Multi-part cross-intersecting families

Abstract

Let ${\mathcal {A}}\subseteq {[n]\atopwithdelims ()a}$ and ${\mathcal {B}}\subseteq {[n]\atopwithdelims ()b}$ be two families of subsets of [n], we say ${\mathcal {A}}$ and ${\mathcal {B}}$ are cross-intersecting if $A\cap B\ne \emptyset$ for all $A\in {\mathcal {A}}$, $B\in {\mathcal {B}}$. In this paper, we study cross-intersecting families in the multi-part setting. By characterizing the independent sets of vertex-transitive graphs and their direct products, we determine the sizes and structures of maximum-sized multi-part cross-intersecting families. This generalizes the results of Hilton’s (J Lond Math Soc 15(2):369–376, 1977) and Frankl–Tohushige’s (J Comb Theory Ser A 61(1):87–97, 1992) on cross-intersecting families in the single-part setting.

1 Introduction

Let [n] be the standard n-element set. For an integer $1\le k\le n$, denote ${[n]\atopwithdelims ()k}$ as the family of all k-element subsets of [n]. A family ${\mathcal {F}}$ is said to be intersecting if $A\cap B\ne \emptyset$ for any $A,B\in {\mathcal {F}}$. The celebrated Erdős–Ko–Rado theorem [7] says that if ${\mathcal {F}}\subseteq {[n]\atopwithdelims ()k}$ is an intersecting family with $k\le \frac{n}{2}$, then

$$\begin{aligned} |{\mathcal {F}}|\le {{n-1}\atopwithdelims (){k-1}}, \end{aligned}$$

and if $n>2k$, the equality holds if and only if every subset in ${\mathcal {F}}$ contains a fixed element.

Due to its fundamental role in extremal set theory, the Erdős–Ko–Rado theorem has numerous extensions that have been explored in various ways. One of the extensions is the study of cross-t-intersecting families: Let n and m be two positive integers. Denote $2^{[n]}$ as the power set of [n] and for $1\le i\le m$, let ${\mathcal {A}}_i\subseteq 2^{[n]}$. Then, families ${\mathcal {A}}_1,{\mathcal {A}}_2,\ldots ,{\mathcal {A}}_m$ are called cross-t-intersecting, if $|A\cap B|\ge t$ for any $A\in {\mathcal {A}}_i$ and $B\in {\mathcal {A}}_j$, $i\ne j$. Especially, we say ${\mathcal {A}}_1,{\mathcal {A}}_2,\ldots ,{\mathcal {A}}_m$ are cross-intersecting if $t=1$.

Hilton [12] investigated cross-intersecting families in ${[n]\atopwithdelims ()k}$ and proved the following inequality.

Theorem 1.1

([12]) Let ${\mathcal {A}}_1,{\mathcal {A}}_2,\ldots , {\mathcal {A}}_m$ be cross-intersecting families in ${[n]\atopwithdelims ()k}$ with $n\ge 2k$. Then,

$$\begin{aligned} \sum \limits _{i=1}^{m}|{\mathcal {A}}_i|\le \left\{ \begin{array}{ll}{n\atopwithdelims ()k},&{}~~~\text {if}~m\le \frac{n}{k};\\ m\cdot {{n-1}\atopwithdelims (){k-1}},&{}~~~\text {if}~m\ge \frac{n}{k}.\end{array}\right. \end{aligned}$$

In [12], Hilton also determined the structures of ${\mathcal {A}}_i$’s when the equality holds. Since then, there have been many extensions about Theorem 1.1. For examples, Borg [3] gave a simple proof of Theorem 1.1, and obtained similar results for labeled sets [2], signed sets [5] and permutations [4]. Based on results of the independent number of vertex-transitive graphs, Wang and Zhang [18] extended Theorem 1.1 to general symmetric systems, which comprise finite sets, finite vector spaces and permutations, etc.

As a variant of Theorem 1.1, Hilton and Milner [13] dealt with pairs of non-empty cross-intersecting families in ${[n]\atopwithdelims ()k}$ and proved the following theorem.

Theorem 1.2

([13]) Let ${\mathcal {A}},{\mathcal {B}}\subseteq {[n]\atopwithdelims ()k}$ be non-empty cross-intersecting families with $n\ge 2k$. Then, $|{\mathcal {A}}|+|{\mathcal {B}}|\le {n\atopwithdelims ()k}-{{n-k}\atopwithdelims ()k}+1$.

This result was generalized by Frankl and Tokushige [10] to the case when ${\mathcal {A}}$ and ${\mathcal {B}}$ are not necessarily in the same k-uniform subfamily of $2^{[n]}$:

Theorem 1.3

([10]) Let ${\mathcal {A}}\subseteq {[n]\atopwithdelims ()a}$ and ${\mathcal {B}}\subseteq {[n]\atopwithdelims ()b}$ be non-empty cross-intersecting families with $n\ge a+b$, $a\le b$. Then $|{\mathcal {A}}|+|{\mathcal {B}}|\le {n\atopwithdelims ()b}-{{n-a}\atopwithdelims ()b}+1$.

In [19], Wang and Zhang generalized Theorem 1.3 to cross-t-intersecting families. Recently, using shifting techniques, Frankl and Kupavskii [9] gave another proof of the result of Wang and Zhang for the case when ${\mathcal {A}},{\mathcal {B}}\subseteq {[n]\atopwithdelims ()k}$.

As another extension, the multi-part Erdős–Ko–Rado theorem was introduced by Frankl [8], in connection with a similar result of Sali [17]. For positive integers p and $n_1,\ldots , n_p$, let $[\sum _{i\in [p]}n_i]$ be the ground set. Then, this ground set can be viewed as the disjoint union of p parts, denoted as $\bigsqcup _{i\in [p]}S_i$,¹ where $S_1=[n_1]$ and $S_i=\{1+\sum _{j\in [i-1]}n_j,\ldots ,\sum _{j\in [i]}n_j\}$ for $2\le i\le p$. Denote $2^{S_i}$ as the power set of $S_i$, for $A_1\in 2^{S_1},\ldots , A_p\in 2^{S_p}$, let $\bigsqcup _{i\in [p]} A_i\subseteq \bigsqcup _{i\in [p]}S_i$ be the disjoint union of $A_1,\ldots ,A_p$. For families ${\mathcal {F}}_1\subseteq 2^{S_1},\ldots ,{\mathcal {F}}_p\subseteq 2^{S_p}$, let $\prod _{i\in [p]}{\mathcal {F}}_i=\{\bigsqcup _{i\in [p]} A_i:~A_i\in {\mathcal {F}}_i\}$. For integers $1\le k_1\le n_1,\ldots , 1\le k_p\le n_p$, we denote $\prod _{i\in [p]}{[n_i]\atopwithdelims ()k_i}$ as the family of all subsets of $\bigsqcup _{i\in [p]}S_i$ which have exactly $k_i$ elements in the i-th part. Therefore, a family ${\mathcal {F}}\subseteq \prod _{i\in [p]}{[n_i]\atopwithdelims ()k_i}$ can be viewed as the natural generalization of k-uniform families to the multi-part setting. We call this family ${\mathcal {F}}$ a multi-part family with p parts. Moreover, a multi-part family is intersecting if any two sets of this family intersect in at least one of the p parts.

Frankl [8] proved that for positive integers $p\ge 1$, $n_1,\ldots ,n_p$ and $k_1,\ldots ,k_p$ satisfying $\frac{k_1}{n_1}\le \ldots \le \frac{k_p}{n_p}\le \frac{1}{2}$, if ${\mathcal {F}}\subseteq \prod _{i\in [p]}{[n_i]\atopwithdelims ()k_i}$ is a multi-part intersecting family, then

$$\begin{aligned} |{\mathcal {F}}|\le \frac{k_p}{n_p}\cdot \prod _{i\in [p]}{n_i\atopwithdelims ()k_i}. \end{aligned}$$

(1)

This bound is sharp; for example, it is attained by the following family:

$$\begin{aligned} \left\{ A\in {[n_p]\atopwithdelims ()k_p}:~i\in A,\text { for some }i\in [n_p]\right\} \times \prod _{i\in [p-1]}{[n_i]\atopwithdelims ()k_i}. \end{aligned}$$

Recall that the Kneser graph $KG_{n,k}$ is the graph on the vertex set ${[n]\atopwithdelims ()k}$, with $A,B\in {[n]\atopwithdelims ()k}$ forming an edge if and only if $A\cap B=\emptyset$. Hence, an intersecting family in ${[n]\atopwithdelims ()k}$ is an independent set in $KG_{n,k}$. Similarly, an intersecting family in $\prod _{i\in [p]}{[n_i]\atopwithdelims ()k_i}$ is an independent set in the graph (direct) product $KG_{n_1,k_1}\times \ldots \times KG_{n_p,k_p}$. Therefore, inequality (1) can be viewed as a consequence of the result stating that $\alpha (G\times H)=\max \{\alpha (G)|H|,\alpha (H)|G|\}$ for general vertex-transitive graphs G and H, which was proved by Zhang in [21].

For a subset $S\subseteq [n]$, denote ${\bar{S}}$ as the complementary set of S in [n]. In this paper, we consider cross-intersecting families under the multi-part setting and our main results are as follows.

Theorem 1.4

Given a positive integer p, let $n_1,n_2,\ldots ,n_p$ and $k_1,k_2,\ldots ,k_p$ be positive integers satisfying $n_i\ge 2k_i\ge 4$ for all $i\in [p]$. Let $X=\prod _{i\in [p]}{[n_i]\atopwithdelims ()k_i}$ and ${\mathcal {A}}_1,{\mathcal {A}}_2,\ldots ,{\mathcal {A}}_m$ be cross-intersecting families over X with ${\mathcal {A}}_1\ne \emptyset$. Then,

$$\begin{aligned} \sum \limits _{i=1}^{m}|{\mathcal {A}}_i|\le \left\{ \begin{array}{ll}|X|,&{}~~~\text {if}~m\le \min \limits _{i\in [p]} \frac{n_i}{k_i};\\ m\cdot M,&{}~~~\text {if}~m\ge \min \limits _{i\in [p]}\frac{n_i}{k_i},\end{array}\right. \end{aligned}$$

(2)

where $M=\max _{i\in [p]}{n_i-1\atopwithdelims ()k_i-1}\prod _{j\ne i}{n_j\atopwithdelims ()k_j}$. Furthermore, the bound is attained if and only if one of the following holds:

• (i) $m<\min _{i\in [p]}\frac{n_i}{k_i}$ and ${\mathcal {A}}_1=X$, ${\mathcal {A}}_2=\cdots ={\mathcal {A}}_m=\emptyset$;

• (ii) $m>\min _{i\in [p]}\frac{n_i}{k_i}$ and ${\mathcal {A}}_1=\cdots ={\mathcal {A}}_m=I$, where I is a maximum intersecting family in X;

• (iii) $m=\min _{i\in [p]}\frac{n_i}{k_i}$ and ${\mathcal {A}}_1,\ldots ,{\mathcal {A}}_m$ are as in (i) or (ii), or $m=2$ and $S=\{s\in [p]:~\frac{n_s}{k_s}=2\}$ and ${\mathcal {F}}=\prod _{s\in S}{[n_s]\atopwithdelims ()k_s}$ such that

  $$\begin{aligned} \begin{array}{l} {\mathcal {A}}_1=({\mathcal {A}}\sqcup {\mathcal {A}}_1')\times \prod \limits _{s\in [p]\setminus S}{[n_s]\atopwithdelims ()k_s}~\text {and}~ {\mathcal {A}}_2=({\mathcal {A}}\sqcup {\mathcal {A}}_2')\times \prod \limits _{s\in [p]\setminus S}{[n_s]\atopwithdelims ()k_s} \end{array} \end{aligned}$$

  (3)

  where ${\mathcal {A}}\subseteq {\mathcal {F}}$ is an intersecting family of size $|{\mathcal {A}}|<\frac{|{\mathcal {F}}|}{2}$, ${\mathcal {A}}_1'=\{E_1,{\bar{E}}_1,\ldots , E_a,{\bar{E}}_{a}\}\subseteq {\mathcal {F}}$ and ${\mathcal {A}}_2'=\{F_1,{\bar{F}}_1,\ldots , F_b,{\bar{F}}_{b}\}\subseteq {\mathcal {F}}$ satisfying ${\mathcal {A}}_1'\cap {\mathcal {A}}_2'=\emptyset$ and $2(|{\mathcal {A}}|+a+b)=|{\mathcal {F}}|$.

Remark 1.5

In [18], Wang and Zhang proved a similar result (Theorem 2.5 in [18]) for general connected symmetric systems. Therefore, Theorem 1.4 can be viewed as an application of the method introduced in [18]. However, different from the general case, Theorem 1.4 determines all the extremal families that attain the bound in (2).

Theorem 1.6

For any integer $p\ge 2$, let $n_i, t_i, s_i$ be positive integers satisfying $n_i\ge s_i+t_i+1$, $2\le s_i,t_i\le \frac{n_i}{2}$ for every $i\in [p]$ and $n_i\le \frac{7}{4}n_j$ for all distinct $i,j\in [p]$. If $\prod _{i\in [p]}{n_i\atopwithdelims ()s_i}\ge \prod _{i\in [p]}{n_i\atopwithdelims ()t_i}$ and ${\mathcal {A}}\subseteq \prod _{i\in [p]}{[n_i]\atopwithdelims ()t_i},~{\mathcal {B}}\subseteq \prod _{i\in [p]}{[n_i]\atopwithdelims ()s_i}$ are non-empty cross-intersecting families, then

$$\begin{aligned} |{\mathcal {A}}|+|{\mathcal {B}}|\le \prod _{i\in [p]}{n_i\atopwithdelims ()s_i}-\prod _{i\in [p]}{{n_i-t_i}\atopwithdelims ()s_i}+1, \end{aligned}$$

(4)

and the bound is attained if and only if the following holds:

• (i) $\prod _{i\in [p]}{n_i\atopwithdelims ()s_i}\ge \prod _{i\in [p]}{n_i\atopwithdelims ()t_i}$, ${\mathcal {A}}=\{A\}$ and ${\mathcal {B}}=\{B\in \prod _{i\in [p]}{[n_i]\atopwithdelims ()s_i}:B\cap A\ne \emptyset \}$ for some $A\in \prod _{i\in [p]}{[n_i]\atopwithdelims ()t_i}$;

• (ii) $\prod _{i\in [p]}{n_i\atopwithdelims ()s_i}= \prod _{i\in [p]}{n_i\atopwithdelims ()t_i}$, ${\mathcal {B}}=\{B\}$ and ${\mathcal {A}}=\{A\in \prod _{i\in [p]}{[n_i]\atopwithdelims ()t_i}:B\cap A\ne \emptyset \}$ for some $B\in \prod _{i\in [p]}{[n_i]\atopwithdelims ()s_i}$.

Remark 1.7

The constraints $s_i, t_i\le \frac{n_i}{2}$ for every $i\in [p]$ and $n_i\le \frac{7}{4}n_j$ for all distinct $i,j\in [p]$ in Theorem 1.6 are necessary.

When $s_i, t_i\le \frac{n_i}{2}$ is violated, for example, let $n_1=n_2=18$, $(s_1,t_1)=(15,2)$ and $(s_2,t_2)=(2,3)$, we have $s_1>\frac{n_1}{2}$ and ${n_1\atopwithdelims ()s_1}\cdot {n_2\atopwithdelims ()s_2}={n_1\atopwithdelims ()t_1}\cdot {n_2\atopwithdelims ()t_2}$. Set ${\mathcal {A}}=\{A_1\}\times {[n_2]\atopwithdelims ()t_2}$ for some 2-subset $A_1\subseteq [n_1]$ and ${\mathcal {B}}=\{B_1\in {[n_1]\atopwithdelims ()s_1}:B_1\cap A_1\ne \emptyset \}\times {[n_2]\atopwithdelims ()s_2}$. Then, we have $|{\mathcal {A}}|+|{\mathcal {B}}|>{n_1\atopwithdelims ()s_1}\cdot {n_2\atopwithdelims ()s_2}-{{n_1-t_1}\atopwithdelims ()s_1}\cdot {{n_2-t_2}\atopwithdelims ()s_2}+1$.

As for the constraint $n_i\le \frac{7}{4}n_j$, the constant $\frac{7}{4}$ here might not be tight, but the quantities of $n_i,n_j$ for distinct $i,j\in [p]$ need to be close. For example, let $n_1=5,n_2=12$ and $(s_1,t_1)=(s_2,t_2)=(2,2)$. Then, we have $n_2>\frac{7}{4}n_1$ and ${n_1\atopwithdelims ()s_1}\cdot {n_2\atopwithdelims ()s_2}={n_1\atopwithdelims ()t_1}\cdot {n_2\atopwithdelims ()t_2}$. Similarly, set ${\mathcal {A}}=\{A_1\}\times {[n_2]\atopwithdelims ()t_2}$ for some 2-subset $A_1\subseteq [n_1]$ and ${\mathcal {B}}=\{B_1\in {[n_1]\atopwithdelims ()s_1}:B_1\cap A_1\ne \emptyset \}\times {[n_2]\atopwithdelims ()s_2}$. Then, we have $|{\mathcal {A}}|+|{\mathcal {B}}|>{n_1\atopwithdelims ()s_1}\cdot {n_2\atopwithdelims ()s_2}-{{n_1-t_1}\atopwithdelims ()s_1}\cdot {{n_2-t_2}\atopwithdelims ()s_2}+1$.

We shall introduce some results about the independent sets of vertex-transitive graphs and their direct products in the next section. Then, we present the proof of Theorem 1.4 in Sect. 3 and the proof of Theorem 1.6 in Sect. 4. In Sect. 5, we will conclude the paper and discuss some remaining problems. For the convenience of the proof, if there is no confusion, we will denote $\prod _{i\in [p]}A_i$ as the disjoint union $\bigsqcup _{i\in [p]}A_i$ in the rest of the paper.

2 Preliminary results

2.1 Independent sets of vertex-transitive graphs

Given a finite set X, for $A\subseteq X$, denote ${\bar{A}}=X\setminus A$. For a simple graph $G=G(V,E)$, denote $\alpha (G)$ as the independent number of G and I(G) as the family of all maximum independent sets of G. For a vertex $v\in V(G)$, define its neighborhood as $N_G(v)=\{u\in V(G):(u,v)\in E(G)\}$. For a subset $A\subseteq V(G)$, denote $N_G(A)=\{b\in V(G):(a,b)\in E(G)$ for some $a\in A\}$ and $N_G[A]=A\cup N_G(A)$. If G is clear from the context, it will be omitted from the notation, so we use N(A) and N[A] to denote $N_G(A)$ and $N_G[A]$, respectively.

A graph G is said to be vertex-transitive, if its automorphism group $\text {Aut}(G)$ acts transitively upon its vertices. As a corollary of the “No-Homomorphism” lemma for vertex-transitive graphs in [1], Cameron and Ku [6] proved the following theorem.

Theorem 2.1

([6]) Let G be a vertex-transitive graph and B a subset of V(G). Denote G[B] as the induced subgraph of G on B. Then, any independent set S in G satisfies that $\frac{|S|}{|V(G)|}\le \frac{\alpha (G[B])}{|B|}$, equality implies that $|S\cap B|=\alpha (G[B])$.

Using the above theorem, Zhang [20] proved the following result.

Lemma 2.2

([20]) Let G be a vertex-transitive graph and A be an independent set of G. Then, $\frac{|A|}{|N[A]|}\le \frac{\alpha (G)}{|G|}$. Moreover, the equality implies that $|S\cap N[A]|=|A|$ for every $S\in I(G)$, and in particular $A\subseteq S$ for some $S\in I(G)$.

An independent set A in G is said to be imprimitive if $|A|<\alpha (G)$ and $\frac{|A|}{|N[A]|}=\frac{\alpha (G)}{|G|}$, and G is called IS-imprimitive if G has an imprimitive independent set. Otherwise, G is called IS-primitive. Note that a disconnected vertex-transitive graph G is always IS-imprimitive. Hence, IS-primitive vertex-transitive graphs are all connected.

For the proof of Theorem 1.4, we need the following result about the relationship between $\alpha (G)$ and the size of an independent set A and its non-neighbors ${\bar{N}}[A]$.

Lemma 2.3

Let G be a vertex transitive graph, and let A be an independent set of G. Then,

$$\begin{aligned} |A|+\frac{\alpha (G)}{|G|}|{\bar{N}}[A]|\le \alpha (G). \end{aligned}$$

Equality holds if and only if $A=\emptyset$ or $|A|=\alpha (G)$ or A is an imprimitive independent set.

For the integrity of the paper, we include the proof here. In [18], Wang and Zhang proved the same inequality for a more generalized combinatorial structure called symmetric system (see [18], Corollary 2.4).

Proof

If $A=\emptyset$ or $A=\alpha (G)$, the equality trivially holds. Suppose $0<|A|<\alpha (G)$, and let B be a maximum independent set in ${\bar{N}}[A]$. Then, $|B|=\alpha ({\bar{N}}[A])$. Clearly, $A\cup B$ is also an independent set of G, thus we have $|A|+|B|\le \alpha (G)$. By Theorem 2.1, we obtain that $\frac{|B|}{|{\bar{N}}[A]|}\ge \frac{\alpha (G)}{|G|}$. Therefore,

$$\begin{aligned} |A|+\frac{\alpha (G)}{|G|}|{\bar{N}}[A]|\le |A|+|B|\le \alpha (G), \end{aligned}$$

the equality holds when $\alpha (G)=|A|+\frac{\alpha (G)}{|G|}|{\bar{N}}[A]|=|A|+\frac{\alpha (G)}{|G|}(|G|-|N[A]|)$, which leads to $\frac{|A|}{|N[A]|}=\frac{\alpha (G)}{|G|}$, i.e., A is an imprimitive independent set. $\square$

Let X be a finite set, and $\Gamma$ a group acting transitively on X. Then, $\Gamma$ is said to be primitive on X if it preserves no non-trivial partition of X. A vertex-transitive graph G is called primitive if $\text {Aut}(G)$ is primitive on V(G). In [20], Zhang (see Proposition 2.4 in [20]) proved that if a vertex-transitive graph G is primitive, then, it must be IS-primitive. As a consequence of this result, Wang and Zhang [18] derived the IS-primitivity of the Kneser graph.

Proposition 2.4

( [18]) The Kneser graph $KG_{n,k}$ is IS-primitive except for $n=2k\ge 4$.

In order to deal with the multi-part case, we also need the results about the independent sets in direct products of vertex-transitive graphs. Let G and H be two graphs, the direct product $G\times H$ of G and H is defined by

$$\begin{aligned} V(G\times H)=V(G)\times V(H), \end{aligned}$$

and

$$\begin{aligned} E(G\times H)=\{[(u_1,v_1),(u_2,v_2)]:(u_1,u_2)\in E(G)\text { and }(v_1,v_2)\in E(H)\}. \end{aligned}$$

Clearly, $G\times H$ is a graph with $\text {Aut}(G)\times \text {Aut}(H)$ as its automorphism group. Moreover, if G, H are vertex-transitive, then, $G\times H$ is also vertex-transitive under the actions of $\text {Aut}(G)\times \text {Aut}(H)$. We say the direct product $G\times H$ is MIS-normal (maximum-independent-set-normal) if every maximum independent set of $G\times H$ is a preimage of an independent set of one factor under projections.

In [21], Zhang obtained the exact structure of the maximal independent set of $G\times H$.

Theorem 2.5

([21]) Let G and H be two vertex-transitive graphs with $\frac{\alpha (G)}{|G|}\ge \frac{\alpha (H)}{|H|}$. Then,

$$\begin{aligned} \alpha (G\times H)=\alpha (G)|H|, \end{aligned}$$

and exactly one of the following holds:

• (i) $G\times H$ is MIS-normal;

• (ii) $\frac{\alpha (G)}{|G|}=\frac{\alpha (H)}{|H|}$ and one of G or H is IS-imprimitive;

• (iii) $\frac{\alpha (G)}{|G|}>\frac{\alpha (H)}{|H|}$ and H is disconnected.

Let G and H be two non-empty vertex-transitive graphs with $\frac{\alpha (G)}{|G|}\ge \frac{\alpha (H)}{|H|}$. By Theorem 2.5, we have $\alpha (G\times H)=\alpha (G)|H|$. For each imprimitive independent set A in G, we have $\frac{|A|}{|N[A]|}=\frac{\alpha (G)}{|G|}$. Then, for an edge $(u,v)\in E(H)$, this implies that

$$\begin{aligned} \frac{|A\times \{u,v\}|}{|N[A\times \{u,v\}]|}=\frac{|A|}{|N[A]|}=\frac{\alpha (G)}{|G|}=\frac{\alpha (G\times H)}{|G\times H|}. \end{aligned}$$

Therefore, $A\times \{u,v\}$ is an imprimitive independent set in $G\times H$. This indicates that if G is IS-imprimitive, then, $G\times H$ is also IS-imprimitive. Moreover, if $\frac{\alpha (G)}{|G|}=\frac{\alpha (H)}{|H|}$ and A is an imprimitive independent set of G, then, $S=(A\times V(H))\cup ({\bar{N}}[A]\times I)$ is an independent set of $G\times H$ of size $\alpha (G)|H|$ for every $I\in I(H)$.

When $G\times H$ is MIS-normal, Zhang [20] also investigated the relationship between the primitivity of $G\times H$ and the primitivities of G and H.

Theorem 2.6

([20]) Suppose $G\times H$ is MIS-normal and $\frac{\alpha (H)}{|H|}\le \frac{\alpha (G)}{|G|}$. If $G\times H$ is IS-imprimitive, then one of the following two possible cases holds:

• (i) $\frac{\alpha (G)}{|G|}=\frac{\alpha (H)}{|H|}$ and one of them is IS-imprimitive or both G and H are bipartite;

• (ii) $\frac{\alpha (G)}{|G|}>\frac{\alpha (H)}{|H|}$ and G is IS-imprimitive.

As an application of Theorems 2.5 and 2.6, Geng et al. [11] showed the MIS-normality of the direct products of Kneser graphs.

Theorem 2.7

([11]) Given a positive integer p, let $n_1,n_2,\ldots ,n_p$ and $k_1,k_2,\ldots ,k_p$ be 2p positive integers with $n_i\ge 2k_i$ for $1\le i\le p$. Then the direct product of the Kneser graph

$$\begin{aligned} \text {KG}_{n_1,k_1}\times \text {KG}_{n_2,k_2}\times \cdots \times \text {KG}_{n_p,k_p} \end{aligned}$$

is MIS-normal except that there exist i, j and $\ell$ with $n_i=2k_i\ge 4$ and $n_j=2k_j$, or $n_i=n_j=n_{\ell }=2$.

2.2 Non-trivial independent sets of part-transitive bipartite graphs

For a bipartite graph G(X, Y) with two parts X and Y, an independent set A is said to be non-trivial if $A\nsubseteq X$ and $A\nsubseteq Y$. G(X, Y) is said to be part-transitive if there is a group $\Gamma$ acting transitively upon each part and preserving its adjacency relations. Clearly, if G(X, Y) is part-transitive, then every vertex of X (Y) has the same degree, written as d(X) (d(Y)). We use $\alpha (X,Y)$ and I(X, Y) to denote the size and the set of the maximum-sized non-trivial independent sets of G(X, Y), respectively.

Let G(X, Y) be a non-complete bipartite graph and let $A\cup B$ be a non-trivial independent set of G(X, Y), where $A\subseteq X$ and $B\subseteq Y$. Then, $A\subseteq X{\setminus } N(B)$ and $B\subseteq Y{\setminus } N(A)$, which implies

$$\begin{aligned} |A|+|B|\le \max {\{|A|+|Y|-|N(A)|, |B|+|X|-|N(B)|\}}. \end{aligned}$$

So we have

$$\begin{aligned} \alpha (X,Y)=\max {\{|Y|-\epsilon (X), |X|-\epsilon (Y)\}}, \end{aligned}$$

(5)

where $\epsilon (X)=\min \{|N(A)|-|A|: A\subseteq X, N(A)\ne Y\}$ and $\epsilon (Y)=\min \{|N(B)|-|B|: B\subseteq Y, N(B)\ne X\}$.

We call $A\subseteq X$ a fragment of G(X, Y) in X if $N(A)\ne Y$ and $|N(A)|-|A|=\epsilon (X)$, and we denote ${\mathcal {F}}(X)$ as the set of all fragments in X. Similarly, we can define ${\mathcal {F}}(Y)$. Denote ${\mathcal {F}}(X,Y)={\mathcal {F}}(X)\cup {\mathcal {F}}(Y)$ and we call the fragment $A\in {\mathcal {F}}(X,Y)$ a k-fragment if $|A|=k$. Moreover, we call a fragment $A\in {\mathcal {F}}(X)$ trivial if $|A|=1$ or $A=X{\setminus } N(b)$ for some $b\in Y$. For each $A\in {\mathcal {F}}(X)$, by Lemma 2.11 below, $Y{\setminus } N(A)$ is a fragment in ${\mathcal {F}}(Y)$. Hence, once we know ${\mathcal {F}}(X)$, ${\mathcal {F}}(Y)$ can also be determined.

Let X be a finite set, and $\Gamma$ a group acting transitively on X. If $\Gamma$ is imprimitive on X, then it preserves a non-trivial partition of X, called a block system, each element of which is called a block. Clearly, if $\Gamma$ is both transitive and imprimitive, there must be a subset $B\subseteq X$ such that $1<|B|<|X|$ and $\gamma (B)\cap B=B$ or $\emptyset$ for every $\gamma \in \Gamma$. In this case, B is called an imprimitive set in X. Furthermore, a subset $B\subseteq X$ is said to be semi-imprimitive if $1<|B|<|X|$ and for each $\gamma \in \Gamma$ we have $\gamma (B)\cap B=B$, $\emptyset$ or $\{b\}$ for some $b\in B$.

The following theorem (cf. [14, Theorem 1.12]) is a classical result on the primitivity of group actions.

Theorem 2.8

([14]) Suppose that a group $\Gamma$ transitively acts on X. Then $\Gamma$ is primitive on X if and only if for each $a\in X$, $\Gamma _a$ is a maximal subgroup of $\Gamma$. Here $\Gamma _a=\{\gamma \in \Gamma :\gamma (a)=a\}$, the stabilizer of $a\in X$.

Noticing the similarities about families that are cross-t-intersecting or cross-Sperner, Wang and Zhang [19] proved the following theorem about $\alpha (X,Y)$ and I(X, Y) of a special kind of part-transitive bipartite graphs.

Theorem 2.9

([19]) Let G(X, Y) be a non-complete bipartite graph with $|X|\le |Y|$. If G(X, Y) is part-transitive and every fragment of G(X, Y) is primitive under the action of a group $\Gamma$. Then $\alpha (X,Y)=|Y|-d(X)+1$. Moreover,

• (1) If $|X|<|Y|$, then X has only 1-fragments;

• (2) If $|X|=|Y|$, then each fragment in X has size 1 or $|X|-d(X)$ unless there is a semi-imprimitive fragment in X or Y.

To deal with multi-part cross-intersecting families, we introduce the following variant of Theorem 2.9.

Theorem 2.10

Let G(X, Y) be a non-complete bipartite graph with $|X|\le |Y|$. If G(X, Y) is part-transitive under the action of a group $\Gamma$, then

$$\begin{aligned} \alpha (X,Y)=\max {\{|Y|-d(X)+1, |A'|+|Y|-|N(A')|, |B'|+|X|-|N(B')|\}},\nonumber \\ \end{aligned}$$

(6)

where $A'$ and $B'$ are imprimitive subsets of X and Y, respectively, such that

$$\begin{aligned} |N(A')|-|A'|&= \min {\{|N(A)|-|A|: A\subseteq X~\text {is imprimitive and}~N(A)\ne Y\}},\nonumber \\ |N(B')|-|B'|&= \min {\{|N(B)|-|B|: B\subseteq Y~\text {is imprimitive and}~N(B)\ne X\}}. \end{aligned}$$

(7)

For the proof of Theorem 2.10, we need the following two lemmas from [19].

Lemma 2.11

([19]) Let G(X, Y) be a non-complete bipartite graph. Then, $|Y|-\epsilon (X)=|X|-\epsilon (Y)$, and

• (i) $A\in {\mathcal {F}}(X)$ if and only if $(Y{\setminus } N(A))\in {\mathcal {F}}(Y)$ and $N(Y{\setminus } N(A))=X{\setminus } A$;

• (ii) $A\cap B$ and $A\cup B$ are both in ${\mathcal {F}}(X)$ if A, $B\in {\mathcal {F}}(X)$, $A\cap B\ne \emptyset$ and $N(A\cup B)\ne Y$.

Lemma 2.12

([19]) Let G(X, Y) be a non-complete and part-transitive bipartite graph under the action of a group $\Gamma$. Suppose that $A\in {\mathcal {F}}(X,Y)$ such that $\emptyset \ne \gamma (A)\cap A\ne A$ for some $\gamma \in \Gamma$. Define $\phi : {\mathcal {F}}(X,Y)\rightarrow {\mathcal {F}}(X,Y)$,

$$\begin{aligned} \phi (A)=\left\{ \begin{array}{ll}Y\setminus N(A),&{}~~~\text {if}~A\in {\mathcal {F}}(X);\\ X\setminus N(A),&{}~~~\text {if}~A\in {\mathcal {F}}(Y).\end{array}\right. \end{aligned}$$

If $|A|\le |\phi (A)|$, then $A\cup \gamma (A)$ and $A\cap \gamma (A)$ are both in ${\mathcal {F}}(X,Y)$.

Remark 2.13

As a direct consequence of Lemma 2.11, a maximum-sized non-trivial independent set in G(X, Y) is of the form $A\sqcup (Y{\setminus } N(A))$ for some $A\in {\mathcal {F}}(X)$ or $B\sqcup (X{\setminus } N(B))$ for some $B\in {\mathcal {F}}(Y)$. Therefore, in order to address our problems, it suffices to determine ${\mathcal {F}}(X)$ $(\text {or }{\mathcal {F}}(Y))$.

Meanwhile, for the mapping $\phi$ in Lemma 2.12, we have $\phi ^{-1}=\phi$ and $|A|+|\phi (A)|=\alpha (X,Y)$. When $|A|=|\phi (A)|$, we call the fragment A balanced. Thus, all balanced fragments have size $\frac{1}{2}\alpha (X,Y)$.

Proof of Theorem 2.10

The same as the original proof of Theorem 2.9 in [19], we apply Lemma 2.12 repeatedly. For any $A_0\in {\mathcal {F}}(X,Y)$ satisfying $|A_0|\le |\phi (A_0)|$, we have the following two cases: (1) there exists $\gamma \in \Gamma$ such that $\emptyset \ne \gamma (A_0)\cap A_0\ne A_0$; (2) $\gamma (A_0)\cap A_0=\emptyset$ or $\gamma (A_0)\cap A_0=A_0$ for any $\gamma \in \Gamma$.

For case (1), by Lemma 2.12, we have $A_0\cap \gamma (A_0)\in {\mathcal {F}}(X,Y)$. Denote

$$\begin{aligned} A_1=\left\{ \begin{array}{ll}A_0\cap \gamma (A_0), &{}~\text {if}~|A_0\cap \gamma (A_0)|\le |\phi (A_0\cap \gamma (A_0))|; \\ \phi (A_0\cap \gamma (A_0)), &{}~~~~~~~~~~~~~~~~\text {otherwise}.\end{array}\right. \end{aligned}$$

Clearly, $A_1\in {\mathcal {F}}(X,Y)$. Next, we consider the primitivity of $A_1$, i.e., whether there is a $\gamma '\in \Gamma$ such that $\emptyset \ne \gamma '(A_1)\cap A_1\ne A_1$ or not.

For case (2), if $|A_0|\ne 1$, then, $A_0$ is an imprimitive set of X (or Y). Otherwise, $|A_0|=1$, which means ${\mathcal {F}}(X,Y)$ contains a singleton.

By doing the above procedures repeatedly, after r $(0\le r\le |A_0|-1)$ steps, we can obtain a fragment $A_r\in {\mathcal {F}}(X,Y)$ such that $A_r$ is either a singleton or an imprimitive set. When $A_r$ is a singleton in X, we have $\epsilon (X)=|N(A_r)|-|A_r|=d(X)-1$. When $A_r$ is an imprimitive set in X, we have $\epsilon (X)=|N(A_r)|-|A_r|$. By the definition of $\epsilon (X)$, this implies that

$$\begin{aligned} |N(A_r)|-|A_r|=\min \left\{ |N(A)|-|A|: A\subseteq X,~N(A)\ne Y\right\} . \end{aligned}$$

Thus, we have $|N(A_r)|-|A_r|= \min {\{|N(A)|-|A|: A\subseteq X~\text {is imprimitive and}} {N(A)\ne Y\}}$. Therefore, by (5), we have

$$\begin{aligned} \alpha (X,Y)=\max {\left\{ |Y|-d(X)+1, |A'|+|Y|-|N(A')|, |B'|+|X|-|N(B')|\right\} }, \end{aligned}$$

(8)

where $A'$ and $B'$ are imprimitive subsets of X and Y, respectively, such that

$$\begin{aligned} |N(A')|-|A'|&= \min {\left\{ |N(A)|-|A|: A\subseteq X~\text {is imprimitive and}~N(A)\ne Y\right\} },\\ |N(B')|-|B'|&= \min {\left\{ |N(B)|-|B|: B\subseteq Y~\text {is imprimitive and}~N(B)\ne X\right\} }. \end{aligned}$$

Note that $|Y|\ge |X|$ and $d(X)|X|=d(Y)|Y|$. Thus, we have $d(X)=d(Y)|Y|/|X|\ge d(Y)$. This leads to

$$\begin{aligned} |Y|-|X|=d(X)|X|/d(Y)-|X|=(d(X)-d(Y))|X|/d(Y)\ge d(X)-d(Y), \end{aligned}$$

which implies that $|X|-d(Y)+1\le |Y|-d(X)+1$. Then, (6) follows directly from (8) and $|X|-d(Y)+1\le |Y|-d(X)+1$. $\square$

3 Proof of Theorem 1.4

Throughout this section, for any non-empty subset $S\subseteq [p]$ and $A=\prod _{i\in S}A_i\in \prod _{i\in S}{[n_i]\atopwithdelims ()k_i}$, denote ${\bar{A}}=\prod _{i\in S}{\bar{A}}_i$. Before we start the proof of Theorem 1.4, we introduce the following proposition about the direct product of Kneser graphs.

Proposition 3.1

Given a positive integer $p\ge 2$, let $n_1,n_2,\ldots ,n_p$ and $k_1,k_2,\ldots ,k_p$ be positive integers with $n_i\ge 2k_i\ge 4$ for $1\le i\le p$. Let $G=\prod _{i\in [p]}{KG_{n_i,k_i}}$. Then, G is IS-imprimitive if and only if there exists an $i\in [p]$ such that $n_i=2k_i\ge 4$.

Proof

Note that $KG_{n,k}$ is disconnected only if $n=2k\ge 4$. Thus, by Proposition 2.4, a disconnected Kneser graph $KG_{n,k}$ is IS-imprimitive for all $k\ge 2$. Moreover, since $\chi ({KG_{n,k}})=n-2k+2$ for all $n\ge 2k$ (Lovász–Kneser Theorem, see [15]), if $KG_{n,k}$ is bipartite, then we must have $n=2k\ge 2$.

To start with, we show that G is IS-imprimitive if there is an $i\in [p]$ such that $n_i=2k_i\ge 4$. Assume that $n_i=2k_i\ge 4$ for some $i\in [p]$. By Proposition 2.4, the i-th factor $KG_{n_i,k_i}$ is IS-imprimitive. Since direct products of Kneser graphs are vertex transitive graphs, by Theorem 2.5, G is also IS-imprimitive.

Next, assume that G is IS-imprimitive. To prove the other direction, we use induction on the number of factors p.

When $p=2$, let $G_1=KG_{n_1,k_1}$, $G_2=KG_{n_2,k_2}$, and $G=G_1\times G_2$. W.l.o.g., assume that $\frac{\alpha (G_1)}{|G_1|}\ge \frac{\alpha (G_2)}{|G_2|}$. By Theorem 2.5, exactly one of the following holds:

• (i) $G_1\times G_2$ is MIS-normal,

• (ii) $\frac{\alpha (G_1)}{|G_1|}=\frac{\alpha (G_2)}{|G_2|}$ and one of $G_1$ and $G_2$ is IS-imprimitive,

• (iii) $\frac{\alpha (G_1)}{|G_1|}>\frac{\alpha (G_2)}{|G_2|}$ and $G_2$ is disconnected.

For case (i), by Theorem 2.6, at least one factor of G is IS-imprimitive or both $G_1$ and $G_2$ are bipartite. Therefore, either there exists an $i\in [2]$ such that $n_i=2k_i\ge 4$ or both $G_1$ and $G_2$ are IS-primitive and bipartite. Note that $KG_{2,1}$ is the only IS-primitive bipartite Kneser graph. Thus, both $G_1$ and $G_2$ are isomorphic to $KG_{2,1}$. This contradicts the assumption that $n_i\ge 2k_i\ge 4$. Note that G is not MIS-normal for cases (ii) and (iii). Therefore, by Theorem 2.7, there exists an $i\in [2]$ such that $n_i=2k_i\ge 4$. Thus, the result holds when $p=2$.

Now, suppose the result holds when the number of factors is $p-1$. W.l.o.g., assume that

$$\begin{aligned} \frac{\alpha (KG_{n_1,k_1})}{|KG_{n_1,k_1}|}\ge \frac{\alpha (KG_{n_2,k_2})}{|KG_{n_2,k_2}|}\ge \cdots \ge \frac{\alpha (KG_{n_p,k_p})}{|KG_{n_p,k_p}|}. \end{aligned}$$

Set $G'_1=\prod _{i=1}^{p-1}KG_{n_i,k_i}$ and $G'_2=KG_{n_p,k_p}$. By Theorem 2.5, we have $\frac{\alpha (G'_1)}{|G'|}=\frac{\alpha (KG_{n_1,k_1})}{|KG_{n_1,k_1}|}$ and exactly one of the following holds:

• (i) $G'_1\times G'_2$ is MIS-normal,

• (ii) $\frac{\alpha (G'_1)}{|G'_1|}=\frac{\alpha (G'_2)}{|G'_2|}$ and one of $G'_1$ and $G'_2$ is IS-imprimitive,

• (iii) $\frac{\alpha (G'_1)}{|G'_1|}>\frac{\alpha (G'_2)}{|G'_2|}$ and $G'_2$ is disconnected.

For case (i), by Theorem 2.6, at least one factor of $G'_1$ and $G'_2$ is IS-imprimitive or both $G'_1$ and $G'_2$ are bipartite. If $G_1'$ is IS-imprimitive, by the induction hypothesis, there exists an $i'\in [p-1]$ such that $n_{i'}=2k_{i'}\ge 4$. If $G_2'$ is IS-imprimitive, then $n_p=2k_p\ge 4$. Otherwise, both $G'_1$ and $G'_2$ are IS-primitive and bipartite. Thus, for $G'_2$, we have $n_p=2k_p=2$, a contradiction. Similarly, note that G is not MIS-normal for cases (ii) and (iii). Therefore, by Theorem  2.7, there exists an $i\in [2]$ such that $n_i=2k_i\ge 4$.

This completes the proof. $\square$

The idea of the proof for Theorem 1.4 is similar to that for general connected symmetric systems in [18]. Since $\prod _{i=1}^{p}\text {KG}_{n_i,k_i}$ is a vertex transitive graph, by Lemma 2.3, we can prove the bound (2). Then, through a careful analysis, we can obtain the structure of all imprimitive independent sets of this graph. This leads to the unique structures of ${\mathcal {A}}_1$ and ${\mathcal {A}}_2$ in (3).

Proof of Theorem 1.4

Define a graph G on the vertex set $X=\prod _{s\in [p]}{[n_s]\atopwithdelims ()k_s}$ with $A,B\in X$ forming an edge in G if and only if $A\cap B=\emptyset$. Therefore, G is the direct product of Kneser graphs $\text {KG}_{n_1,k_1}\times \cdots \times \text {KG}_{n_p,k_p}$.

Assume that $2\le \frac{n_1}{k_1}\le \frac{n_2}{k_2}\le \ldots \le \frac{n_p}{k_p}$, then $\frac{|G|}{\alpha (G)}=\frac{n_1}{k_1}$ by Theorem 2.5. Following the notations of Borg in [3,4,5], write ${\mathcal {A}}^*_i=\{A\in {\mathcal {A}}_i|A\cap B\ne \emptyset \text { for any }B\in {\mathcal {A}}_i\}$, $\hat{{\mathcal {A}}}_i={\mathcal {A}}_i\setminus {\mathcal {A}}_i^*$, ${\mathcal {A}}^*=\bigcup _{i=1}^m{\mathcal {A}}_i^{*}$, $\hat{{\mathcal {A}}}=\bigcup _{i=1}^m\hat{{\mathcal {A}}}_i$. Thus, ${\mathcal {A}}^*$ is an intersecting family. Moreover, note that ${\bar{N}}_{G}[{\mathcal {A}}]=\{B\in X{\setminus } {\mathcal {A}}|A\cap B\ne \emptyset \text { for any }A\in {\mathcal {A}}\}$ for any ${\mathcal {A}}\subseteq X$. Thus, $\hat{{\mathcal {A}}}\subseteq {\bar{N}}_{G}[{\mathcal {A}}^*]$. It follows that ${\mathcal {A}}_i\cap {\mathcal {A}}_j\subseteq {\mathcal {A}}_i^*\cap {\mathcal {A}}_j^*$ from the definition, therefore $\hat{{\mathcal {A}}}_i\cap \hat{{\mathcal {A}}}_j=\emptyset$ for $i\ne j$, and $|\hat{{\mathcal {A}}}|=\sum _{i=1}^m|\hat{{\mathcal {A}}}_i|$. Thus, by Lemma 2.3 we have

$$\begin{aligned} \sum \limits _{i=1}^m|{\mathcal {A}}_i|&=\sum \limits _{i=1}^m|\hat{{\mathcal {A}}}_i| +\sum \limits _{i=1}^m|{\mathcal {A}}_i^*|\le |\hat{{\mathcal {A}}}| +m|{\mathcal {A}}^*|\le |{\bar{N}}_{G}[{\mathcal {A}}^*]| +m|{\mathcal {A}}^*|\\&=\frac{|G|}{\alpha (G)}\left( \frac{\alpha (G)}{|G|}|{\bar{N}}_{G}[{\mathcal {A}}^*]| +|{\mathcal {A}}^*|\right) +(m-\frac{|G|}{\alpha (G)})|{\mathcal {A}}^*|\\&\le |G|+\left( m-\frac{|G|}{\alpha (G)}\right) |{\mathcal {A}}^*|=|G|+(m-\frac{n_1}{k_1}) |{\mathcal {A}}^*|. \end{aligned}$$

If $m<\frac{n_1}{k_1}$, then $\sum _{i=1}^m|{\mathcal {A}}_i|\le |G|$, and the equality implies ${\mathcal {A}}^*=\emptyset$. Thus, ${\mathcal {A}}_i=\hat{{\mathcal {A}}_i}$ for every $i\in [m]$, and this yields that the graph G is a disjoint union of the induced subgraph $G[{\mathcal {A}}_i]'s$. Note that for each $i\in [m]$ and $A\in {\mathcal {A}}_i$, there exists some $B\in {\mathcal {A}}_i$ such that $A\cap B=\emptyset$. Thus, each $G[{\mathcal {A}}_i]$ is a connected component of G. Since G is connected when $\frac{n_s}{k_s}>2$ for all $s\in [p]$ and $m\ge 2$, we know that one of ${\mathcal {A}}_i$ is X and the rest are empty sets, as case (i).

If $m>\frac{n_1}{k_1}$, then $\sum _{i=1}^m|{\mathcal {A}}_i|\le m\alpha (G)$, and the equality implies that ${\mathcal {A}}_1^*=\ldots ={\mathcal {A}}_m^*={\mathcal {A}}^*$, $|{\mathcal {A}}^*|=\alpha (G)$, as case (ii).

If $m=\frac{n_1}{k_1}$, then $\sum _{i=1}^m|{\mathcal {A}}_i|\le |X|$, and the equality implies that ${\mathcal {A}}_1^*=\ldots ={\mathcal {A}}_m^*={\mathcal {A}}^{*}$ and $\frac{\alpha (G)}{|G|}|{\bar{N}}_{G}[{\mathcal {A}}^*]|+|{\mathcal {A}}^*|=\alpha (G)$. By Lemma 2.3, we know that $|{\mathcal {A}}^*|=0$, or $|{\mathcal {A}}^*|=\alpha (G)$, or ${\mathcal {A}}^*$ is an imprimitive independent set of G. In the last case, $\hat{{\mathcal {A}}}_1,\ldots ,\hat{{\mathcal {A}}}_m$ are cross-intersecting families and form a partition of ${\bar{N}}_{G}[{\mathcal {A}}^*]$. In order to determine the structures of the maximum-sized cross-intersecting families in this case, we shall characterize the imprimitive independent set of G.

Claim 3.2

Let ${\mathcal {F}}=\prod _{s\in S}{[n_s]\atopwithdelims ()k_s}$ and $X'=\prod _{s\in [p]\setminus S}{[n_s]\atopwithdelims ()k_s}$, where $S=\{s\in [p]:~\frac{n_s}{k_s}=2\}$. If ${\mathcal {A}}^*$ is an imprimitive independent set of G, then $S\ne \emptyset$ and ${\mathcal {A}}^*={\mathcal {A}}\times X'$, where ${\mathcal {A}}\subseteq {\mathcal {F}}$ is a non-maximum intersecting family.

According to Proposition 3.1, G is IS-imprimitive if and only if there exists an $i\in S$ such that $n_i=2k_i\ge 4$. Thus, with the assumptions in this claim, $S\ne \emptyset$ and $i_0\in S$ if and only if $n_{i_0}=2k_{i_0}\ge 4$ for some $i_0\in [p]$. W.l.o.g., assume that $S=[s_0]$, where $s_0=|S|$. Under this circumstance, $m=\frac{n_1}{k_1}=2$.

Divide ${\mathcal {A}}^*$ into u disjoint parts $\{C_i\times {\mathcal {D}}_i\}_{i=1}^{u}$, where $C_i=C_{i,1}\times \ldots \times C_{i,s_0}\in {\mathcal {F}}$, ${\mathcal {D}}_i\subseteq X'$ for all $i\in [u]$ and $C_i\ne C_j$ for any $i\ne j\in [u]$. Since $N_G(C_i\times {\mathcal {D}}_i)=\bar{C_i}\times {\mathcal {D}}_i'$, where ${\mathcal {D}}'_i=\{A\in X': A\cap D_i=\emptyset \text { for some } D_i\in {\mathcal {D}}_i\}$, we know that $N_G[C_i\times {\mathcal {D}}_i]\cap N_G[C_j\times {\mathcal {D}}_j]=\emptyset$ for all $i\ne j\in [u]$. Meanwhile, $(C_i\times {\mathcal {D}}_i)\cap N_G(C_j\times {\mathcal {D}}_j)=\emptyset$ for all $i\ne j\in [u]$. Otherwise, assume that there exists $T_1\times T_2\in (C_i\times {\mathcal {D}}_i)\cap N_G(C_j\times {\mathcal {D}}_j)$, for some $T_1\in {\mathcal {F}}$ and $T_2\in X'$. Thus, we have $(T_1\times T_2)\cap (C_j\times D_j)=\emptyset$, for some $D_j\in {\mathcal {D}}_j$, which contradicts the fact that ${\mathcal {A}}^*$ is an intersecting family.

By projecting G onto the last $p-s_0$ factors, we obtain a graph $G'$ with vertex set $X'$ such that $A,B\in X'$ form an edge in $G'$ if and only if A, B are disjoint. Thus, we have ${\bar{N}}_{G'}({\mathcal {A}})=\{B\in X'|A\cap B\ne \emptyset \text { for any }A\in {\mathcal {A}}\}$ for any ${\mathcal {A}}\subseteq X'$. Consider the cross-intersecting families $\{{\mathcal {D}}_i,{\bar{N}}_{G'}({\mathcal {D}}_i)\}$ in $X'$, since $|\{{\mathcal {D}}_i,{\bar{N}}_{G'}({\mathcal {D}}_i)\}|=2<\frac{n_{s_0+1}}{k_{s_0+1}}$, by case (i), we know that

$$\begin{aligned} |{\mathcal {D}}_i|+|{\bar{N}}_{G'}({\mathcal {D}}_i)|=|{\mathcal {D}}_i| +|X'|-|N_{G'}({\mathcal {D}}_i)|\le |X'|, \end{aligned}$$

thus we have $|{\mathcal {D}}_i|\le |N_{G'}({\mathcal {D}}_i)|$, and $|C_i\times {\mathcal {D}}_i|=|{\mathcal {D}}_i|\le |N_{G'}({\mathcal {D}}_i)|=|N_G(C_i\times {\mathcal {D}}_i)|$. Therefore,

$$\begin{aligned} \frac{|{\mathcal {A}}^*|}{|N_G[{\mathcal {A}}^*]|} =\frac{\sum _{i\in [u]}|C_i\times {\mathcal {D}}_i|}{\sum _{i\in [u]}|N_G[C_i\times {\mathcal {D}}_i]|} \le \frac{1}{2}=\frac{\alpha (G)}{|G|}=\frac{k_1}{n_1}, \end{aligned}$$

and the equality holds if and only if for all $i\in [u]$, ${\mathcal {D}}_i=X'$ or ${\bar{N}}_{G'}({\mathcal {D}}_i)=X'$. Since ${\mathcal {D}}_i\ne \emptyset$, we have ${\mathcal {A}}^*=\bigsqcup _{i=1}^{u}C_i\times X'={\mathcal {A}}\times X'$. Recall that $\frac{n_s}{k_s}>2$ for all $s>s_0$, hence $C_i\cap C_j\ne \emptyset$ for any $i\ne j\in [u]$. Therefore, by the imprimitivity of ${\mathcal {A}}^*$, ${\mathcal {A}}^*$ is a non-maximum independent set of G, thus ${\mathcal {A}}\subseteq {\mathcal {F}}$ is a non-maximum intersecting family and the claim holds.

Recall that by Claim 3.2, $S\ne \emptyset$, thus $|{\mathcal {F}}|=\prod _{s\in S}{n_s\atopwithdelims ()k_s}$ is an even number. Let $w=\frac{|{\mathcal {F}}|}{2}$. Then, we can assume that

$$\begin{aligned} {\mathcal {F}}=\{A_1,{\bar{A}}_1,\ldots , A_{w},{\bar{A}}_{w}\}. \end{aligned}$$

For every intersecting family ${\mathcal {B}}\subseteq {\mathcal {F}}$, by (1), we have $|{\mathcal {B}}|\le w$. Thus, as a non-maximum intersecting family in ${\mathcal {F}}$, w.l.o.g., we can assume that ${\mathcal {A}}=\{A_1,A_2,\ldots ,A_{w_0}\}$ for some positive integer $w_0<w$. Therefore,

$$\begin{aligned} {\mathcal {A}}^*&=\{A_1,A_2,\ldots ,A_{w_0}\}\times X',\\ N_G({\mathcal {A}}^*)&=\{{\bar{A}}_1,{\bar{A}}_2,\ldots ,{\bar{A}}_{w_0}\}\times X'. \end{aligned}$$

This further implies that

$$\begin{aligned} {\bar{N}}_G[{\mathcal {A}}^*]=\{E_1,{\bar{E}}_1,E_2,{\bar{E}}_2, \ldots ,E_{w-w_0},{\bar{E}}_{w-w_0}\}\times X', \end{aligned}$$

where $E_{i}=A_{w_0+i}$ for $1\le i\le w-w_0$.

Next, we claim that for every $1\le i\le w-w_0$, either $E_i\times X'\subset \hat{{\mathcal {A}}_1}$ or $E_i\times X'\subset \hat{{\mathcal {A}}_2}$. W.l.o.g., assume that $E_1\times B_1\in \hat{{\mathcal {A}}_1}$ for some $B_1\in X'$. Since $\hat{{\mathcal {A}}_1}$ and $\hat{{\mathcal {A}}_2}$ are cross-intersecting and $\hat{{\mathcal {A}}_1}$ and $\hat{{\mathcal {A}}_2}$ form a partition of ${\bar{N}}_G[{\mathcal {A}}^*]$, thus we have ${\bar{E}}_1\times C_1\in \hat{{\mathcal {A}}_1}$ for every $C_1\in X'$ such that $C_1\cap B_1=\emptyset$. Denote $k=\sum _{j\in [p]{\setminus } S}k_j$ and $N=\sum _{j\in [p]\setminus S}n_j$. For $B\in X'$ and $j\in [p]{\setminus } S$, denote $B|_j$ as the projection of B on the j-th factor. Let $B\in X'$ satisfying $|B\cap B_1|=k-1$. Then, for every $j\in [p]\setminus S$, we have $k_j-1\le |(B\cap B_1)|_{j}|\le k_j$. Since $n_j\ge 2k_j+1$, there exists a set $C\in X'$ such that $(C\cap (B\cup B_1))|_{j}=\emptyset$ for every $j\in [p]\setminus S$. Clearly, $C\cap (B\cup B_1)=\emptyset$. Therefore, ${\bar{E}}_1\times C\in \hat{{\mathcal {A}}_1}$. If $E_1\times B\in \hat{{\mathcal {A}}_2}$, then by $C\cap B=\emptyset$, we can also obtain that ${\bar{E}}_1\times C\in \hat{{\mathcal {A}}_2}$. This yields a contradiction. Therefore, $E_1\times B\in \hat{{\mathcal {A}}_1}$ for every $B\in X'$ such that $|B\cap B_1|=k-1$. Note that for any $B'\ne B_1\in X'$ and $|B'\cap B_1|=r<k-1$, there are $B_2,\ldots ,B_{k-r}\in X'$ such that $|B_i\cap B_{i+1}|=k-1$ for $1\le i\le {k-r-1}$ and $|B_{k-r}\cap B'|=k-1$. Therefore, this yields $E_1\times B\in \hat{{\mathcal {A}}_1}$ for any $B\in X'$, which proves the claim.

Moreover, since $\hat{{\mathcal {A}}_1}$ and $\hat{{\mathcal {A}}_2}$ are cross-intersecting, $E_i\times X'$ and ${\bar{E}}_i\times X'$ are both contained in either $\hat{{\mathcal {A}}_1}$ or $\hat{{\mathcal {A}}_2}$. Denote ${\mathcal {A}}_1'= \{E_1,{\bar{E}}_1,\ldots , E_a,{\bar{E}}_{a}\}$ and ${\mathcal {A}}_2'= \{E_{a+1},{\bar{E}}_{a+1},\ldots , E_{w-w_0},{\bar{E}}_{w-w_0}\}$. Therefore, w.l.o.g., we can assume that $\hat{{\mathcal {A}}}_1={\mathcal {A}}_1'\times X'$ and $\hat{{\mathcal {A}}}_2={\mathcal {A}}_2'\times X'$. Finally, by ${\mathcal {A}}_1={\mathcal {A}}^*\sqcup \hat{{\mathcal {A}}}_1$ and ${\mathcal {A}}_2={\mathcal {A}}^*\sqcup \hat{{\mathcal {A}}}_2$, we have

$$\begin{aligned} \begin{array}{l} {\mathcal {A}}_1=({\mathcal {A}}\sqcup {\mathcal {A}}_1')\times X'~\text {and}~ {\mathcal {A}}_2=({\mathcal {A}}\sqcup {\mathcal {A}}_2')\times X'. \end{array} \end{aligned}$$

This completes the proof. $\square$

4 Proof of Theorem 1.6

Throughout this section, we denote $S_n$ as the symmetric group on [n] and $S_C$ as the symmetric group on C for $C\subseteq [n]$. For each $i\in [p]$, let $X_i$ be a finite set, then for each family ${\mathcal {A}}\subseteq \prod _{i\in [p]}X_i$, we denote ${\mathcal {A}}|_i$ as the projection of ${\mathcal {A}}$ onto the i-th factor.

For the proof of Theorem 1.6, we need the following proposition obtained by Wang and Zhang in [19].

Proposition 4.1

([19]) Let G(X, Y) be a non-complete bipartite graph with $|X|=|Y|$ and $\epsilon (X)=d(X)-1$, and let $\Gamma$ be a group part-transitively acting on G(X, Y). If each fragment of G(X, Y) is primitive and there are no 2-fragments in ${\mathcal {F}}(X,Y)$, then every non-trivial fragment $A\in {\mathcal {F}}(X)$ (if there exists) is balanced (see Remark 2.13), and for each $a\in A$, there is a unique non-trivial fragment B such that $A\cap B=\{a\}$.

The proof of Theorem 1.6 is divided into two parts: Firstly, we prove the bound (4). Consider a non-complete bipartite graph defined by the multi-part cross-intersecting family. Through discussions about the primitivity of group $\prod _{i=1}^{p}S_{n_i}$ and careful evaluations about $|{\mathcal {A}}|+|{\mathcal {Y}}|-|N({\mathcal {A}})|$, the bound (4) follows from Theorem 2.10. Secondly, based on a characterization of all non-trivial fragments in this bipartite graph, we determine all the structures of ${\mathcal {A}}$ and ${\mathcal {B}}$ when the bound (4) is attained.

Proof of Theorem 1.6

With the assumptions in the theorem, we define a bipartite graph $G({\mathcal {X}},{\mathcal {Y}})$ with ${\mathcal {X}}=\prod _{i=1}^{p}{[n_i]\atopwithdelims ()t_i}$ and ${\mathcal {Y}}=\prod _{i=1}^{p}{[n_i]\atopwithdelims ()s_i}$. For $A=\prod _{i=1}^{p}{A_i}\in {\mathcal {X}}$ and $B=\prod _{i=1}^{p}{B_i}\in {\mathcal {Y}}$ ($A_i\in {[n_i]\atopwithdelims ()t_i}$ and $B_i\in {[n_i]\atopwithdelims ()s_i}$, for every $1\le i\le p$), (A, B) forms an edge in $G({\mathcal {X}},{\mathcal {Y}})$ if and only if $A\cap B=\emptyset$, i.e., $A_i\cap B_i=\emptyset$ for each $1\le i\le p$.

It can be easily verified that $\prod _{i=1}^{p}S_{n_i}$ acts transitively on ${\mathcal {X}}$ and ${\mathcal {Y}}$, respectively, and preserves the property of cross-intersecting. Thus, we have $d({\mathcal {X}})=|N(A)|$ for each $A\in {\mathcal {X}}$, and $d({\mathcal {Y}})=|N(B)|$ for each $B\in {\mathcal {Y}}$. Since, for each $A=\prod _{i=1}^{p}{A_i}\in {\mathcal {X}}$,

$$\begin{aligned} N(A)=\left\{ B=\prod _{i=1}^{p}{B_i}\in {\mathcal {Y}}:~A_i\cap B_i=\emptyset \text { for each }1\le i\le p\right\} =\prod _{i=1}^{p}{[n_i]\setminus A_i\atopwithdelims ()s_i}, \end{aligned}$$

we have $d({\mathcal {X}})=|N(A)|=\prod _{i=1}^{p}{{n_i-t_i}\atopwithdelims ()s_i}$. Similarly, $d({\mathcal {Y}})=|N(B)|=\prod _{i=1}^{p}{{n_i-s_i}\atopwithdelims ()t_i}$.

By Theorem 2.10, we obtain that

$$\begin{aligned} \alpha ({\mathcal {X}},{\mathcal {Y}})=\max {\{|{\mathcal {Y}}|-d({\mathcal {X}})+1, |{\mathcal {A}}'|+|{\mathcal {Y}}|-|N({\mathcal {A}}')|, |{\mathcal {B}}'|+|{\mathcal {X}}|-|N({\mathcal {B}}')|\}}, \end{aligned}$$

where ${\mathcal {A}}'$ and ${\mathcal {B}}'$ are imprimitive subsets of ${\mathcal {X}}$ and ${\mathcal {Y}}$ satisfying (7), respectively. Therefore, in order to estimate $\alpha ({\mathcal {X}},{\mathcal {Y}})$ accurately, more discussions about the sizes and the structures of the imprimitive subsets of ${\mathcal {X}}$ and ${\mathcal {Y}}$ are necessary. $\square$

Claim 4.2

Let ${\mathcal {A}}$ and ${\mathcal {B}}$ be imprimitive subsets of ${\mathcal {X}}$ and ${\mathcal {Y}}$, respectively, then

$$\begin{aligned} {\mathcal {A}}&=\prod _{i\in T_1}{\{A_i,\bar{A_i}\}}\times \prod _{i\in T_2}{\{A_i\}}\times \prod _{i\in [p]\setminus (T_1\sqcup T_2)}{[n_i]\atopwithdelims ()t_i},~\hbox { for some disjoint}\ T_1, T_2\subseteq [p],\\ {\mathcal {B}}&=\prod _{i\in R_1}{\{B_i,\bar{B_i}\}}\times \prod _{i\in R_2}{\{B_i\}}\times \prod _{i\in [p]\setminus (R_1\sqcup R_2)}{[n_i]\atopwithdelims ()s_i},~\hbox { for some disjoint}\ R_1, R_2\subseteq [p], \end{aligned}$$

where $A_i\in {[n_i]\atopwithdelims ()t_i}$, $B_i\in {[n_i]\atopwithdelims ()s_i}$, $T_1\sqcup T_2\ne \emptyset$, $R_1\sqcup R_2\ne \emptyset$ and $T_2,R_2\ne [p]$. Furthermore, for each $i\in T_1$, $n_i=2t_i$ and for each $i\in R_1$, $n_i=2s_i$.

If $\Gamma =\prod _{i=1}^{p}S_{n_i}$ is imprimitive on ${\mathcal {X}}$, then from the definition we know that $\Gamma$ preserves a non-trivial partition $\{{\mathcal {X}}_j\}_{j=1}^{L}$ of ${\mathcal {X}}$. By projecting ${\mathcal {X}}_j$ to the i-th factor, we can obtain that $\bigsqcup _{j=1}^{L}({\mathcal {X}}_j|_{i})={\mathcal {X}}|_i={[n_i]\atopwithdelims ()t_i}$ and $\Gamma |_{i}=S_{n_i}$ preserving this partition of $[n_i]\atopwithdelims ()t_i$.

It is well known that for each $A_i\in {[n_i]\atopwithdelims ()t_i}$, the stabilizer of $A_i$ is isomorphic to $S_{t_i}\times S_{n_i-t_i}$, which is a maximal subgroup of $S_{n_i}$ if $2t_i\ne n_i$ (see, e.g., [16]). Then, by Theorem 2.8, we obtain that $S_{n_i}$ is primitive on $[n_i]\atopwithdelims ()t_i$ unless $2t_i=n_i$, which means for the factor with $2t_i\le n_i$ the distinct members in $\{{\mathcal {X}}_1|_i,\ldots ,{\mathcal {X}}_L|_i\}$ form a trivial partition of ${[n_i]\atopwithdelims ()t_i}$. Thus, for each $j\in L$, ${\mathcal {X}}_j|_{i}$ is either a singleton in ${[n_i]\atopwithdelims ()t_i}$, or ${\mathcal {X}}_j|_{i}={[n_i]\atopwithdelims ()t_i}$. For convenience, we also use $\bigsqcup _{j=1}^L({\mathcal {X}}_j|_i)$ to denote the partition of ${[n_i]\atopwithdelims ()t_i}$ formed by the distinct members of $\{{\mathcal {X}}_1|_i,\ldots ,{\mathcal {X}}_L|_i\}$.

When $2t_i=n_i$, it can be easily verified that the only imprimitive subset of $[n_i]\atopwithdelims ()t_i$ has the form $\{A_i,\bar{A_i}\}$. Therefore, for the factors with $2t_i=n_i$, the partition $\bigsqcup _{j=1}^{L}({\mathcal {X}}_j|_{i})$ of ${[n_i]\atopwithdelims ()t_i}$ is either a trivial partition, or each partition block has the form ${\mathcal {X}}_j|_{i}=\{A_{i,j},{\bar{A}}_{i,j}\}$ for some $A_{i,j}\in {[n_i]\atopwithdelims ()t_i}$.

Since each imprimitive subset of ${\mathcal {X}}$ can be seen as a block of a non-trivial partition of ${\mathcal {X}}$, we have ${\mathcal {A}}={\mathcal {X}}_j$ for some $j\in [L]$. From the analysis above, we know that ${\mathcal {A}}|_i=\{A_i\}$ or $\{A_i,\bar{A_i}\}$ for some $A_i\in {[n_i]\atopwithdelims ()t_i}$, or ${\mathcal {A}}|_i={[n_i]\atopwithdelims ()t_i}$. Therefore, set $T_1\subseteq [p]$ such that for all $i\in T_1$, $2t_i=n_i$ and ${\mathcal {A}}|_i=\{A_i,\bar{A_i}\}$ for some $A_i\in {[n_i]\atopwithdelims ()t_i}$; set $T_2\subseteq [p]$ such that for all $i\in T_1$, ${\mathcal {A}}|_i$ is a singleton, finally, we have

$$\begin{aligned} {\mathcal {A}}=\prod _{i\in T_1}{\{A_i,\bar{A_i}\}}\times \prod _{i\in T_2}{\{A_i\}}\times \prod _{i\in [p]\setminus (T_1\sqcup T_2)}{[n_i]\atopwithdelims ()t_i}. \end{aligned}$$

The proof for the imprimitive subsets of ${\mathcal {Y}}$ is the same as that of ${\mathcal {X}}$. Thus, the claim holds.

By Claim 4.2, we know that for the imprimitive subsets ${\mathcal {A}}$ and ${\mathcal {B}}$ above

$$\begin{aligned} |{\mathcal {A}}|=2^{|T_1|}\cdot \prod _{i\in [p]\setminus (T_1\sqcup T_2)}{n_i \atopwithdelims ()t_i}~\text {and}~|{\mathcal {B}}|=2^{|R_1|}\cdot \prod _{i\in [p]\setminus (R_1\sqcup R_2)}{n_i \atopwithdelims ()s_i}. \end{aligned}$$

Moreover, since

$$\begin{aligned} N({\mathcal {A}})&=\{B\in {\mathcal {Y}}:~A\cap B=\emptyset \text { for some }A\in {\mathcal {A}}\}\\&=\prod _{i\in T_1}{\left( {A_i\atopwithdelims ()s_i}\sqcup {{\bar{A}}_i\atopwithdelims ()s_i}\right) }\times \prod _{i\in T_2}{{[n_i]\setminus A_i}\atopwithdelims ()s_i}\times \prod _{i\in [p]\setminus (T_1\sqcup T_2)}{[n_i]\atopwithdelims ()s_i},\\ N({\mathcal {B}})&=\{A\in {\mathcal {X}}:~A\cap B=\emptyset \text { for some }B\in {\mathcal {B}}\}\\&=\prod _{i\in R_1}{\left( {B_i\atopwithdelims ()t_i}\sqcup {{\bar{B}}_i\atopwithdelims ()t_i}\right) }\times \prod _{i\in R_2}{{[n_i]\setminus B_i}\atopwithdelims ()t_i}\times \prod _{i\in [p]\setminus (R_1\sqcup R_2)}{[n_i]\atopwithdelims ()t_i},\\ \end{aligned}$$

we have

$$\begin{aligned} |N({\mathcal {A}})|=2^{|T_1|}\cdot \prod _{i\in T_1}{{\frac{n_i}{2}}\atopwithdelims ()s_i}\cdot \prod _{i\in T_2}{{n_i-t_i}\atopwithdelims ()s_i}\cdot \prod _{i\in [p]\setminus (T_1\sqcup T_2)}{n_i\atopwithdelims ()s_i},\\ |N({\mathcal {B}})|=2^{|R_1|}\cdot \prod _{i\in R_1}{{\frac{n_i}{2}}\atopwithdelims ()t_i}\cdot \prod _{i\in R_2}{{n_i-s_i}\atopwithdelims ()t_i}\cdot \prod _{i\in [p]\setminus (R_1\sqcup R_2)}{n_i\atopwithdelims ()t_i}. \end{aligned}$$

Now we can estimate quantities $|{\mathcal {A}}'|+|{\mathcal {Y}}|-|N({\mathcal {A}}')|$ and $|{\mathcal {B}}'|+|{\mathcal {X}}|-|N({\mathcal {B}}')|$.

Claim 4.3

With the assumptions in the theorem, for all imprimitive subsets ${\mathcal {A}}\subseteq {\mathcal {X}}$ and ${\mathcal {B}}\subseteq {\mathcal {Y}}$, $|{\mathcal {Y}}|-d({\mathcal {X}})+1> |{\mathcal {A}}|+|{\mathcal {Y}}|-|N({\mathcal {A}})|$, and $|{\mathcal {Y}}|-d({\mathcal {X}})+1> |{\mathcal {B}}|+|{\mathcal {X}}|-|N({\mathcal {B}})|$.

We prove the claim by estimating the difference directly. Denote

$$\begin{aligned}&~D_1=|N({\mathcal {A}})|-|{\mathcal {A}}|-d({\mathcal {X}})+1\text { and}\\ D_2&=|{\mathcal {Y}}|-|{\mathcal {X}}|+|N({\mathcal {B}})|-|{\mathcal {B}}| -d({\mathcal {X}})+1 \end{aligned}$$

to be the differences between $|{\mathcal {Y}}|-d({\mathcal {X}})+1$ and, respectively, $|{\mathcal {A}}|+|{\mathcal {Y}}|-|N({\mathcal {A}})|$ and $|{\mathcal {B}}|+|{\mathcal {X}}|-|N({\mathcal {B}})|$. Set $d_1=\frac{D_1}{|N({\mathcal {A}})|}$, $d_2=\frac{D_2}{|{\mathcal {X}}|}$. Then, we have $d_1=1-\beta _1-\beta _2+\theta$, $d_2=\delta +\eta _0\cdot (1-\eta _1-\eta _2)+\theta '$, where $\theta =|N({\mathcal {A}})|^{-1}$, $\delta =\frac{|{\mathcal {Y}}|-|{\mathcal {X}}|}{|{\mathcal {X}}|}$, $\eta _0=\frac{|N({\mathcal {B}})|}{|{\mathcal {X}}|}$, $\theta '=|{\mathcal {X}}|^{-1}$, $\beta _1=\frac{|{\mathcal {A}}|}{|N({\mathcal {A}})|}$, $\beta _2=\frac{d({\mathcal {X}})}{|N({\mathcal {A}})|}$, $\eta _1=\frac{|{\mathcal {B}}|}{|N({\mathcal {B}})|}$, and $\eta _2=\frac{d({\mathcal {X}})}{|N({\mathcal {B}})|}$.

Since ${n_i\atopwithdelims ()t_i}\cdot {{n_i-t_i}\atopwithdelims ()s_i}={n_i\atopwithdelims ()s_i}\cdot {{n_i-s_i}\atopwithdelims ()t_i}$ for each $i\in [p]$, we have $1/{{n_i-t_i}\atopwithdelims ()s_i}={n_i\atopwithdelims ()t_i}/({n_i\atopwithdelims ()s_i}\cdot {{n_i-s_i}\atopwithdelims ()t_i})$ for each $i\in [p]$. This yields that

$$\begin{aligned}&~~~~\beta _1=\prod _{i\in [p]}\frac{{n_i \atopwithdelims ()t_i}}{{n_i\atopwithdelims ()s_i}}\cdot \prod _{i\in T_1\sqcup T_2}\frac{1}{{{n_i-s_i}\atopwithdelims ()t_i}},~~\beta _2=\frac{1}{2^{|T_1|}}\cdot \prod _{i\in [p]\setminus (T_1\sqcup T_2)}\prod _{j=0}^{s_i-1}\left( 1-\frac{t_i}{n_i-j}\right) ,\\ \eta _1&=\prod _{i\in [p]}\frac{{n_i \atopwithdelims ()s_i}}{{n_i\atopwithdelims ()t_i}}\cdot \prod _{i\in R_1\sqcup R_2}\frac{1}{{{n_i-t_i}\atopwithdelims ()s_i}},~~\eta _2=\prod _{i\in [p]}\frac{{n_i \atopwithdelims ()s_i}}{{n_i\atopwithdelims ()t_i}}\cdot \frac{1}{2^{|R_1|}}\cdot \prod _{i\in [p]\setminus (R_1\sqcup R_2)}\prod _{j=0}^{t_i-1}\left( 1-\frac{s_i}{n_i-j}\right) . \end{aligned}$$

Moreover, by $s_i,t_i\ge 2$, we have

$$\begin{aligned} \prod _{j=0}^{s_i-1}\left( 1-\frac{t_i}{n_i-j}\right)&\le \left( 1-\frac{t_i}{n_i}\right) \left( 1-\frac{t_i}{n_i-1}\right) \nonumber \\&\le \left( 1-\frac{2}{n_i}\right) \left( 1-\frac{2}{n_i-1}\right) =1-\frac{4n_i-6}{n_i(n_i-1)}. \end{aligned}$$

(9)

By the assumptions, we know that $n_i\ge s_i+t_i+1\ge 5$, $\prod _{i\in [p]}\frac{{n_i \atopwithdelims ()t_i}}{{n_i\atopwithdelims ()s_i}}\le 1$ and ${{n_i-s_i}\atopwithdelims ()t_i}\ge {\lceil \frac{n_i}{2}\rceil \atopwithdelims ()t_i}\ge \frac{n_i}{2}$.

Note that for every $i\in T_1$, $n_i=2t_i$. Thus, by $n_i\ge s_i+t_i+1$, we have $n_i-s_i\ge t_i+1=\frac{n_i}{2}+1$ for every $i\in T_1$. This implies that ${{n_i-s_i}\atopwithdelims ()t_i}\ge t_i+1$ for every $i\in T_1$. Therefore, we have

$$\begin{aligned} \beta _1&\le \prod _{i\in T_1\sqcup T_2}\frac{1}{{{n_i-s_i}\atopwithdelims ()t_i}}\le \prod _{i\in T_1}\frac{1}{t_i+1}\cdot \prod _{i\in T_2}\frac{2}{n_i}\\&\le \max \limits _{i\in T_1\sqcup T_2}{\left\{ \left( \frac{2}{n_i+2}\right) ^{|T_1|}\cdot \left( \frac{2}{n_i}\right) ^{|T_2|}\right\} },\\ \beta _2&\le \frac{1}{2^{|T_1|}}\cdot \max \limits _{i\in [p]\setminus (T_1\sqcup T_2)}\left\{ \left( 1-\frac{4n_i-6}{n_i(n_i-1)}\right) ^{p-(|T_1|+|T_2|)}\right\} . \end{aligned}$$

and

$$\begin{aligned} \eta _1&\le (1+\delta )\cdot \prod _{i\in R_1\sqcup R_2}\frac{1}{{{n_i-t_i}\atopwithdelims ()s_i}}\le (1+\delta )\cdot \max \limits _{i\in R_1\sqcup R_2}{\left\{ \left( \frac{2}{n_i+2}\right) ^{|R_1|} \cdot \left( \frac{2}{n_i}\right) ^{|R_2|}\right\} },\\&~~~~~~~~~\eta _2\le (1+\delta )\cdot \frac{1}{2^{|R_1|}}\cdot \max \limits _{i\in [p]\setminus (R_1\sqcup R_2)}\left\{ \left( 1-\frac{4n_i-6}{n_i(n_i-1)} \right) ^{p-(|R_1|+|R_2|)}\right\} . \end{aligned}$$

Since $T_1\sqcup T_2\ne \emptyset$, $R_1\sqcup R_2\ne \emptyset$ and $T_2,R_2\ne [p]$, we can obtain

$$\begin{aligned} \left\{ \begin{array}{ll} \beta _1\le \max \limits _{i\in T_1}{\left\{ \frac{2}{n_i+2}\right\} }~\text {and}~\beta _2\le \frac{1}{2}, &{}~\text {if}~T_1\ne \emptyset ;\\ \beta _1\le \max \limits _{i\in T_2}{\left\{ \frac{2}{n_i}\right\} }~\text {and}~\beta _2\le \max \limits _{i\in [p]\setminus T_2}\left\{ 1-\frac{4n_i-6}{n_i(n_i-1)}\right\} , &{}~\text {otherwise};\end{array}\right. \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{array}{ll} \frac{\eta _1}{1+\delta }\le \max \limits _{i\in R_1}{\left\{ \frac{2}{n_i+2}\right\} }~\text {and}~\frac{\eta _2}{1+\delta }\le \frac{1}{2}, &{}~\text {if}~R_1\ne \emptyset ;\\ \frac{\eta _1}{1+\delta }\le \max \limits _{i\in R_2}{\left\{ \frac{2}{n_i}\right\} }~\text {and}~\frac{\eta _2}{1+\delta }\le \max \limits _{i\in [p]\setminus R_2}\left\{ 1-\frac{4n_i-6}{n_i(n_i-1)}\right\} , &{}~\text {otherwise}.\end{array}\right. \end{aligned}$$

Since $5\le n_i\le \frac{7}{4} n_j$ for all distinct $i,j\in [p]$, we have $\frac{2}{n_i+2}\le \frac{2}{7}$ and

$$\begin{aligned} \frac{2}{n_j}+1-\frac{4n_i-6}{n_i(n_i-1)}&\le \frac{7}{2n_i}+1-\frac{4n_i-6}{n_i(n_i-1)}\nonumber \\&\le 1-\frac{n_i-5}{2n_i(n_i-1)}\le 1 \end{aligned}$$

(10)

for any $i\ne j\in [p]$. This leads to

$$\begin{aligned} \beta _1+\beta _2\le \left\{ \begin{array}{ll} \max \limits _{i\in T_1}{\left\{ \frac{2}{n_i+2}\right\} }+\frac{1}{2} \le \frac{11}{14}, &{}~\text {if}~T_1\ne \emptyset ;\\ \max \limits _{i\in T_2}{\left\{ \frac{2}{n_i}\right\} }+\max \limits _{i\in [p]\setminus T_2}\left\{ 1-\frac{4n_i-6}{n_i(n_i-1)}\right\} < 1 &{}~\text {otherwise};\end{array}\right. \end{aligned}$$

and

$$\begin{aligned} \frac{\eta _1+\eta _2}{1+\delta }\le \left\{ \begin{array}{ll} \max \limits _{i\in R_1}{\left\{ \frac{2}{n_i+2}\right\} }+\frac{1}{2}\le \frac{11}{14}, &{}~\text {if}~R_1\ne \emptyset ;\\ \max \limits _{i\in R_2}{\left\{ \frac{2}{n_i}\right\} }+\max \limits _{i\in [p]\setminus R_2}\left\{ 1-\frac{4n_i-6}{n_i(n_i-1)}\right\} < 1 &{}~\text {otherwise}.\end{array}\right. \end{aligned}$$

Thus, we have both $\beta _1+\beta _2\le 1$ and $\frac{\eta _1+\eta _2}{1+\delta }\le 1$. Therefore,

$$\begin{aligned} ~~~~~~~~~~~~~~~~~~~~~~~~~d_1&=1-\beta _1-\beta _2+\theta> 1-\beta _1-\beta _2>0,\\ d_2&=\delta +\eta _0\cdot (1-\eta _1-\eta _2)+\theta ' \\ {}&=\delta \cdot \left( 1-\eta _0\cdot \frac{\eta _1+\eta _2}{1+\delta }\right) +\eta _0\cdot \left( 1-\frac{\eta _1+\eta _2}{1+\delta }\right) +\theta '>0. \end{aligned}$$

Thus, the claim holds.

For each pair of non-empty cross-intersecting families $({\mathcal {A}},{\mathcal {B}})\in 2^{{\mathcal {X}}}\times 2^{{\mathcal {Y}}}$, ${\mathcal {A}}\cup {\mathcal {B}}$ forms a non-trivial independent set of $G({\mathcal {X}},{\mathcal {Y}})$. Therefore, by Claim 4.3, the inequality (4) holds.

To complete the proof, assume that the equality in (4) holds, we need to characterize all the non-trivial fragments in ${\mathcal {F}}({\mathcal {X}})$. As a direct consequence of Claim 4.3, every fragment of $G({\mathcal {X}},{\mathcal {Y}})$ is primitive. Hence, by Theorem 2.9, when $\prod _{i\in [p]}{n_i \atopwithdelims ()t_i}<\prod _{i\in [p]}{n_i\atopwithdelims ()s_i}$, ${\mathcal {X}}$ has only 1-fragments.

When $\prod _{i\in [p]}{n_i \atopwithdelims ()t_i}=\prod _{i\in [p]}{n_i\atopwithdelims ()s_i}$, suppose there are non-trivial fragments in ${\mathcal {F}}({\mathcal {X}})$. W.l.o.g., assume that ${\mathcal {S}}$ is a minimal-sized non-trivial fragment in ${\mathcal {X}}$. By Theorem 2.9, ${\mathcal {S}}$ is semi-imprimitive. Since for any two different elements $A,B\in {\mathcal {X}}$, $|N(A)\cap N(B)|<\prod _{i\in [p]}{{n_i-t_i}\atopwithdelims ()s_i}-1$. Therefore, there are no 2-fragments in $\mathcal {F({\mathcal {X}})}$. By Proposition 4.1, ${\mathcal {S}}$ is balanced.

Now we are going to prove the non-existence of such ${\mathcal {S}}$ by analyzing its size and structure, which will yield that ${\mathcal {X}}$ also has only 1-fragments when $\prod _{i\in [p]}{n_i \atopwithdelims ()t_i}=\prod _{i\in [p]}{n_i\atopwithdelims ()s_i}$.

For each $A=\prod _{i\in [p]}A_i\in {\mathcal {S}}$, let $\Gamma _A=\prod _{i\in [p]}(S_{A_i}\times S_{{\bar{A}}_i})$, $\Gamma _{{\mathcal {S}}}=\{\sigma \in \Gamma :~\sigma ({\mathcal {S}})={\mathcal {S}}\}$ and $\Gamma _{A,{\mathcal {S}}}=\{\sigma \in \Gamma _A:~\sigma ({\mathcal {S}})={\mathcal {S}}\}$. We claim that there exists a subset $C\in {\mathcal {S}}$ such that $\Gamma _{C}\ne \Gamma _{C,{\mathcal {S}}}$. Otherwise, for any two different subsets $B,B'\in {\mathcal {S}}$, we have $\Gamma _{B}=\Gamma _{B,{\mathcal {S}}}$ and $\Gamma _{B'}= \Gamma _{B',{\mathcal {S}}}$. Since $\Gamma _{B,{\mathcal {S}}}$ and $\Gamma _{B',{\mathcal {S}}}$ are both subgroups of $\Gamma _{{\mathcal {S}}}$, we have $\langle \Gamma _{B},\Gamma _{B'}\rangle$ is a subgroup of $\Gamma _{{\mathcal {S}}}$. Let $T\subseteq [p]$ be the factors where $B'_i=B_i~(\text {or}~{\bar{B}}_i~\text {if}~2t_i=n_i)$, write

$$\begin{aligned} \Gamma _B=\prod _{i\in T}(S_{B_i}\times S_{{\bar{B}}_i})\times \prod _{i\in [p]\setminus T}(S_{B_i}\times S_{{\bar{B}}_i}), \end{aligned}$$

then we have,

$$\begin{aligned} \Gamma _{B'}=\prod _{i\in T}(S_{B_i}\times S_{{\bar{B}}_i})\times \prod _{i\in [p]\setminus T}(S_{B'_i}\times S_{{\bar{B}}'_i}). \end{aligned}$$

Since $\langle S_{B_i}\times S_{{\bar{B}}_i}, S_{B'_i}\times S_{{\bar{B}}_i'}\rangle =S_{n_i}$ for each $B'_i\ne B_i~(\text {and}~B'_i\ne {\bar{B}}_i~\text {if}~2t_i=n_i)$, we have

$$\begin{aligned} \langle \Gamma _{B},\Gamma _{B'}\rangle =\prod _{i\in T}(S_{B_i}\times S_{{\bar{B}}_i})\times \prod _{i\in [p]\setminus T}S_{n_i}. \end{aligned}$$

Therefore, for some fixed $B\in {\mathcal {S}}$, $\Gamma _{{\mathcal {S}}}$ contains $\prod _{i\in T'}(S_{B_i}\times S_{{\bar{B}}_i})\times \prod _{i\in [p]{\setminus } T'}S_{n_i}$ as a subgroup, where

$$\begin{aligned} T'=\{i|i\in [p],\text { such that }A_i=B_i~(\text {or }{\bar{B}}_i\text { if }2t_i=n_i)\text { for all }A\in {\mathcal {S}}\}. \end{aligned}$$

When $T'=\emptyset$, we have $\Gamma _{{\mathcal {S}}}=\prod _{i\in [p]}S_{n_i}$, thus ${\mathcal {S}}={\mathcal {X}}$, yielding a contradiction. When $T'\ne \emptyset$, if $|T'|=1$, w.l.o.g., taking $T'=\{1\}$, we have $(S_{B_1}\times S_{{\bar{B}}_1})\times \prod _{i\in [p]{\setminus } \{1\}}S_{n_i}\subseteq \Gamma _{{\mathcal {S}}}$. Therefore, since ${\mathcal {S}}\ne {\mathcal {X}}$, from the definition of $T'$ we have

$$\begin{aligned} {\mathcal {S}}=\{B_1\}\times \prod _{i\in [p]\setminus \{1\}}{[n_i]\atopwithdelims ()t_i}, \hbox { or } S=\{B_1,{\bar{B}}_1\}\times \prod _{i\in [p]\setminus \{1\}}{[n_i]\atopwithdelims ()t_i}\text { when}\ 2t_1=n_1. \end{aligned}$$

In both cases, $|{\mathcal {S}}|<\frac{\alpha ({\mathcal {X}},{\mathcal {Y}})}{2}$. If $|T'|\ge 2$, we have

$$\begin{aligned} {\mathcal {S}}\subseteq \{B_{i_0}\}\times \prod _{i\in [p]\setminus \{i_0\}}{[n_i]\atopwithdelims ()t_i}, \hbox { or } S\subseteq \{B_{i_0},{\bar{B}}_{i_0}\}\times \prod _{i\in [p]\setminus \{i_0\}}{[n_i]\atopwithdelims ()t_i}\text { when}\ 2t_{i_0}=n_{i_0}, \end{aligned}$$

for some $i_0\in T'$. Therefore, when $T'\ne \emptyset$, we always have $|{\mathcal {S}}|<\frac{\alpha ({\mathcal {X}},{\mathcal {Y}})}{2}$, which contradicts the fact that ${\mathcal {S}}$ is balanced. Hence, the existence of C is guaranteed.

By Proposition 4.1 we have that $[\Gamma _{C}:\Gamma _{C,{\mathcal {S}}}]$, the index of $\Gamma _{C,{\mathcal {S}}}$ in $\Gamma _{C}$, equals 2. Now let $\Gamma _{C,{\mathcal {S}}}[C_i]$ be the projection of $\Gamma _{C,{\mathcal {S}}}$ onto $S_{C_i}$ and let $\Gamma _{C,{\mathcal {S}}}|_{S_{C_{i}}\times S_{{\bar{C}}_{i}}}$ be the projection of $\Gamma _{C,{\mathcal {S}}}$ onto $S_{C_{i}}\times S_{{\bar{C}}_{i}}$. $\Gamma _{C,{\mathcal {S}}}[C_i]$ must be a subgroup of $S_{C_i}$ of index no greater than 2. Thus, $\Gamma _{C,{\mathcal {S}}}[C_i]=S_{C_i}$ or $A_{C_i}$. Since $\Gamma _C=\prod _{i\in [p]}(S_{C_i}\times S_{{\bar{C}}_i})$, we know that $\Gamma _{C,{\mathcal {S}}}=\prod _{i\in [p]{\setminus }\{j\}}(S_{C_i}\times S_{{\bar{C}}_i})\times (A_{C_{j}}\times S_{{\bar{C}}_{j}})$ or $\prod _{i\in [p]{\setminus }\{j\}}(S_{C_i}\times S_{{\bar{C}}_i})\times (S_{C_{j}}\times A_{{\bar{C}}_{j}}$), for some $j\in [p]$.

Since for all $i\in [p]$, $t_i=|B_i\cap C_i|+|B_i\cap {\bar{C}}_i|$ for each pair $B,C\in {\mathcal {S}}$. If $|B_i\cap C_i|>1$, let $s,t\in B_i\cap C_i$, then the transposition (s t) fixes both $C_i$ and $B_i$. Taking $i=j$, the semi-imprimitivity of ${\mathcal {S}}$ implies that $(s~t)\in \Gamma _{C,{\mathcal {S}}}|_{S_{C_{j}}\times S_{{\bar{C}}_{j}}}$. This yields $\Gamma _{C,{\mathcal {S}}}|_{S_{C_{j}}\times S_{{\bar{C}}_{j}}}=S_{C_{j}}\times A_{{\bar{C}}_{j}}$. From this process it follows that, for each $B\in {\mathcal {S}}$, there exists at most one of $|B_j\cap C_j|$ and $|B_j\cap {\bar{C}}_j|$ to be greater than 1. Note that if $B_j\subseteq {\bar{C}}_j$, then $S_{C_j}$ and $S_{B_j}$ fix both $C_j$ and $B_j$, i.e., $S_{C_j}\times S_{B_j}\subseteq \Gamma _{C,{\mathcal {S}}}|_{S_{C_{j}}\times S_{{\bar{C}}_{j}}}$. Since $\Gamma _{C,{\mathcal {S}}}|_{S_{C_{j}}\times S_{{\bar{C}}_{j}}}=A_{C_{j}}\times S_{{\bar{C}}_{j}}$ or $S_{C_{j}}\times A_{{\bar{C}}_{j}}$, and neither $A_{C_{j}}\times S_{{\bar{C}}_{j}}$ nor $S_{C_{j}}\times A_{{\bar{C}}_{j}}$ contains $S_{C_j}\times S_{B_j}$. Therefore, we obtain that $|B_j\cap C_j|=1$ for each $B\in {\mathcal {S}}{\setminus }\{C\}$, or $|B_j\cap C_j|=t_j-1$ for each $B\in {\mathcal {S}}{\setminus }\{C\}$.

We claim that for both cases, ${\mathcal {S}}$ cannot be balanced.

Suppose $|B_j\cap C_j|=1$ for each $B\in {\mathcal {S}}$. W.l.o.g., assume $B_j\cap C_j=\{1\}$ for some $B\in {\mathcal {S}}$. From the semi-imprimitivity of ${\mathcal {S}}$, we know that for all $\gamma \in \Gamma ,~\gamma ({\mathcal {S}})\cap {\mathcal {S}}=\emptyset , ~{\mathcal {S}}$ or $\{A\}$ for some $A\in {\mathcal {S}}$. Thus, $(\gamma ({\mathcal {S}})\cap {\mathcal {S}})|_j=\emptyset ,~{\mathcal {S}}|_j$ or $\{A_j\}$ for some $A_j\in {[n_j]\atopwithdelims ()t_j}$. If $t_j>2$, then $|B_j\cap {\bar{C}}_j|\ge 2$, so $\Gamma _{C,{\mathcal {S}}}|_{S_{C_{j}}\times S_{{\bar{C}}_{j}}}=A_{C_{j}}\times S_{{\bar{C}}_{j}}$. On the other hand, since $(1~s~t)\in A_{C_j}$, we can find distinct $s,t\in C_j$ such that $(1\,s~t)(B_j)=(B_j{\setminus }\{1\})\cup \{s\}\in {\mathcal {S}}|_j$. Then, $(1\,s)({\mathcal {S}}|_j)$ has more than one element of ${\mathcal {S}}|_j$, therefore $(1\,s)\in \Gamma _{C,{\mathcal {S}}}|_{S_{C_{j}}\times S_{{\bar{C}}_{j}}}$. This contradiction proves that $t_j=2$. Thus, we can conclude that ${\mathcal {S}}|_j={\mathcal {C}}=\{A_j\in {[n_j]\atopwithdelims ()2}:~1\in A_j\}$. Otherwise, w.l.o.g., assume $C_j=\{1,2\}$ and there exists another $B'\in {\mathcal {S}}$ such that $B'_j\cap C_j=\{2\}$. Since $\Gamma _{C,{\mathcal {S}}}|_{S_{C_{j}}\times S_{{\bar{C}}_{j}}}=A_{C_{j}}\times S_{{\bar{C}}_{j}}$ or $S_{C_{j}}\times A_{{\bar{C}}_{j}}$ and $A_{C_{j}}=\{(1)(2)\}$, thus we have ${\mathcal {C}}\subseteq {\mathcal {S}}|_j$ and $\mathcal {C'}=\{A_j\in {[n_j]\atopwithdelims ()2}:~2\in A_j\}\subseteq {\mathcal {S}}|_j$. Therefore, ${\mathcal {S}}|_j={\mathcal {C}}\cup \mathcal {C'}$. This yields $\Gamma _{C,{\mathcal {S}}}|_{S_{C_{j}}\times S_{{\bar{C}}_{j}}}=S_{C_{j}}\times S_{{\bar{C}}_{j}}$, a contradiction.

Now, suppose $|B_j\cap C_j|=t_j-1>1$ for each $B\in {\mathcal {S}}$. Similarly, we can prove that $n_j-t_j=2$, which contradicts $n_j\ge s_j+t_j+1$ and $2\le s_j$, $t_j\le \frac{n}{2}$. Therefore, for each $B\in {\mathcal {S}}$, $|B_j\cap C_j|=1$.

From the analysis above, w.l.o.g., we can conclude that $t_j=2$ and for each $B\in {\mathcal {S}}$, $B_j=\{1,b\}$ for some $b\in [n_j]$. Thus, for each $B\in {\mathcal {S}}$, we have $\Gamma _{B,{\mathcal {S}}}|_{S_{B_{j}}\times S_{{\bar{B}}_{j}}}=A_{B_{j}}\times S_{{\bar{B}}_{j}}$, and $\Gamma _{B,{\mathcal {S}}}=\prod _{i\in [p]{\setminus } \{j\}}(S_{B_i}\times S_{{\bar{B}}_i})\times (A_{B_{j}}\times S_{{\bar{B}}_{j}})$. Therefore, $\Gamma _{{\mathcal {S}}}$ contains

$$\begin{aligned} \langle \Gamma _{B,{\mathcal {S}}},\text { for all }B\in {\mathcal {S}}\rangle =\prod _{i\in T''}(S_{C_i}\times S_{{\bar{C}}_i})\times \prod _{i\in [p]\setminus (T''\cup \{j\})}S_{n_i}\times S_{[n_j]\setminus \{1\}} \end{aligned}$$

as a subgroup, where $T''=\{i|i\in [p],\text { such that }B_i=C_i~(\text {or }{\bar{C}}_i\text { if }2t_i=n_i)\text { for all }B\in {\mathcal {S}}\}$. Similarly, by arguing the structure of ${\mathcal {S}}$, if $T''\ne \emptyset$, we can prove that $|{\mathcal {S}}|<\frac{\alpha ({\mathcal {X}},{\mathcal {Y}})}{2}$. Thus, we have $T''=\emptyset$ and

$$\begin{aligned} \prod _{i\in [p]\setminus \{j\}}S_{n_i}\times S_{[n_j]\setminus \{1\}}\subseteq \Gamma _{{\mathcal {S}}}. \end{aligned}$$

Note that for each $B\in {\mathcal {S}}$, $B_j=\{1,b\}$ for some $b\in [n_j]$. Thus, ${\mathcal {S}}=\prod _{i\in [p]{\setminus }\{j\}}{[n_i]\atopwithdelims ()t_i}\times {\mathcal {C}}$.

Since ${\mathcal {S}}$ is assumed to be balanced, we have ${\mathcal {S}}=\frac{\alpha ({\mathcal {X}},{\mathcal {Y}})}{2}$. Note that $\prod _{i\in [p]}{n_i \atopwithdelims ()t_i}=\prod _{i\in [p]}{n_i\atopwithdelims ()s_i}$ and $|{\mathcal {S}}|=\prod _{i\in [p]{\setminus }\{j\}}{n_i\atopwithdelims ()t_i}\cdot (n_j-1)$. Thus, by the equality in (4), we have

$$\begin{aligned} 2\prod _{i\in [p]\setminus \{j\}}{n_i\atopwithdelims ()t_i}\cdot (n_j-1)&=\prod _{i\in [p]\setminus \{j\}}{n_i\atopwithdelims ()t_i}\cdot {n_j\atopwithdelims ()2}- \prod _{i\in [p]\setminus \{j\}}{{n_i-s_i}\atopwithdelims ()t_i}\cdot {{n_j-s_j}\atopwithdelims ()2}+1. \end{aligned}$$

(11)

Dividing both sides of (11) by $b_0=\prod _{i\in [p]\setminus \{j\}}{n_i\atopwithdelims ()t_i}$, we can obtain

$$\begin{aligned} 2(n_j-1)={n_j\atopwithdelims ()2}-a_0{{n_j-s_j}\atopwithdelims ()2}+\frac{1}{b_0}, \end{aligned}$$

where $a_0=\prod _{i\in [p]{\setminus } \{j\}}\frac{{{n_i-s_i}\atopwithdelims ()t_i}}{{n_i\atopwithdelims ()t_i}}$. This indicates that

$$\begin{aligned} 1=\frac{n_j}{4}-a_0\frac{(n_j-s_j)(n_j-s_j-1)}{4(n_j-1)}+\frac{1}{2b_0(n_j-1)}. \end{aligned}$$

(12)

Next, we claim that this is impossible under the assumption that $n_i\ge 3+s_i$ and $2\le s_j\le \frac{n_j}{2}$. Note that

$$\begin{aligned} a_0=\prod _{i\in [p]\setminus \{j\}}\frac{{{n_i-s_i}\atopwithdelims ()t_i}}{{n_i\atopwithdelims ()t_i}}=\prod _{i\in [p]\setminus \{j\}}\prod _{j=0}^{t_i-1}\left( 1-\frac{s_i}{n_i-j}\right) . \end{aligned}$$

Thus, by similar arguments as (9), we have $a_0\le 1-\frac{4n_i-6}{n_i(n_i-1)}$ for some $i\in [p]{\setminus }\{j\}$. Thus, by $b_0>0$, the RHS of (12) is larger than

$$\begin{aligned} \frac{n_j}{4}-\left( 1-\frac{4n_i-6}{n_i(n_i-1)} \right) \cdot \frac{(n_j-s_j)(n_j-s_j-1)}{4(n_j-1)}. \end{aligned}$$

(13)

Since $(n_j-s_j)(n_j-s_j-1)=s_j^2-(2n_j-1)s_j+n_j(n_j-1)$ is an decreasing function of $s_j$ when $2\le s_j\le \frac{n_j}{2}$, we have $(n_j-s_j)(n_j-s_j-1)\le (n_j-2)(n_j-3)$. Moreover, by $n_i\ge 5$, we also have $\frac{4n_i-6}{n_i(n_i-1)}\ge \frac{3}{n_i}$. Therefore,

$$\begin{aligned} (13)&\ge \frac{n_j}{4}-\left( 1-\frac{3}{n_i}\right) \frac{(n_j-2)(n_j-3)}{4(n_j-1)}\\&= \frac{n_j}{4}-\frac{(n_j-2)(n_j-3)}{4(n_j-1)} +\frac{3}{n_i}\cdot \frac{(n_j-2)(n_j-3)}{4(n_j-1)}\\&= \frac{4n_j-6}{4(n_j-1)}+\frac{3}{n_i}\cdot \frac{(n_j-2)(n_j-3)}{4(n_j-1)}. \end{aligned}$$

Note that $n_i\le \frac{7}{4}n_j$. Thus, we have

$$\begin{aligned} \frac{3}{n_i}(n_j-2)(n_j-3)&\ge \frac{12}{7n_j}(n_j-2)(n_j-3)\\&= \frac{12}{7}\left( 1-\frac{2}{n_j}\right) (n_j-3)\ge \frac{72}{35}, \end{aligned}$$

where the last inequality follows from $n_j\ge 5$. This implies that $\frac{3}{n_i}\cdot \frac{(n_j-2)(n_j-3)}{4(n_j-1)}>\frac{2}{4(n_j-1)}$, which leads to $(13)>1$. Thus, the RHS of (12) is strictly larger than 1, which yields a contradiction. Therefore, there is no such $n_j$ satisfying (12) and (11), which implies that S is not balanced.

This completes the proof. $\square$

5 Concluding remarks

In this paper, we investigate two multi-part generalizations of the cross-intersecting theorems. Our main contribution is determining the maximal size and the corresponding structures of the families for both trivially and non-trivially (with the non-empty restriction) cross-intersecting cases.

The method we used for the proof was originally introduced by Wang and Zhang in [18], which was further generalized to the bipartite case in [19]. This method can deal with set systems, finite vector spaces and permutations uniformly. It is natural to ask whether we can extend the single-part cross-intersecting theorems for finite vector spaces and permutations to the multi-part case. It is possible for permutations when considering the case without the non-empty restriction, and we believe it is also possible for finite vector spaces. But when it comes to the case where the families are non-empty, as far as we know, there is still no result for finite vector spaces and permutations.

For single-part families ${\mathcal {A}}$ and ${\mathcal {B}}$, it is natural to define cross-t-intersecting as $|A\cap B|\ge t$ for each pair of $A\in {\mathcal {A}}$ and $B\in {\mathcal {B}}$. But for multi-part families, when defining cross-t-intersecting between two families, the simple extension of the definition for single-part case can be confusing. Therefore, a reasonable definition and related problems for multi-part cross-t-intersecting families are also worth considering.