数理解析研究所講究録
991 巻 1997 年 80-92

80

On Association Schemes of Balanced
Property with $m_{1}--4$
E\"uchi Bannai

Attila Sali*

Department of
Kyushu University
Fukuoka, Higashi-ku
Hakozaki 6-10-1 Japan.812$\mathrm{M}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{C}\mathrm{s}\backslash$February 18, 1997

Abstract

The work of classification of association schemes began in [2] is continued.
The assumption of $m_{1}=4$ allows considein geometrical representation on
the unit sphere. Using the balanced property $[6, 7]$ of Terwilliger aclassification of all possible local structures
is given.
$\mathrm{g}$$S^{3}\subset R^{4}$$\Gamma_{1}(x)$*Permanent address: Mathematical Institute of the Hungarian Academy of
Sciences, Budapest P.O.B. 127 H-1364 HUNGARY

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1

Introduction

The classification of association schemes is one of the most important tasksof Algebraic Combinatorics. In the present paper we continue the work
began by Bannai [2] and consider the case $m_{1}=4$ . (For definitions and basic
properties of association schemes the reader is referred to the book of Bannai
and Ito [1].)We restrict our attention to primitive symmetric association schemes. Let
be a symmetric association scheme. Let $A_{i}(0\leq i\leq d)$
be the adjacency matrix with respect to the relation $R_{i}(0\leq i\leq d)$ on$X$ and let $A=$be the Bose-Mesner algebra of . Let$E_{i}(0\leq i\leq d)$ be the primitive idempotents
be theo.f $A$ . Let
be the dual subdegrees of .
subdegrees of and let$\Gamma_{i}(x)=\{w:(x,w\rangle\inR_{i}\}$
be the -neighborhood of .
Let
By renumbering the relations if neccessary, we may assume without loss
. Let us setof generality that $q_{1}(1)\geq q_{1}(i)$ for all$\mathcal{X}=(X, \{R_{i}\}_{0\leq i}\leq d)$$(A_{0}, A_{1}, \ldots , A_{d})$$\mathcal{X}$$k_{\dot{f}}=p_{ii}^{0}$$m_{i}(=q_{ii}^{0}=\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(E_{i}))$$\mathcal{X}$$\mathcal{X}$$x$$\dot{i}$$1\leq\dot{i}\leq d$$E_{1}= \frac{1}{|X|}.\sum_{i=0}^{d}q1(i)A\dot{.}$.

Note that $q_{1}(0)=m_{1}=4$ . We consider the spherical embedding of $X$ in$S^{3}=\{(x, y, z,v)\in R^{4}:x^{2}+y^{2}+z^{2}+v^{2}=1\}$ , with respect to. That is,$X$ is embedded inwith its Gramm matrix given by$E_{1}$$G$$S^{3}$$G= \frac{|X|}{4}E_{1}$.

By the primitivity of $X$ this embedding is injective. For the sake of simplicity,of the above embedding.
we identify elements of $X$ with the vectors of
The concept of balanced sets was introduced by Terwilliger in $[6, 7]$ .
$R^{4}$

Definition 1.1 $LetX$ be set of vectors in $S^{N-1}$ , andbe the set of scalar products of elements of X. $X$ is called balanced$letA=\{\alpha_{0}, \alpha_{1}, \ldots, \alpha_{d}\}$(i) for all $x\in X$ and all $i(0\leq i\leq d)_{f}$ the vector
multiple of$x$$\sum_{y\in x,():^{y}}x,y=\alpha$$\dot{i}f$

is a scalar

82(ii) for all$x,$ $y\inX$and all $i,j(0\leq i,j\leq d)$ , the vector $\Sigma z\in X$,$\langle x, z\rangle=\alpha_{i}$

$z-$

,

$(y,$ $z\rangle=\alpha_{j}$$\sum w\in X$$w$,$\langle x, w\rangle=\alpha_{j}$is a scalar multiple of $x-y$ .

,$(y,w\rangle=\alpha_{i}$An association scheme
is said to have the balanced propis a balanced set. It was proved inerty if its embedding with respect to[7] that a -polynomial association scheme has the balanced property. However, there are examples of association schemes of balanced property, which
are not Q-polynomial.
A general method of classification of association schemes was introducedin [2]. That is, consider the embedding with respect toin $S^{n-1}$ andsuppose, that $\alpha_{1}=q_{1}(0)$ is maximal amongst the. Then an upperbound exists for by the so called kissing problem of spheres of equal radii.
That is, the points ofare on a sphere in $R^{n-1}$ whose radius is smaller
than the minimum distancepoints of $X$ . Then one should try tocharacterize the possible geometric configurations on that sphere and “lift itup” to $S^{n-1}$ This was successfully done in [2] in the case of $m_{1}=3$ , that is$n=3$ . The aim of this paper is to investigate $m_{1}=4$ case and determine all
possible geometric configurations- ofprovided the association scheme
is of the balanced property.
Section 2 contains general observations, the case-by-case analysis is contained in Section 3. In many cases the tedious proofs are omitted, or just a
brief sketch is included. Finally, Section 4 contains some concluding remarks.
$\mathcal{X}=(X, \{R_{i}\}_{0\leq i}\leq d)$$E_{1}$$Q$$E_{1}$$\alpha_{i}’ \mathrm{s}$$k_{1}$$\Gamma_{1}(x)$$\mathrm{a}\mathrm{m}\mathrm{o}\mathrm{n}\mathrm{g}\mathrm{s}\mathrm{t}\wedge$$\Gamma_{1}(x)$2General observations

Rom now on, throughout the paper we assume thatis
an association scheme of balanced property such thatand $(x, y)\in$. The following Proposition is a direct consequence ofthe solution of the kissing problem in the three dimensional Euclidean space,see Leech [5].
$\mathcal{X}=(X, \{R_{i}\}_{0\leq i}\leq d)$$X\subset S^{3}$$R_{i}\Leftrightarrow(x,$$y\rangle$$=\alpha_{i}$Proposition 2.1 By the above settings, $k_{1}\leq 12$ holds.

I83

is an ordinary three dimenLet $\gamma_{i}(x)=\{y\in S^{3}:(x, y\rangle=\alpha_{i}\}$ . Thensional sphere. On this sphere, the smallest distance between points of $X$ is. Thus, applying Leech’s ideas [5] it can be proved that thelarger thanwhose distance is lessgraph obtained by connecting points of $X$ on, is planar.
thenby
Let us denote the scalar products of vectors of $X$ on. That is corresponds to the smallest distancebe the distance corresponding to ,. Furthermore, letoccuring on
two
be the degree of the regular graph whose vertex set isand letvertices are connected iff their distance is . We use the ambiguity in thismay denote angular or Euclidean distance, depending on
notation, thatthe environment. Then, applying again Leech’s ideas$\gamma_{1}(x)$$60^{\mathrm{O}}$$\gamma_{1}(x)$$90^{\mathrm{o}}$$\gamma_{1}(x)$$\alpha_{i_{2}}=\sqrt 2>\ldots>\alpha_{\dot{\mathrm{z}}_{\iota}}=\beta_{s}$$\alpha_{i_{1}}=\beta_{1}>$$\beta_{1}$$\gamma_{1}(x)$$d_{i}$$\sqrt i$$\Gamma_{1}(x)$$deg_{i}$$d_{i}$$d_{i}$$k_{1} \leq 7+.\sum_{\mathrm{o}d.<90}$

(1)

degi

follows. Indeed, there can be at most 6 points of $X$ on a half sphere.
The balanced property is used for the following two Lemmas.
Lemma 2.2 Let as assume that $deg_{i}=1$ for some . Then there exists an $N$
positive integer such that is the length of the shortest diagonal of a regularN-gon whose side is .
$\dot{i}$$d_{i}$$d_{1}$

Proof of Lemma 2.2 By the assumption, $p_{1j:}^{1}=1$ . Let $y\in\Gamma_{1}(x)$ andand $\{v\}=\Gamma_{j-}.(y)\cap\Gamma_{1}(x)$ . By (ii) of the definition ofare coplanar and form a symmetricbalanced sets, the four points $x,$
playing the same role as $x,$ before,
trapezoid. Now considering the pair
another coplanar point is obtained, say . Continuing this process, the newly
obtained point has to coincide with after a while, otherwise we would get.
a shorter distance than onI
Next case is degree 2.
$\{z\}=\Gamma_{j}\dot{.}(x)\cap\Gamma_{1}(y)$$y,$ $z,$ $v$$y,$ $z$

$y$

$w$

$v$

$d_{1}$$\gamma_{1}(x)$Lemma 2.3 Let as suppose that $deg_{i}=2$ for some . Then $i=1_{f}$ furthermore, $j_{i}=1$ .
$i$and suppose that $deg_{i}=2$ . We may
Proof of Lemma 2.3 Let
assume without loss of generality that $x=(0,0, a, b)$ and $y=(0,0, a, -b)$ ,where $a^{2}+b^{2}=1$ and $\alpha_{1}=a^{2}-b^{2}$ . Let $z_{k}=(z_{k1}, z_{k2}, zk3, Zk4)k=1,2$$\langle x,y\rangle=\alpha_{1}$84, while $u_{k}=(u_{k1}, u_{k2}, u_{k3}, uk4)$ be those of
be the elements of. If, thenor. The distance relations canbe expressed in the following set of equations.
$\Gamma_{j}.\cdot(y)\cap\Gamma_{1}(x)$$\Gamma_{j}.\cdot(x)\cap\Gamma_{1}(y)$$\dot{i}\neq 1$$j_{i}\neq 1$$z_{k}\neq u_{k}$

(2)
(3)
(4)
(5)
(6)
(7)
(8)

$z_{13}a+z14b=a^{2}-b^{2}$$z_{23}a+Z24b=a^{2}-b^{2}$$u_{13}a-u_{14}b=a^{2}-b^{2}$$u_{23}a-u_{24}b=a^{2}-b^{2}$$u_{13}a+u_{14}b=z_{13}a-z14b$$u_{23}a+u_{24}b=z_{23}a-z24b$$u_{13}a+u_{14}b=z_{23}a-z24b$

The (2) $-(5)$ express the fact thatand, respectively.
The (6) $-(8)$ are for$(k=1,2l=1,2)$ . From these equations
it can be easily inferred that$u_{k}\in\Gamma_{1}(y)$$z_{k}\in\Gamma_{1}(x)$$\langle x, u_{k}\rangle=(y,$$z_{l}\rangle$

(9)
(10)
(11)
(12)

$u_{13}$$=$$z_{13}$$u_{23}$$=$$z_{23}$$u_{14}$$=$$-z_{14}$$u_{14}$$=$$-z_{14}$$z_{11}+z_{21}$$=$$u_{11}+u_{21}$$z_{12}+z_{22}$$=u_{12}+u_{22}$.

The balanced property implies that

Furthermore, because$z_{k}$

and$u_{k}$

(13)
(14)

.

are unit vectors,(15)$z_{1}^{2}+1Z12z^{2}=221^{+=}Z222u^{2}11+u=u21221^{+}2u^{2}22$

holds, as well. Thus,
andare 2-dimensional
vectors of same length, and their pairwise sum is the same, hence they are
the same pair of vectors. Consequently,
and , furthermore
andare reflections of each other, respectively, to the hyperplane othogonal to$(0,0,0,1)$ . However, this implies thatand cannot be inat thesame time, a contradiction.
$(z_{11}, z_{12}),$ $(z_{21}, z_{22})$$(u_{11}, u_{12}),$ $(u_{21}, u_{22})$$z_{1}$$z_{1},$$z_{2}$$u_{1}$$y$$z_{2}$$u_{2}$$\Gamma_{1}(x)$

1

85

3

Cases according to$k_{1}$In this section several cases are considered according to possible values of .
If the appropriate geometric embedding exists, then a-regular planargraph of vertices exists. Many cases can be ruled out by Euler’s formula.
If the planar graph exists, then geometric considerations rule out many cases,
i.e. the planar graph cannot be represented onby uniform length edges.
As usual, denotes the number of faces, the number of vertices, thenumber of edges of a planar graph, respectively. Similarly,
denotes thenumber of -sided faces. The main formulas used are$k_{1}$$deg_{1}$$k_{1}$$S^{2}$

$f$

$v$

$e$

$f_{i}$$\dot{i}$

$v+f$

$=$

$\sum f_{f}..=$

(16)
(17)

$e+2$
$f$

$=2e$

$\sum if_{i}$$\sum(i-3)f*\cdot$$\geq$

(18)

(19)

$0$

For the sake of convenience, let denote an -sided face, thus a planar graphcan be represented by the the symbol $f_{3}*F_{3}+f_{4}*F_{4}+\ldots$ .
$i$$F_{i}$

3.1$k_{1}=12$This is an extreme case. By (1)$de\geq 5$ holds. Thus, connecting
those points ofwhose distance is less than
, we obtain a regularplanar graph on 12 vertices of degree at least 5. Easy application of Euler’s
formula shows, that this graph has degree 5, indeed, and all its faces aretriangles. Thus, it is the edge-graph of the icosahedron. By Lemma 2.2$d_{i}<90$ implies $deg_{i}\neq 1$ furthermore, if, then $deg_{i}\neq 2$ by Lemma 2.3.
Thus,
can only happen if either $j_{1}=1$ and $deg_{1}=2$ and$deg_{2}=3$ or $deg_{1}=5$ . In the first case, or in the second case if $j_{1}\neq 1$ ,we have a distance onthat is very close to the shortest one, say the
distance corresponding to(eitheror ). Thus, onandtogeher we have at least 15 points, which can be shown to be impossible byroutine calculations.
So, if $k_{1}=12$ , thenis an icosahedron, with edge length equal to theshortest distance among points of $X$ . Considering the graph with vertex set$X$ , edge set determined by
the shortest ditance, a locally icosahedron regulargraph is obtained, such graphs are characterized by Blokhuis et. al [3].
$\sum_{\mathrm{A}<90gi}\mathrm{o}$$\Gamma_{1}(x)$$90^{\mathrm{o}}$$j_{i}\neq 1$$\Sigma_{d\dot{.}<90gi}\circ de=5$$\gamma_{1}(x)$$\alpha_{i}$$\Gamma_{1}(x)$$d_{2}$$d_{1}$$\gamma_{1}(x)$$\gamma_{i}(x)$86

Figure 13.2$k_{1}=11$

contradicts to Euler Formula.
are even.
In this case allfollows, aholds, so
If $deg_{1}=2$ , then by (1)$degi=6$ . This could only occur as
contradiction. Thus, $deg_{1}=4$ and$3+3,5+1,4+1+1,3+\cdot 1+1+1$ , because of Lemma 2.3. However, in eachcase an odd degree should exist, a contradiction.
$\sum_{d\dot{.}<90}\circ deg_{i}\geq 6$$deg_{i}’ \mathrm{s}$$\sum_{d\dot{.}<9}0\circ deg_{i}\geq 6$$d_{2}<90^{\mathrm{o}}$$\sum_{i>1}$

3.3$k_{1}=10$$degi\geq 5$ , which contradicts to Euler Formula.
If $deg_{1}<3$ then by (1)is if, then the only way to obtain 10 points on
If $deg_{1}=3$ andtwo points of distance do not have common neighbour of distance from$f_{3}=0$ .
each. Thus, considering the planar graph determined by distanceOn the other hand, $v=10,$ $e=15,$ $f=7,$ $\sum(i-4)f_{i}=2$ . This allows
two possibilities: $5*F_{4}+2*F_{5}$ or $6*F_{4}+1*F_{6}$ . The second possibility
is easyly seen to be unrealizable. Also, it is not hard to see that the only
realization (as a planar graph) of the first possibility is the graph of Figure
1. Routine, but very tedious case-by-case analysis shows that the only way
to geometrically represent this graph is if the two pentagons are in paralell, a contradiction.
planes. But in this case $deg_{i}=2$ for someIf $deg_{1}=4$ , then by Euler Formula we have the following possibilities.
$1*F_{7}+11*F_{3}$ : clearly impossible.
$1*F_{6}+1*F_{4}+10*F_{3}$ : If there is a point surrounded by triangles only, thengeometrically it means a square pyramid with uniform edge lengths. That$\Sigma_{d:<90}\mathrm{o}$$\gamma_{1}(x)$$d_{2}\geq 90^{\mathrm{o}}$$d_{1}$$d_{1}$$d_{1},$$\dot{i}>1$

87

Figure 2
already determines the sphere, and easy to see, that there is not enoughroom left for the other five points. If there is no such vertex, then each one
is incident to the quadrangle or the hexagon, i.e. these two faces have nocommon vertex, which is impossible to draw.
$1*F_{5}+2*F_{4}+9*F_{3}:$ Again, either we get a point surrounded by triangles,
or cannot finish the drawing of the graph.
there is no vertex surronded by triangles, then the only
possibility is that the two pentagons are disjoint, which leads to the graphobtained from the icosaahedron by removing an antipodal point pair. The
only geometric representation of this graph having no with $deg_{i}=2$ is 10
points of an icosahedron, so that the missing two points are antipodal. The
same geometric consideration can be applied as in thecase to show that $j_{1}=1$ . Then the graph on $X$ whose edges are the pairsof distanceis locallyis the graph obtained from thegraph, where
icosahedron by removing an antipodal point pair. This can be eliminated by
the method of Blokhuis et. al. [3].
$4*F_{4}+8*F_{3}$ : Again either there exists a vertex surrounded by triangles
or the graph of Figure 2 is
. Consider the following transformation$\phi=(8,1)(7, 5)(10, 6)(4,3)(9,2)$ . It is not hard to see tharpreserves distances amongst points of, i.e. it can be extended to an orthogonaltransformation of
the property $\phi^{2}=Id$ . Thus is eithera reflection to plane or 180 degree rotation around a line or reflection to apoint. Now, if is not reflection to a plane, then all $(j, \phi(j))$ pairs should
be antipodal, but $(3, \phi(3))$ is the shortest distance, a contradiction. So is$2*F_{5}+10*F_{3}:\mathrm{I}\mathrm{f}$$\dot{i}$$\Gamma_{1}(x)=\mathrm{i}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{a}\mathrm{h}\mathrm{e}\mathrm{d}\mathrm{r}\mathrm{o}\mathrm{n}$$P_{10}$$d_{1}$$P_{10}$$\mathrm{o}\mathrm{b}\mathrm{t}\mathrm{a}\mathrm{i}\mathrm{n}\mathrm{e}\dot{\mathrm{d}}$$\phi$$\Gamma_{1}(x)$$\gamma_{1}(x)\mathrm{s}\mathrm{a}\mathrm{t}\mathrm{i}_{\mathrm{S}}\mathfrak{h}\mathrm{i}\mathrm{n}\mathrm{g}$$\phi$$\phi$$\phi$

88$\mathrm{d}$$\mathrm{e}$$\mathrm{h}$Figure 3
reflection to a plane. In that case, lines 1-8, 7-5, 10-6, 4-3, and 9-2 are allparalell. However, the orientation is different, a contradiction.

3.4$k_{1}=9$All
must be even, and they sum up to 8. There cannot be two of themequal to 2, so the only possibility is $deg_{1}=deg_{2}=4$ . However, two-distanceset on 9 points does not exist.
$deg_{i}’ \mathrm{s}$3.5$k_{1}=8$

If $deg_{1}=2$ , then $deg_{2}=5$ or $deg_{2}=4$ and $deg_{3}=1$ . In both cases easygeometric considerations yield contradictions.
If $deg_{1}=3$ , then Euler Formula yield $f=6$ and $\sum(\dot{i}-3)f_{i}=6$ . It is
immediate to rule out the $f_{8}>0$ and $f_{7}>0$ cases. If $f_{6}>0$ , then the graphshown on Figure 3 is possible only. In the geometric representation of thegraph of Figure 3 the two tetrahedrons abcd and $efgh$ are isomorphic. By (i)of balanced property, these two tetrahedrons must be in antipodal position.
However, that would result in $deg_{i}=2$ for some $i>1$ .
If $f_{5}>0$ , then there is only one possible graph, shown on Figure 4. Again,using (i) of balanced property, the bisector plane of the angle formed by two
normal vectors of the planes of the triangular faces must contain the twovertices not incident with the triangles. This results in again $deg_{i}=2$ for

89

Figure 4

Figure 5some $i>1$ .
If $f_{8}=f_{7}=f_{6}=f_{5}=0$ , then the graph is the edge graph of the cube,
and the only possible balanced geometric realization is the cube itself.
If $deg_{1}=4$ , then the only possible planar graoh is shown on Figure 5.
Since the points are on a sphere, tetrahedrons (4835) and (2367) are isomorphic, hence distances 35 and 36 are equal. This implies by Lemma 2.3
that $deg_{2}=3$ . Thus, any two points not joined by adistance edge areof distance . In particular, tetrahedrons (4578) and (4378) are isomorphic,
which implies distances 47 and 43 are equal, i.e. $d_{1}=d_{2}$ , a contradiction.
$deg_{1}\geq 5$ is clearly impossible geometrically.
$d_{1}$$d_{2}$

90

Figure 6

3.6$k_{1}=7$

There exists no -regular planar graph on 7 vertices if $a\geq 3$ . Thus $deg_{1}=2$
and $deg_{2}=4$ . Let
be of distance . Then they have at least
three common neighbours of distancethat possible only if and are
antipodal, a Contradiction.
$a$$a,$ $b\in\Gamma_{1}(x)$$d_{1}$$d_{2}$$a$$b$$\cdot$3.7$k_{1}=6$

If $deg_{1}=4$ then $deg_{2}=1$ . The graph is the edge graph of the octahedron,and the only geometrical representation is the octahedron itself, because itis an antipodal 2-distance set of 6 points, which must be a tight spherical
3-design, see [4].
If $deg_{1}=3$ , then the only possible planar graph is shown on Figure 6.
By (i) of balanced property, the two triangles must lay in paralell planes
so that their centers are mirror images of each other to the center of the
Elementary geometric considerations yield that either $deg_{2}=2$ or$deg_{2}=1$ and $deg_{3}=1$ but one ofor is less thanthat contradicts
to Lemma 2.2.
If $deg_{1}=2$ , then the graph of the shortest distance on
is eithera union of two triangles or a hexagon. The other degrees must be $deg_{2}=$$deg_{3}=deg_{4}=1$ in both cases. Using Lemma 2.2 elementary geometriccalculations show that the required polyhedra do not exist.
$\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\vee\cdot$$d_{2}$$d_{3}$$\sqrt{d_{1}}$$\Gamma_{1}(x)$

91

3.8

$k_{1}=5$This case is impossible because allare even and their sum is 4, but$deg_{1}=4$
then either, which is clearly impossible, or $deg_{1}=deg_{2}=2$ that
contradicts to Lemma 2.3.
$deg_{i}’ \mathrm{s}$3.9$k_{1}=4$

If $deg_{1}=3$ , thenis the tetrahedron.
$deg_{1}=2$$deg_{2}=1$
If
then
and
is a square on a great circle ofby (i) of balanced property. This implies using Lemma 2.3 that $d_{1}=90^{0}$ on. However, then the configuration is antipodal, that is is imprimitive.
$\Gamma_{1}(x)$$\Gamma_{1}(x)$$S^{3}$

3.10$\gamma_{1}(x)$$\mathcal{X}$$k_{1}=3$

In this caseis a regular triangle on a great circle ofby (i) ofbalanced property. This implies using Lemma 2.3 that $d_{1}=120^{0}$ on . It iseasy to see that this case.results in an ordinary tetrahedron, that is $m_{1}=3$ ,a contradiction.
$\Gamma_{1}(x)$$\gamma_{1}(x)$$S^{3}$

3.11

The Main Theorem

According to the above analysis, the following is true.

Theorem 3.1 Let
be an association scheme of balanced$m_{1}=4$
property with. Then in the geometric representation of the neighborhoodof a point $x\in X$ is one of the following regular polyhedra:
Icosahedronf Cube, Octahedron or Tetrahedron.
$\mathcal{X}=(X, \{R\}_{0\leq i\leq}d)$$\mathcal{X}$$\Gamma_{1}(x)$4Some Remarks

The case of Icoshedron is completely settled by Blokhuis et. al. in [3]. The
regular polytopes provide examples for the other cases. If the edges of the
four-dimensional polytope $X$ are only those of the shortest distance, thenthe dual polytope
is regular faced polytope. Those are well known to beclassified, which classification gives results for our case by duality.
$X^{*}$

92

References[1] E. Bannai and T. Ito, Algebraic $Comb\dot{i}nator\dot{i}CsI$: Association Schemes,
Benjamin Cummings , Menlo Park, Cal. 1984
[2] E. Bannai, On Primitive Symmetric Association Schemes with $m_{1}=3$ ,, preprint.
$\mathrm{K}\mathrm{y}\mathrm{u}\mathrm{s}\mathrm{h}\mathrm{u}- \mathrm{M}\mathrm{P}\mathrm{s}_{-}1996- 3$[3] A. Blokhuis, A.E. Brouwer, D. Buset and A.M. Cohen, The LocallyIcosahedral Graphs, 19-22.
[4] P. Delsarte, J.M. Goethals and J.J. Seidel, Spherical Codes and Designs,
Dedicata, 6 (1977), 363-388.
$Geomet7\dot{\mathrm{B}}a$[5] J. Leech, The problem of the thirteen spheres, Math. Gazette 40 (1956),22-23.

[6] P. Terwilliger, A Characterization of P- and -Polynomial Association
Schemes, J.
Th. (A)45, (1987) 8-26.
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Combin..

[7] P. Terwilliger, Balanced Sets and -Polynomial Association Schemes,
Graphs and Combin. 4, (1988) 87-94.
$Q$