Ramsey-Turán numbers for intersecting odd cliques

Guantao Chen and Min Liu
Department of Mathematics and Statistics, Georgia State University, Atlanta, GA 30303
College of Economics, Northwest University of Politics and Law, Xi’an, Shaanxi, China 710063
Supported in part by NSF grant DMS-1855716Supported in part by the Research Foundation of College of Economics, Northwest University of Politics and Law (No. 19XYKY07), and by China Scholarship Council (No. 201808610115).

Abstract

Given a graph $H$ and a function $f:\mathbb{Z}^{+}\longrightarrow\mathbb{Z}^{+}$ , the Ramsey-Turán number of $H$ and $f$ , denoted by $RT(n,H,f(n))$ , is the maximum number of edges a graph $G$ on $n$ vertices can have, which does not contain $H$ as a subgraph and also does not contain a set of $f(n)$ independent vertices. Let $r$ be a positive integer. In 1969, Erdős and Sós proved that $RT(n,K_{\_}{2r+1},o(n))=\frac{n^{2}}{2}(1-\frac{1}{r})+o(n^{2})$ . Let $F_{\_}k(2r+1)$ denote the graph consisting of $k$ copies of complete graphs $K_{\_}{2r+1}$ sharing exactly one vertex. In this paper, we show that $RT(n,F_{\_}k(2r+1),o(n))=\frac{n^{2}}{2}(1-\frac{1}{r})+o(n^{2})$ , which is of the same magnitude with $RT(n,K_{\_}{2r+1},o(n))$ .

Keywords: Turán number, Ramsey-Turán number, odd cliques, independent set

1 Introduction

Let $H$ be a graph. A graph $G$ is called $H$ -free if $G$ does not contain a copy of $H$ as a subgraph. For a positive integer $n$ , let $ex(n,H)$ denote the maximum number of edges of an $H$ -free graph of order $n$ , and call such a graph an extremal graph for $H$ . For any positive integer $r$ , let $K_{\_}r$ denote the $n$ -vertex complete graph. Turán’s classical result states that the balanced complete $n$ -vertex $r$ -partite graph, denoted by $T_{\_}r(n)$ , is the unique $K_{\_}{r+1}$ -extremal graph of order $n$ . A graph on $2k+1$ vertices consisting of $k$ triangles which intersect in exactly one common vertex is called a $k$ -fan and denoted by $F_{\_}k$ . Erdős, Füredi, Gould, and Gunderson [2] determined $ex(n,F_{\_}k)$ for $n\geq 50k^{2}$ . A graph on $(r-1)k+1$ vertices consisting of $k$ copies of $K_{\_}r$ sharing exactly one common vertex is called a $(k,r)$ -fan and denoted by $F_{\_}k(r)$ . Chen, Gould, Pfender, and Wei [1] determined $ex(n,F_{\_}k(r))$ for $n\geq 16k^{3}r^{8}$ .

Notice that the independence number $\alpha(T_{\_}r(n))$ is at least $\frac{n}{r}$ . Given a graph $H$ and a function $f:\ \mathbb{Z}^{+}\longrightarrow\mathbb{Z}^{+}$ , the Ramsey-Turán number for a large positive integer $n$ with regard to $H$ and $f(n)$ , denoted by $RT(n,H,f(n))$ , is the maximum number of edges of an $n$ -vertex $H$ -free graph $G$ with $\alpha(G)\leq f(n)$ . Erdős and Sós in 1969 [3] studied of Ramsey-Turán numbers and showed that $RT(n,K_{\_}{2r+1},o(n))=\frac{n^{2}}{2}(1-\frac{1}{r})+o(n^{2})$ , which is approximately the value of $ex(n,K_{\_}{r+1})$ . The study of Ramsey-Turán numbers has drawn a great deal of attention over the last 40 years; see the survey by Simonovits and Sós [7] and [4, 5, 6] for more results on various Ramsey-Turán problems. Inspired by the study of extremal numbers $ex(n,F_{\_}k)$ and $ex(n,F_{\_}k(r))$ in [2] and [1], respectively, we consider Ramsey-Turán numbers for $F_{\_}k(r)$ and obtained the following result.

Theorem 1.

For any two fixed positive integers $k$ and $r$ ,

RT(n,F_{\_}{k}(2r+1),o(n))=\frac{n^{2}}{2}\left(1-\frac{1}{r}\right)+o(n^{2}).

In the next section we will first give a short proof of the case $r=1$ , i.e., $RT(n,F_{\_}{k}(3),o(n))=o(n^{2})$ , and then prove the general case. We now introduce notations and terminology that will be used in the proof.

All graphs considered in this paper are simple graphs. Let $G=(V(G),E(G))$ be a graph with vertex set $V(G)$ and edge set $E(G)$ . Denote by $|G|$ and $||G||$ for $|V(G)|$ and $|E(G)|$ , respectively; and call them the order and the size of $G$ , respectively. Denote by $G(n)$ a graph of order $n$ and denote by $G(n,m)$ a graph of order $n$ and size $m$ . For a vertex $x\in V(G)$ , the neighborhood of $x$ in $G$ , denoted by $N_{\_}G(x)$ , is the set of vertices adjacent to $x$ , i.e., $N_{\_}{G}(x)=\{y\in V(G):xy\in E(G)\}$ . We use $N(x)$ for $N_{\_}G(x)$ when the referred graph $G$ is clear. The degree of $x$ in $G$ , denoted by $d_{\_}G(x)$ , or $d(x)$ , is the cardinality of $N_{\_}{G}(x)$ , i.e., $d_{\_}G(x)=|N_{\_}G(x)|$ . Denote by $\delta(G)$ the minimum degree in $G$ . For a subset $X\subseteq V(G)$ , let $G[X]$ denote the subgraph of $G$ induced by $X$ , i.e., a subgraph of $G$ with vertex set $X$ and the set of all edges with two ends in $X$ . For a vertex set $X=\{x_{\_}1,x_{\_}2,\dots,x_{\_}r\}$ , denote by $G[x_{\_}1,x_{\_}2,\dots,x_{\_}r]$ for $G[\{x_{\_}1,x_{\_}2,\dots,x_{\_}r\}]$ .

The matching number of a graph $G$ , denoted by $\nu(G)$ , is the maximum number of edges in a matching in $G$ . Let $\alpha(G)$ denote the independence number of $G$ . Clearly, if $G$ is an $n$ -vertex graph, then $\alpha(G)\geq n-2\nu(G)$ .

2 Proof of Theorem 1

2.1 Prilimilary

We first consider triangles sharing a common vertex and prove the following simple fact.

Lemma 1.

For any fixed positive integer $k$ , $RT(n,F_{\_}{k}(3),o(n))=o(n^{2})$ .

Proof.

Let $G$ be an $n$ -vertex $F_{\_}{k}(3)$ -free graph with independence number $\alpha(G)\leq\epsilon n$ for some small positive real number $\epsilon$ . For any $x\in V(G)$ , since $G$ does not contain $k$ edge-disjoint triangles intersecting in the vertex $x$ , the neighborhood of $x$ contains at most $k-1$ independent edges, i.e., $\nu(G[N(x)])\leq k-1$ . Thus, $\alpha(G[N(x)])\geq d(x)-2(k-1)$ . Since $\alpha(G[N(x)])\leq\alpha(G)\leq\epsilon n$ , we have $d(x)\leq\epsilon n+2(k-1)$ . Counting the total degree of $G$ , we have the following inequality: $|E(G)|\leq\frac{1}{2}(\epsilon+2(k-1)/n)n^{2}$ . Since $\lim_{\_}{n\rightarrow\infty}\frac{1}{2}(\epsilon+2(k-1)/n)=\frac{1}{2}\epsilon$ , we complete the proof of Lemma 1. ∎

The following result from [3] is needed in our proof and for completeness we give its proof here.

Lemma 2 (Erdős and Sós).

Let $\beta$ and $\epsilon$ be two positive numbers with $0<\beta<\frac{1}{2}$ and $0<\epsilon<1$ . Then for every positive number $c$ with $0<c<\sqrt{\beta\epsilon}$ and any graph $G(n)$ with $|E(G)|\geq\beta n^{2}(1+\epsilon)$ , there is a subgraph $G(m)\subseteq G(n)$ with $m>cn$ and $\delta(G(m))>2\beta(1+\epsilon/2)m$ .

Proof.

To avoid cumbersome notation, we assume without loss of generality that $cn$ is an integer. Suppose on the contrary a such graph $G(m)$ does not exist. Denote the graph $G(n)$ by $G$ with $|E(G)|\geq\beta n^{2}(1+\epsilon)$ . Then the vertices of $G$ can be written in a sequence $x_{\_}1,x_{\_}2,\dots,x_{\_}n$ so that for every $i\leq(1-c)n$ the degree of $x_{\_}i$ in $G[x_{\_}i,x_{\_}{i+1},\dots,x_{\_}n]$ is less than $2\beta(1+\epsilon/2)(n-i+1)$ .

$\displaystyle\beta n^{2}(1+\epsilon)\leq\|E(G)\|$	$\displaystyle\leq$	$\displaystyle 2\beta\left(1+\frac{\epsilon}{2}\right)(n+(n-1)+\dots+(cn+1))+{cn\choose 2}$
	$\displaystyle<$	$\displaystyle 2\beta\left(1+\frac{\epsilon}{2}\right)\left(\frac{n^{2}}{2}\right)+\frac{c^{2}n^{2}}{2}$
	$\displaystyle<$	$\displaystyle\beta n^{2}(1+\epsilon)\qquad\mbox{ (since $c<\sqrt{\beta\epsilon}$), }$

which gives a contradiction. ∎

Lemma 3.

Let $r$ be a positive integer. For any $\epsilon>0$ , there exists positive integer $N=N(\epsilon)$ such that for a clique $D$ of a graph $G(n)$ with $n>N$ and $|D|\leq r$ , if $d(v)\geq\left(1-\frac{1}{r}+\frac{\epsilon}{3}\right)n$ for each $v\in D$ , then there exists a clique $D^{*}$ that contains $D$ as a proper subset.

Proof.

Let $m$ denote the number of vertices in $V(G)-D$ that are adjacent to all vertices in $D$ . It is sufficient to show that $m>0$ . By counting the edges between $D$ and $G-D$ , we have the following inequality.

\displaystyle|D|m+(|D|-1)(n-|D|-m)\geq\sum_{\_}{v\in D}(d(v)-(|D|-1))\geq|D|\left(\left(1-\frac{1}{r}+\frac{\epsilon}{3}\right)n-(|D|-1)\right)

Thus, $m\geq|D|\left(1-\frac{1}{r}+\frac{\epsilon}{3}\right)n-(|D|-1)n=n-\left(\frac{1}{r}-\frac{\epsilon}{3}\right)|D|n>0$ , where in the last inequality we used $|D|\leq r$ . ∎

Lemma 4.

Let $r$ be a positive integer. For any $\epsilon>0$ , there exist $\delta=\frac{\epsilon}{4}$ and $N=N(\epsilon)$ such that for any clique $D$ of a graph $G(n)$ with $n>N$ and $\alpha(G(n))\leq\delta n$ , if $|D|\leq 2r$ and $d(v)\geq(1-\frac{1}{r}+\frac{\epsilon}{3})n$ for each $v\in D$ , then there exists a clique $D^{*}$ with $|D^{*}|=|D|+1$ such that $|D^{*}\cap D|\geq|D|-1$ .

Proof.

We set $|D|=s\leq 2r$ and $G=G(n)$ . If there is a vertex $v\in V(G)-D$ that is adjacent to all vertices in $D$ , then $D\cup\{v\}$ is the desired clique. So, we assume that $d_{\_}D(v)\leq s-1$ for all $v\in V(G)-D$ . We divide $V(G-D)$ into two vertex-disjoint sets $U$ and $W$ such that

	$\displaystyle U$	$\displaystyle=$	$\displaystyle\{v\in V(G)-D\ \|\ d_{\_}D(v)\leq s-2\},\mbox{ and}$
	$\displaystyle W$	$\displaystyle=$	$\displaystyle\{v\in V(G)-D)\ \|\ d_{\_}D(v)=s-1\}.$

Computing the edges between $D$ and $V(G)-D$ , we get the following inequality.

\displaystyle(s-2)|U|+(s-1)|W|\geq\sum_{\_}{v\in D}d_{\_}{G-D}(v)\geq s\left(\left(1-\frac{1}{r}+\frac{\epsilon}{3}\right)n-(s-1)\right)

By assuming $n\geq N>\frac{3}{\epsilon}$ and applying the inequality $s\leq 2r$ , we get

|W|\geq\left(2-\frac{s}{r}+s\frac{\epsilon}{3}-\frac{s}{n}\right)n>\left(s\frac{\epsilon}{3}-\frac{s}{n}\right)n>0.

Let $D=\{v_{\_}1,v_{\_}2,\dots,v_{\_}s\}$ and divide $W$ into $s$ vertex-disjoint sets as $W_{\_}1,W_{\_}2,\dots W_{\_}s$ such that $W_{\_}i=\{w\in W\ |\ v_{\_}iw\notin E(G)\}$ for each $1\leq i\leq s$ . We claim $W_{\_}i$ is an independent set. Otherwise, let $x$ and $y$ be two adjacent vertices in $W_{\_}i$ . Then, $D^{*}=D\backslash\{v_{\_}i\}\cup\{x,y\}$ is the desired clique. Hence,

\alpha(G)\geq max\{|W_{\_}i|,{1\leq i\leq s}\}>\left(\frac{\epsilon}{3}-\frac{1}{n}\right)n>\delta n,

giving a contradiction. ∎

2.2 Proof of Theorem 1

Recall the graph $F_{\_}{k}(2r+1)$ is the union of $k$ copies of $K_{\_}{2r+1}$ sharing a common vertex. We call the common vertex the center of $F_{\_}{k}(2r+1)$ . For convention, if $k=0$ , $F_{\_}{k}(2r+1)$ is a single vertex graph, which is the center. For any $m$ nonnegative integers $k_{\_}1$ , $k_{\_}2$ , $\dots$ , $k_{\_}m$ , let $F_{\_}{k_{\_}1,k_{\_}2,\dots,k_{\_}m}(2r+1)$ be the graph consisting of a clique $B$ with $m$ vertices and $m$ vertex-disjoint copies of $F_{\_}{k_{\_}i}(2r+1)$ with its center in $B$ . In the above definition, we call $B$ the base of $F_{\_}{k_{\_}1,k_{\_}2,\dots,k_{\_}m}(2r+1)$ .

We now show $RT(n,F_{\_}{k}(2r+1),o(n))=\frac{n^{2}}{2}(1-\frac{1}{r})+o(n^{2})$ , where $r\geq 2$ , as the theorem holds when $r=1$ by Lemma 1. Since $RT(n,K_{\_}{2r+1},o(n))=\frac{n^{2}}{2}(1-\frac{1}{r})+o(n^{2})$ , we only need to show that $RT(n,F_{\_}{k}(2r+1),o(n))\leq\frac{n^{2}}{2}(1-\frac{1}{r})+o(n^{2})$ . Let $\epsilon>0$ be a small positive real number. We will show that there exists a positive number $\delta:=\delta(\epsilon)$ and a large positive integer $N:=N(\epsilon)$ such that all graphs $G(n)$ with $n>N$ , if $|E(G)|\geq(1-\frac{1}{r}+\epsilon)n^{2}/2$ and $\alpha(G(n))\leq\delta n$ , then there exists a $F_{\_}k(2r+1)$ as a subgraph of $G(n)$ . In the proof, we will take $\delta=\frac{\sqrt{2}}{10}\epsilon^{2}$ and we will not specify the value of $N(\epsilon)$ by only assuming $n$ is large enough to ensure our counting work.

Since $|E(G)|\geq\left(1-\frac{1}{r}+\epsilon\right)n^{2}/2$ , by Lemma 2, $G$ contains a subgraph $G_{\_}1$ of order $n_{\_}1\geq\sqrt{\epsilon/2}n$ such that the minimum degree $\delta(G_{\_}1)\geq(1-\frac{1}{r}-\frac{\epsilon}{2})n_{\_}1$ . Since $\alpha(G_{\_}1)\leq\alpha(G)\leq\frac{\sqrt{2}}{10}\epsilon^{2}n\leq\frac{\epsilon}{5}n_{\_}1$ . So, we assume without loss of generality that $\delta(G)\geq(1-\frac{1}{r}-\frac{\epsilon}{2})n$ and $\alpha(G)\leq\frac{\sqrt{\epsilon}}{5}n$ .

For each integer $m$ with $1\leq m\leq r+1$ , let $\mathcal{F}_{\_}{m}$ be the set of $m$ -tuples $(k_{\_}1,k_{\_}2,\dots,k_{\_}m)$ of $m$ nonnegative integers $k_{\_}1$ , $k_{\_}2$ , $\dots$ , $k_{\_}m$ listed with non-increasing order such that $G$ contains a copy of $F_{\_}{k_{\_}1,k_{\_}2,\dots,k_{\_}m}(2r+1)$ . By Lemma 3, $G$ contains a copy of $K_{\_}{m}$ for every $1\leq m\leq r+1$ . Thus, $\mathcal{F}_{\_}m\neq\emptyset$ . We also notice that if there exists an $m$ -tuple $(k_{\_}1,k_{\_}2,\dots,k_{\_}m)\in\mathcal{F}_{\_}m$ with $k_{\_}1\geq k$ , then $G$ contains a copy of $F_{\_}k(2r+1)$ . So, we assume $k_{\_}1\leq k-1$ for all $(k_{\_}1,k_{\_}2,\dots,k_{\_}m)\in\mathcal{F}_{\_}m$ . Consequently, $|\mathcal{F}_{\_}m|\leq m^{k}$ .

For each $\mathcal{F}_{\_}m$ , we define a lexicographic order $\prec$ such that $(k_{\_}1,k_{\_}2,\dots,k_{\_}m)\prec(\ell_{\_}1,\ell_{\_}2,\dots,\ell_{\_}m)$ if there exist $s$ with $1\leq s\leq m$ such that $k_{\_}i=\ell_{\_}i$ for $i<s$ and $k_{\_}i<\ell_{\_}i$ . Clearly, $(0,0,\dots,0)$ is the minimum element in $\mathcal{F}_{\_}m$ following this order.

Let $(k_{\_}1,k_{\_}2,\dots,k_{\_}{r+1})$ be the maximum element in $\mathcal{F}_{\_}{r+1}$ . We will lead a contradiction by showing that either $G$ contains a copy of $F_{\_}{k}(2r+1)$ or $(k_{\_}1,k_{\_}2,\dots,k_{\_}{r+1})$ is not maximum in $\mathcal{F}_{\_}{r+1}$ . Let $F$ be a copy of $F_{\_}{k_{\_}1,k_{\_}2,\dots,k_{\_}{r+1}}(2r+1)$ in $G$ and $B$ be the base. Denote $B$ by $\{v_{\_}1,v_{\_}2,\dots,v_{\_}{r+1}\}$ and assume $v_{\_}i$ is the center of the corresponding $F_{\_}{k_{\_}i}(2r+1)$ in $F$ . Let $H$ be obtained from $G_{\_}1$ by deleting all vertices in $V(F)-B$ , i.e., $H=G-(V(F)-B)$ . Since $|V(F)|\leq 2r(k-1)(r+1)+(r+1)$ and $\delta(G)\geq(1-\frac{1}{r}+\frac{\epsilon}{2})n$ , we may assume the minimum degree $\delta(H)\geq(1-\frac{1}{r}-\frac{\epsilon}{3})|H|$ and $\alpha(H)\leq\frac{\epsilon}{4}|H|$ provided $n$ is large enough.

Starting with clique $B_{\_}0=B$ in $H$ , we apply Lemma 4 repeatedly $r$ times, we get a sequence of cliques $B_{\_}1$ , $B_{\_}2$ , $\dots$ , $B_{\_}r$ such that $|B_{\_}i|=|B_{\_}{i-1}|+1$ and $|B_{\_}i\cap B_{\_}{i-1}|\geq|B_{\_}{i-1}|-1$ . At the end, we have $|B_{\_}r|=|B_{\_}0|+r=(r+1)+r=2r+1$ and $|B_{\_}r\cap B_{\_}0|\geq|B_{\_}0|-r\geq 1$ . Let $s\geq 1$ be the smallest integer such that $v_{\_}s\in B_{\_}r\cap B_{\_}0$ . Then, we get subgraph $F^{*}$ which is a copy of $F_{\_}{k_{\_}1,k_{\_}2,\dots,k_{\_}s+1}(2r+1)$ with base $B^{*}=\{v_{\_}1,v_{\_}2,\dots,v_{\_}s\}$ . Let $H^{*}=G-(V(F^{*})-B^{*})$ .

Applying Lemma 3 repeatedy $(r+1-s)$ times, we get a clique $B^{\prime}\supseteq B^{*}$ with $r+1$ vertices in $H^{*}$ . Combining $B^{\prime}$ with $F^{*}$ , we get a copy of $F_{\_}{k_{\_}1,\dots,k_{\_}s+1,0,\dots,0}(2r+1)$ with base $B^{\prime}$ . Let $\ell_{\_}1,\ell_{\_}2,\dots,\ell_{\_}s$ be an new list of $k_{\_}1,k_{\_}2,\dots,k_{\_}s+1$ in non-increasing order. Then, $(r+1)$ -tuple $(\ell_{\_}1,\ell_{\_}2,\dots,\ell_{\_}s,0,\dots,0)\in\mathcal{F}_{\_}{r+1}$ . But, $(k_{\_}1,k_{\_}2,\dots,k_{\_}{r+1})\prec(\ell_{\_}1,\ell_{\_}2,\dots,\ell_{\_}s,0,\dots,0)$ , giving a contradiction to the maximality of $(k_{\_}1,k_{\_}2,\dots,k_{\_}{r+1})$ .

3 Acknowledgement

We would like to thank Yan Cao for the helpful communication.

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