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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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Since a face 2-colourable embedding is sought, it can be assumed that, for

i = 1,2,...,7, the triangles ?0,2i−1,2i? are coloured black and the triangles

?0,2i,2i + 1? (with “15” replaced by “1”) are coloured white. We look for

bi-embeddings where the white and black systems are isomorphic to systems

#1 and #2 respectively.

It was shown in [1] that there are precisely 480 differently labelled copies

of system #1 on the point set Z15 and containing the seven white triples

{0,2i,2i + 1}; it was also explained there how these 480 copies may be ob-

tained. In that paper we sought bi-embeddings of system #1 with itself.

There are also 480 differently labelled copies of system #1 on the point set

Z15and containing the seven black triples {0,2i − 1,2i}, and these may be

obtained from the 480 white systems by applying the permutation (0)(14

13 12 . . . 1). In the case being considered in the current paper, we seek

bi-embeddings of system #1 with system #2. The strategy employed is to

take the 480 black systems just identified and to apply permutations and

Pasch-switches which yield all the differently labelled copies of system #2 on

the point set Z15and which contain the seven black triples {0,2i − 1,2i}.

Given a realisation of system #2 there are precisely 15×(14×12×10×

... × 2) ways of mapping the blocks through a single point onto the seven

black triples {0,2i−1,2i}. However, system #2 has an automorphism group

of order 192 [8]. Consequently the number of differently labelled copies of

system #2 on the point set Z15and containing the seven specified triples is

15.27.7!/192 = 105 × 480. All such systems may be obtained in one of two

ways from the 480 copies of system #1 containing the same black triples.

The first of these is by switching any Pasch configuration which does

not involve the seven specified triples. There are 7 × 6 = 42 Pasch config-

urations in system #1 which involve triples through a specified point, and

consequently there are 105−42 = 63 which do not. Thus we obtain 63×480

copies of system #2 containing the seven specified triples. We show below

that these are distinct, and we refer to them as Type I copies.

The second possibility is that a copy of system #2 containing the seven

specified triples results from a Pasch switch on a copy of system #1 which

does not contain all the specified triples. The Pasch configuration involved

in such a switch which lies in system #2 must contain two of the speci-

fied triples, say {0,2i − 1,2i} and {0,2j − 1,2j} (i ?= j) together with a

pair of other triples which may either be {{x,2i − 1,2j − 1},{x,2i,2j}}

or {{x,2i − 1,2j},{x,2i,2j − 1}}. The corresponding copy of system #1

will contain five of the specified triples together with either {{0,2i − 1,2j −

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1},{0,2i,2j},{x,2i − 1,2i},{x,2j − 1,2j}} or {{0,2i − 1,2j},{0,2i,2j −

1},{x,2i − 1,2i},{x,2j − 1,2j}}. If we apply the permutation (2i − 1 2j)

(alternatively (2i 2j−1)) to the former case or (2i−1 2j−1) (alternatively

(2i 2j)) to the latter case we obtain a copy of system #1 containing all seven

of the specified triples. The process is reversible; we may start with any of

the 480 copies of system #1 containing all seven of the specified triples, ap-

ply an appropriate permutation, carry out the corresponding Pasch-switch,

and obtain a copy of system #2 containing the seven specified triples. There

are 2 × 7 × 6 = 84 transformations to consider, leading to 84 × 480 copies

of system #2 containing the seven specified triples. We will refer to these

as Type II copies. We show below that these are distinct from the Type I

copies and that there are precisely 42×480 distinct Type II copies, each copy

being generated precisely twice by the procedure described above. Thus we

are able to construct all (63+42)×480 = 105×480 distinct copies of system

#2 containing the seven specified black triples.

Lemma 2.1 (a) The 63 × 480 Type I copies are all distinct.

(b) The Type I copies are all distinct from the Type II copies.

(c) The Type II copies form 42 × 480 distinct pairs of identical systems.

Proof. Parts (a) and (b) follow immediately from the fact that in a copy of

system #2, there is precisely one Pasch configuration which may be switched

to give a copy of system #1 [4]. To establish part (c), note firstly that

a copy of system #1 containing five of the specified triples together with

{0,2i−1,2j −1} and {0,2i,2j} may be obtained from a copy containing all

seven of the specified triples by means of either the permutation (2i−1 2j)

or the permutation (2i 2j − 1). A similar duplication occurs in respect of

a copy of system #1 containing five of the specified triples together with

{0,2i − 1,2j} and {0,2i,2j − 1}. Thus there are at most 42 × 480 distinct

Type II systems. If two Type II systems are identical then they arise from

identical copies of system #1 as described and only two transpositions of the

forms described are capable of producing such a copy of system #1 from a

copy containing all seven of the specified triples.

Putting together a white system #1 and a black system #2, the assumed

rotation at 0 together with the lists of black and white triples determines a

potential rotation scheme. As a consequence, there are 480 × (105 × 480)

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potential bi-embeddings of system #1 with system #2. Each of these was

examined to check firstly that the potential rotation at each vertex indeed

comprises a single 14-cycle and, in such cases, secondly that the whole scheme

satisfies Rule R. The rotation schemes so identified were then further tested

against Rule R* to determine those which are orientable.

The procedure just described leads to the conclusion that there is no bi-

embedding of system #1 with system #2 in an orientable surface. However

non-orientable bi-embeddings were obtained. Isomorphisms between these

bi-embeddings may be determined in the manner given below. The same

approach can also be used to determine the automorphism groups. Since the

black and the white systems are themselves non-isomorphic, mappings which

reverse the colours cannot form isomorphisms between (or automorphisms of)

the bi-embeddings obtained.

Consider two rotation schemes, R1and R2, defined on the points of Z15

and representing bi-embeddings of system #1 (white) and system #2 (black).

To determine those mappings (if any) φ : Z15→ Z15which take R1to R2

we only need consider 15 × 14 = 210 possibilities. For suppose two points

x and y are fixed in R1, then once their images φ(x) and φ(y) are chosen

in R2, the circularly ordered rotations at x in R1and at φ(x) in R2must

correspond. Since y corresponds to φ(y), the images of the remaining points

are determined up to a reversal of one of these rotations. However, only one

of the two orientations is possible because colour reversals are not allowed.

Thus there are 210 possible mappings which might provide an isomorphism

and, similarly, there are 210 possible mappings of an embedding which might

provide an automorphism.

3 Results

From the 480×(105×480) possibilities described above, 1050 bi-embeddings

of system #1 with system #2 were identified and these fall into just five

isomorphism classes. A representative of each class is given in Table 2. None

of these bi-embeddings can be oriented to satisfy Rule R* and so there is no

orientable bi-embedding of these systems. Each isomorphism class contains

210 bi-embeddings satisfying Rule R. Consequently all the bi-embeddings

have only the trivial automorphism. These computational results have been

verified by two independently written computer programs. We summarise

the results as follows.

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Theorem 3.1 Up to isomorphism, there are five non-orientable bi-

embeddings of system #1 with system #2. There is no orientable bi-

embedding of these systems.

Class #1 Representative

56

57 10

58

118

18 11

82 14

11 1410

9 1411

41 12

1 1214

814

6 14

37

710

611

0:

1:

2:

3:

4:

5:

6:

7:

8:

9:

10:

11:

12:

13:

14:

1

2

0

4

0

6

0

8

0

2

0

1

0

3

0

5

0

7

0

9

0

34

3

789

6

9

10

13

11

11 12

12

10

1314

1411984

34

2

14

13

6

7

7 12 13

1210

13

69

6

3

5

9

9

8

5

1 14

142712 10

11

5

4 10127

9

1

4

13

131

2

3

4

6

7

3

4

1227

3

9

4

6

3

8

4

13 101 12

1311

13

2

2

5 10 14

101153

7

5

6

1

4

6

1

9

4

9

3

05

1

6

7

5

2

7

122

3

1

5

5

13

13

11

12108

9

2

0112 10 148

6

2

13

140 1243

9

11

12

8

10 138 10

(Continued on the next page)

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Class #2 Representative

56

58 12

68

106

131

93

28

10 1411

3 11

126

8 13

136

48

79

122

Class #3 Representative

56

58 12

57

108

1 13

93

10 13

101411

310

122

83 13

138

48

79

126 10

0:

1:

2:

3:

4:

5:

6:

7:

8:

9:

10:

11:

12:

13:

14:

1

2

0

4

0

6

0

8

0

2

0

1

0

3

0

5

0

7

0

9

0

34

3

9

789

7

1011

11

10

1213

13

13

14

1410694

34

2

7

8

2

1

3

5 14

12

11

11

14

12

12

11

14

12 11

13117

7

95

8

7

1 14

1469

8

1210

13

5

4 10 142 12

129

1

6

104

3

4

1

7

7

1311

13

13

1

5

7

2

9

5

6

2

8

5

4

1

4

6

9

9

5

2

3

1

1

1

5

10

13

14

101473

2

8

5

04

2

7

5

9

112 11

12 104

2

6

7

145

6

3

4

3

0 11310 14

11

10

9

2

6

13

140 124 108

10 138 113

0:

1:

2:

3:

4:

5:

6:

7:

8:

9:

10:

11:

12:

13:

14:

1

2

0

4

0

6

0

8

0

2

0

1

0

3

0

5

0

7

0

9

0

34

3

789

7

9

10 11

11

10

12 13

13

13

14

14 10694

34

2

14

13

8

6

6

1

4

61211

1112

11

11

7

7

95

8

7

1 14

1429

8

9

4

12 10

13

5

4 10

14

142

3

9

12

1228

3

5

1

7

4

9

11

13

13

1

5

7

2

9

5

6

2

8

5

1

6

12

4

6

121 11

13

14

10144

6

11

14

73

2

6

5

04

2

7

5

9

1

9

1211

12 105

1

1

14

10

7

2

6

1

6

2

4

3

01131413

140124

7

103

2

118

10 138 1153

(Continued on the next page)

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Class #4 Representative

56

133

1210 13

98

108 11

17

81 10

51 11

4 10

146 13

5 14

714

36

8 11

7 11

Class #5 Representative

56

97 13

144

105

142

137

1 11

123

145 12

14 12

12 13

5213

149

32 11

129

0:

1:

2:

3:

4:

5:

6:

7:

8:

9:

10:

11:

12:

13:

14:

1

2

0

4

0

6

0

8

0

2

0

1

0

3

0

5

0

7

0

9

0

34

4

7

8

8

6

4

9 10

11

11 1213

12

14

10

12

11

14

14 107

5

6

1

5

8

9

3

7

5

3

7

9

2

7

4

3

6

11

14

67

51 13 1112

1496

8

2 13

14

14

12

12

2 1013

11

9

4

9

1

13

14

92

2 10

14

11

12

34

6

1

12 13

13 1152

5

3

4

123

7

1

9

7

9

4

9

4 108

4

3

8

1

2

2

12

07

6

8

3

6

136

5

9

6

11

12108

4

4

3

133

1

2

0 112 10 145

2

2

13

140 1271

9

105

10 135 10812

0:

1:

2:

3:

4:

5:

6:

7:

8:

9:

10:

11:

12:

13:

14:

1

2

0

4

0

6

0

8

0

2

0

1

0

3

0

5

0

7

0

9

0

34

3

9

78

4

9 10 11

12

1213

10

13

14

14 11

12

11

6

6

8

58

3

6

5

2

7

8

2

6

4

2

6

7

7

7

3

9

5

10

12

11

10

5 11

13

10

1419

91

1

4

1

13

12

10

8

8

8

12

14

13

911

1412

14

3

139

6

10

10

2

1

3

4

7

2

4

8

4

2

1

11

113

2

2

4

5

5

9

8

9

4

13

13

9

7 108

7

4

3

116

6

4

6

1

5

01

6

5

5

8

1

1

7

14 11

12101483

0113741013

140 12108

3

6

10 13672 1011

Table 2. Isomorphism class representatives.

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References

[1] G. K. Bennett, M. J. Grannell and T. S. Griggs, Bi-embeddings of the

projective space PG(3,2), J. Statist. Plann. Inference 86 (2000), 321-329.

[2] G. K. Bennett, M. J. Grannell and T. S. Griggs, Bi-embeddings of Steiner

triple systems of order 15, Graphs Combin. 17 (2001), 193-197.

[3] C. P. Bonnington, M. J. Grannell, T. S. Griggs and J.ˇSir´ aˇ n, Exponential

families of non-isomorphic triangulations of complete graphs, J. Combin.

Theory Ser. B 78 (2000), 169-184.

[4] M. J. Grannell, T. S. Griggs and J. P. Murphy, Switching cycles in

Steiner triple systems, Util. Math. 56 (1999), 3-21.

[5] M. J. Grannell, T. S. Griggs and J.ˇSir´ aˇ n, Face 2-colourable triangular

embeddings of complete graphs, J. Combin. Theory Ser. B 74 (1998),

8-19.

[6] M. J. Grannell, T. S. Griggs and J.ˇSir´ aˇ n, Surface embeddings of Steiner

triple systems, J. Combin. Des. 6 (1998), 325-336.

[7] M. J. Grannell, T. S. Griggs and J.ˇSir´ aˇ n, Recursive constructions for

triangulations, J. Graph Theory, to appear.

[8] R. A. Mathon, K. T. Phelps and A. Rosa, Small Steiner triple systems

and their properties, Ars Combin. 15 (1983), 3-110.

[9] G. Ringel, Map Color Theorem, Springer, 1974.

[10] R. M. Wilson, Nonisomorphic Steiner triple systems, Math. Z. 135

(1974), 303-313.

[11] J. W. T. Youngs, The mystery of the Heawood conjecture, in: “Graph

Theory and its Applications” (ed. B. Harris), Academic Press, 1970,

17-50.

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