Geometriae Dedicata 69: 287–289, 1998. c 1998 Kluwer Academic Publishers. Printed in the Netherlands.
287
64 Lines from a Quaternionic Polytope STUART G. HOGGAR Department of Mathematics, University of Glasgow, Glasgow G12 8QW, Scotland e-mail:
[email protected] (Received: 4 December 1996) Abstract. An earlier computer calculation produced the vertices of a polytope in quaternionic 4-space, and determined their mutual angles all to be arccos(1/3). This note establishes a computer-independent verification. Mathematics Subject Classifications (1991): 05B30, 51F15, 51M20, 52B11, 68R99. Key words: quaternion, polytope, lines.
This note is to verify a computer calculation used in [2, 3] which at the time appeared unmanageable by human hand, but whose veracity has resumed importance [4]. The assertion is that an explicit set of 64 lines through the origin in 4-space H 4 over the quaternions, when complexified to 64 lines in complex 8-space, becomes equiangular, with cos2 = 19 for any pair of distinct lines. The angle ; 0 6 6 =2, between two lines is defined via representative vectors u; v by cos2 = ju:v j2 =(juj jv j)2 , where (ui ):(vi ) = ui vi and jv j2 = v:v , the star denoting conjugation. Some details. The original vectors were computer-generated in an orbit of vertices of a polytope in H 4 [2] and calculated as satisfying cos2 = 19 or 13 , thus meeting the upper bound on size of such a set in H 4 [3]. The complexified vectors meet a similar bound for C 8 [1, 3]; we shall verify their value of cos2 in a way that is easily checked by hand. Let
r=
p
2;
q
i = (,1); s = (1 + i)=r; t = s = (1 , i)=r:
(1)
Thus
jsj = jtj = 1 = st;
and
s2 = i; t2 = ,i; s + t = r; s , t = ri;
(2)
and the complex numbers
s t; i(s t); s(1 i); t(1 i) have modulus r:
(3)
It is important to note that the angle between two lines is unchanged if we multiply their representative vectors by scalars; also, multiplying a vector v by a unit scalar
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STUART G. HOGGAR
Table I. Matrix of complex 2-vectors (a, b) giving the 64 lines Row
O
D
1 2 3 4
(0; 0)
(s; t)
(s;
,
(0; r )
(0; 0)
(t;
(s; s)
(r; 0)
(0; 0)
(t; s)
(t; t)
(0; ri)
(0; 0)
(s;
Table II.
S
,
s)
,) t
Matrix with
R
(t;
s)
,)
i; j
t
Type 1: Type 2: Type 3: Type 4:
(ri; 0)
ODSR DORS SROD RSDO
entry the triple
(Di :Dj ; Si :Sj ; Ri :Rj )
1 2 3 4
1
2
3
2; 2; 2 i; 0; 0 0; 0; 2i 0; 2i; 0
2i; 0; 0 2; 2; 2 0; 2i; 0 0; 0; 2i
0; 0; 2i 0; 2i; 0 2; 2; 2 2i; 0; 0
,2
,
,
4
,
0; 2i; 0 0; 0; 2i 2i; 0; 0 2; 2; 2
,
,
leaves its modulus unchanged. In particular, j iv j = jv j. Our vectors v will satisfy jvj2 = 6; we must show that ju:vj2 = 4 when u; v represent distinct lines (they represent the same lines if and only if u:v = 6). Our verification actually holds for a range of sets of representative vectors that includes the original set of [2]. Let O, D, S, R be the columns in order of the matrix of complex 2-vectors shown in Table I, with rows numbered 1 to 4. Thus for example D = [D1 D2 D3 D4 ]T ; D3 = (t; s). A given row exhibits a complex 8-vector v , with jv j2 = 6 by (1), (2); coordinate sign changes yield vectors in four (but no more) distinct lines. Inserting signs in each row against any chosen two of its nonzero 2-vectors, we obtain a total of 4 4 = 16 vectors spanning distinct lines through the origin in C 8 . The same set of lines is obtained for all such choices. Call them Type 1 lines. Now, for each n = 2; 3; 4 we obtain 16 lines said to be of Type n by interchanging columns 1 and n in the matrix, and similarly the remaining two columns. These types are listed to the right of the matrix. We now check the inner products, aided first by Table II. EXAMPLE. Constructing Table II: D1 :D2 = (s; t):(t; ,s) = st ,ts = s2 ,t2 2i. Using Table II: (0; D1 ; ,S1 ; R1 ):(0; D2 ; S2 ; R2 ) = D1 :D2 , S1 :S2 + R1 :R2 2i.
= =
For lines of the same type we require for 1 6 i; j 6 4: jDi :Dj Si :Sj Ri :Rj ] = 2 (excluding case i = j and all signs positive). This is verified by inspecting Table II, with i; j entry the triple Di :Dj ; Si :Sj ; Ri :Rj . For lines of different type we must check that for each pair X; Y chosen from S; R; D we have: jXi :Yj Yi :Xj j = 2. For a table with i; j entry mij = (1=r )Xi :Yj (1 6 i; j 6 4) this is equivalent to
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64 LINES FROM A QUATERNIONIC POLYTOPE
Table III. i; j entry is (1=r )Si :Rj . 1 2 3 4
1
2
3
s
s
si
s
s
t
t
t
t
,
,
, ,
si ti
ti
Table IV.
entry is
i; j
(1=r )Di :Rj .
4
1
2
si
t
s
s
t
s
t
t
s
, , , ,
si ti ti
,
,
3
,
ti
si
,
si ti
Table V. i; j entry is (1=r )Di :Sj .
4
1
2
3
si
s
t
s
s
t
, , , ,
ti ti
si
, ,
t
,
s
t
t
s
4
,
t
s
s
t
t
s
,
jmij mjij = r. The required tables are given in Tables III–V. With the aid of (1) to (3), the verification may be done by inspection. References 1. Delsarte, P., Goethals, J.M. and Seidel, J.J.: Bounds for systems of lines, and Jacobi polynomials, Philips Res. Rep. 30 (1975), 91–105 (Bouwkamp Volume). 2. Hoggar, S.G.: Two quaternionic 4-polytopes, in The Geometric Vein, Coxeter Festschrift, SpringerVerlag, New York, 1981, pp. 219–230. 3. Hoggar, S.G.: t-designs with general angle set, European J. Combin. 13 (1992), 257–271. 4. Seidel, J.J.: Private communication.
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