combinatorial uniqueness of Hesse Configuration

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

2

)

=

36

elements. Each
element of

L

is a subset of

P

with three elements,
hence has

(

3

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

∈

P

, there exist exactly four elements

ℓ

∈

L

such that

p

∈

ℓ

.

Proof.

To every

q

∈

P

such that

q

≠

p

, there exists
exactly one

ℓ

∈

L

such that

p

∈

ℓ

and

q

∈

ℓ

. Furthermore, for every

ℓ

∈

L

such
that

p

∈

ℓ

, there will be exactly two elements
of

ℓ

other than

p

. Hence, there exist

(

9

-

1

)

/

2

=

4

elements

ℓ

∈

L

such that

p

∈

ℓ

.
∎

Theorem 3

.

If

(

P

,

L

)

is an abstract Hesse configuration and

ℓ

∈

L

,
then there exist

m

,

n

∈

L

such that

ℓ

∩

m

=

ℓ

∩

n

=

m

∩

n

=

∅

.

Proof.

By the foregoing result, given

p

∈

ℓ

, there are four
elements of

L

to which

p

belongs. One of these, of course,
is

ℓ

itself, and the other three are distinct from

ℓ

.
Since

ℓ

has three elements, this means that there are at
most

3

⋅

3

+

1

=

10

elements

k

∈

L

such that

P

∩

L

≠

∅

. Because

L

has 12 elements. there
must exist

m

,

n

∈

L

such that

ℓ

∩

m

=

ℓ

∩

n

=

∅

.

It remains to show that

m

∩

n

=

∅

. Suppose to the
contrary that there exists a

p

such that

p

∈

m

and

p

∈

n

. Since

m

∩

ℓ

=

∅

, it follows that

p

∉

ℓ

, hence there will exist three distinct elements
of

L

containing

p

and an element of

ℓ

. Because

ℓ

∩

m

=

ℓ

∩

n

=

∅

, 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

∩

n

=

∅

.
∎

Theorem 4

.

If

m

,

n

,

k

are elements of

L

such that

m

∩

n

=

n

∩

k

=

k

∩

m

=

∅

and

ℓ

∈

L

∖

{

m

,

n

,

k

}

,
then

ℓ

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

ℓ

must belong to one of

m

,

n

,

k

. Were
two elements of

ℓ

to belong to the same element of

{

m

,

n

,

k

}

then, by the third

defining property

, that element
would have to equal

ℓ

, contrary to its definition. Hence,
each element of

ℓ

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

∈

L

such that

a

∩

b

=

b

∩

c

=

c

∩

a

=

∅

. 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

∈

L

such that

d

∩

e

=

e

∩

f

=

f

∩

d

=

∅

.

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

ℓ

∈

L

∖

{

a

,

b

,

c

,

d

,

e

,

f

}

, then

ℓ

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

∈

ℓ

. It could not be the case that

D

∈

ℓ

because then

ℓ

would have two elements in
common with

d

. Since

ℓ

must have one element in common
with

b

, that means that either

E

∈

ℓ

or

F

∈

ℓ

.
If

A

,

E

∈

ℓ

, then the element

ℓ

has in common with

c

cannot be

G

because

ℓ

would have both

A

and

G

in common with

c

and it cannot be

H

because

ℓ

and

e

would have both

E

and

H

in common, hence the only
possibility is to have

I

∈

ℓ

, i.e.

ℓ

=

{

A

,

E

,

I

}

.
Likewise, if

A

,

F

∈

ℓ

, it follows that

H

∈

ℓ

.

Summarrizing the last few

sentences

, if

A

∈

ℓ

, then
either

ℓ

=

{

A

,

E

,

I

}

or

ℓ

=

{

A

,

F

,

H

}

. By a

similar

line of reasoning, if

B

∈

ℓ

, then either

ℓ

=

{

B

,

D

,

I

}

or

ℓ

=

{

B

,

F

,

G

}

and, if

B

∈

ℓ

, then
either

ℓ

=

{

C

,

D

,

H

}

or

ℓ

=

{

C

,

E

,

G

}

. Since

ℓ

must contain one of

A

,

B

,

C

, it follows that there are omnly
the following six possibilities for

ℓ

:

{

D

,

H

,

C

}

,

{

A

,

E

,

I

}

,

{

B

,

F

,

G

}

,

{

B

,

D

,

I

}

,

{

C

,

E

,

G

}

,

{

A

,

F

,

H

}

.

However, since

L

∖

{

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

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