combinatorial uniqueness of Hesse Configuration

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

Definition 1.

An abstract Hesse configuration is a pair ofsets $\mathrm{(}P\mathrm{,}L\mathrm{)}$ 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 $P$, thereexists 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. Eachelement 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 $D$ mustbe a subset of exactly one element of $L$. For thisto be possible, $L$ must have cardinality $36/3=12$.∎

Theorem 2.

If $\mathrm{(}P\mathrm{,}L\mathrm{)}$ is an abstract Hesse configuration then, forevery $p\mathrm{\in }P$, 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 existsexactly one $\mathrm{\ell }\in L$ such that $p\in \mathrm{\ell }$ and$q\in \mathrm{\ell }$. Furthermore, for every $\mathrm{\ell }\in L$ suchthat $p\in \mathrm{\ell }$, there will be exactly two elementsof $\mathrm{\ell }$ 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 fourelements 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 atmost $3\cdot 3+1=10$ elements $k\in L$ such that$P\cap L\ne \mathrm{\varnothing }$. Because $L$ has 12 elements. theremust 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 thecontrary 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 elementsof $L$ containing $p$ and an element of $\mathrm{\ell }$. Because$\mathrm{\ell }\cap m=\mathrm{\ell }\cap n=\mathrm{\varnothing }$, these three elementsare distinct from $m$ and $n$. That makes for a total offive distinct elements of $L$ containing $p$, whichcontradicts 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 elementsand $m,n,k$ are pairwise disjoint but $P$ only has nineelements, it follows that every element of $P$ must belongto exactly one of $m,n,k$. In particular, this means thatevery element of $\mathrm{\ell }$ must belong to one of $m,n,k$. Weretwo elements of $\mathrm{\ell }$ to belong to the same element of$\{m,n,k\}$ then, by the third defining property, that elementwould 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 canlabel the elements of $P$ as A,B,C,D,E,F,G,H,I in such a waythat 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 $L$ hastwelve elements, there must exist an a elemetn of $L$ distinctfrom $a,b,c$. Pick such an element and call it $d$. Byanother 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 witheach 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 theelement it has in common with $f$. Likewise, $b$ must haveexactly 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 theelement $b$ has in common with $e$, $F$ be the element $b$has in common with $f$, $G$ be the element $c$ has in commonwith $d$, $H$ 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 assignedlabels $A,B,C,D,E,F,G,H,I$ to the elements of $P$ in sucha 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 labelthe 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 incommon with $d$. Since $\mathrm{\ell }$ must have one element in commonwith $b$, 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 onlypossibility 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 $A\in \mathrm{\ell }$, theneither $\mathrm{\ell }=\{A,E,I\}$ or $\mathrm{\ell }=\{A,F,H\}$. By a similarline of reasoning, if $B\in \mathrm{\ell }$, then either $\mathrm{\ell }=\{B,D,I\}$ or $\mathrm{\ell }=\{B,F,G\}$ and, if $B\in \mathrm{\ell }$, theneither $\mathrm{\ell }=\{C,D,H\}$ or $\mathrm{\ell }=\{C,E,G\}$. Since $\mathrm{\ell }$must contain one of $A,B,C$, it follows that there are omnlythe 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 cardinalitysix, all these possibilities must be actual members of the set.∎