Balanced Subdivisions of Cliques in Graphs

Abstract

Given a graph H, a balanced subdivision of H is a graph obtained from H by subdividing every edge the same number of times. In 1984, Thomassen conjectured that for each integer $k\ge 1$, high average degree is sufficient to guarantee a balanced subdivision of $K_k$. Recently, Liu and Montgomery resolved this conjecture.We give an optimal estimate up to an absolute constant factor by showing that there exists $c>0$ such that for sufficiently large d, every graph with average degree at least d contains a balanced subdivision of a clique with at least $cd^{1/2}$ vertices. It also confirms a conjecture from Verstraëte: every graph of average degree $cd^2$, for some absolute constant $c>0$, contains a pair of disjoint isomorphic subdivisions of the complete graph $K_d$. We also prove that there exists some absolute $c>0$ such that for sufficiently large d, every $C_4$-free graph with average degree at least d contains a balanced subdivision of the complete graph $K_{cd}$, which extends a result of Balogh, Liu and Sharifzadeh.

A Dependent random choice

The following lemma can be regarded as a bipartite version of dependent random choice, of which the proof follows from that of Fox and Sudakov [7].

Lemma A.1

Given integers $a, t, n_1,n_2, c, r$ and a constant $\alpha >0$, let $G=(V_1,V_2,E)$ be a bipartite graph such that $|V_1|=n_1$, $|V_2|=n_2$ and $|E|\ge \alpha n_1n_2$. If it holds that

$$\begin{aligned} \alpha ^tn_1-\left( {\begin{array}{c}n_1\\ r\end{array}}\right) \left( \frac{c}{n_2}\right) ^t\ge a, \end{aligned}$$

then there exists a set $A_0\subseteq V_1$ of size at least a such that every r-subset of $A_0$ has at least c common neighbours in $V_2$.

Proof

Pick a set of t vertices of $V_2$ uniformly at random with repetition, say $b_1,b_2,\cdots ,b_t$. Let A denote the set of common neighbours for all vertices $b_i$ and $X=|A|$. By linearity of expectation,

$$\begin{aligned} {\mathbb {E}}(X)=\sum _{v\in V_1}{\mathbb {P}}(b_1,\cdots ,b_t\in N(v))=\sum _{v\in V_1}\left( \frac{d(v)}{n_2}\right) ^t. \end{aligned}$$

By the convexity of the function $f(x)=x^t$, we have

$$\begin{aligned} {\mathbb {E}}(X)\ge \frac{n_1(\sum _{v\in V_1}d(v)/n_1)^t}{n_2^t}\ge \frac{n_1(\alpha n_2)^t}{n_2^t}\ge \alpha ^tn_1. \end{aligned}$$

Let Y denote the random variable counting the number of r-subsets in A with fewer than c common neighbours in $V_2$. Therefore, the probability that a randomly chosen $b_i$ is one of the common neighbours of such an r-set is at most $\frac{c}{n_2}$. Hence, since we made random choices of $b_i$ uniformly and independently, the probability that such an r-tuple be contained in A is at most $\left( \frac{c}{n_2}\right) ^t$. As there are at most $\left( {\begin{array}{c}|V_1|\\ r\end{array}}\right)$ subsets of size r, it follows that

$$\begin{aligned} {\mathbb {E}}(Y)\le \left( {\begin{array}{c}|V_1|\\ r\end{array}}\right) \left( \frac{c}{n_2}\right) ^t. \end{aligned}$$

Again, by linearity of expectation, it holds that

$$\begin{aligned} {\mathbb {E}}(X-Y)\ge \alpha ^tn_1-\left( {\begin{array}{c}n_1\\ r\end{array}}\right) \left( \frac{c}{n_2}\right) ^t\ge a. \end{aligned}$$

Hence there exists a choice of A for which $X-Y\ge a$. Delete one vertex from each subset r-subset of A with fewer than c common neighbours and let $A_0$ be the remaining subset of A. Thus, $|A_0|\ge a$ and every r-subset of $A_0$ has at least c common neighbours. $\square$