Zero divisor Example:
To find the zerodivisor, Z,
of S := e2 -
2 e4 - e7 +
2 e9 - e11
+ e13 .
First, choose the following octonion
multiplication rule for the basal elements, ij
.
ij ik
= - djk
+ S(jk)
ejkl im
,
where the summation is over j and
k for a fixed m, and the structure constant, ejkm
, is determined by the following cycles,
(1,2,3) , (1,4,5) , (1,7,6) , (2,4,6)
, (2,5,7) , (3,4,7) , (3,6,5) ,
and is given by the usual permutation
rule for a 3-indexed antisymmetric tensor with values 0 or ± 1,
i.e.,
ijkm
= { |
1 , if
(j,k,m) is an even permutation
-1, if (j,k,m)
is an odd permutation
0 , otherwise. |
} of the octonion cycle. |
Writing S in terms of octonions
A and B, i.e., S := ( A ; B ) , where
A = i
2
- 2 i 4 - i 7
, B = 2 i 1
- i 3 + i 5
we see that S satisfies the zerodivisor
vector-closure conditions as well as the following conditions discussed
in our publication
A .
B = 0,
n(A) = n(B), w
:= ( n (S) )2 / n (A,B) = 1 .
Choosing x 1
, x 5 , x 6
, x 7 , as the free parameters, we obtain
X = 18x 1i
1
+ 6 ( 14x
1 - 4x 5
- 4x 6 + 3 x 7
) i 2 + 18 ( 2x 1
+ x 5 ) i 3
+
6 ( 7x 1 - 2x 5
- 2x 6 ) i 4
+ 18x 5 i 5
+ 18x 6 i 6
+ 18x 7 i 7
,
and
Y = (
- 6x1 - 6x5 + 3x6
) i 1 + ( 2x1 - 16x5
- 7x6 - 12x7
) i 2 + ( 38x1
- 16x5 + 2x6
+ 6x7 ) i 3
+( -34x1+2x5
+ 11x6 - 12x7
) i 4 + ( 50x 1
- 4x 5 - 4x 6
+ 6x 7 ) i 5
+ ( -20x 1 - 20x 5
- 2x 6 + 12x 7
) i 6 + ( 70x 1
- 20x 5 - 299x 6
+ 12x7 ) i 7
|
... |