A remark on the paper "Properties of intersecting families of ordered sets" by O. Einstein

TL;DR

This paper discusses a correction to a previous theorem on intersecting families of ordered sets and subspaces, providing two weaker variants due to an identified mistake in the original proof.

Contribution

It identifies a mistake in Einstein's original proof and offers two alternative weaker theorems for intersecting families of ordered subspaces.

Findings

Original theorem's proof contains a mistake.

Two weaker variants of the theorem are proven.

Clarifies the limitations of the original results.

Abstract

O. Einstein (2008) proved Bollob\'as-type theorems on intersecting families of ordered sets of finite sets and subspaces. Unfortunately, we report that the proof of a theorem on ordered sets of subspaces had a mistake. We prove two weaker variants.

Equations

$$\begin{array}{cccc}{U}_{11}& U_{12}& \cdots & U_{1k}\\ U_{21}& U_{22}& \cdots & U_{2k}\\ \cdots & \cdots & ⋮& \cdots \\ U_{m1}& U_{m2}& \cdots & U_{mk}\end{array}$$

$$m\le \frac{({\sum }_{r=1}^k{\ell }_r)!}{{\prod }_{r=1}^k{\ell }_r!}.$$

$$\begin{array}{cccc}{U}_{11}& U_{12}& \cdots & U_{1k}\\ U_{21}& U_{22}& \cdots & U_{2k}\\ \cdots & \cdots & ⋮& \cdots \\ U_{m1}& U_{m2}& \cdots & U_{mk}\end{array}$$

$$m\le \frac{({\sum }_{r=1}^k{\ell }_r)!}{{\prod }_{r=1}^k{\ell }_r!}.$$

$$\begin{array}{cccc}{A}_{11}& A_{12}& \cdots & A_{1k}\\ A_{21}& A_{22}& \cdots & A_{2k}\\ \cdots & \cdots & ⋮& \cdots \\ A_{m1}& A_{m2}& \cdots & A_{mk}\end{array}$$

$$m\le \frac{({\sum }_{r=1}^k{\ell }_r)!}{{\prod }_{r=1}^k{\ell }_r!}.$$

$$m\le \frac{{\prod }_{1\le a<b\le k}({\ell }_a+{\ell }_b)!}{({\prod }_{r=1}^k{\ell }_r!{)}^{k-1}}.$$

$$\begin{array}{cccc}{U}_{11}& U_{12}& \cdots & U_{1k}\\ U_{21}& U_{22}& \cdots & U_{2k}\\ \cdots & \cdots & ⋮& \cdots \\ U_{m1}& U_{m2}& \cdots & U_{mk}\end{array}$$

$$m\le \frac{[({\sum }_{r=1}^k{\ell }_r)-kt]!}{{\prod }_{r=1}^k({\ell }_r-t)!}.$$

$$\begin{array}{cccc}{A}_{11}& A_{12}& \cdots & A_{1k}\\ A_{21}& A_{22}& \cdots & A_{2k}\\ \cdots & \cdots & ⋮& \cdots \\ A_{m1}& A_{m2}& \cdots & A_{mk}\end{array}$$

$$m\le \frac{[({\sum }_{r=1}^k{\ell }_r)-kt]!}{{\prod }_{r=1}^k({\ell }_r-t)!}.$$

$$m\le \frac{{\prod }_{1\le a<b\le k}({\ell }_a+{\ell }_b-2t)!}{({\prod }_{r=1}^k({\ell }_r-t)!{)}^{k-1}}.$$

Full text

1 Introduction

The following theorem generalizing the theorem of Bollobás [2] is well known and proved by using the wedge product method (see [1]).

Theorem 1** (Lovász [6]; skew version).**

Let $a$, $b$ be positive integers. Let $U_{1},U_{2},\ldots,U_{m}$, $V_{1},V_{2},\ldots,V_{m}$ be subspaces satisfying the following:

1. (i)

$\dim U_{i}\leq a$* and $\dim V_{i}\leq b$ for all $i=1,2,\ldots,m$.* 2. (ii)

$U_{i}\cap V_{i}=\{0\}$* for all $i=1,2,\ldots,m$.* 3. (iii)

$U_{i}\cap V_{j}\neq\{0\}$* for all $1\leq i<j\leq m$.*

Then $m\leq\binom{a+b}{a}$.

Ori Einstein [3] published a paper on a generalization of the above theorem and its consequence on finite sets by Frankl [4]. We will show that his proof of Theorem 2.7 in [3] is incorrect and so we state it as a conjecture.

Conjecture 1** (Theorem 2.7 of [3]).**

Let $\ell_{1}$, $\ell_{2}$, $\ldots$, $\ell_{k}$ be positive integers. Let $U$ be a linear space over a field $\mathbb{F}$. Consider the following matrix of subspaces:

[TABLE]

If these subspaces satisfy:

1. (i)

for every $1\leq j\leq k$, $1\leq i\leq m$, $\dim U_{ij}\leq\ell_{j}$; 2. (ii)

for every fixed $i$, all subspaces $U_{ij}$ are pairwise disjoint; 3. (iii)

for each $i<i^{\prime}$, there exist some $j<j^{\prime}$ such that $U_{ij}\cap U_{i^{\prime}j^{\prime}}\neq\{0\}$;

then

[TABLE]

Here is the overview of this note. In the next section, we will sketch the reason why the proof of Theorem 2.7 in [3] is incorrect and present a weaker theorem (Theorem 3) obtained by tightening condition (ii). In Section 3, we prove another weaker theorem (Theorem 4), by providing a weaker upper bound for $m$ instead of modifying any assumptions. Section 4 will discuss the threshold versions.

2 The mistake and its first remedy

Let us first point out the mistake in the proof of Conjecture 1 in [3]. As it is typical in the wedge product method, we take $v_{i}=\bigwedge_{j=1}^{k-1}\wedge T_{j}(U_{ij})$ and $w_{i}=\bigwedge_{j=1}^{k-1}\bigwedge_{r=j+1}^{k}\wedge T_{j}(U_{ir})$ for some linear transformations $T_{1}$, $T_{2}$, $\ldots$, $T_{k-1}$. Then the following claim is made:

Claim** (Page 41 in [3]).**

For every $i\leq i^{\prime}$, $v_{i}\wedge w_{i^{\prime}}\neq 0$ if and only if $i=i^{\prime}$.

This claim is false in general. For instance, if $U_{11}\cap(U_{12}+U_{13}+\cdots+U_{1,k-1})\neq\{0\}$, then $\wedge T_{1}(U_{11})\wedge\bigwedge_{r=2}^{k}\wedge T_{1}(U_{1r})=0$ and therefore $v_{1}\wedge w_{1}^{\prime}=0$. The crucial mistake is that condition (ii) in Conjecture 1 does not imply that $\dim(U_{i1}+U_{i2}+\cdots+U_{ik})=\sum_{j=1}^{k}\dim U_{ij}$. (For instance the spans of $\left(\begin{smallmatrix}1\\ 0\end{smallmatrix}\right)$, $\left(\begin{smallmatrix}0\\ 1\end{smallmatrix}\right)$, and $\left(\begin{smallmatrix}1\\ 1\end{smallmatrix}\right)$ are pairwise disjoint and yet their sum has dimension $2$ only.)

If $\dim(U_{i1}+U_{i2}+\cdots+U_{ik})=\sum_{j=1}^{k}\dim U_{ij}$, then the claim is true and so we can recover the following weaker theorem by the proof in [3].

Theorem 2**.**

Let $\ell_{1}$, $\ell_{2}$, $\ldots$, $\ell_{k}$ be positive integers. Let $U$ be a linear space over a field $\mathbb{F}$. Consider the following matrix of subspaces:

[TABLE]

If these subspaces satisfy:

1. (i)

for every $1\leq j\leq k$, $1\leq i\leq m$, $\dim U_{ij}\leq\ell_{j}$; 2. (ii)

for every fixed $i$, $\dim(\sum_{j=1}^{k}U_{ij})=\sum_{j=1}^{k}\dim U_{ij}$; 3. (iii)

for each $i<i^{\prime}$, there exist some $j<j^{\prime}$ such that $U_{ij}\cap U_{i^{\prime}j^{\prime}}\neq\{0\}$;

then

[TABLE]

Though Theorem 2 is weaker than Conjecture 1, it allows us to recover Theorem 2.8 of [3].

Theorem 3** (Theorem 2.8 of [3]).**

Let $\ell_{1}$, $\ell_{2}$, $\ldots$, $\ell_{k}$ be positive integers. Consider the following matrix of sets:

[TABLE]

If these sets satisfy:

1. (i)

for every $1\leq j\leq k$, $1\leq i\leq m$, $\lvert A_{ij}\rvert\leq\ell_{j}$; 2. (ii)

for every fixed $i$, all sets $A_{ij}$ are pairwise disjoint; 3. (iii)

for each $i<i^{\prime}$, there exist some $j<j^{\prime}$ such that $A_{ij}\cap A_{i^{\prime}j^{\prime}}\neq\emptyset$;

then

[TABLE]

Note that Theorem 3 implies that Conjecture 1 is true when $\ell_{1}=\ell_{2}=\cdots=\ell_{k}=1$.

3 Second remedy

Naturally we ask whether Conjecture 1 can be proven with some upper bound on $m$. Here we show that this is possible, while generalizing Theorem 1.

Theorem 4**.**

Under the same assumptions of Conjecture 1, we have

[TABLE]

Proof.

We may assume that $\dim U_{ij}=\ell_{j}$ for all $i$, $j$ and $\mathbb{F}$ is infinite. Let $V=V_{1,2}\oplus V_{1,3}\oplus\cdots\oplus V_{1,k}\oplus\cdots\oplus V_{2,3}\oplus\cdots\oplus V_{k-1,k}=\bigoplus_{a=1}^{k-1}\bigoplus_{b=a+1}^{k}V_{a,b}$ be a $\sum_{a=1}^{k-1}\sum_{b=a+1}^{k}(\ell_{a}+\ell_{b})$-dimensional vector space over $\mathbb{F}$, decomposed into the direct sum of subspaces $V_{a,b}$, each of dimension $\ell_{a}+\ell_{b}$. By Corollary 3.14 of [1], for all $i<j$, there exists a linear transformation $T_{ab}:U\to V_{a,b}$ such that for all $1\leq i\leq m$, $\dim T_{ab}(U_{ia})=\ell_{a}$, $\dim T_{ab}(U_{ib})=\ell_{b}$, and $\dim T_{ab}(U_{ia})\cap T_{ab}(U_{jb})=\dim U_{ia}\cap U_{jb}$ for all $1\leq i,j\leq m$. Finally, for each $1\leq i\leq m$, let $v_{i}=\bigwedge_{a=1}^{k-1}\bigwedge_{b=a+1}^{k}\wedge T_{ab}(U_{ia})$ and $w_{i}=\bigwedge_{a=1}^{k-1}\bigwedge_{b=a+1}^{k}\wedge T_{ab}(U_{ib})$.

We claim that for $i\leq i^{\prime}$, $v_{i}\wedge w_{i^{\prime}}\neq 0$ if and only if $i=i^{\prime}$. If $i<i^{\prime}$, then there exist $1\leq j<j^{\prime}\leq k$ such that $U_{ij}\cap U_{i^{\prime}j^{\prime}}\neq\{0\}$. By the choice of $T_{jj^{\prime}}$, $T_{jj^{\prime}}(U_{ij})\cap T_{jj^{\prime}}(U_{i^{\prime}j^{\prime}})\neq\{0\}$ and so $(\wedge T_{jj^{\prime}}(U_{ij}))\wedge(\wedge T_{jj^{\prime}}(U_{i^{\prime}j^{\prime}}))=0$, which implies that $v_{i}\wedge w_{i^{\prime}}=0$. If $i=i^{\prime}$, then $v_{i}\wedge w_{i^{\prime}}$ is the wedge product of disjoint subspaces and so $v_{i}\wedge w_{i^{\prime}}\neq 0$.

Therefore $v_{1},v_{2},\ldots,v_{m}$ are linearly independent in the space $\bigwedge_{a=1}^{k-1}\bigwedge_{b=a+1}^{k}\bigwedge^{\ell_{a}}V_{a,b}$, whose dimension is $\prod_{a=1}^{k-1}\prod_{b=a+1}^{k}\binom{\ell_{a}+\ell_{b}}{\ell_{a}}=\frac{\prod_{1\leq a<b\leq k}(\ell_{a}+\ell_{b})!}{(\prod_{i=1}^{k}\ell_{i}!)^{k-1}}$. This proves that $m\leq\frac{\prod_{1\leq a<b\leq k}(\ell_{a}+\ell_{b})!}{(\prod_{i=1}^{k}\ell_{i}!)^{k-1}}$. ∎

4 Threshold versions

The paper [3] uses Conjecture 1 to deduce the threshold versions (Lemma 2.9 and Theorem 2.10) to generalize a result of Füredi [5]. We do not know how to prove Lemma 2.9 and Theorem 2.10 of [3] and so we leave them as conjectures. It is not clear how one can relax conditions in Lemma 2.9 and Theorem 2.10 of [3], while avoiding ugly conditions from (ii) of Theorem 3. (A necessary condition $\ell_{i}\geq t$ was missing in [3].)

Conjecture 2** (Lemma 2.9 of [3]).**

Let $\ell_{1}$, $\ell_{2}$, $\ldots$, $\ell_{k}$ be positive integers such that $\ell_{i}\geq t$ for all $i$. Let $U$ be a linear space over a field $\mathbb{F}$. Consider the following matrix of subspaces:

[TABLE]

If these subspaces satisfy:

1. (i)

for every $1\leq j\leq k$, $1\leq i\leq m$, $\dim U_{ij}\leq\ell_{j}$; 2. (ii)

for every fixed $i$, $\dim(U_{ij}\cap U_{ij^{\prime}})\leq t$; 3. (iii)

for each $i<i^{\prime}$, there exists some $j<j^{\prime}$ such that $\dim(U_{ij}\cap U_{i^{\prime}j^{\prime}})>t$;

then

[TABLE]

Conjecture 3** (Theorem 2.10 of [3]).**

Let $\ell_{1}$, $\ell_{2}$, $\ldots$, $\ell_{k}$ be positive integers such that $\ell_{i}\geq t$ for all $i$. Consider the following matrix of sets:

[TABLE]

If these sets satisfy:

1. (i)

for every $1\leq j\leq k$, $1\leq i\leq m$, $\lvert A_{ij}\rvert\leq\ell_{j}$; 2. (ii)

for every $i$, $j$ and $j^{\prime}$, $\lvert A_{ij}\cap A_{ij^{\prime}}\rvert\leq t$; 3. (iii)

for each $i<i^{\prime}$, there exists some $j<j^{\prime}$ such that $\lvert A_{ij}\cap A_{i^{\prime}j^{\prime}}\rvert>t$;

then

[TABLE]

By using Theorem 4, we can prove the following weaker variants of Conjectures 2 and 3 by the same reduction in [3].

Theorem 5**.**

Under the same assumptions of Conjecture 2, we have

[TABLE]

Theorem 6**.**

Under the same assumptions of Conjecture 3, we have

[TABLE]

Acknowledgement

The author would like to thank students attending the graduate course on combinatorics in the spring semester of 2017; they pointed out the difficulty, when they were given Conjecture 1 for $k=3$ as a homework problem. We also like to thank the anonymous referees for their helpful comments.