COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.

310 C. M. Glennie
n = 4. Take d = 11. L is spanned by the twelve monomials upqr, ap(qr),
a(pq)r (see proof of Lemma 1). These are subject to the relation
aP@, c, 4 = aQ@, c, 4
and the word-relation matrix has rank 1. So by Lemma 5 (with U = (0))
dim L 4 11 = d. From Lemma 4 we have that Mf [W] = N where w = 1.
So dimMa dimN-dim [FYI * dimN-w== 12-l = 11 = d.
n = 5. Take d = 55. Since pqr(st) = stR(pqr) = stQ(p, q, r)-stS(p, q, r)
we have that VC U and U = L. Then dim L = dim Us 5 dim L(4) =
=Sxll=55=d. From Lemma 4, M+[W]=N with w=5. So
dim M 3 dim N - w = 60- 5 = 55 = d.
IZ = 6. Take d = 330. From Lemma 2, dim US 5 dim L(5) = 275. V is
spanned by (i) 60 elements apqr(st), (ii) 30 elements a(pq)r(st), (iii) 30 elements
ap(qr)(st). From (1), (8), (9), (lo), (11) we have
&W) - 4mW) - 4wXqr) E u.
Defining T(p, q, r, s, t) as
[Qtq, r, P) - S(q, r, pNJW + [Qh t, P) - S(s, t, P)lNqr)
we have
dqr)W - aT@, 4, r, s, 0 E U. (12)
Also, from (5):
apqptr, s, t> E U (13)
4&W-, 4 t) E u. (14)
and from (1) :
apti, q, r)R(st) - aQCp, q, r)R(st) E U. (15)
(12) to (15’) give respectively 30, 20, 10, 10 relations. Setting up the wordrelation
matrix for the 120 spanning elements of V and these 70 relations we
get a 70x 120 matrix of which the rank is 65. Then by Lemma 5,
dim (U+ V) 6 dim U+(l20-65). So
dim L == dim (U+ V) =s 275f 55 = 330 = d.
From Lemma 4, M+ [W] = N with w = 45. Now let W’ be the subset
- - -
of W consisting of the 30 elements pqrstufpqrsut, pqrs(tu), and let N’ =
= M+ [ wl]. We have 45 relations amongst elements of W- w’ obtained
from
- ~ - - - -~abcdef
- abcdfe + bcdefa - bcdeaf f cdefab - cdefba + acdfeb- acdfbe E N’
(16)
by permuting a, b, c, d, e, f and using Lemma 3. We have a further 6 relations
obtained from
---caefab
- cdefba f defbac- defbca f efbcad- efbcda
----
+- fbcdae -fbcdea i- bcdeaf - bcdefa E N (17)
I
Identities in Jordan algebras
by permuting a, b, c, d, e, f and using Lemma 3. (16) is the linearized
form of
abcdab-abcdba E N’
which comes from
-- __ ~
acdb2a - cdb2aa + cdb2a2 - bdca2b + dca26b - dca2b2 = 0
using Lemma 3 and
pe______pqrst
= qrstp-rstpqfstpqr- tpqrs+pqrst. (18)
~-
(17) comes from c (cdefub- cdef(ab)) = 0 where the sum is taken over
the cyclic permutations of b, c, d, e, f and Lemma 3 is used where necessary.
The rank of the word-relation matrix for the 15 elements in W- W
and the 51 relations above is 15. So dim N = dim (N’f [W- W’]) 4
dim N’+l5-15 = dim N’. Whence N = N’. So dim M==dim N-30 =
360-30 = 330 = d.
n = 7. Take d = 2345. From Lemma 2, dim S == 7 dim L(6) = 7 x 330 =
2310. V is spanned by elements of types (i) upqrs(tu), (ii) a(pq)rs(tu),
&) ap(qr)s(tu), (iv) apq(rs)(tu), (v) a(pq)(rs)(tu). Now tuR(apqrs),
tuR(a(pq)rs), tuR(apq(rs)), and tuR(a(pq)(rs)) are in S. This follows at
once on expanding the operator R using (3) and then using (3) again where
necessary. So L = Sf V is spanned by S and the set of 180 elements
up(qr)s(tu). Now let X be the set of the 48 elements of type (iii) in which
q = b and t = c or q = c and t = b. Consider the following table, in which
each element is to represent the set of elements obtained from it by
replacing p, q, r, s by all permutations of d, e, & g:
aptkM4 w Wr(W
aptqrP(cs) ~pt&Mrs) dcq)bW dqrW4
4kMcs) ap(qr)sW aptWq(rs) a+MW
4cMrs) %sMrd adz4btrs) &vMrs)
Each element in the table can be expressed modulo S as a linear combination
of elements in higher rows. Thus, for example:
ap(qr)b(cs) = -up(bq)r(cs)-ap(br)q(cs) (mod S)
since apQ(q, r, b)R(cs) = apP(q, r, b)R(cs) and the elements in this last
expression are all of type (iv) and so in S. The expression for ab(cp)q(rs)
arises from
cpQ@, b, q)R(rs) + rsQ(a, c, p)R(W + bqQ(a, r, s)R(cp) - aQ@q, cp, rs) E S.
So we now have that S+ [X] = L. But there are further relations modulo
S amongst the elements of X. These are:
CPA 21
311
1 MsM4 E s (1%
c ~p(cdr@s) E s, (20)