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On the bi-embeddability of certain Steiner

triple systems of order 15

G. K. Bennett, M. J. Grannell and T. S. Griggs,

Department of Pure Mathematics,

The Open University,

Walton Hall, Milton Keynes, MK7 6AA,

United Kingdom.

September 2001

This is a preprint of an article accepted for publication in the European

Journal of Combinatorics c ?2001 (copyright owner as specified in the jour-

nal).

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Running head:

Bi-embeddings of STS(15)s

Corresponding author:

M. J. Grannell, 17 Royal Avenue, Leyland, Lancashire, PR25 1BQ, England.

Tel: 01772-433430email: m.j.grannell@open.ac.uk

AMS classifications:

05B07, 05C10.

Abstract

There are 80 non-isomorphic Steiner triple systems of order 15. A standard

listing of these is given in [8]. We prove that systems #1 and #2 have no bi-

embedding together in an orientable surface. This is the first known example

of a pair of Steiner triple systems of order n, satisfying the admissibility

condition n ≡ 3 or 7 (mod 12), which admits no orientable bi-embedding.

We also show that the same pair has five non-isomorphic bi-embeddings in a

non-orientable surface.

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1 Introduction

The background to this paper lies in the result of Ringel and Youngs [9, 11]

that, for all n ≡ 0, 3, 4 or 7 (mod 12), there exists a triangulation of the

complete graph Knin an orientable surface of appropriate genus. Here we

give a brief summary of those aspects required for the purpose of the present

paper; further details may be found in [1, 3, 5, 6] as well as in Ringel’s book

[9].

For n ≡ 3 or 7 (mod 12) results of Ringel and Youngs establish the ex-

istence of a triangulation of Kn, in an orientable surface, and having the

additional property that the faces may be properly 2-coloured. The triangu-

lar faces in each of the two colour classes of such a triangulation necessarily

form a Steiner triple system of order n (STS(n)); that is a set of triples from

a point set of cardinality n such that every pair of points (corresponding to

the edges of Kn) lies in a unique triple (the face, in the relevant colour class,

which contains that edge). In such a triangulation we will say that the two

STS(n)s, are embedded together in the surface.

Given a pair of STS(n)s, say A and B, one may ask whether there exists an

embedding of A together with B. The answer to this question will sometimes

be no, for example if A and B have a triple in common. However, the question

may be refined to ask if A and an isomorphic copy of B can be embedded

together. With this question in mind, we define a bi-embedding of A and B

to be an embedding of A with an isomorphic copy of B.

It is not known whether every STS(n) with n ≡ 3 or 7 (mod 12) has a bi-

embedding with some other STS(n) in an orientable surface. An affirmative

answer would entail the existence of nO(n2)non-isomorphic face 2-colourable

triangulations of Knin an orientable surface, since there are nn2/6−o(n2)non-

isomorphic STS(n)s [10]. However, the best existing lower bounds for the

numbers of such triangulations all have the form 2O(n2)[3, 7]. The lowest

non-trivial specific value of n for which one might investigate the question is

n = 15.

There are 80 non-isomorphic STS(15)s and it is known that at least three

of these have bi-embeddings in an orientable surface. In each of these three

cases the bi-embedding is of a system with an isomorphic copy of itself. Using

the standard listing of the STS(15)s given in [8], the three systems involved

are #1 (which is the point-line design of the projective geometry PG(3,2)),

#76 and #80. The bi-embedding of system #80 was given by Ringel [9], that

of #1 was given in [1], and that of #76 together with current graphs which

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generate all three bi-embeddings was given in [2]. It seems to be a difficult

problem to determine whether or not all the remaining 77 STS(15)s admit a

bi-embedding in an orientable surface. However, a more tractable question

is whether particular pairs of STS(15)s may be bi-embedded together.

The rich structure of PG(3,2) facilitated computer analysis which resulted

in the construction of the unique (up to isomorphism) bi-embedding of sys-

tem #1 with itself in an orientable surface. None of the other STS(15)s

possesses a comparable degree of symmetry. However, system #2 may be

obtained from #1 by means of a Pasch-switch. A Pasch configuration also

known as a quadrilateral in a Steiner triple system is a set of four triples

whose union has cardinality six. Such a configuration is isomorphic to

{{a,b,c},{a,y,z},{x,b,z},{x,y,c}}. A Pasch-switch is the operation of re-

placing this set of four triples by {{x,y,z},{x,b,c},{a,y,c},{a,b,z}} which

covers the same pairs. System #1 has 105 Pasch configurations and it was

shown in [4] that switching any one of these results in a copy of system #2.

Using this fact, we show (Theorem 3.1) that there is no bi-embedding of sys-

tem #1 with system #2 in an orientable surface. This is the first example of

a pair of STS(n)s (with n ≡ 3 or 7 (mod 12)) which cannot be bi-embedded

in an orientable surface.

Ringel and Youngs’ work also dealt with triangulations of Kn in non-

orientable surfaces and one may ask questions, similar to those given above,

for bi-embeddings of STS(n)s in non-orientable surfaces. Here the necessary

conditions are n ≡ 1 or 3 (mod 6). In the course of the investigation we

prove that there are precisely five non-isomorphic bi-embeddings of system

#1 with system #2 in a non-orientable surface.

2 Method

In a bi-embedding of two STS(15)s there will be 15 vertices, 105 edges and

70 triangular faces. The genus of the surface may be determined from Euler’s

formula. In the orientable case the surface is S11, the sphere with 11 handles,

and in the non-orientable case it is¯S22, the sphere with 22 crosscaps. We

will refer to the colour classes for the faces as black and white.

A triangulation of Knmay be described by means of a rotation scheme.

This comprises a set of circularly ordered lists, one for each vertex of Kn. The

list corresponding to the vertex x, the rotation at x, gives the remaining n−1

vertices in the order in which they appear around x in the given embedding.

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If the embedding is in an orientable surface then a consistent orientation, say

clockwise, may be selected for the entire rotation scheme. As an example,

Table 1 gives a rotation scheme for an embedding of K7in a torus. In fact

this embedding is unique up to isomorphism. The vertices of K7are taken

to be the points of Z7.

0:

1:

2:

3:

4:

5:

6:

1

2

3

4

5

6

0

3

4

5

6

0

1

2

2

3

4

5

6

0

1

6

0

1

2

3

4

5

4

5

6

0

1

2

3

5

6

0

1

2

3

4

Table 1. Rotation scheme for embedding K7.

Given a triangular embedding of Kn, by considering each pair of adjacent

triangular faces, ?i,j,k? and ?i,k,l?, it is easy to see that the rotation scheme

must satisfy the following:

Rule R. If the rotation at i contains ...jkl... then the rotation at k contains

either ...lij ... or ...jil....

The converse is also true (see for example [9], p76), namely a rotation

scheme on n points (with the rotation at each point x containing all the n−1

points apart from x) which satisfies Rule R represents a triangular embedding

of Knin some surface. The surface may or may not be orientable. It will be

orientable if it is possible to orient the rotations at the vertices consistently,

i.e. to satisfy the following:

Rule R*. If the rotation at i contains ...jkl... then the rotation at k

contains ...lij ....

The necessity of Rule R* may be seen in a similar fashion to that of Rule

R. A proof of its sufficiency is given in [9].

We take the vertices of K15to be the elements of Z15. Without loss of

generality, the rotation at 0 can be taken as:

0 :1 2 3 4 5 6 7 8 9 10 11 12 13 14

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