\( \def\<#1>{\left<#1\right>} \def\f{\mathcal{F}} \newcommand{\R}{\mathcal{R}} \def\aveDeg#1#2{d_{#2}(#1)} \def\minDeg#1#2{\delta_{#2}(#1)} \def\proj#1#2{{#1}_{#2}} \def\fact#1#2{f_{#1}\left(#2\right)} \newcommand{\ffact}[1]{f_{#1}} \def\euler{\mathrm{e}} \def\rect{{\cal R}} \def\dens#1{\alpha_{#1}} \)

Two extremal properties of sets of vectors

Here I present two properties of sets of vectors, much in the spirit of Sects. 2.4, 2.5 and Ex. 2.7 in the book (2nd edition).

Thinkness lemma

Suppose that the set $E$ of edges of a bipartite $n\times n$ graph has the property that, in average, an edge $e\in E$ shares a common left vertex as well as a common right vertex with at least $\alpha n$ edges of $E$. Then there exists a subset $E'\subseteq E$ of $|E'|\geq |E|/4$ edges such that every edge of $E'$ shares a common left vertex as well as a common right vertex with at least $\alpha n/4$ edges of $E'$. This is a direct consequence of the following more general "thinkness lemma" (Corollary 1).

Let $A\subseteq [n]^r$ a set of vectors $a=(a_1,,\ldots,a_r)$ with all $a_i\in[n]=\{1,\ldots,n\}$. The projection of a vector $a=(a_1,\ldots,a_r)$ onto a set of coordinates $S=\{i_1, < \ldots < i_k\}$ is the vector $\proj{a}{S}= (a_{i_1},\ldots,a_{i_k})$. The projection of a set of vectors $A$ is the set $\proj{A}{S}=\left\{\proj{a}{S}\,:\, a\in A\right\}$. Note that \[ |\proj{A}{S}|\leq |A|\leq n^{r-|S|}\cdot |\proj{A}{S}| \] holds for all $S\subset[n]$. For a vector $x\in \proj{A}{S}$, the extensions of $x$ in $A$ are all vectors $a\in A$ with $\proj{a}{S}=x$. Note that $x$ has at least one and at most $n^{r-|S|}$ extensions in $A$. The average number of extensions of a vector of $\proj{A}{S}$ in $A$ is \[ \aveDeg{A}{S}=\frac{|A|}{|\proj{A}{S}|} \] The minimum number, $\minDeg{A}{S}$, of extensions of vectors in $\proj{A}{S}$ is the minimum, over all vectors $x\in\proj{A}{S}$, of the number of extensions of $x$ in $A$. Note that \[ 1\leq \minDeg{A}{S}\leq \aveDeg{A}{S}\leq n^{r-|S|} \] holds for every $S\subseteq[r]$. For a family $\f\subseteq 2^{[r]}$ of sets of positions, let $\minDeg{A}{\f}=\min\{\minDeg{A}{S}\colon S\in\f\}$.

The following lemma is a sight generalization of a more specific fact (Corollary 2 below) proved by Ran Raz and Pierre McKenzie [1]; a simpler argument given below is due to Mike Saks.

Thinkness Lemma: Let $A\subseteq[n]^r$ be a set of vectors, $0<\alpha <1$ a real number, and $\f\subseteq 2^{[r]}$ a family of sets of positions, each of size $k$. If $\aveDeg{A}{S}\geq \alpha\cdot n^{r-k}$ holds for all $S\in\f$, then for every $\beta\geq 0$ there is a subset $B\subseteq A$ of $|B|\geq (1-\beta)|A|$ vectors such that $\minDeg{B}{}\geq \gamma\cdot n^{r-k}$ for $\gamma=\alpha\beta/|\f|$.

Proof: The lemma is trivial for $\beta=0$; so, assume that $\beta>0$. Define a sequence $A=A^0\supset A^1\supset \ldots$ of subsets of $A$ by the following procedure.

  1. If $|A^j|<(1-\beta)|A|$, then stop.
  2. If $\minDeg{A}{}\geq \Delta:=\alpha\beta n^{r-k}/|\f|$, then stop.
  3. Otherwise, there is some set $A\in\f$ and a vector $x$ in the projection of $A^j$ onto $S$ such that $x$ has $ < \Delta$ extensions in $A^j$. To obtain the next set $A^{j+1}$, we remove from $A^j$ all extensions of the vector $x$ in $A^j$.
The last set, $A^l$, in this sequence is the desired set $B$. To show that $A^l$ satisfies the requirements, it is enough to show that the procedure will always stop because of the second condition (2), and never because of the first condition (1) (hence, $B=A^l$ has both required properties).

Note that when going from $A^j$ to $A^{j+1}$, at least one vector $x$ is removed from the projection of $A^j$ onto the set $S$. So, the total number $l$ of steps satisfies \[ l\leq \sum_{S\in\f} |\proj{A}{S}| =\sum_{S\in\f}\frac{|A|}{\aveDeg{A}{S}}\,, \] and since $\aveDeg{A}{S}\geq \alpha n^{r-k}$ holds for every $i$, we can conclude that the procedure will (definitely) stop after \[ l\leq \frac{|\f|\cdot|A|}{\alpha n^{r-k}} \] steps. On the other hand, $A^{j+1}$ is obtained from $A^j$ by removing at most $\Delta$ vectors. So, the total number of vectors removed from $A$ in all $l$ steps to get $A^l$ is at most \[ l\cdot \Delta\leq \frac{|\f|\cdot |A|}{\alpha n^{r-k}}\cdot\Delta =\frac{|\f|\cdot |A|}{\alpha n^{r-k}}\cdot \frac{\alpha\beta n^{r-k}}{|\f|}=\beta|A|\,. \] Thus, when the procedure stops after $l$ steps, $|A^l|\geq (1-\beta)|A|$ still holds (the first stopping condition (1) is not fulfilled). So, since the procedure stopped, the second condition $\minDeg{A^l}{}\geq \Delta$ is fulfilled. $\Box$

Corollary 1: Let $G=(V_1\cup V_2,E)$ be a bipartite $n\times n$ graph such that the average degrees of non-isolated vertices in $V_1$ and in $V_2$ is at least $\alpha n$. Then for every $\beta\geq 0$, there exists a subset $E'\subseteq E$ of $|E'|\geq (1-\beta)|E|$ edges such that the the degree of every non-isolated vertex of $G'=(V_1\cup V_2,E')$ is at least $\alpha\beta n/2$.

Proof: Apply the thinkness lemma with $r=2$ and $\f=\{\{1\},\{2\}\}$. $\Box$

Let $A\subseteq[n]^r$. Given a position $1\leq i\leq n$, the $i$-th neighbor of a vector $a\in A$ is a vector $b\in A$ such that $b_j=a_j$ for all $j\neq i$.

Corollary 2: Let $A\subseteq[n]^r$ and $0<\alpha <1$. If for every position $i\in [r]$ the average number of $i$-th neighbors of a vector of $A$ is at least $\alpha n$, then there is a subset $B\subseteq A$ of $|B|\geq (1-\beta)|A|$ vectors such that, for every position $i\in [r]$, every vector of $B$ has at least $ \alpha\beta n/r$ $i$-th neighbors in $B$.
Proof: Apply the thinkness lemma with $\f=\{[r]-\{i\}\colon i=1,\ldots,r\}$. $\Box$

Multidimensional Zarankiewicz problem

An $(r,k)$-rectangle (or simply, $k$-rectangle if its dimension $r$ is clear from the context) is a Cartesian product of $r$ (not necessarily distinct) sets, each of cardinality $k$. Let $1\leq k\leq m$ and $r\geq 1$ be integers, and $V_1,\ldots,V_r$ be finite sets of size $|V_1|=\ldots=|V_r|=n$. Note that the total number of $k$-rectangles in $V_1\times \cdots\times V_r$ is $\binom{n}{k}^r$: just pick $k$-element subsets in each of the blocks $V_1,\ldots,V_r$. For a real number $0\leq \alpha\leq 1$, define $\fact{0}{\alpha}:=\alpha$ and \[ \fact{i+1}{\alpha}:= \binom{\fact{i}{\alpha}\cdot n}{k} \Big/\binom{n}{k}\,. \] Since $(a/b)^b\leq \binom{a}{b}\leq (\euler a/b)^b$, we have $\fact{r}{\alpha}\ge (\alpha/\euler)^{k^r}$. The following lower bound on the number of rectangles was proved by Paul Erdös,  and Joel Spencer [2].
Rectangle Lemma: Every subset $X\subseteq V_1\times \cdots\times V_r$ of $|X|\geq \alpha n^r$ vectors contains at least $\fact{r}{\alpha}\cdot \binom{n}{k}^r$ $k$-rectangles.
The proof generalizes the double-counting argument of Kövari, Sós and Turán for bipartite graphs (see Sect. 2.4 in the book): the idea is to use (in a clever way) the convexity of the function $g(x)=\binom{x}{k}$ (with $\binom{x}{k}=0$ for $ x< k-1$) together with the Jensen inequality for convex functions $g(x)$: \begin{equation}\label{eq:1} \sum_{i=1}^m g(x_i)\geq m\cdot g\Big(\frac{1}{m}\sum_{i=1}^m x_i\Big)\,. \end{equation} Proof: We proceed by induction on the dimension $r$. The basis case $r=1$ is trivial. Assume the theorem holds for $r$, and take a set $X\subseteq V_1\times \cdots\times V_{r+1}$ of $|X|\geq \alpha n^{r+1}$ vectors. For $v\in V_{r+1}$, let $\dens{v}:=|X_v|/n^r$ be the density of the set of vectors \[ X_v=\{(v_1,\ldots,v_{r})\colon (v_1,\ldots,v_{r},v)\in X\}\,. \] Then we have \[ \sum_{v\in V_{r+1}}\dens{v}\cdot n^r=|X|\geq \alpha n^{r+1}\,, \] and hence, the average density is \begin{equation}\label{eq:2} \frac{1}{n}\sum_{v\in V_{r+1}}\dens{v}\geq \alpha\,. \end{equation} Let $\rect$ be the set of all $|\rect|=\binom{n}{k}^r$ $k$-rectangles in $V_1\times\cdots\times V_r$, and consider the bipartite graph $G\subseteq \rect\times V_{r+1}$ such that $(R,v)\in G$ if and only if $R\times \{v\}\subseteq X$. Since $|X_v|\geq \dens{v}\cdot n^r$ holds for every point $v\in V_{r+1}$, the induction hypothesis implies that each set $X_v$ must contain at least \begin{equation}\label{eq:3} d_v:=\fact{r}{\dens{v}}\cdot\binom{n}{k}^r=\fact{r}{\dens{v}}\cdot|\rect| \end{equation} $k$-rectangles, and hence, at least so many edges of $G$ must be incident with the vertex $v$. Since the function $h(x)=\fact{i}{x}$ is convex, the inequalities \eqref{eq:1}-\eqref{eq:3} (with $g(x):= \fact{n}{x}$, and $x_v:=\dens{v}$) yield the following lower bound on the number $|G|$ of edges in $G$: \begin{equation}\label{eq:4} |G|\geq \sum_{v\in V_{r+1}} d_v\overset{\eqref{eq:3}}{=} |\rect|\sum_{v\in V_{r+1}}\fact{r}{\dens{v}}\overset{\eqref{eq:1}}{\geq} | \rect|\cdot n\cdot \ffact{r}\bigg(\frac{1}{n}\sum_{v\in V_{r+1}}\dens{v}\bigg) \overset{\eqref{eq:2}}{\geq} |\rect|\cdot n\cdot \fact{r}{\alpha}\,. \end{equation} For a rectangle $R\in\rect$, let $\dens{R}$ denote its degree in the graph $G$; hence, $\sum_{R\in\rect}\dens{R}=|G|$. By the definition of the graph $G$, for every rectangle $R\in\rect$ and for every subset $S\subseteq V_{r+1}$ of $|S|=k$ its neighbors in $G$, the Cartesian product $R\times S$ is a $k$-rectangle lying in our set $X$. So, we can again use the Jensen inequality \eqref{eq:1} with $g(x):=\binom{x}{k}$ and $m:=|\rect|=\binom{n}{k}^{r}$ to conclude that the number of $k$-rectangles in $X$ is at least \begin{align*} \sum_{R\in\rect}\binom{\dens{R}}{k}\overset{\eqref{eq:1}}{\geq} |\rect|\binom{\tfrac{1}{|\rect|}\sum_{R\in\rect}\dens{R}}{k} \overset{\eqref{eq:4}}{\geq} |\rect| \binom{n\cdot \fact{r}{\alpha}}{k} =\binom{n}{k}^r\binom{n\cdot \fact{r}{\alpha}}{k}\,, %\qedhere \end{align*} which is exactly $\fact{r+1}{\alpha}\cdot \binom{n}{k}^{r+1}$, by the definition of functions $\ffact{i}$. $\Box$

Corollary 3: If $\log n\geq k^r+\log k$, then any subset $A\subseteq [n]^r$ of $|A|\geq 4\left(\frac{n}{2^k}\right)^r$ vectors contains at least one $(r,k)$-rectangle.
Proof: We have $|A|\geq \alpha n^r$ for $\alpha:=2^{2-kr}$. We also have $\fact{n}{\alpha}\ge (\alpha/\euler)^{k^r}\geq 2^{-krk^r}$. By the rectangle lemma, the set $A$ contains at least $\fact{r}{\alpha}\cdot \binom{n}{k}^r\geq \fact{r}{\alpha}\cdot (n/k)^k$ $(n,k)$-rectangles. So, it is enough to ensure that $(n/k)^{kr}\geq (\euler/\alpha)^{k^r}$. By taking logarithms, this turns into the inequality \[ kr\log n\geq k^r\log(\euler/\alpha)+kr\log k\,. \] After the division by $kr$, we obtain that $\log n$ needs only be at least $\gamma k^r+\log k$ for $\gamma= \log(\euler/\alpha)/kr$. By our condition on $\alpha$, $\gamma$ is at most $1$, so the inequality holds. $\Box$

References:

  1. R. Raz and P. McKenzie, Separation of the Monotone NC Hierarchy, Combinatorica, 19 (1999) 403�435.
  2. P. Erdös,  and J. Spencer, Probabilistic Methods in Combinatorics, Akadmiai Kiad�o, 1974.

S. Jukna, October 16, 2020