## You are here

Homecombinatorial uniqueness of Hesse Configuration

## Primary tabs

# 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 $(P,L)$ 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 $(P,L)$ 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 ${9\choose 2}=36$ elements. Each element of $L$ is a subset of $P$ with three elements, hence has ${3\choose 2}=3$ 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 $(P,L)$ is an abstract Hesse configuration then, for every $p\in P$, there exist exactly four elements $\ell\in L$ such that $p\in\ell$.

###### Proof.

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

###### Theorem 3.

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

###### Proof.

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

It remains to show that $m\cap n=\emptyset$. Suppose to the contrary that there exists a $p$ such that $p\in m$ and $p\in n$. Since $m\cap\ell=\emptyset$, it follows that $p\notin\ell$, hence there will exist three distinct elements of $L$ containing $p$ and an element of $\ell$. Because $\ell\cap m=\ell\cap n=\emptyset$, 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 $m\cap n=\emptyset$. ∎

###### Theorem 4.

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

###### 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 $\ell$ must belong to one of $m,n,k$. Were two elements of $\ell$ to belong to the same element of $\{m,n,k\}$ then, by the third defining property, that element would have to equal $\ell$, contrary to its definition. Hence, each element of $\ell$ must belong to a distinct element of $\{m,n,k\}$. ∎

###### Theorem 5.

If $(P,L)$ 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 $a\cap b=b\cap c=c\cap a=\emptyset$. 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 $d\cap e=e\cap f=f\cap d=\emptyset$.

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

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

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

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

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

## Mathematics Subject Classification

51E20*no label found*51A45

*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections