\leq i\leq k$ and any $i$-edge subgraph $\{A_{r_{1}},\ldots,A_{r_{i}}\}\subset F_{m}^{d_{1},\ldots,d_{k}}$, the size of the common intersection $a_{i}:=|A_{r_{1}}\cap\cdots\cap A_{r_{i}}|$ is given by

[TABLE]

Proof of Claim 4.4. Suppose $G=\{A_{r_{1}},\ldots,A_{r_{i}}\}\subset F_{m}^{d_{1},\ldots,d_{k}}$. We shall now count $|A_{r_{1}}\cap\cdots\cap A_{r_{i}}|$. For a fixed hypergraph $F_{m}^{d_{1},\ldots,d_{k}}\supseteq G^{\prime}\supseteq G$, let $U_{G^{\prime}}$ denote the set of all vertices of $F_{m}^{d_{1},\ldots,d_{k}}$ which are in all the edges of $G^{\prime}$ but none of the edges of $F_{m}^{d_{1},\ldots,d_{k}}\setminus G^{\prime}$. Notice that $A_{r_{1}}\cap\cdots\cap A_{r_{i}}$ is a disjoint union of $U_{G^{\prime}}$’s, $G^{\prime}\supseteq G$. Therefore,

[TABLE]

Fix a $G^{\prime}\supseteq G$. Let $G^{\prime}=\{A_{r_{1}},\ldots,A_{r_{i}},A_{s_{1}},\ldots,A_{s_{|G^{\prime}|-i}}\}$. We observe that,

• •

For $i\leq|G^{\prime}|<k$, $U_{G^{\prime}}$ consists exactly of the vertices $\left\{v_{j}^{\{r_{1},\ldots,r_{i},s_{1},\ldots,s_{|G^{\prime}|-1}\}}:1\leq j\leq d_{|G^{\prime}|}\right\}$.

• •

For $k\leq|G^{\prime}|<m$, $\bigcap_{X\in G^{\prime}}X=\{w_{1},\ldots,w_{d_{k}}\}\subseteq\bigcup_{X\not\in G^{\prime}}X$, thus $U_{G^{\prime}}=\varnothing$.

• •

For $|G^{\prime}|=m$, $U_{G^{\prime}}=\bigcap_{X\in G^{\prime}}X=\{w_{1},\ldots,w_{d_{k}}\}$.

Therefore,

[TABLE]

Plugging back these values into (8), we get

[TABLE]

for every $1\leq i\leq k$. $\blacksquare$

Now we return to the proof of Lemma 4.3. Given a length $k$ vector $\vec{b}\geq 0$ which satisfies (6) for $1\leq i\leq k-1$, let $d_{i}$ be the left hand side of (6), i.e.,

[TABLE]

and let $d_{k}=b_{k}$. Now, we look at the construction $F_{m}=F_{m}^{d_{1},\ldots,d_{k}}$, and pick a $k$-edge subgraph $G\subset F_{m}$. Observe that $G\in\mathbf{EIP}_{k}$, and therefore there is a length $k$ vector $\vec{g}$ such that $G=H(\vec{g})$. It suffices to check that $\vec{g}=\vec{b}$.

Suppose $G=\{A_{1},\ldots,A_{k}\}$. For $1\leq i\leq k$, let $a_{i}:=|A_{1}\cap\cdots\cap A_{i}|$. Recall that Lemma 4.2 gave us a way of computing $\vec{g}$ in terms of $\vec{a}$, and Claim 4.4 computes $\vec{a}$ in terms of $\vec{d}$. In order to precisely write down these relations, we introduce a few matrices.

Notation**.**

• •

Let $a^{(m)}_{ij}=\binom{m-i}{j-i}$ and $b^{(m)}_{ij}=(-1)^{j-i}\binom{m-i}{j-i}$.‡‡‡ By our convention, $\binom{x}{y}=0$ if $y<0$. Thus $a^{(m)}_{ij}=b^{(m)}_{ij}=0$ whenever $j<i$. Then, we denote by $A_{k,m}$ and $B_{k,m}$ the upper triangular matrices

[TABLE]

• •

Let $\vec{\mathbf{1}}$ denote the all-one vector, and $\vec{\mathbf{0}}$ the all-zero vector.

• •

Define $D:=\begin{bmatrix}A_{k-1,m}&\vec{\bf 1}\\ \vec{\mathbf{0}}^{\intercal}&1\end{bmatrix}$.

• •

Let $W_{k-1,m}$ be the $(k-1)\times(k-1)$ matrix given by $W_{k-1,m}=(w^{(m)}_{ij})_{1\leq i,j\leq k-1}$, and $w^{(m)}_{ij}=(-1)^{j-i}\binom{m-k+j-i-1}{j-i}$. Define $W^{\prime}:=\begin{bmatrix}W_{k-1,m}&\vec{\mathbf{0}}\\ \vec{\mathbf{0}}^{\intercal}&1\end{bmatrix}$.

First, we observe that the assertion of Lemma 4.2 can be rephrased as,

[TABLE]

Next, in terms of matrices, equality (7) reads

[TABLE]

Finally, by the definition of $\vec{d}$, we have

[TABLE]

Putting together Equations (9,10,11), we obtain:

[TABLE]

By Proposition 8.2 from the Appendix, we know that the product matrix $B_{k,k}DW^{\prime}$ is $I_{k}$, and this concludes the proof of Lemma 4.3. $\blacksquare$

Proof of Theorem 2.7.

We now have gathered all the equipment required to complete the proof of Theorem 2.7. Recall that $\alpha=\min\limits_{1\leq i\leq k-2}\left(\frac{b_{i}(m)}{mb_{i+1}(m)}\right)$, and we wish to prove that

[TABLE]

Note that this bound is trivial if $\frac{k(k-1)}{\alpha}\geq m$, therefore we may assume that $\alpha m>k(k-1)$. From the definition of $\alpha$, note that $b_{i}\geq\alpha mb_{i+1}$ for each $1\leq i\leq k-2$. By successively applying these inequalities we obtain $b_{i}\geq\alpha mb_{i+1}\geq\alpha^{2}m^{2}b_{i+2}\geq\cdots\geq\alpha^{k-i-1}m^{k-i-1}b_{k-1}$. Thus,

[TABLE]

The last inequality follows from $X^{t}\geq\binom{\lfloor X\rfloor}{t}$. Observe that the assumption $\frac{\alpha m}{k}>k-1$ implies $\left\lceil\frac{\alpha m}{k}\right\rceil\geq k$. Therefore, for $1\leq i\leq k-2$ and $i+1\leq r\leq k-1$, we have

[TABLE]

Thus, (12) gives us

[TABLE]

implying

[TABLE]

This is exactly the condition (6), with $m$ replaced by $\left\lceil\frac{\alpha m}{k}\right\rceil$, so Lemma 4.3 gives us a hypergraph $K$ on $\left\lceil\frac{\alpha m}{k}\right\rceil$ edges such that every $k$ sets of $K$ are isomorphic to $H(\vec{\bf b})$.

Now, consider a $\left\lceil\frac{k}{\alpha}\right\rceil$-fold disjoint union of $K$’s. This hypergraph $F_{m}$ has $\left\lceil\frac{k}{\alpha}\right\rceil\cdot\left\lceil\frac{\alpha m}{k}\right\rceil\geq m$ edges, and note that as long as we pick $1+\left\lceil\frac{k}{\alpha}\right\rceil\cdot(k-1)$ edges, some $k$ of them fall in the same copy of $K$. These $k$ edges create a $H(\vec{\bf b})$ by construction of $K$. This shows $f(m,H(\vec{\bf b}))\leq\left\lceil\frac{k}{\alpha}\right\rceil\cdot(k-1)$, completing the proof of the upper bound.

Now we prove the lower bound. Recall that we are aiming to prove

[TABLE]

Suppose $F$ is a hypergraph on $m$ edges. Either $F$ has a subgraph $F_{1}$ of size $\frac{m}{2}$ which is of the same uniformity as $H(\vec{b})$, or it has a subgraph of size $\frac{m}{2}$ which is not of the this uniformity. If the latter is true, then $\mbox{ex}(F,H(\vec{b}))\geq\frac{m}{2}$. Otherwise, we focus on the subgraph $F_{1}$. Let $T$ be a $H(\vec{b})$-free subgraph in $F_{1}$ of maximum size, say $|T|=t$. Then, for every $S\in F_{1}\setminus T$, there exist distinct $A_{1},\ldots,A_{k-1}\in T$ such that $\{A_{1},\ldots,A_{k-1},S\}$ forms a $H(\vec{b})$. Therefore, there are fixed $A_{1},\ldots,A_{k-1}\in T$ such that $\{A_{1},\ldots,A_{k-1},S\}$ forms a $H(\vec{b})$ for every $S\in F_{2}$, where

[TABLE]

Further, note that $|A_{1}\cap\cdots\cap A_{k-1}\cap S|=b_{k}$ for every $S\in F_{2}$, therefore there is a subgraph $F_{3}\subseteq F_{2}$ such that every element $S\in F_{3}$ intersects $A_{1}\cap\cdots\cap A_{k-1}$ in the exact same set, and

[TABLE]

Finally, for any $1\leq i\leq k-2$, let

[TABLE]

Let $D:=\max\limits_{x\in V(F_{3})}\deg_{F_{3}}(x)$. As $\{A_{1},\ldots,A_{k-1},B\}$ is an $H(\vec{b})$,

[TABLE]

Now, for a fixed $S\in F_{3}$, let $X:=A_{1}\cap\cdots\cap A_{i}\setminus(A_{i+1}\cup\cdots\cup A_{k-1})$. Note that,

[TABLE]

Therefore (14) implies

[TABLE]

Clearly the sets in $F_{3}$ that achieve the maximum degree $D$ is $H(\vec{b})$-free, leading to the inequality

[TABLE]

i.e.,

[TABLE]

Since $m\geq 6$, note that if $t\geq\frac{m}{4}$, then $t\geq\left(\frac{m}{2}\right)^{\frac{1}{3}}\geq\left(\frac{m}{2}\right)^{\frac{1}{k}}$, which is larger than the right side of (13). So we may assume $t<\frac{m}{4}$, which would lead us to

[TABLE]

As (15) holds for every $1\leq i\leq k-2$, this gives the bound that we seek. ∎

5 Proof of Theorem 2.8

In this section we prove Theorem 2.8. The proof is by induction on $b_{k}$, starting from $b_{k}=0$. Notice that the lower bound of Theorem 2.7 gives us the following corollary, which serves as the base case for our induction argument:

Corollary 5.1**.**

For $m\geq 6$,

[TABLE]

Further, one can asymptotically improve this bound when $k=3$:

Proposition 5.2**.**

For $m\geq 4$,

[TABLE]

Proof.

Let $|F|=m$ and $H=H(b_{1},b_{2},0)$. Either $F$ has a $(b_{1}+2b_{2})$-uniform subgraph $F_{1}$ of size $\frac{m}{2}$, or it has a subgraph of size $\frac{m}{2}$ in which none of the edges have size $(b_{1}+2b_{2})$. If the latter is true, then $\mbox{ex}(F,H)\geq\frac{m}{2}$. Otherwise let us focus on $F_{1}$. Let $T$ be an $H$-free subset of maximum size in $F_{1}$, and suppose $|T|=t$. Note that for any $B\in F_{1}\setminus T$, there are sets $A_{1},A_{2}\in T$ such that $(B,A_{1},A_{2})$ is a $H(b_{1},b_{2},0)$. Suppose $V=\bigcup_{A\in T}A$, then we have $|B\cap V|\geq 2b_{2}$, and $|V|\leq t(b_{1}+2b_{2})$. Let $D=\max\limits_{x\in V}\deg_{F_{1}}(x)$. Then,

[TABLE]

and

[TABLE]

Let $x\in V$ have the maximum degree in $F$. Since the subgraph of size $D$ containing $x$ is $H$-free, we obtain

[TABLE]

If $t\geq\frac{m}{4}$, then $t\geq\frac{1}{2}\sqrt{m}\geq\sqrt{\frac{mb_{2}}{2(b_{1}+2b_{2})}}$. So assume $t<\frac{m}{4}$, and therefore $t^{2}\geq\frac{mb_{2}}{2(b_{1}+2b_{2})}$, as desired. ∎

Before we prove Theorem 2.8 we require the following lemma from [13]:

Lemma 5.3**.**

Let $H=(V,E)$ be a $k$-graph on $m$ vertices, and let $\alpha(H)$ denote the independence number of $H$. Then,

[TABLE]

Now we are prepared to prove Theorem 2.8.

Proof of Theorem 2.8.

Fix $k$ and $\vec{\bf b}$. Recall that $b_{k}$ is fixed, and we wish to show that for $m\geq 6$,

[TABLE]

Suppose $|F|=m$. Then, either $F$ has a subgraph $F_{1}$ of size at least $\frac{m}{2}$ which has uniformity the same as that of $H(\vec{b})$, or it does not. When the latter is true, we have $\mbox{ex}(F,H(\vec{b}))\geq\frac{m}{2}$. Since $\frac{m}{2}\geq m^{\frac{1}{4}}\cdot\left(\frac{1}{8}\right)^{\frac{1}{4}}$ and $\frac{m}{2}\geq m^{\frac{1}{2}}\cdot(\frac{1}{8})^{\frac{1}{2}}$, we may assume that the former is true. We wish to show that $F_{1}$ contains a $H(\vec{b})$-free subgraph of large size.

We proceed by induction on $b_{k}$. Notice that we already established the results for $b_{k}=0$ in Corollary 5.1 (using $b_{k-1}\leq 2b_{k-1}$) and Proposition 5.2.

Construct a $k$-graph $G$ with vertex set $F_{1}$ and call $\{A_{1},\ldots,A_{k}\}$ an edge in $G$ iff $\{A_{1},\ldots,A_{k}\}\cong H(\vec{b})$. Clearly, $t=\alpha(G)$ is a lower bound to our problem. By Lemma 5.3,

[TABLE]

Given $1\leq i\leq k$ and $B_{1},\ldots,B_{i}\in F_{1}$, denote by $\deg_{G}(B_{1},\ldots,B_{i})$ the number of edges of $G$ containing $\{B_{1},\ldots,B_{i}\}$. As $\sum_{A_{1},\ldots,A_{k-2}\in F_{1}}\deg_{G}(A_{1},\ldots,A_{k-2})=\binom{k}{2}|E(G)|$, we have

[TABLE]

The sum on the left side has at most $\binom{m/2}{k-2}\leq\frac{(m/2)^{k-2}}{(k-2)!}$ terms, therefore there exist distinct $A_{1},\ldots,A_{k-2}\in F_{1}$ such that

[TABLE]

Note that $\frac{(k-2)!(k-1)^{k}}{2k^{k-1}}>\frac{1}{4}$ for every $k\geq 3$. Let $\mathcal{B}$ denote the set of all edges $B\in F_{1}$ which are covered by an edge through $\{A_{1},\ldots,A_{k-2}\}$. Then, $|\mathcal{B}|^{2}\geq\deg_{G}(A_{1},\ldots,A_{k-2})$, and so

[TABLE]

As $\{A_{1},\ldots,A_{k-2}\}$ is a subgraph of $H(\vec{b})$, we have

[TABLE]

Also, for every $B\in\mathcal{B}$, $\{A_{1},\ldots,A_{k-2},B\}$ is a subgraph of $H(\vec{b})$. Thus,

[TABLE]

Now,

[TABLE]

Let $D$ be the maximum degree of a vertex in $F_{1}$. Then, by (17),

[TABLE]

Also, note that

[TABLE]

Therefore (18) gives us,

[TABLE]

Now, we notice that if $x$ is a vertex of degree $D$, then deleting it from the edges through $x$ gives us a family of uniformity one less than that of $F_{1}$. By induction on $b_{k}$, this subfamily already contains a $H(b_{1},\ldots,b_{k-1},b_{k}-1)$-free family of size $f(D,H(b_{1},\ldots,b_{k-1},b_{k}-1))$, which is a natural lower bound to our problem. Therefore,

[TABLE]

We now split into two cases.

• •

Case I: $k\geq 4$. Now we use the inductive lower bound given by (16):

[TABLE]

Combining this bound with (19), we get

[TABLE]

Which, on invoking ${t^{\frac{k-1}{2}}}\leq{t^{k}}$, leads us to

[TABLE]

finishing off the induction step.

• •

Case II: $k=3$. In this case we use the inductive lower bound in (16) of

[TABLE]

Again, combining this bound with (19), we obtain

[TABLE]

This implies $t\geq m^{\frac{1}{b_{3}+2}}\left(\frac{b_{2}}{4(b_{1}+2b_{2})}\right)^{\frac{b_{3}+1}{b_{3}+2}}$, completing the induction step.

∎

6 Proof of Theorem 2.10

In this section, we prove Theorem 2.10. For the proof, we rely upon the incidence structure of Miquelian inversive planes $\mathbf{M}(q)$ of order $q$. An inversive plane consists of a set of points $\mathcal{P}$ and a set of circles $\mathcal{C}$ satisfying three axioms [14]:

• •

Any three distinct points are contained in exactly one circle.

• •

If $P\neq Q$ are points and $c$ is a circle containing $P$ but not $Q$, then there is a unique circle $b$ through $P,Q$ and satisfying $b\cap c=\{P\}$.

• •

$\mathcal{P}$ contains at least four points not on the same circle.

Every inversive plane is a $3$-$(n^{2}+1,n+1,1)$-design for some integer $n$, which is called its order. An inversive plane is called Miquelian if it satisfies Miquel’s theorem [14]. The usefulness of Miquelian inversive planes lies in the fact that their automorphism groups are sharply 3-transitive (cf. pp 274-275, Section 6.4 of [15]). There are a few known constructions of $\mathbf{M}(q)$, one such construction is outlined here. The points of $\mathbf{M}(q)$ are elements of $\mathbb{F}_{q}^{2}$ and a special point at infinity, denoted by $\infty$. The circles are the images of the set $K=\mathbb{F}_{q}\cup\{\infty\}$ under the permutation group $PGL_{2}(q^{2})$, given by

[TABLE]

For further information on inversive planes and their constructions, the reader is referred to [15, 16, 17].

Now, we prove Theorem 2.10.

Proof of Theorem 2.10..

Recall that for every odd prime power $q$, we are required to demonstrate a hypergraph on $q^{2}+1$ edges with the property that every three edges form an $H(q^{2}-q-1,q,1)$. Let $\mathbf{M}(q)$ be a Miquelian inversive plane, with points labelled $\{1,2,\ldots,q^{2}+1\}$. Then, we consider the $(q^{2}+q)$-graph $F=\{A_{1},\ldots,A_{q^{2}+1}\}$, whose vertex set $V(F)$ is the circles of $\mathbf{M}(q)$, and $A_{i}$ is the collection of circles containing $i$. By the inversive plane axiom, any three distinct points have a unique circle through them. It suffices to show that any two distinct points $P,Q$ in $\mathbf{M}(q)$ have $q+1$ distinct circles through them. By 2-transitivity of the Automorphism group, we know that any two points have the same number $a_{2}$ of circles through them. Now, for any $P\neq Q$,

[TABLE]

Thus $a_{2}=q+1.$ So, $F$ is $(q^{2}+q)$-uniform, every two edges of $F$ have an intersection of size $q+1$, and every three edges of $F$ have an intersection of size $1$. By inclusion-exclusion, they form a $H(q^{2}-q-1,q,1)$. ∎

Now, we prove Corollary 2.11.

Proof of Corollary 2.11.

Recall that we wish to extend the result of the Theorem 2.10. We wish to show that when $b_{1}\geq b_{2}^{\,2}-b_{2}-1$ and $b_{2}^{\,2}\geq m$, $f(m,H(b_{1},b_{2},1))=2$.

Notice that the inversive plane construction asymptotically gives us $b_{2}^{\,2}+1$ sets such that any three of them are an isomorphic copy of $H(b_{2}^{\,2}-b_{2}-1,b_{2},1)$. As long as $b_{2}^{\,2}+1\geq m$, we can take a subgraph of the construction and still obtain $m$ sets satisfying the same property. Also note that as long as $b_{1}\geq b_{2}^{\,2}-b_{2}-1$, we can create a construction by first creating an inversive plane $F$ satisfying $b_{1}=b_{2}^{\,2}-b_{2}-1$, and then adding $(b_{1}-b_{2}^{\,2}+b_{2}+1)$ new distinct points to each set in $F$. ∎

7 Further Problems

We discuss a few further problems that are interesting. Of course, the main open question is (2), which asks to characterize all sequences of $k$-edge hypergraphs $H_{m}$ for which $f(m,H_{m})$ is bounded. As we discussed, even the case $k=3$ turns out to be interesting.

Let us focus on the case $k=3$ and $\vec{\bf b}=(b_{1},b_{2},1)$. The current state of affairs was summarized in Figure 3. An interesting observation is that all the upper bounds in the light regions are actually upper bounds of $2$. Therefore, the following question may be interesting to ask:

Problem 7.1**.**

Characterize all values of $(b_{1},b_{2})$ such that $f(m,H_{3}(b_{1},b_{2},1))=2$.

We cannot solve this problem completely. However, a straightforward counting argument estimating $\sum d_{i}$, $\sum d_{i}^{2}$ and $\sum d_{i}^{3}$, along with Cauchy-Schwarz yields the following necessary condition.

Theorem 7.2**.**

Suppose $f(m,H_{3}(b_{1},b_{2},b_{3}))=2$. Then, for large enough $m$, $b_{1},b_{2},b_{3}$ must satisfy

[TABLE]

In particular, when $b_{3}=1$, these numbers satisfy

[TABLE]

Theorem 7.2 gives more insight into Figure 3. Basically, there are two cases to consider. When $b_{1}$ is asymptotically larger than $\frac{b_{1}b_{2}}{m}$, ie when $b_{2}=o(m)$, this means that $b_{1}\geq b_{2}^{\,2}$ is necessary for $f=2$. When $b_{2}\geq m$, this gives us $b_{1}\geq mb_{2}$, which is exactly the construction in Lemma 4.3. Further, note that this transition occurs exactly at the intersection of the line $b_{1}=mb_{2}$ and the parabola $b_{1}=b_{2}^{\,2}$.

As a further special case of Problem 7.1, one can look at $\vec{\bf b}=(m,b_{2},1)$ where $1\ll b_{2}\ll\sqrt{m}$. We expect this range to be solvable via a construction, since there are constructions for $b_{2}=1$ (Theorem 2.7) and $b_{2}=\sqrt{m}$ (Theorem 2.10). The problem is equivalent to constructing bipartite graphs with certain properties, as stated below.

Problem 7.3**.**

Suppose $1\ll b_{2}\ll\sqrt{m}$. Is there a bipartite graph $G$ with parts $A$, $B$, such that $|A|=m$, the degree of every vertex in $A$ is asymptotic to $m$, the size of the common neighborhood of every pair in $A$ is asymptotic to $b_{2}$, and every three vertices in $A$ have a unique common neighbor in $B$?

If such a bipartite graph can be constructed, then we can let $F=\{N_{G}(u):u\in A\}$. This hypergraph will testify for $f(m,H(m,b_{2},1))=2$. From the proof of Theorem 7.2, we know that if such a bipartite graph exists, it cannot be regular from $B$. A regular construction from $B$ implies equality in the Cauchy-Schwarz inequality, which would imply $b_{2}=\Theta(\sqrt{m})$. Therefore if such a graph is constructed, $B$ needs to have vertices of different degrees.

Notice also, that if the answer to Problem 7.3 is affirmative, then we can shade the small triangle in Figure 3 light. This is courtesy the fact that given any $(b_{1},b_{2})$ in that region, we can write it as a sum of $(x,y)+(m,z)$, with $x\geq my$. We can then take a $H_{3}(x,y,0)$-construction $\{A_{1},\ldots,A_{m}\}$ and a $H_{3}(m,z,1)$-construction $\{A^{\prime}_{1},\ldots,A^{\prime}_{m}\}$, and merge them together to obtain the $H_{3}(b_{1},b_{2},1)$-construction $\{A_{1}\cup A^{\prime}_{1},\ldots,A_{m}\cup A^{\prime}_{m}\}$.

8 Appendix

Our goal in this section is to prove the matrix identity asserted in Proposition 8.2. Before seeing the proof, we mention an useful binomial identity in Lemma 8.1.

Lemma 8.1**.**

For integers $x\geq 0,y\geq z\geq 0$, we have

[TABLE]

Proof.

One can prove this identity using induction on $y$. Note that when $y=z$, the identity becomes

[TABLE]

which follows from applying Pascal’s identity $\binom{x}{t}=\binom{x-1}{t}+\binom{x-1}{t-1}$ to each term and telescoping.

Now suppose that (20) holds for some $y$. Then,

[TABLE]

By induction hypothesis, the first term is $(-1)^{z}\binom{x-y+z-1}{z}$ and the second term is $(-1)^{z-1}\binom{x-y+z-2}{z-1}$. Their sum is $(-1)^{z}\binom{x-y+z-2}{z}$, as desired. ∎

Notice that in some cases $x-y+z-1$ could be negative, whence we interpret the binomial coefficient $\binom{-a}{s}$ as $(-1)^{s}\binom{a+s-1}{s}$. Also observe that the generalized binomial coefficients satisfy Pascal’s identity $\binom{a}{s}=\binom{a-1}{s}+\binom{a-1}{s-1}$.

We are now going to state and prove Proposition 8.2. Recall the following notation: $a^{(m)}_{ij}=\binom{m-i}{j-i}$, $b^{(m)}_{ij}=(-1)^{j-i}\binom{m-i}{j-i}$, $w^{(m)}_{ij}=(-1)^{j-i}\binom{m-k+j-i-1}{j-i}$,

[TABLE]

and,

[TABLE]

Proposition 8.2**.**

[TABLE]

Proof.

Note that $B_{k,k}=\begin{bmatrix}B_{k-1,k}&\vec{v}\\ \vec{\bf 0}^{\intercal}&1\end{bmatrix}$, where $v_{i}=(-1)^{k-i}$, and therefore

[TABLE]

We verify that $B_{k-1,k}\vec{\bf 1}+v=\vec{\bf 0}$ and $B_{k-1,k}A_{k-1,m}W_{k-1,m}=I_{k-1}$ in Claims 8.3 and 8.4, respectively.

Claim 8.3**.**

$B_{k-1,k}\vec{\bf 1}+v=\vec{\bf 0}.$

Proof of Claim 8.3.

Note that the $i$’th row of $B_{k-1,k}\vec{\bf 1}$ is

[TABLE]

as desired. ∎

Claim 8.4**.**

$B_{k-1,k}A_{k-1,m}W_{k-1,m}=I_{k-1}$.

Proof of Claim 8.4.

Note that the $(i,j)$th entry of the product matrix is given by

[TABLE]

Observe that, using Lemma 8.1 for $x=k-i,y=m-i,z=s-i$, we get

[TABLE]

Plugging this back into (21), we get that the $(i,j)$th entry of the product matrix is

[TABLE]

Notice that the sum in (22) only runs from $s=i$ to $s=j$, and therefore after the change of variable $s\mapsto s+i$, the expression reduces to

[TABLE]

Note that $\binom{s-(m-k)-1}{s}=(-1)^{s}\binom{m-k}{s}$, so (23) is the sum

[TABLE]

which, on invoking Lemma 8.1 for $x=m-k,y=m-k+j-i-1,z=j-i$, reduces to

[TABLE]

Clearly, this is [math] when $j\neq i$ and $1$ when $j=i$. ∎

This completes the proof of Proposition 8.2. ∎