Note on the diversity of intersecting families

TL;DR

This paper constructs two intersecting families with higher diversity than a certain sum, disproving Huang's conjecture for odd n=2k+1 with k≥3.

Contribution

It provides explicit examples of intersecting families with greater diversity, challenging previous conjectures in combinatorics.

Findings

Disproved Huang's conjecture for odd n=2k+1 with k≥3.

Constructed two intersecting families with higher diversity than the specified sum.

Enhanced understanding of the diversity properties of intersecting families.

Abstract

Let $n=2k+1$ be odd with $k\geq3$. In this note, we give two intersecting families with diversity larger than $\sum_{i=k+1}^{2k} \binom {2k}{i}$, which disprove a conjecture of Huang.

Equations

$$div(\mathcal{F})\le div({\mathcal{Q}}_k)=\sum _{i=k+1}^{2k}\left(\genfrac{}{}{0ex}{}{2k}{i}\right)$$

$$\sigma (i)\equiv i+1\text{ mod }n,\text{ for all entries }i=1,2,\dots ,n,$$

$${\mathcal{F}}_k=\{{\sigma }^i(L_k):i=0,1,\dots ,n-1\}$$

$${\mathcal{P}}_k={\mathcal{F}}_k\cup ({\mathcal{Q}}_k\setminus \overline{{\mathcal{F}}_k})$$

$$div({\mathcal{P}}_k)=|{\mathcal{P}}_k|-\sigma ({\mathcal{F}}_k)-(\sigma ({\mathcal{Q}}_k)-\sigma (\overline{{\mathcal{F}}_k}))=\sum _{i=k+1}^{2k}\left(\genfrac{}{}{0ex}{}{2k}{i}\right)+1=div({\mathcal{Q}}_k)+1$$

$${\mathcal{R}}_k=\mathcal{A}\cup ({\mathcal{Q}}_k\setminus \bar{\mathcal{A}}).$$

$$div({\mathcal{R}}_k)-div({\mathcal{Q}}_k)>\sum _{i=q+1}^k\frac{n-2i}{2}\left(\genfrac{}{}{0ex}{}{n-q-1}{i-q-1}\right)$$

$${\mathcal{A}}_i\ge n\left(\genfrac{}{}{0ex}{}{n-q-1}{i-q-1}\right)-\left(\genfrac{}{}{0ex}{}{n}{2}\right)\left(\genfrac{}{}{0ex}{}{n-2q-1}{i-2q-1}\right)\ge \frac{n}{2}\left(\genfrac{}{}{0ex}{}{n-q-1}{i-q-1}\right)$$

$$div({\mathcal{R}}_k)$$

$$=div({\mathcal{Q}}_k)+\sum _{i=q+1}^k(|{\mathcal{A}}_i|-2\sigma ({\mathcal{A}}_i)),$$

$$div({\mathcal{R}}_k)-div({\mathcal{Q}}_k)$$

$$>\sum _{i=q+1}^k\frac{n-2i}{2}\left(\genfrac{}{}{0ex}{}{n-q-1}{i-q-1}\right).$$

Full text

1 Introduction

Throughout the paper, we denote $[n]:=\{1,2,\ldots,n\}$, and $2^{[n]}:=\{X:X\subset[n]\}$. A family $\mathcal{F}\subset 2^{[n]}$ is called intersecting if any two of its elements intersect. The degree $\delta(x)$ of $x\in[n]$ is defined to be the number of members of $\mathcal{F}$ that contain $x$, and $\mathcal{F}$ is called regular if all of its elements have the same degree. If $\mathcal{F}$ is regular, we also denote $\sigma(\mathcal{F})$ the degree of elements of $\mathcal{F}$. Given an intersecting family $\mathcal{F}$, the diversity of $\mathcal{F}$ is defined to be $div(\mathcal{F}):=|\mathcal{F}|-\mathrm{max}_{x\in{[n]}}|\mathcal{F}(x)|$, where $\mathcal{F}(x)=\{F:x\in F\in\mathcal{F}\}$. Let $n=2k+1$ and $\mathcal{Q}_{k}={\{A:A\subset[2k+1],|A|\geq k+1\}}$. About the maximal diversity among families included in $2^{[n]}$, Huang[1] gave the following conjecture.

Conjecture 1.1**.**

For $n=2k+1$, suppose $\mathcal{F}\subset 2^{[n]}$ is intersecting. Then

[TABLE]

In this note, we will give two regular intersecting families with diversity larger than $\mathcal{Q}_{k}$, which disprove the above conjecture.

2 Counterexamples

The first example is constructed from a regular intersecting family in $\binom{[n]}{k}$. The circular shift operation $\sigma$ on $[n]$ is defined as

[TABLE]

and we denote $\sigma(A)=\{\sigma(i):i\in A\}$ for $A\subset[n]$.

Let $L_{3}=\{1,2,4\}$ and $L_{k}=\{1,2,4\}\cup\{7,9,\ldots,2k-1\}$ for $k\geq 4$. For $k\geq 3$, we define $\mathcal{F}_{k}$ to be the family obtained by repeatedly applying $\sigma$ to $L_{k}$, i.e.

[TABLE]

For example, $\mathcal{F}_{3}$ consists of the 7 lines of the Fano plane.

Lemma 2.1**.**

For $k\geq 3$ and $n=2k+1$, $\mathcal{F}_{k}\subset 2^{[n]}$ is a regular intersecting family with degree equal to $k$.

Proof.

$\mathcal{F}_{3}$ consists of the 7 lines of the Fano plane, and thus is a regular intersecting family with degree 3. For $k>3$, it is clear that $\mathcal{F}_{k}$ is regular with degree $|\mathcal{F}_{k}|*k/n=k$. To prove that $\mathcal{F}_{k}$ is intersecting, we just need to show that $L_{k}$ and $\sigma^{i}(L_{k})(1\leq i\leq n-1)$ intersect, since $\sigma^{n-i}(\sigma^{i}(L_{k}))=L_{k}$ and $\sigma^{n-i}(\sigma^{j}(L_{k}))=\sigma^{j-i}(L_{k})$ for $1\leq i<j\leq n-1$. Assume that $L_{k}$ and $\sigma^{i}(L_{k})(1\leq i\leq n-1)$ do not intersect for some $i\in[n-1]$, then we must have $\sigma^{i}(1)=5$ or $\sigma^{i}(1)=2k$. In the first case, we obtain $\sigma^{i}(2k-1)=2$, and $\sigma^{i}(4)=2$ in the second one. Both of the two cases contradict the assumption, thus $\mathcal{F}_{k}$ is intersecting. ∎

Given a set $A\subset[n]$, We denote $\bar{A}$ the complement of $A$ in $[n]$, and denote $\bar{\mathcal{F}}:=\{\bar{A}:A\in\mathcal{F}\}$ for $\mathcal{F}\subset 2^{[n]}$. The family $\mathcal{P}_{k}$ is defined as

[TABLE]

theorem 2.2**.**

For $n=2k+1$ with $k\geq 3$, $\mathcal{P}_{k}$ is a regular intersecting family and $div(\mathcal{P}_{k})=div(\mathcal{Q}_{k})+1$.

Proof.

By Lemma 2.1, the family $\mathcal{F}_{k}$ is intersecting. If $\mathcal{P}_{k}$ is not intersecting, then there exist disjoint sets $F$ and $Q$ with $F\in\mathcal{F}_{k}$ and $Q\in\mathcal{Q}_{k}\setminus\bar{\mathcal{F}_{k}}$. Since $F\cap Q=\emptyset$ and $|Q|\geq|\bar{F}|=k+1$, we must have $Q=\bar{F}$, which contradicts the choice of $F$ and $Q$. Therefore $\mathcal{P}_{k}$ is intersecting.

Since $\mathcal{F}_{k}$, $\bar{\mathcal{F}_{k}}$ and $\mathcal{Q}_{k}$ are regular with degree equal to $k$, $k+1$ and $\sum_{i=k}^{2k}\binom{2k}{i}$ respectively, and $|\mathcal{P}_{k}|=|\mathcal{Q}_{k}|=\sum_{i=k+1}^{2k+1}\binom{2k+1}{i}$, we have

[TABLE]

∎

The second example is related to finite projective planes. Let $q$ be an odd prime power, and let $\mathds{P}$ be the transitive projective plane $\mathds{P}^{2}(\mathds{F}_{q})$ over the finite field $\mathds{F}_{q}$. We take $n=q^{2}+q+1$, and identify the set of points of $\mathds{P}$ with $[n]$. For $q+1\leq i\leq\frac{q^{2}+q}{2}$, we define $\mathcal{A}_{i}$ to be the family of all $i-$element subsets consisting of the points of $\mathds{P}$ that contain a line of $\mathds{P}$. Since $\mathds{P}$ is transitive, $\mathcal{A}_{i}$ is a regular intersecting family for $i\geq q+1$. See [2] for more details about the family $\mathcal{A}_{i}$.

Let $k=\frac{q^{2}+q}{2}$ and $\mathcal{A}=\bigcup_{i=q+1}^{k}\mathcal{A}_{i}$. We define the family $R_{k}$ as

[TABLE]

Then the following result will show that $\mathcal{R}_{k}$ is a family in $2^{[n]}$ with larger diversity than $\mathcal{Q}_{k}$.

theorem 2.3**.**

For an odd prime power $q$, let $n=q^{2}+q+1$ and $k=\frac{q^{2}+q}{2}$. Then $\mathcal{R}_{k}$ is a regular intersecting family, and we have

[TABLE]

Proof.

If $\mathcal{R}_{k}$ is not intersecting, then there exist disjoint sets $A$ and $Q$ with $A\in\mathcal{A}$ and $Q\in\mathcal{Q}_{k}\setminus\mathcal{\bar{A}}$. Since $A\cap Q=\emptyset$ and $|Q|\leq k$, we have $A\subset\bar{Q}\in 2^{[n]}\setminus\mathcal{Q}_{k}$. On the other hand, we know that $\mathcal{R}_{k}$ is an upset by its definition , i.e. is closed under taking supersets. Thus we must have $\bar{Q}\in\mathcal{A}$, which contradicts the choice of $Q$. Therefore $\mathcal{R}_{k}$ is an intersecting family.

It is clear that $\mathcal{A}_{i}$ and $\bar{\mathcal{A}_{i}}$ are regular, so $\mathcal{R}_{k}$ is also regular. To obtain the diversity of $\mathcal{R}_{k}$, we firstly compute the size of $\mathcal{A}_{i}$. Since $\mathcal{A}_{q+1}$ consists of all lines of the projective plane, we have $|\mathcal{A}_{q+1}|=n$. For $q+1<i\leq k$, by using the Bonferroni inequalities, we have

[TABLE]

Since $|\mathcal{R}_{k}|=|\mathcal{Q}_{k}|,\sigma(\mathcal{A}_{i})+\sigma(\bar{\mathcal{A}}_{i})=|\mathcal{A}_{i}|$, we have

[TABLE]

Combining (1) and (2) together, we have

[TABLE]

∎

Remark 2.4**.**

By the construction of the above example, to get the maximum diversity among families included in $2^{[n]}$, it will be meaningful to study the existence of regular intersecting families included in $\binom{[n]}{k}$ and the maximum size of such families when $k$ is relatively small compared to $[n]$. Readers could refer to [3] for more information about regular intersecting families.