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Algebraic approach to stability results for Erdős-Ko-Rado theorem

Gennian Ge School of Mathematical Sciences, Capital Normal University, Beijing, China. Emails: [email protected], [email protected]. Gennian Ge was supported by the National Key Research and Development Program of China under Grant 2020YFA0712100, the National Natural Science Foundation of China under Grant 12231014, and Beijing Scholars Program.    Zixiang Xu Extremal Combinatorics and Probability Group (ECOPRO), Institute for Basic Science (IBS), Daejeon, South Korea. Email: [email protected]. Supported by IBS-R029-C4.    Xiaochen Zhao11footnotemark: 1
Abstract

Celebrated results often unfold like episodes in a long-running series. In the field of extremal set thoery, Erdős, Ko, and Rado in 1961 established that any kk-uniform intersecting family on [n][n] has a maximum size of (n1k1)\binom{n-1}{k-1}, with the unique extremal structure being a star. In 1967, Hilton and Milner followed up with a pivotal result, showing that if such a family is not a star, its size is at most (n1k1)(nk1k1)+1\binom{n-1}{k-1}-\binom{n-k-1}{k-1}+1, and they identified the corresponding extremal structures. In recent years, Han and Kohayakawa, Kostochka and Mubayi, and Huang and Peng have provided the second and third levels of stability results in this line of research.

In this paper, we provide a unified approach to proving the stability result for the Erdős-Ko-Rado theorem at any level. Our framework primarily relies on a robust linear algebra method, which leverages appropriate non-shadows to effectively handle the structural complexities of these intersecting families.

1 Introduction

1.1 Overview

We say that a set system 2[n]\mathcal{F}\subseteq 2^{[n]} is intersecting if F1F2F_{1}\cap F_{2}\neq\emptyset for every pair of sets F1,F2F_{1},F_{2}\in\mathcal{F}. A fundamental result in extremal set theory, due to Erdős, Ko, and Rado [7], determines the maximum size of a kk-uniform intersecting family \mathcal{F}.

Theorem 1.1 (Erdős-Ko-Rado [7]).

Let n,kn,k be positive integers with n2kn\geq 2k. If ([n]k)\mathcal{F}\subseteq\binom{[n]}{k} is an intersecting family, then we have

||(n1k1).|\mathcal{F}|\leq\binom{n-1}{k-1}.

The equality holds if and only if ={F([n]k):pF}\mathcal{F}=\{F\in\binom{[n]}{k}:p\in F\} for some p[n]p\in[n].

Note that the only extremal family in the Erdős-Ko-Rado theorem is the star, also known as a trivially intersecting family. Hilton and Milner [21] established a stability result, showing that the size of non-trivially intersecting families is significantly smaller than the maximum given by the Erdős-Ko-Rado theorem. Before presenting a series of stability results, we first introduce some necessary notations. For a set system 2[n]\mathcal{F}\subseteq 2^{[n]}, we define the maximum degree of \mathcal{F} as

dmax():=maxi[n]|{F:iF}|.d_{\textup{max}}(\mathcal{F}):=\max\limits_{i\in[n]}|\{F\in\mathcal{F}:i\in F\}|.

We use [i,j][i,j] to denote the set {i,i+1,,j}\{i,i+1,\ldots,j\}. Let k3k\geq 3 be a positive integer, we will use ([n]k)\binom{[n]}{k} to denote the family of all subsets of [n][n] with size kk. For i[3,k+1]i\in[3,k+1], we define

i:={F([n]k):1F,F[2,i]}{F([n]k):1F,[2,i]F}.\mathcal{M}_{i}:=\bigg{\{}F\in\binom{[n]}{k}:1\in F,F\cap[2,i]\neq\emptyset\bigg{\}}\cup\bigg{\{}F\in\binom{[n]}{k}:1\notin F,[2,i]\subseteq F\bigg{\}}.

It is not hard to check that i\mathcal{M}_{i} is intersecting and |i|=j=2i(njk2)+(niki+1)|\mathcal{M}_{i}|=\sum\limits_{j=2}^{i}\binom{n-j}{k-2}+\binom{n-i}{k-i+1}. In particular, we can see that |3|=(n2k2)+2(n3k2)=(n2k2)+(n3k2)+(n4k2)+(n4k3)=|4||\mathcal{M}_{3}|=\binom{n-2}{k-2}+2\binom{n-3}{k-2}=\binom{n-2}{k-2}+\binom{n-3}{k-2}+\binom{n-4}{k-2}+\binom{n-4}{k-3}=|\mathcal{M}_{4}|. Moreover, we can see dmax(i)=j=2i(njk2)d_{\textup{max}}(\mathcal{M}_{i})=\sum\limits_{j=2}^{i}\binom{n-j}{k-2}. For j[1,nk]j\in[1,n-k] we further define

k,j:={F([n]k):1F,F[2,k]}{F([n]k):1F,F[2,k]=,[k+1,k+j]F}\displaystyle\mathcal{M}_{k,j}:=\bigg{\{}F\in\binom{[n]}{k}:1\in F,F\cap[2,k]\neq\emptyset\bigg{\}}\cup\bigg{\{}F\in\binom{[n]}{k}:1\in F,F\cap[2,k]=\emptyset,[k+1,k+j]\subseteq F\bigg{\}}
{F([n]k):1F,[2,k]F,|F[k+1,k+j]|=1}.\displaystyle\cup\bigg{\{}F\in\binom{[n]}{k}:1\notin F,[2,k]\subseteq F,|F\cap[k+1,k+j]|=1\bigg{\}}.

In particular, one can easily check that k,nk=k,k,1=k+1\mathcal{M}_{k,n-k}=\mathcal{M}_{k},\mathcal{M}_{k,1}=\mathcal{M}_{k+1}.

For two distinct sets E1,E2([n]k)E_{1},E_{2}\in\binom{[n]}{k} with |E1E2|=k2\left|E_{1}\cap E_{2}\right|=k-2, and an element x0[n](E1E2)x_{0}\in[n]\setminus\left(E_{1}\cup E_{2}\right), we define

𝒦(E1,E2,x0):={G([n]k):x0G,GE1,GE2}{E1,E2},\mathcal{K}\left(E_{1},E_{2},x_{0}\right):=\left\{G\in\binom{[n]}{k}:x_{0}\in G,G\cap E_{1}\neq\emptyset,G\cap E_{2}\neq\emptyset\right\}\cup\left\{E_{1},E_{2}\right\},

and we write 𝒦2\mathcal{K}_{2} for any family isomorphic to 𝒦(E1,E2,x0)\mathcal{K}\left(E_{1},E_{2},x_{0}\right).

Theorem 1.2 (Hilton-Milner [21]).

Let n,kn,k be positive integers with n>2kn>2k. If ([n]k)\mathcal{F}\subseteq\binom{[n]}{k} is an intersecting family and FF=\bigcap_{F\in\mathcal{F}}F=\emptyset, then we have

||(n1k1)(nk1k1)+1.|\mathcal{F}|\leq\binom{n-1}{k-1}-\binom{n-k-1}{k-1}+1.

For k=3k=3, the equality holds if and only if \mathcal{F} is isomorphic to 3\mathcal{M}_{3} or 4\mathcal{M}_{4}; for k4k\geq 4, the equality holds if and only if \mathcal{F} is isomoprhic to k+1\mathcal{M}_{k+1}.

If the intersecting family ([n]k)\mathcal{F}\subseteq\binom{[n]}{k} is neither extremal Erdős-Ko-Rado family nor extremal Hilton-Milner family, what is the maximum size of \mathcal{F}? Han and Kohayakawa [20] resolved this problem. Here we use EKR\mathcal{F}_{\textup{EKR}} and HM\mathcal{F}_{\textup{HM}} to denote the corresponding extremal families in Theorems 1.1 and 1.2 respectively.

Theorem 1.3 (Han-Kohayakawa [20]).

Suppose that k3k\geq 3 and n>2kn>2k, let ([n]k)\mathcal{F}\subseteq\binom{[n]}{k} be an intersecting family. Assume that \mathcal{F} is neither a sub-family of EKR\mathcal{F}_{\textup{EKR}} nor HM\mathcal{F}_{\textup{HM}}, then we have

||(n1k1)(nk1k1)(nk2k2)+2.|\mathcal{F}|\leq\binom{n-1}{k-1}-\binom{n-k-1}{k-1}-\binom{n-k-2}{k-2}+2.

Moreover, when k=4k=4, the equality holds if and only if \mathcal{F} is isomorphic to 4,2,3\mathcal{M}_{4,2},\mathcal{M}_{3} or 4\mathcal{M}_{4}; when k5k\geq 5 or k=3k=3, the equality holds if and only if \mathcal{F} is isomorphic to k,2\mathcal{M}_{k,2}.

Han and Kohayakawa [20] further asked what the next maximum intersecting kk-uniform families on [n][n] are? Kostochka and Mubayi [25] answered this question when nn is sufficiently large. Recently, Huang and Peng [22] completely answered this question for any n2k+1n\geq 2k+1 as follows.

Theorem 1.4 (Huang-Peng [22]).

Let k4k\geq 4 and ([n]k)\mathcal{F}\subseteq\binom{[n]}{k} be an intersecting family which is neither a sub-family of EKR\mathcal{F}_{\textup{EKR}} nor HM\mathcal{F}_{\textup{HM}}. Furthermore, k,2\mathcal{F}\not\subseteq\mathcal{M}_{k,2}, in addition 3\mathcal{F}\not\subseteq\mathcal{M}_{3} and 4\mathcal{F}\not\subseteq\mathcal{M}_{4} if k=4k=4. Then the followings hold.

  1. (1)

    If 2k+1n3k32k+1\leq n\leq 3k-3, then ||(n1k1)2(nk1k1)+(nk3k1)+2|\mathcal{F}|\leq\binom{n-1}{k-1}-2\binom{n-k-1}{k-1}+\binom{n-k-3}{k-1}+2. Moreover, when k5k\geq 5, the equality holds if and only if \mathcal{F} is isomorphic to 𝒦2\mathcal{K}_{2}. When k=4k=4, the equality holds if and only if \mathcal{F} is isomorphic to 𝒦2\mathcal{K}_{2} or 4,3\mathcal{M}_{4,3}.

  2. (2)

    If n3k2n\geq 3k-2, then ||(n1k1)(nk1k1)(nk2k2)(nk3k3)+3|\mathcal{F}|\leq\binom{n-1}{k-1}-\binom{n-k-1}{k-1}-\binom{n-k-2}{k-2}-\binom{n-k-3}{k-3}+3. Moreover, when k=5k=5, the equality holds if and only if \mathcal{F} is isomorphic to 5,3\mathcal{M}_{5,3} or 5\mathcal{M}_{5}. For every other kk, the equality holds if and only if \mathcal{F} is isomorphic to k,3\mathcal{M}_{k,3}.

The theorems above can be viewed as the 1st, 2nd, and 3rd-level full stability results for the Erdős-Ko-Rado theorem, respectively. Currently, there are several different proofs for these stability results, particularly regarding the Hilton-Milner theorem [9, 11, 15, 23, 28, 29] and some interesting variants [12, 27]. However, as far as we know, no work has emerged that approaches proving these stability results from multilinear polynomial methods. We aim to fill this gap.

The main contribution of this paper is to provide an algebraic approach to proving the stability result at the tt-th level of the Erdős-Ko-Rado theorem for arbitrary positive integer tt. To more clearly demonstrate our method, we will present a slightly weaker version of the stability result under the following conditions. Specifically, we make two assumptions: first, for the stability result at the tt-th level, we assume kt+2k\geq t+2; second, we assume n>(5+5)k723.618kn>\frac{(5+\sqrt{5})k-7}{2}\approx 3.618k instead of n>2kn>2k.

Theorem 1.5 (tt-th level stability).

Let t4t\geq 4 be a positive integer. For any positive integer kt+2k\geq t+2, let n>(5+5)k72n>\frac{(5+\sqrt{5})k-7}{2} and ([n]k)\mathcal{F}\subseteq\binom{[n]}{k} be a non-trivial intersecting family. Suppose k,t01\mathcal{F}\not\subseteq\mathcal{M}_{k,t_{0}-1} for any 2t0t2\leq t_{0}\leq t, then

||(n1k1)j=1t(nkjkj)+t.|\mathcal{F}|\leq\binom{n-1}{k-1}-\sum\limits_{j=1}^{t}\binom{n-k-j}{k-j}+t.

When k=t+2k=t+2, the equality holds if and only if \mathcal{F} is isomorphic to t+2,t\mathcal{M}_{t+2,t} or t+2\mathcal{M}_{t+2}, and when k>t+2k>t+2, the equality holds if and only if \mathcal{F} is isomorphic to k,t\mathcal{M}_{k,t}.

Note that with our framework, combined with some simple and appropriate structural analysis, we can also obtain stability results in Theorems 1.2, 1.3 and 1.4 mentioned earlier, as well as higher-layer stability results. However, since the main purpose of this paper is to present a new framework and method, we will not fully expand on these full proofs. For interested readers, we will provide an alternative proof of Theorem 1.3 in the Appendix using our unified framework.

We remark that in Theorem 1.5, we provide the upper bound for |||\mathcal{F}| under the assumption that we prohibit all of the extremal structures up to the first t1t-1 levels. Moreover, we can characterize the corresponding extremal structures. After completing this draft, we are informed by Jian Wang that recent work by Kupavskii, and by Frankl and Wang [14, 16, 26] established a similar upper bound under the condition that the diversity γ()=||dmax()\gamma(\mathcal{F})=|\mathcal{F}|-d_{\textup{max}}(\mathcal{F}) is large. Although we do not use the concept of diversity here, it has proven valuable in the study of kk-uniform intersecting families. We recommend the interested readers to [10, 13, 14, 26, 27] and the references therein. Instead, we will introduce our framework and give the self-contained and elementary proofs in Section 3.2. When we finally try to analyze the extremal structures in Section 3.3 and Appendix, for convenience, we will use a celebrated result of Frankl [8], see Theorem 2.3.

2 Robust linear algebra methods

2.1 Some useful lemmas

The following triangular criterion is useful when we want to prove a sequence of polynomials to be linearly independent.

Proposition 2.1.

Let f1,f2,,fmf_{1},f_{2},\ldots,f_{m} be functions in a linear space. If 𝐯(1),𝐯(2),,𝐯(m)\bm{v}^{(1)},\bm{v}^{(2)},\ldots,\bm{v}^{(m)} are vectors such that fi(𝐯(i))0f_{i}(\bm{v}^{(i)})\neq 0 for 1im1\leq i\leq m and fi(𝐯(j))=0f_{i}(\bm{v}^{(j)})=0 for i>ji>j, then f1,f2,,fmf_{1},f_{2},\ldots,f_{m} are linearly independent.

We follow the notations on a proof of Erdős-Ko-Rado theorem via multilinear polynomials [18]. Suppose that :={F1,F2,,Fm}\mathcal{F}:=\{F_{1},F_{2},\ldots,F_{m}\}, where Fi[n]F_{i}\subseteq[n] for each 1im1\leq i\leq m. For a set P[n]P\subseteq[n] and a non-negative integer β\beta, we say the set FF satisfies the property (P,β)(P,\beta)-intersection if |FP|=β|F\cap P|=\beta. Now suppose for each FiF_{i}\in\mathcal{F}, we write a collection of ss intersection properties (allow repetition) as

Ri={(Pi1,βi1),(Pi2,βi2)),,(Pis,βis)}.R_{i}=\{(P_{i_{1}},\beta_{i_{1}}),(P_{i_{2}},\beta_{i_{2}})),\ldots,(P_{i_{s}},\beta_{i_{s}})\}.

In [18], the authors built the relation between the multilinear polynomials and the certain collections of intersection properties, here we introduce the following key lemma and the proof in details.

Lemma 2.2.

Let ={F1,F2,,Fm}2[n]\mathcal{F}=\{F_{1},F_{2},\ldots,F_{m}\}\subseteq 2^{[n]}. Suppose that for each FiF_{i}\in\mathcal{F}, one can find a set Xi[n]X_{i}\subseteq[n] and a collection of ss intersection properties RiR_{i} such that

  1. (1)

    XiX_{i} does not satisfy any of the conditions in RiR_{i};

  2. (2)

    XiX_{i} satisfies at least one condition in RjR_{j} for all j>ij>i.

Then we have ||h=0s(nh)|\mathcal{F}|\leq\sum\limits_{h=0}^{s}\binom{n}{h}.

Proof of Lemma 2.2.

For 𝒙=(x1,x2,,xn)\bm{x}=(x_{1},x_{2},\ldots,x_{n}), we define a sequence of nn-variate real polynomials fi(𝒙)f_{i}(\bm{x}) for 1im1\leq i\leq m as

fi(𝒙)=a=1s(bPiaxbβia).f_{i}(\bm{x})=\prod\limits_{a=1}^{s}\bigg{(}\sum\limits_{b\in P_{i_{a}}}x_{b}-\beta_{i_{a}}\bigg{)}.

For a subset A[n]A\subseteq[n], we will use 𝒂=(a1,a2,,an)\bm{a}=(a_{1},a_{2},\ldots,a_{n}) to represent its characteristic vector, that is, for each 1in1\leq i\leq n, ai=1a_{i}=1 if iAi\in A and ai=0a_{i}=0 otherwise. Observe that the scale product 𝒂𝒃=|AB|\bm{a}\cdot\bm{b}=|A\cap B| for any A,B[n]A,B\subseteq[n]. Thus, we can write fi(𝒙)f_{i}(\bm{x}) as

fi(𝒙)=a=1s(|XPia|βia).f_{i}(\bm{x})=\prod\limits_{a=1}^{s}(|X\cap P_{i_{a}}|-\beta_{i_{a}}).

Let 𝒙(i)\bm{x}^{(i)} be the characteristic vector of set XiX_{i}. By condition (1), we can see fi(𝒙(i))0f_{i}(\bm{x}^{(i)})\neq 0. By condition (2), fj(𝒙(i))=0f_{j}(\bm{x}^{(i)})=0 for all j>ij>i. Then by Proposition 2.1, {fi}i=1m\{f_{i}\}_{i=1}^{m} are linearly independent. Moreover, as each polynomial contains nn variables and the degree of each polynomial is at most ss, thus we have mh=0s(nh)m\leq\sum\limits_{h=0}^{s}\binom{n}{h}, as claimed. ∎

When we analyze the structural properties of set systems in proofs of Theorem 1.3 and Theorem 1.5, we will take advantage of the following result of Frankl [8].

Theorem 2.3.

Suppose that n>2kn>2k, 3ik+13\leq i\leq k+1, ([n]k)\mathcal{F}\subseteq\binom{[n]}{k} is an intersecting family with dmax()dmax(i)d_{\textup{max}}(\mathcal{F})\leq d_{\textup{max}}(\mathcal{M}_{i}), then |||i||\mathcal{F}|\leq|\mathcal{M}_{i}|. Moreover if ||=|i||\mathcal{F}|=|\mathcal{M}_{i}|, then either \mathcal{F} is isomorphic to i\mathcal{M}_{i}, or when i=4i=4, \mathcal{F} is isomorphic to 3\mathcal{M}_{3}.

2.2 Overview of the robust linear algebra methods

Our main approach is based on a robust linear algebra method developed in recent work of Gao, Liu and the second author [19]. Here we first show an example and explain how the standard linear algebra method works and then summarize some interesting tricks. Furthermore, we will briefly introduce the main ideas on the robust linear algebra method. For more on the linear algebra methods in combinatorics, we recommend the interested readers to the great textbook [3] and a recent note [32].

Suppose that 2[n]\mathcal{F}\subseteq 2^{[n]} is an LL-intersecting family for some subset L[n]L\subseteq[n] with |L|=s|L|=s, Frankl and Wilson [17] showed that ||i=0s(ni)|\mathcal{F}|\leq\sum\limits_{i=0}^{s}\binom{n}{i}. We sketch the shorter proof of the above result by Babai [2] as follows. Let ={F1,F2,,Fm}\mathcal{F}=\{F_{1},F_{2},\ldots,F_{m}\}, indexed so that |F1||Fm||F_{1}|\leq\cdots\leq|F_{m}| and L={1,,s}L=\{\ell_{1},\ldots,\ell_{s}\}. For each ii let vi\vec{v}_{i} be the incidence vector of FiF_{i}. Define polynomials f1,f2,,fmf_{1},f_{2},\ldots,f_{m} by fi(x)=k<|Fi|(xvik)f_{i}(\vec{x})=\prod\limits_{\ell_{k}<|F_{i}|}(\vec{x}\cdot\vec{v}_{i}-\ell_{k}). Then one can easily prove that {fi}i=1m\{f_{i}\}_{i=1}^{m} are linearly independent via Proposition 2.1. Thus |||\mathcal{F}| can be upper bounded by the number of all possible polynomials. The first trick one can use is multilinear reduction, that is, we can replace each xitx_{i}^{t} term by xix_{i} for any 1in1\leq i\leq n and t2t\geq 2 because xi{0,1}x_{i}\in\{0,1\}. Using this trick, one can efficiently give the upper bound i=0s(ni)\sum\limits_{i=0}^{s}\binom{n}{i} for |||\mathcal{F}| as the total degree of each monomial is at most ss.

The second trick is that, one can further add more associated polynomials. For example, we can add i=0s1(n1i)\sum_{i=0}^{s-1}\binom{n-1}{i} many extra polynomials in the following way. Label the sets in ([n1]s)\binom{[n-1]}{\leq s} with label BiB_{i} for 1iq=i=0s1(n1i)1\leq i\leq q=\sum_{i=0}^{s-1}\binom{n-1}{i} such that |Bi||Bj||B_{i}|\leq|B_{j}| when i<ji<j. Let wi\vec{w}_{i} be the characteristic vector of BiB_{i}, and let hBi(x)=jBixjh_{B_{i}}(\vec{x})=\prod_{j\in B_{i}}x_{j} for i>1i>1. Then define a multilinear polynomial gBig_{B_{i}} in nn variables as follows. gB1=xn1g_{B_{1}}=x_{n}-1 and gBi=(xn1)hBi(x)g_{B_{i}}=(x_{n}-1)h_{B_{i}}(\vec{x}) for i>1i>1. In [33], Snevily proved that {fi}i=1m\{f_{i}\}_{i=1}^{m} and {gBj}j=1q\{g_{B_{j}}\}_{j=1}^{q} are linearly independent. Thus the upper bound for |||\mathcal{F}| can be improved to i=0s(ni)i=0s1(n1i)=i=0s(n1i)\sum_{i=0}^{s}\binom{n}{i}-\sum_{i=0}^{s-1}\binom{n-1}{i}=\sum_{i=0}^{s}\binom{n-1}{i}. This trick has been applied to several problems, for example, see [1, 4, 5, 6, 24, 30, 31]. In a word, this trick consists of two parts, the first step is to choose some appropriate extra polynomials, one can see that in many previous works involved with this trick, the extra polynomials usually are clear and natural, which associate some explicit family of subsets. For instance in the above famous case, the extra polynomials associate to ([n1]s)\binom{[n-1]}{\leq s} exactly. The second part is that we need to show the union of the original polynomials and the extra polynomials are linearly independent. Usually the second part is much more complicated and difficult. Once we prove the linear independence, we can immediately see the improvement on the bounds for size of the family.

In the proofs of stability results of Kleitman’s isodiametric inequality, the authors in [19] mainly focus on another direction about the second trick. More precisely, one can first carefully choose a family of subsets which satisfies some appropriate properties and then associate each subset of the chosen family with the extra polynomial one by one. Then it will be easier to prove the linear independence. At the cost, one cannot know all of information about the chosen family, but sometimes one can ignore it, e.g., see the proof in [19, Theorem 1.10]. While in some cases, then the main task in the robust linear algebra method is to dig out the structural properties of the family one chooses, to achieve this, usually one can apply some structural analysis, e.g., see the proof in [19, Theorem 1.8].

There are several advantages in this robust linear algebra method. The first is that the linear independence usually will be easier to show. The second is that, usually the previous linear algebra methods just provide the bound of the size, while the new method can be used to prove some stability results, that means, we can not only capture the size of the family, but also obtain some structural information. We believe that this method has the potential to be applied to a wider range of problems.

3 A unified framework and proof of Theorem 1.5

3.1 High level overview of our framework

Although our proofs are relatively simple, it might be helpful to briefly outline the main ideas.

  1. 1.

    Following the ideas in [18], we partition the family \mathcal{F} into two sub-families, 0\mathcal{F}_{0} and 1\mathcal{F}_{1}, where the sets in 1\mathcal{F}_{1} contain the element that attains the maximum degree in \mathcal{F}. We then introduce two auxiliary families, \mathcal{H} and 𝒢\mathcal{G}. At this stage, we can associate these families with certain intersection conditions, which leads to an algebraic proof of [18].

  2. 2.

    The key ingredient involves finding one more auxiliary family 𝒮\mathcal{S}, which is a sub-family of the non-shadows of 1\mathcal{F}_{1}. We carefully associate 𝒮\mathcal{S} with appropriate intersection conditions. The first crucial point occurs at 3.1, after proving that the families 1\mathcal{F}_{1}, 𝒢\mathcal{G}, \mathcal{H}, and 𝒮\mathcal{S} satisfy certain properties analogous to linear independence in normal linear algebra method, we obtain an important quantitative relation: |1|(n1k1)|𝒮||\mathcal{F}_{1}|\leq\binom{n-1}{k-1}-|\mathcal{S}|, leading to the inequality ||(n1k1)|𝒮|+|0||\mathcal{F}|\leq\binom{n-1}{k-1}-|\mathcal{S}|+|\mathcal{F}_{0}|.

  3. 3.

    To show the upper bound on |||\mathcal{F}|, it is sufficient to analyze the lower bound on |𝒮||0||\mathcal{S}|-|\mathcal{F}_{0}|. We establish a general lower bound on |𝒮||\mathcal{S}| in 3.3, and later, through a stability argument, refine this bound in 3.4, which is essential for determining the extremal structures.

  4. 4.

    The final task is to carefully compare the specific sizes of several numbers. To do this, we establish inequalities between variables, explore monotonicity, and analyze the structure at extreme values in 3.5 and 3.6. Additionally, we will use Theorem 2.3 of Frankl. These steps are relatively straightforward and follow from natural analytical considerations.

3.2 Our framework

For positive integers n2k+1n\geq 2k+1, without loss of generality, we can assume that \mathcal{F} is a maximal non-trivial intersecting family of ([n]k)\binom{[n]}{k}, that is, if we add any new member of ([n]k)\binom{[n]}{k} to \mathcal{F}, then \mathcal{F} is not non-trivial intersecting. For a family ([n]k)\mathcal{F}\in\binom{[n]}{k}, we denote the shadow of \mathcal{F} as k1:={T([n]k1):TFfor some F}\partial_{k-1}{\mathcal{F}}:=\{T\in\binom{[n]}{k-1}:T\subseteq F\ \text{for\ some\ }F\in\mathcal{F}\}. Let p[n]p\in[n] be the element which attains the maximum degree in \mathcal{F}. Consider the following families:

  • 0:={F:pF}\mathcal{F}_{0}:=\{F\in\mathcal{F}:p\notin F\};

  • :={H[n]:pH,0|H|k2}\mathcal{H}:=\{H\subseteq[n]:p\notin H,0\leq|H|\leq k-2\};

  • 1:={F:pF}\mathcal{F}_{1}:=\{F\in\mathcal{F}:p\in F\};

  • 𝒢:={G[n]:pG,1|G|k1}\mathcal{G}:=\{G\subseteq[n]:p\in G,1\leq|G|\leq k-1\};

  • 𝒮:={S([n]k1)k11:pS,F0 such that SF=}\mathcal{S}:=\{S\subseteq\binom{[n]}{k-1}\setminus\partial_{k-1}{\mathcal{F}_{1}}:p\notin S,\exists F\in\mathcal{F}_{0}\textup{ such\ that\ }S\cap F=\emptyset\}.

Let 𝒜:=1𝒢𝒮={A1,,Am}\mathcal{A}:=\mathcal{F}_{1}\sqcup\mathcal{H}\sqcup\mathcal{G}\sqcup\mathcal{S}=\{A_{1},\ldots,A_{m}\}. We define an ordering \prec on the sets, and for two families 𝒜,\mathcal{A},\mathcal{B}, denote 𝒜\mathcal{A}\prec\mathcal{B} if and only if for any A𝒜A\in\mathcal{A} and BB\in\mathcal{B}, we have ABA\prec B. We first arrange the sets in a linear order as follows: 1𝒢𝒮\mathcal{H}\prec\mathcal{F}_{1}\prec\mathcal{G}\prec\mathcal{S}. We put the members of 1\mathcal{F}_{1} and 𝒮\mathcal{S} in arbitrary order and the members of \mathcal{H} and 𝒢\mathcal{G} in order increasing by size, for example, HiHjH_{i}\prec H_{j} if |Hi||Hj||H_{i}|\leq|H_{j}|. To apply Lemma 2.2, we need to associate each member A𝒜A\in\mathcal{A} with a set XX and at most k1k-1 many intersection conditions as follows.

  • For HH\in\mathcal{H}, we can set X:=HX:=H with intersection conditions ({h},0)(\{h\},0) for each hHh\in H and ([n],nk1)([n],n-k-1).

  • For F1F\in\mathcal{F}_{1}, we can set X:=F{p}X:=F\setminus\{p\} with intersection conditions (F{p},β)(F\setminus\{p\},\beta) for 0βk20\leq\beta\leq k-2.

  • For G𝒢G\in\mathcal{G}, we can set X:=GX:=G with intersection conditions ({g},0)(\{g\},0) for each gGg\in G.

  • For S𝒮S\in\mathcal{S}, we can set X:=SX:=S with intersection conditions ({s},0)(\{s\},0) for each sSs\in S.

We claim that the system (Ai,Xi,Ri)(A_{i},X_{i},R_{i}), in which Ai𝒜A_{i}\in\mathcal{A}, XiX_{i} and the intersections conditions RiR_{i} are defined as above, satisfies the conditions in Lemma 2.2 with s=k1s=k-1.

Claim 3.1.

For each Ai𝒜A_{i}\in\mathcal{A}, XiX_{i} does not satisfy any of the conditions in RiR_{i}, and XiX_{i} satisfies at least one condition in RjR_{j} for all j>ij>i. In particular, |𝒜|=1k1(n)|\mathcal{A}|\leq\sum\limits_{\ell=1}^{k-1}\binom{n}{\ell}.

Proof of claim.

Recall that 1𝒢𝒮\mathcal{H}\prec\mathcal{F}_{1}\prec\mathcal{G}\prec\mathcal{S}, and we put the elements of 1\mathcal{F}_{1} and 𝒮\mathcal{S} in arbitrary order and the members of \mathcal{H} and 𝒢\mathcal{G} in order increasing by size. For distinct sets AiAj𝒜A_{i}\preceq A_{j}\in\mathcal{A}, we write AiAjA_{i}\rightarrow A_{j} if XiX_{i} satisfies at least one condition in RjR_{j}, and in particular, AiAiA_{i}\rightarrow A_{i} if XiX_{i} does not satisfy any of the conditions in RiR_{i}. For sub-families 𝒳𝒴\mathcal{X}\preceq\mathcal{Y} of 𝒜\mathcal{A}, we write 𝒳𝒴\mathcal{X}\rightarrow\mathcal{Y} if for every X𝒳X\in\mathcal{X} and Y𝒴Y\in\mathcal{Y} with XYX\preceq Y, we have XYX\rightarrow Y. Then it suffices to check the following 1010 situations:

  1. 1.

    \mathcal{H}\rightarrow\mathcal{H}: For each HiH_{i}\in\mathcal{H}, obviously it does not satisfy ({h},0)(\{h\},0) for each hHih\in H_{i} and ([n],nk1)([n],n-k-1) since |H|k2|H|\leq k-2, moreover, for any other HjH_{j} with |Hj||Hi||H_{j}|\geq|H_{i}|, there exists some hHjh\in H_{j} such that hHih\notin H_{i}. Then \mathcal{H}\rightarrow\mathcal{H} is checked.

  2. 2.

    1\mathcal{H}\rightarrow\mathcal{F}_{1}: For any HH\in\mathcal{H} and any F1F\in\mathcal{F}_{1}, we have 0|HF|k20\leq|H\cap F|\leq k-2, then 1\mathcal{H}\rightarrow\mathcal{F}_{1} is checked.

  3. 3.

    𝒢\mathcal{H}\rightarrow\mathcal{G}: Since for any HH\in\mathcal{H} and G𝒢G\in\mathcal{G}, we have pHp\notin H and pGp\in G, then 𝒢\mathcal{H}\rightarrow\mathcal{G} is checked.

  4. 4.

    𝒮\mathcal{H}\rightarrow\mathcal{S}: Since for any HH\in\mathcal{H} and S𝒮S\in\mathcal{S}, |S|>|H||S|>|H|, there exists some sSs\in S such that sHs\notin H, then 𝒮\mathcal{H}\rightarrow\mathcal{S} is checked.

  5. 5.

    11\mathcal{F}_{1}\rightarrow\mathcal{F}_{1}: Since 1\mathcal{F}_{1} is kk-uniform, for any distinct FiFj1F_{i}\prec F_{j}\in\mathcal{F}_{1}, we have |FiFj|k1|F_{i}\cap F_{j}|\leq k-1 and pFiFjp\in F_{i}\cap F_{j}. Therefore there exists some xFj{p}x\in F_{j}\setminus\{p\} and xFi{p}x\notin F_{i}\setminus\{p\}, then 11\mathcal{F}_{1}\rightarrow\mathcal{F}_{1} is checked.

  6. 6.

    1𝒢\mathcal{F}_{1}\rightarrow\mathcal{G}: Since for any F1F\in\mathcal{F}_{1} and G𝒢G\in\mathcal{G}, we can see pGp\in G and pF{p}p\notin F\setminus\{p\}, then 1𝒢\mathcal{F}_{1}\rightarrow\mathcal{G} is checked.

  7. 7.

    1𝒮\mathcal{F}_{1}\rightarrow\mathcal{S}: Since for any F1F\in\mathcal{F}_{1} and S𝒮S\in\mathcal{S}, S([n]k1)k11S\subseteq\binom{[n]}{k-1}\setminus\partial_{k-1}\mathcal{F}_{1}, then there exists some xSx\in S such that xF{p}x\notin F\setminus\{p\}. Then 1𝒮\mathcal{F}_{1}\rightarrow\mathcal{S} is checked.

  8. 8.

    𝒢𝒢\mathcal{G}\rightarrow\mathcal{G}: For any distinct GiGj𝒢G_{i}\prec G_{j}\in\mathcal{G} with |Gi||Gj||G_{i}|\leq|G_{j}|, obviously there exists some gGjg\in G_{j} such that GGiG\notin G_{i}, then 𝒢𝒢\mathcal{G}\rightarrow\mathcal{G} is checked.

  9. 9.

    𝒢𝒮\mathcal{G}\rightarrow\mathcal{S}: Since for any G𝒢G\in\mathcal{G} and S𝒮S\in\mathcal{S}, we have pGp\in G and pSp\notin S. Moreover, notice that |G|k1|G|\leq k-1 and |S|=k1|S|=k-1, therefore there exists some element sSs\in S such that sGs\notin G. Then 𝒢𝒮\mathcal{G}\rightarrow\mathcal{S} is checked.

  10. 10.

    𝒮𝒮\mathcal{S}\rightarrow\mathcal{S}: Since for any distinct SiSj𝒮([n]k1)S_{i}\prec S_{j}\in\mathcal{S}\subseteq\binom{[n]}{k-1} with SiSjS_{i}\prec S_{j}, there exists some sSjs\in S_{j} such that sSis\notin S_{i}, then 𝒮𝒮\mathcal{S}\rightarrow\mathcal{S} is checked.

This finishes the proof. ∎

Then by Lemma 2.2 and 3.1, we have ||+|1|+|𝒢|+|𝒮|i=0k1(ni)|\mathcal{H}|+|\mathcal{F}_{1}|+|\mathcal{G}|+|\mathcal{S}|\leq\sum\limits_{i=0}^{k-1}\binom{n}{i}. Also note that |𝒢|=||=i=0k2(n1i)|\mathcal{G}|=|\mathcal{H}|=\sum\limits_{i=0}^{k-2}\binom{n-1}{i}, which implies

dmax()=|1|i=0k1(ni)2i=0k2(n1i)|𝒮|=(n1k1)|𝒮|.d_{\textup{max}}(\mathcal{F})=|\mathcal{F}_{1}|\leq\sum\limits_{i=0}^{k-1}\binom{n}{i}-2\sum\limits_{i=0}^{k-2}\binom{n-1}{i}-|\mathcal{S}|=\binom{n-1}{k-1}-|\mathcal{S}|.

Moreover, we have

||=|0|+|1|(n1k1)(|𝒮||0|).|\mathcal{F}|=|\mathcal{F}_{0}|+|\mathcal{F}_{1}|\leq\binom{n-1}{k-1}-(|\mathcal{S}|-|\mathcal{F}_{0}|).

We can immediately derive the following results when |0||\mathcal{F}_{0}| is very small.

  1. 1.

    When |0|=0|\mathcal{F}_{0}|=0, suppose that there exists some S𝒮S\in\mathcal{S}, then S{p}S\cup\{p\}\notin\mathcal{F} and (S{p})F(S\cup\{p\})\cap F\neq\emptyset for any FF\in\mathcal{F}, which is a contradiction to that \mathcal{F} is a maximal non-trivial intersecting family. Therefore, when 0=\mathcal{F}_{0}=\emptyset, then 𝒮=\mathcal{S}=\emptyset, which gives ||(n1k1)|\mathcal{F}|\leq\binom{n-1}{k-1}.

  2. 2.

    When |0|=1|\mathcal{F}_{0}|=1, set 0={F0}\mathcal{F}_{0}=\{F_{0}\}. By definitions, for each S𝒮S\in\mathcal{S}, we can see SF0=S\cap F_{0}=\emptyset and pSp\notin S. Moreover, we claim that ([n]({p}F0)k1)𝒮\binom{[n]\setminus(\{p\}\cup F_{0})}{k-1}\subseteq\mathcal{S}. To see this, since \mathcal{F} is an intersecting family, then for any member Pk11P\in\partial_{k-1}\mathcal{F}_{1}, at least one of the events pPp\in P and PF0P\cap F_{0}\neq\emptyset occurs, which yields that k11([n](F0{p})k1)=\partial_{k-1}\mathcal{F}_{1}\cap\binom{[n]\setminus(F_{0}\cup\{p\})}{k-1}=\emptyset. Therefore, when |0|=1|\mathcal{F}_{0}|=1, then |𝒮|(nk1k1)|\mathcal{S}|\geq\binom{n-k-1}{k-1}, which also yields that ||(n1k1)(nk1k1)+1|\mathcal{F}|\leq\binom{n-1}{k-1}-\binom{n-k-1}{k-1}+1.

In general, Let x:=|0|1x:=|\mathcal{F}_{0}|\geq 1, and 0:={F1,F2,,Fx}\mathcal{F}_{0}:=\{F_{1},F_{2},\ldots,F_{x}\}, we have the following crucial claim by extending the above argument in the case of |0|=1|\mathcal{F}_{0}|=1.

Claim 3.2.

𝒮=i=1x([n](Fi{p})k1)\mathcal{S}=\bigcup\limits_{i=1}^{x}\binom{[n]\setminus(F_{i}\cup\{p\})}{k-1}.

Proof of claim.

Set 𝒯:=i=1x([n](Fi{p})k1)\mathcal{T}:=\bigcup\limits_{i=1}^{x}\binom{[n]\setminus(F_{i}\cup\{p\})}{k-1}, we first prove that 𝒯𝒮\mathcal{T}\subseteq\mathcal{S}. Note that for any T𝒯T\in\mathcal{T}, we can see pTp\notin T and by definitions there exists some Fi0F_{i}\in\mathcal{F}_{0} such that TFi=T\cap F_{i}=\emptyset. Then it suffices to show that TT cannot be a shadow of 1\mathcal{F}_{1}. Suppose that there is a T𝒯T\subseteq\mathcal{T} such that TFT\subseteq F for some F1F\in\mathcal{F}_{1}, then T=F{p}T=F\setminus\{p\}. Since \mathcal{F} is an intersecting family, we have (T{p})Fi(T\cup\{p\})\cap F_{i}\neq\emptyset for each i[x]i\in[x]. However, this is impossible, because pFip\notin F_{i}, and there exists some Fi0F_{i}\in\mathcal{F}_{0} such that TFi=T\cap F_{i}=\emptyset, a contradiction to the definition of 𝒯\mathcal{T}.

On the other hand, for any S𝒮S\in\mathcal{S}, there exists some Fi0F_{i}\in\mathcal{F}_{0} such that SFi=S\cap F_{i}=\emptyset. Then S([n](Fi{p})k1)S\in\binom{[n]\setminus(F_{i}\cup\{p\})}{k-1}, which yields 𝒮𝒯\mathcal{S}\subseteq\mathcal{T}. This finishes the proof. ∎

For 0={F1,F2,,Fx}\mathcal{F}_{0}=\{F_{1},F_{2},\ldots,F_{x}\}, let 𝒞1={C([n]k1):pC,CF1=}\mathcal{C}_{1}=\{C\subseteq\binom{[n]}{k-1}:p\notin C,C\cap F_{1}=\emptyset\}. Moreover, for each 2ix2\leq i\leq x, we define

𝒞i:={C([n]k1):pC,CFi=,CFjfor any 1ji1}.\mathcal{C}_{i}:=\bigg{\{}C\subseteq\binom{[n]}{k-1}:p\notin C,C\cap F_{i}=\emptyset,C\cap F_{j}\neq\emptyset\ \textup{for\ any\ }1\leq j\leq i-1\bigg{\}}.

By definitions one can see that 𝒞1,𝒞2,,𝒞x\mathcal{C}_{1},\mathcal{C}_{2},\ldots,\mathcal{C}_{x} are pairwise disjoint, moreover, it is not hard to see that i=1x𝒞i=i=1x([n](Fi{p})k1)\bigcup\limits_{i=1}^{x}\mathcal{C}_{i}=\bigcup\limits_{i=1}^{x}\binom{[n]\setminus(F_{i}\cup\{p\})}{k-1}. When 1ixk1\leq i\leq x\leq k, by definition we have |𝒞i|(nkiki)|\mathcal{C}_{i}|\geq\binom{n-k-i}{k-i}, therefore, we have the following lower bound on |𝒮||\mathcal{S}| by 3.2.

Claim 3.3.

When 1xk1\leq x\leq k, |𝒮|j=1x(nkjkj)|\mathcal{S}|\geq\sum\limits_{j=1}^{x}\binom{n-k-j}{k-j}.

By 3.2 and 3.3, we can see |𝒮|j=1k(nkjkj)|\mathcal{S}|\geq\sum\limits_{j=1}^{k}\binom{n-k-j}{k-j} when x>kx>k. Indeed, we can strengthen 3.3 in the following form, which plays a key role when we analyze the extremal structures. Recall that a family 𝒲\mathcal{W} is called a sunflower with ss common elements, if there is a set S[n]S\subseteq[n] with |S|=s|S|=s such that for any distinct Wi,Wj𝒲W_{i},W_{j}\in\mathcal{W}, we have WiWj=SW_{i}\cap W_{j}=S.

Claim 3.4.

The followings hold.

  1. (1)

    When 1xk1\leq x\leq k, |𝒮|=j=1x(nkjkj)|\mathcal{S}|=\sum\limits_{j=1}^{x}\binom{n-k-j}{k-j} if and only if 0={F1,F2,,Fx}\mathcal{F}_{0}=\{F_{1},F_{2},\ldots,F_{x}\} forms a sunflower with k1k-1 common elements. In particular, if 0\mathcal{F}_{0} is not a sunflower with k1k-1 common elements, then |𝒮|j=1x(nkjkj)+(nk3k2)|\mathcal{S}|\geq\sum\limits_{j=1}^{x}\binom{n-k-j}{k-j}+\binom{n-k-3}{k-2}.

  2. (2)

    When k+1xnkk+1\leq x\leq n-k, |𝒮|=j=1k(nkjkj)|\mathcal{S}|=\sum\limits_{j=1}^{k}\binom{n-k-j}{k-j} if and only if 0={F1,F2,,Fx}\mathcal{F}_{0}=\{F_{1},F_{2},\ldots,F_{x}\} forms a sunflower with k1k-1 common elements. In particular, if 0\mathcal{F}_{0} is not a sunflower with k1k-1 common elements, then |𝒮|j=1k(nkjkj)+(nk3k2)|\mathcal{S}|\geq\sum\limits_{j=1}^{k}\binom{n-k-j}{k-j}+\binom{n-k-3}{k-2}.

Proof of claim.

We focus on the case when 1xk1\leq x\leq k and one can apply the almost identical argument to show the remaining cases. On one hand, if for any distinct i,j[x]i,j\in[x], FiFj=AF_{i}\cap F_{j}=A for some A([n]k1)A\subseteq\binom{[n]}{k-1}, then for each i[x]i\in[x], set Fi=A{ai}F_{i}=A\cup\{a_{i}\}, it is easy to calculate |𝒞|=(nkk)|\mathcal{C}_{\ell}|=\binom{n-k-\ell}{k-\ell} for each [x]\ell\in[x]. This implies that in this case, |𝒮|=j=1x(nkjkj)|\mathcal{S}|=\sum\limits_{j=1}^{x}\binom{n-k-j}{k-j}.

On the other hand, we need to show that if |𝒮|=j=1x(nkjkj)|\mathcal{S}|=\sum\limits_{j=1}^{x}\binom{n-k-j}{k-j}, then 0\mathcal{F}_{0} has to be a sunflower with k1k-1 common elements. First suppose that there exists a pair of sets Fi,Fj0F_{i},F_{j}\in\mathcal{F}_{0} such that |FiFj|k2|F_{i}\cap F_{j}|\leq k-2, by suitable re-labelling, we can assume |F1F2|k2|F_{1}\cap F_{2}|\leq k-2. Then there exist two elements in F1F2F_{1}\setminus F_{2}, denoted by {a1,a2}F1F2\{a_{1},a_{2}\}\subseteq F_{1}\setminus F_{2}. Then we can see the size of 𝒞2\mathcal{C}_{2} is at least (nk2k2)+(nk3k2)\binom{n-k-2}{k-2}+\binom{n-k-3}{k-2}, since one can pick (nk2k2)\binom{n-k-2}{k-2} sets that contain a1a_{1}, and at least (nk3k2)\binom{n-k-3}{k-2} sets that do not contain a1a_{1}. Therefore if |𝒮|=j=1x(nkjkj)|\mathcal{S}|=\sum\limits_{j=1}^{x}\binom{n-k-j}{k-j} for any distinct i,j[x]i,j\in[x], we can assume that |FiFj|=k1|F_{i}\cap F_{j}|=k-1. When x=2x=2, then we are already done. When x3x\geq 3, we set B:=F1F2B:=F_{1}\cap F_{2}, F1=B{b1}F_{1}=B\cup\{b_{1}\} and F2=B{b2}F_{2}=B\cup\{b_{2}\}. Then it suffices to show for any distinct ,m[x]\ell,m\in[x], FFm=BF_{\ell}\cap F_{m}=B.

  1. Case 1.

    If {,m}{1,2}\{\ell,m\}\cap\{1,2\}\neq\emptyset, then by symmetry and suitable re-labelling, it suffices to consider the case that =1\ell=1 and m=3[x]{1,2}m=3\in[x]\setminus\{1,2\}. Suppose that F1F3BF_{1}\cap F_{3}\neq B, then there is some bBb\in B such that bF3b\notin F_{3}, which yields that b(F1F3)(F2F3)b\in(F_{1}\setminus F_{3})\cap(F_{2}\setminus F_{3}). Moreover since |F3|=k|F_{3}|=k and |F3F1|=|F3F2|=k1|F_{3}\cap F_{1}|=|F_{3}\cap F_{2}|=k-1, then we have F3={b1,b2}B{b}F_{3}=\{b_{1},b_{2}\}\cup B\setminus\{b\}. Therefore we can see |𝒞3|(nk2k2)=(nk3k2)+(nk3k3)>(nk3k3)|\mathcal{C}_{3}|\geq\binom{n-k-2}{k-2}=\binom{n-k-3}{k-2}+\binom{n-k-3}{k-3}>\binom{n-k-3}{k-3}, which yields |𝒮|j=1x(nkjkj)+(nk3k2)|\mathcal{S}|\geq\sum\limits_{j=1}^{x}\binom{n-k-j}{k-j}+\binom{n-k-3}{k-2}, a contradiction.

  2. Case 2.

    If {,m}{1,2}=\{\ell,m\}\cap\{1,2\}=\emptyset, by symmetry we can assume that =3\ell=3 and m=4m=4. Suppose that F3F4=DB=F1F2F_{3}\cap F_{4}=D\neq B=F_{1}\cap F_{2}, set F3:=D{d3}F_{3}:=D\cup\{d_{3}\} and F4=D{d4}F_{4}=D\cup\{d_{4}\}. Since |F1F3|=k1|F_{1}\cap F_{3}|=k-1 and BDB\neq D, we have b1=d3b_{1}=d_{3}, similarly since |F1F4|=k1|F_{1}\cap F_{4}|=k-1 and BDB\neq D, we have b1=d4b_{1}=d_{4}. However, this implies that F3=F4F_{3}=F_{4}, a contradiction.

This finishes the proof. One can apply the same argument to show the statement in (2), we omit the repeated details. ∎

Recall that |𝒮|j=1k(nkjkj)|\mathcal{S}|\geq\sum\limits_{j=1}^{k}\binom{n-k-j}{k-j} when x>kx>k. We then define the function g(x)g(x) to be

g(x)={j=1x(nkjkj),if 1xk,j=1k(nkjkj),if k+1xnk.g(x)=\begin{cases}\sum\limits_{j=1}^{x}\binom{n-k-j}{k-j},&\text{if }1\leq x\leq k,\\ \sum\limits_{j=1}^{k}\binom{n-k-j}{k-j},&\text{if }k+1\leq x\leq n-k.\end{cases}

Next we will carefully determine the values at which the function f(x):=g(x)xf(x):=g(x)-x attains its minimum. Recall that |S||0|f(|0|)|S|-|\mathcal{F}_{0}|\geq f(|\mathcal{F}_{0}|), then to obtain the tt-th level of stability result for Erdős-Ko-Rado theorem, we need to understand the exact values of h(t)=mintxnkf(x)h(t)=\min\limits_{t\leq x\leq n-k}f(x) since we have ||(n1k1)h(t)|\mathcal{F}|\leq\binom{n-1}{k-1}-h(t) when txnkt\leq x\leq n-k.

Claim 3.5.

For given 1tnk1\leq t\leq n-k, then the followings hold.

  • If kt+3k\geq t+3, then h(t)=f(t)<minxtf(x)h(t)=f(t)<\min\limits_{x\neq t}f(x).

  • If k=t+2k=t+2, then h(t)=f(t)=f(nk)<minx{t,nk}f(x)h(t)=f(t)=f(n-k)<\min\limits_{x\notin\{t,n-k\}}f(x).

  • If kt+1k\leq t+1, then h(t)=f(nk)<minxnkf(x)h(t)=f(n-k)<\min\limits_{x\neq n-k}f(x).

Proof of claim.

A straightforward calculation shows that f(x)f(x) with x[1,nk]x\in[1,n-k] is monotonically increasing when x[1,k]x\in[1,k] and monotonically decreasing when x[k,nk]x\in[k,n-k]. Therefore h(t)h(t) is either f(t)f(t) or f(nk)f(n-k). Note that f(nk)f(t)=(nktk(t+1))(nkt1)f(n-k)-f(t)=\binom{n-k-t}{k-(t+1)}-\binom{n-k-t}{1}, then the claim follows by computing the exact values directly. ∎

For given positive integer xx, let (x)\mathcal{F}(x) be the largest intersecting family among all ([n]k)\mathcal{F}\subseteq\binom{[n]}{k} with |0|=x|\mathcal{F}_{0}|=x. Let dmax(x)d_{\textup{max}}(x) represent the maximum degree of (x)([n]k)\mathcal{F}(x)\subseteq\binom{[n]}{k}. Observe that for any xx\in\mathbb{N}, |(x)|=x+dmax(x)|\mathcal{F}(x)|=x+d_{\textup{max}}(x). The following observation is crucial.

Claim 3.6.

If x1>x2x_{1}>x_{2}, dmax(x1)dmax(x2)d_{\textup{max}}(x_{1})\leq d_{\textup{max}}(x_{2}).

Proof of claim.

Suppose that x1>x2x_{1}>x_{2} and |(x1)|>|(x2)|+(x1x2)|\mathcal{F}(x_{1})|>|\mathcal{F}(x_{2})|+(x_{1}-x_{2}), then one can remove x1x2x_{1}-x_{2} many sets from (x1)\mathcal{F}(x_{1}) to obtain an intersecting family \mathcal{F} of size larger than |(x2)||\mathcal{F}(x_{2})| with |0|=x2|\mathcal{F}_{0}|=x_{2}, a contradiction to the definition of (x2)\mathcal{F}(x_{2}). ∎

3.3 Stability at arbitrary level: Proof of Theorem 1.5

With the new framework in hand, we then prove the tt-level stability result for Erdős-Ko-Rado theorem. Based on Theorems 1.2, 1.3 and 1.4, we then consider the case of t4t\geq 4. Let ([n]k)\mathcal{F}\subseteq\binom{[n]}{k} be a non-trivial intersecting family and k,t01\mathcal{F}\not\subseteq\mathcal{M}_{k,t_{0}-1} for any 2t0t2\leq t_{0}\leq t. Let 0:={F1,F2,,Fx}\mathcal{F}_{0}:=\{F_{1},F_{2},\ldots,F_{x}\}. Since when |0|1|\mathcal{F}_{0}|\leq 1, \mathcal{F} is a star, or a sub-family of k+1\mathcal{M}_{k+1}, it suffices to consider the case of |0|2|\mathcal{F}_{0}|\geq 2.

Claim 3.7.

min2|0|t1{|𝒮||0|}>mint|0|nk{|𝒮||0|}\min\limits_{2\leq|\mathcal{F}_{0}|\leq t-1}\{|\mathcal{S}|-|\mathcal{F}_{0}|\}>\min\limits_{t\leq|\mathcal{F}_{0}|\leq n-k}\{|\mathcal{S}|-|\mathcal{F}_{0}|\}.

Proof of claim.

When x=|0|t1x=|\mathcal{F}_{0}|\leq t-1, under the assumption that k,t01\mathcal{F}\not\subseteq\mathcal{M}_{k,t_{0}-1} for any 2t0t2\leq t_{0}\leq t, by the definition of k,t01\mathcal{M}_{k,t_{0}-1}, we can see 0\mathcal{F}_{0} is not a sunflower with k1k-1 common elements. Recall that 𝒞1={C([n]k1):pC,CF1=}\mathcal{C}_{1}=\{C\subseteq\binom{[n]}{k-1}:p\notin C,C\cap F_{1}=\emptyset\}, and for each 2i|0|2\leq i\leq|\mathcal{F}_{0}|,

𝒞i:={C([n]k1):pC,CFi=,CFjfor any 1ji1}.\mathcal{C}_{i}:=\bigg{\{}C\subseteq\binom{[n]}{k-1}:p\notin C,C\cap F_{i}=\emptyset,C\cap F_{j}\neq\emptyset\ \textup{for\ any\ }1\leq j\leq i-1\bigg{\}}.

Under the condition that 0\mathcal{F}_{0} is not a sunflower with k1k-1 common elements, we then take advantage of the proof of 3.4. If x=2x=2, then |𝒞2|(nk2k2)+(nk3k2)|\mathcal{C}_{2}|\geq\binom{n-k-2}{k-2}+\binom{n-k-3}{k-2}, which yields that |𝒮||0||𝒞1|+|𝒞2|2(nk1k1)+(nk2k2)+(nk3k2)2|\mathcal{S}|-|\mathcal{F}_{0}|\geq|\mathcal{C}_{1}|+|\mathcal{C}_{2}|-2\geq\binom{n-k-1}{k-1}+\binom{n-k-2}{k-2}+\binom{n-k-3}{k-2}-2. If x3x\geq 3, we can see that |𝒞1|+|𝒞2|+|𝒞3|(nk1k1)+(nk2k2)+(nk3k3)+(nk3k2)|\mathcal{C}_{1}|+|\mathcal{C}_{2}|+|\mathcal{C}_{3}|\geq\binom{n-k-1}{k-1}+\binom{n-k-2}{k-2}+\binom{n-k-3}{k-3}+\binom{n-k-3}{k-2}. Moreover, since for any 3i|0|3\leq i\leq|\mathcal{F}_{0}|, |𝒞i|(nkiki)1|\mathcal{C}_{i}|\geq\binom{n-k-i}{k-i}\geq 1, we then conclude that

min2|0|t1{|𝒮||0|}(nk1k1)+(nk2k2)+(nk3k2)2.\min\limits_{2\leq|\mathcal{F}_{0}|\leq t-1}\left\{|\mathcal{S}|-|\mathcal{F}_{0}|\right\}\geq\binom{n-k-1}{k-1}+\binom{n-k-2}{k-2}+\binom{n-k-3}{k-2}-2.

Since kt+2k\geq t+2, by 3.5, we have mint|0|nk{|𝒮||0|}=j=1t(nkjkj)t\min\limits_{t\leq|\mathcal{F}_{0}|\leq n-k}\{|\mathcal{S}|-|\mathcal{F}_{0}|\}=\sum\limits_{j=1}^{t}\binom{n-k-j}{k-j}-t. Note that

(nk1k1)+\displaystyle\binom{n-k-1}{k-1}+ (nk2k2)+(nk3k2)2(j=1t(nkjkj)t)\displaystyle\binom{n-k-2}{k-2}+\binom{n-k-3}{k-2}-2\quad-\left(\sum\limits_{j=1}^{t}\binom{n-k-j}{k-j}-t\right)
(nk3k2)2((nk2k3)t),\displaystyle\geq\binom{n-k-3}{k-2}-2-\left(\binom{n-k-2}{k-3}-t\right),

where we take advantage of j=3t(nkjkj)j=3k(nkjkj)=(nk2k3)\sum\limits_{j=3}^{t}\binom{n-k-j}{k-j}\leq\sum\limits_{j=3}^{k}\binom{n-k-j}{k-j}=\binom{n-k-2}{k-3}. Note that when n>(5+5)k72n>\frac{(5+\sqrt{5})k-7}{2}, it is easy to check that (nk3k2)2((nk2k3)t)\binom{n-k-3}{k-2}-2-(\binom{n-k-2}{k-3}-t) is strictly larger than 0. This finishes the proof. ∎

By 3.7, it then suffices to consider the case when x=|0|tx=|\mathcal{F}_{0}|\geq t. If xnk+1x\geq n-k+1, by 3.6, we have dmax(x)dmax(nk)=(n1k1)j=1k(nkjkj)=dmax(k)d_{\textup{max}}(x)\leq d_{\textup{max}}(n-k)=\binom{n-1}{k-1}-\sum\limits_{j=1}^{k}\binom{n-k-j}{k-j}=d_{\textup{max}}(\mathcal{M}_{k}), where we take advantage of the formulas (n1k1)=j=1k1(nj1k2)+(nkk1)\binom{n-1}{k-1}=\sum_{j=1}^{k-1}\binom{n-j-1}{k-2}+\binom{n-k}{k-1} and (nkk1)=j=1k(nkjkj).\binom{n-k}{k-1}=\sum_{j=1}^{k}\binom{n-k-j}{k-j}. Therefore by Theorem 2.3 in this case we have |||k|=j=2k(njk1)+nk|\mathcal{F}|\leq|\mathcal{M}_{k}|=\sum\limits_{j=2}^{k}\binom{n-j}{k-1}+n-k. Moreover, Theorem 2.3 states that the equality holds if and only if \mathcal{F} is isomorphic to k\mathcal{M}_{k}. However, in this case we assume that x=|0|>nkx=|\mathcal{F}_{0}|>n-k, therefore \mathcal{F} cannot be isomorphic to k\mathcal{M}_{k}, which yields that ||<|k|=(n1k1)f(nk)|\mathcal{F}|<|\mathcal{M}_{k}|=\binom{n-1}{k-1}-f(n-k).

By definition of h(t)h(t), we can see h(t)f(nk)h(t)\leq f(n-k), it then remains to consider the case when txnkt\leq x\leq n-k. Recall that in this case we have ||(n1k1)h(t)|\mathcal{F}|\leq\binom{n-1}{k-1}-h(t), we then consider the following cases based on 3.5.

  1. Case 1.

    If k=t+2k=t+2, by 3.5 we have h(t)=f(t)=f(nt2)h(t)=f(t)=f(n-t-2). We then determine the extremal structures when ||=(n1k1)h(t)|\mathcal{F}|=\binom{n-1}{k-1}-h(t), according to the size of 0\mathcal{F}_{0}.

    1. Subcase 1.1.

      If 0={F1,F2,,Ft}\mathcal{F}_{0}=\{F_{1},F_{2},\ldots,F_{t}\}, by 3.4, 0\mathcal{F}_{0} is a sunflower with t+1t+1 common elements. We denote A=i=1tFi={a1,a2,,at+1}A=\bigcap\limits_{i=1}^{t}F_{i}=\{a_{1},a_{2},\ldots,a_{t+1}\}, and FA={b}F_{\ell}\setminus A=\{b_{\ell}\} for each [t]\ell\in[t]. Then for any F1F\in\mathcal{F}_{1}, if FA=F\cap A=\emptyset, then {b1,b2,,bt}\{b_{1},b_{2},\ldots,b_{t}\}\subseteq\mathcal{F}, therefore 1={G:([n]t+2):pG,GA}{G([n]t+2):{b1,b2,,bt,p}G}\mathcal{F}_{1}=\{G:\binom{[n]}{t+2}:p\in G,G\cap A\neq\emptyset\}\cup\{G\in\binom{[n]}{t+2}:\{b_{1},b_{2},\ldots,b_{t},p\}\in G\}. Since 0\mathcal{F}_{0} is a sunflower with t+1t+1 common elements, we can see \mathcal{F} is isomorphic to t+2,t\mathcal{M}_{t+2,t} in this case, as desired.

    2. Subcase 1.2.

      If |0|=nt2|\mathcal{F}_{0}|=n-t-2, we can see dmax(nt2)dmax(t+2)d_{\textup{max}}(n-t-2)\leq d_{\textup{max}}(\mathcal{M}_{t+2}), then by Theorem 2.3, |||t+2||\mathcal{F}|\leq|\mathcal{M}_{t+2}|. Moreover, the equality holds if and only if \mathcal{F} is isomorphic to t+2\mathcal{M}_{t+2}, as desired.

  2. Case 2.

    When kt+3k\geq t+3, by 3.5, we have h(t)=f(t)=j=1t(nkjkj)th(t)=f(t)=\sum\limits_{j=1}^{t}\binom{n-k-j}{k-j}-t. Then it suffices to consider the case when |0|=t|\mathcal{F}_{0}|=t. By an almost identical argument as that in Subcase 1.1, we can see \mathcal{F} is isomorphic to k,t\mathcal{M}_{k,t}, we omit the details here.

This finishes the proof.

4 Concluding remarks

In this paper, we focus on developing a unified framework for deriving stability results for the celebrated Erdős-Ko-Rado theorem using a robust linear algebraic approach. To illustrate our main ideas, we first present a slightly weaker version that applies to arbitrary levels, extending beyond previous results in [20, 21, 22, 25]. In Theorem 1.5, we impose two stronger assumptions: kt+2k\geq t+2 and n>(5+5)k72n>\frac{(5+\sqrt{5})k-7}{2}. Indeed, by carefully extending the arguments in the proofs of 3.43.5 and 3.7, the case under the natural conditions n2k+1n\geq 2k+1 and kt+1k\geq t+1 can be readily analyzed. In fact, we provide an alternative proof of Theorem 1.3 in the Appendix, While it is feasible to further extend our methods to give a full proof of Theorem 1.4, we do not pursue it here. It is worth noting that in the Appendix, where we prove Theorem 1.3, the case k=3=2+1k=3=2+1 is relatively more complicated. Similarly, when proving the complete stability of Theorem 1.4 (i.e., for the third level), we find that the case k=4=3+1k=4=3+1 is also more intricate. However, interestingly, when t4t\geq 4 the case k=t+1k=t+1 actually becomes simpler. Although we do not fully present this proof in the paper, we encourage interested readers to explore this phenomenon independently.

We believe that a more interesting direction for research is to continue exploring the potential of this robust linear algebra method to obtain more stability results more efficiently.

Acknowledgement

Zixiang Xu would like to thank Prof. Hao Huang, Dr. Yongtao Li, Prof. Hong Liu, Prof. Benjian Lv, and Prof. Jian Wang for their valuable discussion during the early stages of this work over the past two years. In particular, he thanks Dr. Yongtao Li for providing the reference [18] and Prof. Jian Wang for informing him of the results in [16, 26].

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Appendix: An alternative proof of Theorem 1.3

Since when |0|1|\mathcal{F}_{0}|\leq 1, \mathcal{F} is a star, or a sub-family of k+1\mathcal{M}_{k+1}, it suffices to consider the case when |0|2|\mathcal{F}_{0}|\geq 2. By 3.6, for any x>nkx>n-k, we have dmax(x)dmax(nk)(n1k1)i=1k(nkiki)=dmax(k)d_{\textup{max}}(x)\leq d_{\textup{max}}(n-k)\leq\binom{n-1}{k-1}-\sum_{i=1}^{k}\binom{n-k-i}{k-i}=d_{\textup{max}}(\mathcal{M}_{k}), which yields |||k|=j=2k(njk1)+nk|\mathcal{F}|\leq|\mathcal{M}_{k}|=\sum_{j=2}^{k}\binom{n-j}{k-1}+n-k. Moreover, by Theorem 2.3 the equality holds if and only if \mathcal{F} is isomorphic to k\mathcal{M}_{k}, in particular, when k=4k=4, it could be isomorphic to 3\mathcal{M}_{3}. Therefore, generally in this case except k=4k=4, when x>nkx>n-k, \mathcal{F} cannot be isomorphic to k\mathcal{M}_{k}, which implies ||<|k||\mathcal{F}|<|\mathcal{M}_{k}|. Additionally, when k=4k=4, if ||=|4||\mathcal{F}|=|\mathcal{M}_{4}|, \mathcal{F} can be isomorphic to 3\mathcal{M}_{3}.

When 2xnk2\leq x\leq n-k, note that ||(n1k1)h(2)|\mathcal{F}|\leq\binom{n-1}{k-1}-h(2), we consider the following cases.

  1. Case 1.

    When k=3k=3, by 3.5, h(2)=f(n3)h(2)=f(n-3), then ||(n12)f(n3)(n12)((n42)+(n51)+1)+(n3)=(n12)(n42)+1|\mathcal{F}|\leq\binom{n-1}{2}-f(n-3)\leq\binom{n-1}{2}-(\binom{n-4}{2}+\binom{n-5}{1}+1)+(n-3)=\binom{n-1}{2}-\binom{n-4}{2}+1. Moreover, when |0|=n3|\mathcal{F}_{0}|=n-3 and the above equality holds, then by 3.4 we can see that 0\mathcal{F}_{0} is a sunflower with 22 core elements a1,a2a_{1},a_{2}. Since \mathcal{F} is 33-uniform and n3>3n-3>3, then any F0F\in\mathcal{F}\setminus\mathcal{F}_{0} must contain either a1a_{1} or a2a_{2}, therefore ={F([n]3):|F{a1,a2,p}|2}\mathcal{F}=\{F\in\binom{[n]}{3}:|F\cap\{a_{1},a_{2},p\}|\geq 2\}. Indeed, this configuration is an extremal structure in Theorem 1.2, which is a contradiction to the assumption.

    1. Subcase 1.1.

      When |0|=2|\mathcal{F}_{0}|=2, by 3.3, ||(n12)|𝒮|+2(n12)(n42)(n51)+2=2n2|\mathcal{F}|\leq\binom{n-1}{2}-|\mathcal{S}|+2\leq\binom{n-1}{2}-\binom{n-4}{2}-\binom{n-5}{1}+2=2n-2. By 3.4 the equality holds if and only if 0\mathcal{F}_{0} is a sunflower with 22 common elements, therefore, ||=2n2|\mathcal{F}|=2n-2 if and only if \mathcal{F} is isomorphic to 3,2\mathcal{M}_{3,2}.

    2. Subcase 1.2.

      When 3|0|n33\leq|\mathcal{F}_{0}|\leq n-3, note that \mathcal{F} is not a sub-family of extremal structures in Theorem 1.2, namely 4\mathcal{M}_{4} or 3\mathcal{M}_{3}. Suppose that 0\mathcal{F}_{0} is a sunflower with 22 common elements, say 0={F([n]3):{w1,w2}F}\mathcal{F}_{0}=\{F\in\binom{[n]}{3}:\{w_{1},w_{2}\}\subseteq F\}, then since 1\mathcal{F}_{1} is a star and \mathcal{F} is intersecting, then \mathcal{F} must be a sub-family of 3\mathcal{M}_{3}, a contradiction to the assumption. Therefore 0\mathcal{F}_{0} is not a sunflower with 22 common elements, then by 3.4, ||(n12)|𝒮|+|0|(n12)(n42)(n51)(n61)1+|0||\mathcal{F}|\leq\binom{n-1}{2}-|\mathcal{S}|+|\mathcal{F}_{0}|\leq\binom{n-1}{2}-\binom{n-4}{2}-\binom{n-5}{1}-\binom{n-6}{1}-1+|\mathcal{F}_{0}|, in particular, the equality holds only if |𝒞3|=1|\mathcal{C}_{3}|=1. Then it suffices to consider the case |0|=n3|\mathcal{F}_{0}|=n-3, otherwise ||2n3|\mathcal{F}|\leq 2n-3. Since |0|=n3>3|\mathcal{F}_{0}|=n-3>3, there exists some F30F_{3}\in\mathcal{F}_{0} such that |F1F3|=2|F_{1}\setminus F_{3}|=2 or |F2F3|=2|F_{2}\setminus F_{3}|=2. Suppose that former case occurs, set F1F3:={a1,a2}F_{1}\setminus F_{3}:=\{a_{1},a_{2}\}, and set bF2F3b\in F_{2}\setminus F_{3}, then {a1,b},{a2,b}𝒞3\{a_{1},b\},\{a_{2},b\}\subseteq\mathcal{C}_{3}, which implies that |𝒞3|2|\mathcal{C}_{3}|\geq 2, combining with the proof in 3.4, we can see ||(n12)|𝒮|+n3(n12)(n42)(n51)(n61)2+(n3)=2n3|\mathcal{F}|\leq\binom{n-1}{2}-|\mathcal{S}|+n-3\leq\binom{n-1}{2}-\binom{n-4}{2}-\binom{n-5}{1}-\binom{n-6}{1}-2+(n-3)=2n-3.

    3. Subcase 1.3.

      When |0|n2|\mathcal{F}_{0}|\geq n-2, we first have the following claim.

      Claim 4.1.

      F0F=\bigcap_{F\in\mathcal{F}_{0}}F=\emptyset.

      Proof of claim.

      Suppose that gF0Fg\in\bigcap\limits_{F\in\mathcal{F}_{0}}F, then we consider the sub-family 1,g:={F1:gF}\mathcal{F}_{1,g}:=\{F\in\mathcal{F}_{1}:g\notin F\} of 1\mathcal{F}_{1}. Pick arbitrary set A1,gA\in\mathcal{F}_{1,g}, denoted by A={a1,a2,p}A=\{a_{1},a_{2},p\}. Suppose that there exists some a3ga_{3}\neq g such that {a1,a3,p}1,g\{a_{1},a_{3},p\}\in\mathcal{F}_{1,g}, then the subset in 0\mathcal{F}_{0} that contains a2a_{2} but does not contain a1a_{1} must be {a2,a3,g}\{a_{2},a_{3},g\}. Then 0={a2,a3,g}(a[n]{p,a1,g}{a,a1,g})\mathcal{F}_{0}=\{a_{2},a_{3},g\}\cup\bigg{(}\bigcup\limits_{a\in[n]\setminus\{p,a_{1},g\}}\{a,a_{1},g\}\bigg{)}, which implies 1,g\mathcal{F}_{1,g} must be {a1,a3,p}{a1,a2,p}\{a_{1},a_{3},p\}\cup\{a_{1},a_{2},p\}. By symmetry, we can conclude that in this case, we have |1,g|2|\mathcal{F}_{1,g}|\leq 2, which implies that the degree of gg is larger than the degree of pp, a contradiction. If there is no a3a_{3} such that {a1,a3,p}1,g\{a_{1},a_{3},p\}\in\mathcal{F}_{1,g} or {a2,a3,p}1,g\{a_{2},a_{3},p\}\in\mathcal{F}_{1,g}, then either |1,g|1|\mathcal{F}_{1,g}|\leq 1, or there exist some b1,b2b_{1},b_{2} with {b1,b2}{a1,a2}=\{b_{1},b_{2}\}\cap\{a_{1},a_{2}\}=\emptyset such that {b1,b2,p}1,g\{b_{1},b_{2},p\}\in\mathcal{F}_{1,g}. However, this implies that 0{a1,b1,p}{a1,b2,p}{a2,b1,p}{a2,b2,p}\mathcal{F}_{0}\subseteq\{a_{1},b_{1},p\}\cup\{a_{1},b_{2},p\}\cup\{a_{2},b_{1},p\}\cup\{a_{2},b_{2},p\}, which is a contradiction to |0|n2>4|\mathcal{F}_{0}|\geq n-2>4. ∎

      Now let qq be the element attaining the maximum degree of 0\mathcal{F}_{0}, and denote 0,q={F0,qF}\mathcal{F}_{0,q}=\{F\in\mathcal{F}_{0},q\in F\} and 0,0={F0,qF}\mathcal{F}_{0,0}=\{F\in\mathcal{F}_{0},q\notin F\}. By 4.1, |0,0|>0|\mathcal{F}_{0,0}|>0. Obviously we have |0,q|2|\mathcal{F}_{0,q}|\geq 2, otherwise |0,0|=0|\mathcal{F}_{0,0}|=0, a contradiction. Moreover if |0,q|=2|\mathcal{F}_{0,q}|=2, then 3|0|v[n]{p}|0,q|2(n1)3|\mathcal{F}_{0}|\leq\sum\limits_{v\in[n]\setminus\{p\}}|\mathcal{F}_{0,q}|\leq 2(n-1), which implies |0|2(n1)3<n2|\mathcal{F}_{0}|\leq\frac{2(n-1)}{3}<n-2 when n7n\geq 7, a contradiction to |0|n2|\mathcal{F}_{0}|\geq n-2. Then we divide our argument according to |0,q||\mathcal{F}_{0,q}|:

      1. Subsubcase 1.3.1.

        If |0,q|=3|\mathcal{F}_{0,q}|=3, we arbitrarily pick a set A:={a1,a2,a3}0,0A:=\{a_{1},a_{2},a_{3}\}\in\mathcal{F}_{0,0}, denote 0,q\mathcal{F}_{0,q} to be A1A2A3A_{1}\cup A_{2}\cup A_{3}, observe that the sets in 1,q={F1,qF}\mathcal{F}_{1,q}=\{F\in\mathcal{F}_{1},q\notin F\} are of the form {p,ai,}\{p,a_{i},*\} for some i[3]i\in[3]. If there exists some aia_{i} appearing exactly twice in 0,q\mathcal{F}_{0,q}, assume ai=a1a_{i}=a_{1} by symmetry, then we denote A1={q,a1,b1}A_{1}=\{q,a_{1},b_{1}\}, A2={q,a1,b2}A_{2}=\{q,a_{1},b_{2}\} and A3={q,a2,b3}A_{3}=\{q,a_{2},b_{3}\}, in particular, b3a1b_{3}\neq a_{1}. Then {p,a1,}\{p,a_{1},*\} can only be {p,a1,a2}\{p,a_{1},a_{2}\} or {p,a1,b3}\{p,a_{1},b_{3}\}, and {p,a2,}\{p,a_{2},*\} can only be {p,a1,a2}\{p,a_{1},a_{2}\}. We then consider the possibilities of {p,a3,}\{p,a_{3},*\}.

        • If b1=a3b_{1}=a_{3} or b2=a3b_{2}=a_{3}, without loss of generality, assume b1=a3b_{1}=a_{3}, then if b3=a3b_{3}=a_{3}, then {p,a3,}\{p,a_{3},*\} has to be {p,a3,a1}\{p,a_{3},a_{1}\} or {p,a3,b2}\{p,a_{3},b_{2}\}, in this case {p,a3,a1}={p,a1,b3}\{p,a_{3},a_{1}\}=\{p,a_{1},b_{3}\}, which implies |1,q|3|\mathcal{F}_{1,q}|\leq 3. If b3a3b_{3}\neq a_{3}, then the element * of {p,a3,}\{p,a_{3},*\} belongs to A2A3A_{2}\cap A_{3}, which implies that either {p,a3,}={p,a3,b2}\{p,a_{3},*\}=\{p,a_{3},b_{2}\} or {p,a3,}=\{p,a_{3},*\}=\emptyset. Therefore, in this case we have |1,q|3|\mathcal{F}_{1,q}|\leq 3.

        • If a3{b1,b2}a_{3}\notin\{b_{1},b_{2}\}, then if b3=a3b_{3}=a_{3}, then the element * of {p,a3,}\{p,a_{3},*\} belongs to A1A2A_{1}\cap A_{2}, which yields that {p,a3,}\{p,a_{3},*\} has to be {p,a3,a1}\{p,a_{3},a_{1}\}. Therefore, in this case we have |1,q|3|\mathcal{F}_{1,q}|\leq 3. If b3a3b_{3}\neq a_{3}, it is easy to see that {p,a3,}=\{p,a_{3},*\}=\emptyset or {p,a3,}={p,a3,b1}\{p,a_{3},*\}=\{p,a_{3},b_{1}\} when b1=b2=b3b_{1}=b_{2}=b_{3}, which yields that |1,q|3|\mathcal{F}_{1,q}|\leq 3 in this case.

        In all, we have |1,q|3|\mathcal{F}_{1,q}|\leq 3. If each aia_{i} appears exactly once in 0,q\mathcal{F}_{0,q}, then we denote A1={q,a1,b1}A_{1}=\{q,a_{1},b_{1}\}, A2={q,a2,b2}A_{2}=\{q,a_{2},b_{2}\} and A3={q,a3,b3}A_{3}=\{q,a_{3},b_{3}\} with {b1,b2,b3}{a1,a2,a3}=\{b_{1},b_{2},b_{3}\}\cap\{a_{1},a_{2},a_{3}\}=\emptyset. Observe that if b1,b2,b3b_{1},b_{2},b_{3} are pairwise distinct, then |1,q|=0|\mathcal{F}_{1,q}|=0. By symmetry if b1=b2b3b_{1}=b_{2}\neq b_{3}, then 1,q\mathcal{F}_{1,q} can only consist of {p,a3,b1}\{p,a_{3},b_{1}\}. In particular, if b1=b2=b3b_{1}=b_{2}=b_{3}, then 1,q{p,a1,b1}{p,a2,b1}{p,a3,b1}\mathcal{F}_{1,q}\subseteq\{p,a_{1},b_{1}\}\cup\{p,a_{2},b_{1}\}\cup\{p,a_{3},b_{1}\}. Therefore, |1,q|3|\mathcal{F}_{1,q}|\leq 3. Moreover, we can see 11,q{p,q,a1}{p,q,a2}{p,q,a3}\mathcal{F}_{1}\setminus\mathcal{F}_{1,q}\subseteq\{p,q,a_{1}\}\cup\{p,q,a_{2}\}\cup\{p,q,a_{3}\}, which together implies that |1|6|\mathcal{F}_{1}|\leq 6. Note that 3|0|v[n]{p}|0,q|3(n1)3|\mathcal{F}_{0}|\leq\sum\limits_{v\in[n]\setminus\{p\}}|\mathcal{F}_{0,q}|\leq 3(n-1), therefore, ||=|0|+|1|n+52n3|\mathcal{F}|=|\mathcal{F}_{0}|+|\mathcal{F}_{1}|\leq n+5\leq 2n-3 when n8n\geq 8. In particular, when n=7n=7, if |0,0|=1|\mathcal{F}_{0,0}|=1, then ||1+3+6=1011|\mathcal{F}|\leq 1+3+6=10\leq 11, and if |0,0|2|\mathcal{F}_{0,0}|\geq 2, then |11,q|2|\mathcal{F}_{1}\setminus\mathcal{F}_{1,q}|\leq 2, which implies that |1|5|\mathcal{F}_{1}|\leq 5, then |||0|+56+5=11<12|\mathcal{F}|\leq|\mathcal{F}_{0}|+5\leq 6+5=11<12.

      2. Subsubcase 1.3.2.

        If |0,q|4|\mathcal{F}_{0,q}|\geq 4, then we directly apply the argument in Subsubcase 1.3.1, we can see |1,q|3|\mathcal{F}_{1,q}|\leq 3. Then the degree of element qq is larger than the degree of element pp, a contradiction.

      In all, when |0|n2|\mathcal{F}_{0}|\geq n-2, ||2n3|\mathcal{F}|\leq 2n-3.

    Thus when k=3k=3, ||2n2|\mathcal{F}|\leq 2n-2, the equality holds if and only if \mathcal{F} is isomorphic to 3,2\mathcal{M}_{3,2}.

  2. Case 2.

    When k=4k=4, by 3.5, h(2)=f(2)=f(n4)h(2)=f(2)=f(n-4), then ||(n13)h(2)=(n13)(n53)(n62)+2|\mathcal{F}|\leq\binom{n-1}{3}-h(2)=\binom{n-1}{3}-\binom{n-5}{3}-\binom{n-6}{2}+2. Then there are different types of extremal configurations.

    1. Subcase 2.1.

      When 0={F1,F2}\mathcal{F}_{0}=\{F_{1},F_{2}\}, by 3.4, |F1F2|=3|F_{1}\cap F_{2}|=3. We set F1F2={a1,a2,a3}F_{1}\cap F_{2}=\{a_{1},a_{2},a_{3}\}, F1F2={b1}F_{1}\setminus F_{2}=\{b_{1}\} and F2F1={b2}F_{2}\setminus F_{1}=\{b_{2}\}. Then for any F1F\in\mathcal{F}_{1}, if F{a1,a2,a3}=F\cap\{a_{1},a_{2},a_{3}\}=\emptyset, then {b1,b2}F\{b_{1},b_{2}\}\subseteq F, therefore, 1={G([n]4):{b1,b2,p}G}{G([n]4):pG,G{a1,a2,a3}}\mathcal{F}_{1}=\{G\in\binom{[n]}{4}:\{b_{1},b_{2},p\}\in G\}\cup\{G\in\binom{[n]}{4}:p\in G,G\cap\{a_{1},a_{2},a_{3}\}\neq\emptyset\}. Moreover, 0={G([n]4):{a1,a2,a3}G,G{b1,b2}}\mathcal{F}_{0}=\{G\in\binom{[n]}{4}:\{a_{1},a_{2},a_{3}\}\subseteq G,G\cap\{b_{1},b_{2}\}\neq\emptyset\}. Therefore, \mathcal{F} is isomorphic to k,2\mathcal{M}_{k,2}.

    2. Subcase 2.2.

      |0|=n44|\mathcal{F}_{0}|=n-4\geq 4, dmax(n4)(n13)i=14(n4i4i)=dmax(4)d_{\textup{max}}(n-4)\leq\binom{n-1}{3}-\sum\limits_{i=1}^{4}\binom{n-4-i}{4-i}=d_{\textup{max}}(\mathcal{M}_{4}), then by Theorem 2.3, |||4||\mathcal{F}|\leq|\mathcal{M}_{4}|. Moreover, the equality holds if and only if \mathcal{F} is either isomorphic to 3\mathcal{M}_{3}, or isomorphic to 4\mathcal{M}_{4}. Furthermore, by checking the structure of 3\mathcal{M}_{3}, we can see \mathcal{F} has to be isomorphic to 4\mathcal{M}_{4} in this case.

  3. Case 3.

    When k5k\geq 5, by 3.5, h(2)=f(2)h(2)=f(2), then ||(n1k1)h(2)=(n1k1)(nk1k1)(nk2k2)+2|\mathcal{F}|\leq\binom{n-1}{k-1}-h(2)=\binom{n-1}{k-1}-\binom{n-k-1}{k-1}-\binom{n-k-2}{k-2}+2. Using the almost identical argument as that in Subcase 2.1, we can show the above equality holds if and only if \mathcal{F} is isomorphic to k,2\mathcal{M}_{k,2}, we omit the repeated details here.

In all, when k=4k=4, ||(n13)(n53)(n62)+2|\mathcal{F}|\leq\binom{n-1}{3}-\binom{n-5}{3}-\binom{n-6}{2}+2, and the equality holds if and only if \mathcal{F} is isomorphic to 4,2,3\mathcal{M}_{4,2},\mathcal{M}_{3} or 4\mathcal{M}_{4}. When k5k\geq 5 or k=3k=3, we have ||(n1k1)(nk1k1)(nk2k2)+2|\mathcal{F}|\leq\binom{n-1}{k-1}-\binom{n-k-1}{k-1}-\binom{n-k-2}{k-2}+2, and the equality holds if and only if \mathcal{F} is isomorphic to k,2\mathcal{M}_{k,2}. This finishes the proof.