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 twoelement 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 $(91)/2=4$ 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  20140228 16:18:25 
Last modified on  20140228 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 