combinatorial uniqueness of Hesse Configuration

In this article, we will show that a collection of objects which has the incidence structure of a Hesse configuration is unique up to relabeling.

Definition 1.

An abstract Hesse configuration is a pair of sets $\mathrm{(}P\mathrm{,}L\mathrm{)}$ which satisfies the following conditions:

1. $P$ has nine elements

2. Every element of $L$ is a subset of $P$ with three elements.

3. For any two distinct elements $x,y$ of $P$, there exists exactly one element of $L$ which contains both $x$ and $y$.

Theorem 1.

If $\mathrm{(}P\mathrm{,}L\mathrm{)}$ is an abstract Hesse configuration, then $L$ has 12 elements.

Proof.

Let $D$ be the set of all two-element subsets of $P$. Then $D$ has $\left(\genfrac{}{}{0pt}{}{9}{2}\right)=36$ elements. Each element of $L$ is a subset of $P$ with three elements, hence has $\left(\genfrac{}{}{0pt}{}{3}{2}\right)=3$ subsets of cardinality 2. By the definition above, every element of $D$ must be a subset of exactly one element of $L$. For this to be possible, $L$ must have cardinality $36/3=12$. ∎

Theorem 2.

If $\mathrm{(}P\mathrm{,}L\mathrm{)}$ is an abstract Hesse configuration then, for every $p\mathrm{\in }P$, there exist exactly four elements $\mathrm{\ell }\mathrm{\in }L$ such that $p\mathrm{\in }\mathrm{\ell }$.

Proof.

To every $q\in P$ such that $q\ne p$, there exists exactly one $\mathrm{\ell }\in L$ such that $p\in \mathrm{\ell }$ and $q\in \mathrm{\ell }$. Furthermore, for every $\mathrm{\ell }\in L$ such that $p\in \mathrm{\ell }$, there will be exactly two elements of $\mathrm{\ell }$ other than $p$. Hence, there exist $(9-1)/2=4$ elements $\mathrm{\ell }\in L$ such that $p\in \mathrm{\ell }$. ∎

Theorem 3.

If $\mathrm{(}P\mathrm{,}L\mathrm{)}$ is an abstract Hesse configuration and $\mathrm{\ell }\mathrm{\in }L$, then there exist $m\mathrm{,}n\mathrm{\in }L$ such that $\mathrm{\ell }\mathrm{\cap }m\mathrm{=}\mathrm{\ell }\mathrm{\cap }n\mathrm{=}m\mathrm{\cap }n\mathrm{=}\mathrm{\varnothing }$.

Proof.

By the foregoing result, given $p\in \mathrm{\ell }$, there are four elements of $L$ to which $p$ belongs. One of these, of course, is $\mathrm{\ell }$ itself, and the other three are distinct from $\mathrm{\ell }$. Since $\mathrm{\ell }$ has three elements, this means that there are at most $3\cdot 3+1=10$ elements $k\in L$ such that $P\cap L\ne \mathrm{\varnothing }$. Because $L$ has 12 elements. there must exist $m,n\in L$ such that $\mathrm{\ell }\cap m=\mathrm{\ell }\cap n=\mathrm{\varnothing }$.

It remains to show that $m\cap n=\mathrm{\varnothing }$. Suppose to the contrary that there exists a $p$ such that $p\in m$ and $p\in n$. Since $m\cap \mathrm{\ell }=\mathrm{\varnothing }$, it follows that $p\notin \mathrm{\ell }$, hence there will exist three distinct elements of $L$ containing $p$ and an element of $\mathrm{\ell }$. Because $\mathrm{\ell }\cap m=\mathrm{\ell }\cap n=\mathrm{\varnothing }$, these three elements are distinct from $m$ and $n$. That makes for a total of five distinct elements of $L$ containing $p$, which contradicts the previous theorem, hence $m\cap n=\mathrm{\varnothing }$. ∎

Theorem 4.

If $m\mathrm{,}n\mathrm{,}k$ are elements of $L$ such that $m\mathrm{\cap }n\mathrm{=}n\mathrm{\cap }k\mathrm{=}k\mathrm{\cap }m\mathrm{=}\mathrm{\varnothing }$ and $\mathrm{\ell }\mathrm{\in }L\mathrm{\setminus }\mathrm{\{}m\mathrm{,}n\mathrm{,}k\mathrm{\}}$, then $\mathrm{\ell }$ has exactly one element in common with each of $m\mathrm{,}n\mathrm{,}k$.

Proof.

Since each element of $L$ is a subset of $P$ with three elements and $m,n,k$ are pairwise disjoint but $P$ only has nine elements, it follows that every element of $P$ must belong to exactly one of $m,n,k$. In particular, this means that every element of $\mathrm{\ell }$ must belong to one of $m,n,k$. Were two elements of $\mathrm{\ell }$ to belong to the same element of $\{m,n,k\}$ then, by the third defining property, that element would have to equal $\mathrm{\ell }$, contrary to its definition. Hence, each element of $\mathrm{\ell }$ must belong to a distinct element of $\{m,n,k\}$. ∎

Theorem 5.

If $\mathrm{(}P\mathrm{,}L\mathrm{)}$ is an abstract Hesse configuration, then we can label the elements of $P$ as A,B,C,D,E,F,G,H,I in such a way that the elements of $L$ are

$$\{A,B,C\},\{D,E,F\},\{G,H,I\},$$

$$\{A,D,G\},\{B,E,H\},\{C,F,I\},$$

$$\{D,H,C\},\{A,E,I\},\{B,F,G\},$$

$$\{B,D,I\},\{C,E,G\},\{A,F,H\}.$$

Proof.

By theorem 3, there exist $a,b,c\in L$ such that $a\cap b=b\cap c=c\cap a=\mathrm{\varnothing }$. Since $L$ has twelve elements, there must exist an a elemetn of $L$ distinct from $a,b,c$. Pick such an element and call it $d$. By another application of theorem 3, there must exist $e,f\in L$ such that $d\cap e=e\cap f=f\cap d=\mathrm{\varnothing }$.

By theorem 4, $a$ must have exactly one element in common with each of $d,e,f$; let $A$ the element it has in common with $d$, $B$ be the element it has in common with $e$ and $C$ be the element it has in common with $f$. Likewise, $b$ must have exactly one element in common with each of $d,e,f$, as must $c$. Let $D$ be the element $b$ has in common with $d$, $E$ be the element $b$ has in common with $e$, $F$ be the element $b$ has in common with $f$, $G$ be the element $c$ has in common with $d$, $H$ be the element $c$ has in common with $e$ and $I$ be the element $c$ has in common with $f$.

Summarizing what we just said another way, we have assigned labels $A,B,C,D,E,F,G,H,I$ to the elements of $P$ in such a way that

$$a=\{A,B,C\},b=\{D,E,F\},c=\{G,H,I\},$$

$$d=\{A,D,G\},e=\{B,E,H\},f=\{C,F,I\}.$$

That is half of what we set out to do; we must still label the remaining six elements of $L$.

By theorem 4, if $\mathrm{\ell }\in L\setminus \{a,b,c,d,e,f\}$, then $\mathrm{\ell }$ must have exactly one element in common with each of $a,b,c$ and exactly one element in common with each of $d,e,f$.

Suppose that $A\in \mathrm{\ell }$. It could not be the case that $D\in \mathrm{\ell }$ because then $\mathrm{\ell }$ would have two elements in common with $d$. Since $\mathrm{\ell }$ must have one element in common with $b$, that means that either $E\in \mathrm{\ell }$ or $F\in \mathrm{\ell }$. If $A,E\in \mathrm{\ell }$, then the element $\mathrm{\ell }$ has in common with $c$ cannot be $G$ because $\mathrm{\ell }$ would have both $A$ and $G$ in common with $c$ and it cannot be $H$ because $\mathrm{\ell }$ and $e$ would have both $E$ and $H$ in common, hence the only possibility is to have $I\in \mathrm{\ell }$, i.e. $\mathrm{\ell }=\{A,E,I\}$. Likewise, if $A,F\in \mathrm{\ell }$, it follows that $H\in \mathrm{\ell }$.

Summarrizing the last few sentences, if $A\in \mathrm{\ell }$, then either $\mathrm{\ell }=\{A,E,I\}$ or $\mathrm{\ell }=\{A,F,H\}$. By a similar line of reasoning, if $B\in \mathrm{\ell }$, then either $\mathrm{\ell }=\{B,D,I\}$ or $\mathrm{\ell }=\{B,F,G\}$ and, if $B\in \mathrm{\ell }$, then either $\mathrm{\ell }=\{C,D,H\}$ or $\mathrm{\ell }=\{C,E,G\}$. Since $\mathrm{\ell }$ must contain one of $A,B,C$, it follows that there are omnly the following six possibilities for $\mathrm{\ell }$:

$$\{D,H,C\},\{A,E,I\},\{B,F,G\},$$

$$\{B,D,I\},\{C,E,G\},\{A,F,H\}.$$

However, since $L\setminus \{a,b,c,d,e,f\}$ has cardinality six, all these possibilities must be actual members of the set. ∎

Title	combinatorial uniqueness of Hesse Configuration
Canonical name	CombinatorialUniquenessOfHesseConfiguration
Date of creation	2014-02-28 16:18:25
Last modified on	2014-02-28 16:18:25
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	22
Author	rspuzio (6075)
Entry type	Definition
Classification	msc 51E20
Classification	msc 51A45