Standard Cayley-Dickson Construction of Sedenions

Using the traditional Cayley-Dickson formula (a,b)(c,d) = (ac - db*, a*d + cb) gives as special cases:

(e_i, 0)(0, 1) = (0, e_i*) and (0, 1)(e_i, 0) = (0, e_i)
(0, 1)(0, e_i) = (-e_i, 0) and (0, e_i)(0, 1) = (-e_i*, 0)
(e_i, 0)(0, e_j) = (0, e_i*e_j) and (0, e_j)(e_i, 0) = (0, e_ie_j)
(0, e_i)(0, e_j) = (-e_je_i*, 0)

Using this process to obtain sedenions from octonions requires first choosing a representation of the octonions to act upon. A logical choice is the representation obtained from the right-handed quaternions via the canonical mapping, viz.:

	e₀	e₁	e₂	e₃	e₄	e₅	e₆	e₇
e₀	e₀	e₁	e₂	e₃	e₄	e₅	e₆	e₇
e₁	e₁	-e₀	e₃	-e₂	-e₅	e₄	-e₇	e₆
e₂	e₂	-e₃	-e₀	e₁	-e₆	e₇	e₄	-e₅
e₃	e₃	e₂	-e₁	-e₀	-e₇	-e₆	e₅	e₄
e₄	e₄	e₅	e₆	e₇	-e₀	-e₁	-e₂	-e₃
e₅	e₅	-e₄	-e₇	e₆	e₁	-e₀	-e₃	e₂
e₆	e₆	e₇	-e₄	-e₅	e₂	e₃	-e₀	-e₁
e₇	e₇	-e₆	e₅	-e₄	e₃	-e₂	e₁	-e₀

Applying the Cayley-Dickson construction above and using the canonical mapping gives the following multiplication table containing the triads listed beside it ("T#" is a triad's 0-based position in the lexically ordered list of quaternionic triads; "Handbit" is 0 for righthanded and 1 for lefthanded, giving a signmask value of 29cf9f29e for this multiplication table):

	e₀	e₁	e₂	e₃	e₄	e₅	e₆	e₇	e₈	e₉	e₁₀	e₁₁	e₁₂	e₁₃	e₁₄	e₁₅
e₀	e₀	e₁	e₂	e₃	e₄	e₅	e₆	e₇	e₈	e₉	e₁₀	e₁₁	e₁₂	e₁₃	e₁₄	e₁₅
e₁	e₁	-e₀	e₃	-e₂	-e₅	e₄	-e₇	e₆	-e₉	e₈	-e₁₁	e₁₀	e₁₃	-e₁₂	e₁₅	-e₁₄
e₂	e₂	-e₃	-e₀	e₁	-e₆	e₇	e₄	-e₅	-e₁₀	e₁₁	e₈	-e₉	e₁₄	-e₁₅	-e₁₂	e₁₃
e₃	e₃	e₂	-e₁	-e₀	-e₇	-e₆	e₅	e₄	-e₁₁	-e₁₀	e₉	e₈	e₁₅	e₁₄	-e₁₃	-e₁₂
e₄	e₄	e₅	e₆	e₇	-e₀	-e₁	-e₂	-e₃	-e₁₂	-e₁₃	-e₁₄	-e₁₅	e₈	e₉	e₁₀	e₁₁
e₅	e₅	-e₄	-e₇	e₆	e₁	-e₀	-e₃	e₂	-e₁₃	e₁₂	e₁₅	-e₁₄	-e₉	e₈	e₁₁	-e₁₀
e₆	e₆	e₇	-e₄	-e₅	e₂	e₃	-e₀	-e₁	-e₁₄	-e₁₅	e₁₂	e₁₃	-e₁₀	-e₁₁	e₈	e₉
e₇	e₇	-e₆	e₅	-e₄	e₃	-e₂	e₁	-e₀	-e₁₅	e₁₄	-e₁₃	e₁₂	-e₁₁	e₁₀	-e₉	e₈
e₈	e₈	e₉	e₁₀	e₁₁	e₁₂	e₁₃	e₁₄	e₁₅	-e₀	-e₁	-e₂	-e₃	-e₄	-e₅	-e₆	-e₇
e₉	e₉	-e₈	-e₁₁	e₁₀	e₁₃	-e₁₂	e₁₅	-e₁₄	e₁	-e₀	-e₃	e₂	e₅	-e₄	e₇	-e₆
e₁₀	e₁₀	e₁₁	-e₈	-e₉	e₁₄	-e₁₅	-e₁₂	e₁₃	e₂	e₃	-e₀	-e₁	e₆	-e₇	-e₄	e₅
e₁₁	e₁₁	-e₁₀	e₉	-e₈	e₁₅	e₁₄	-e₁₃	-e₁₂	e₃	-e₂	e₁	-e₀	e₇	e₆	-e₅	-e₄
e₁₂	e₁₂	-e₁₃	-e₁₄	-e₁₅	-e₈	e₉	e₁₀	e₁₁	e₄	-e₅	-e₆	-e₇	-e₀	e₁	e₂	e₃
e₁₃	e₁₃	e₁₂	e₁₅	-e₁₄	-e₉	-e₈	e₁₁	-e₁₀	e₅	e₄	e₇	-e₆	-e₁	-e₀	e₃	-e₂
e₁₄	e₁₄	-e₁₅	e₁₂	e₁₃	-e₁₀	-e₁₁	-e₈	e₉	e₆	-e₇	e₄	e₅	-e₂	-e₃	-e₀	e₁
e₁₅	e₁₅	e₁₄	-e₁₃	e₁₂	-e₁₁	e₁₀	-e₉	-e₈	e₇	e₆	-e₅	e₄	-e₃	e₂	-e₁	-e₀

Heptads contain the triads to the right (as referenced by T#):

H#	T#'s of Member Triads	signmask
H0	0,1,2,7,8,13,14	1101110 = 6e
H1	0,3,4,9,10,15,16	1101110 = 6e
H2	0,5,6,11,12,17,18	0010000 = 10
H3	1,3,5,19,20,23,24	0111011 = 3b
H4	1,4,6,21,22,25,26	1011011 = 5b
H5	2,3,6,27,28,31,32	0111011 = 3b
H6	2,4,5,29,30,33,34	0100011 = 23
H7	7,9,11,19,21,27,29	0111011 = 3b
H8	7,10,12,20,22,28,30	0111101 = 3d
H9	8,9,12,23,25,31,33	1101110 = 6e
H10	8,10,11,24,26,32,34	0010000 = 10
H11	13,15,17,19,22,31,34	0111011 = 3b
H12	13,16,18,20,21,32,33	1011011 = 5b
H13	14,15,18,23,26,27,30	0111011 = 3b
H14	14,16,17,24,25,28,29	0100011 = 23

T#;	Triad Units	Handbit
0	e₁,e₂,e₃	0
1	e₁,e₄,e₅	1
2	e₁,e₆,e₇	1
3	e₁,e₈,e₉	1
4	e₁,e₁₀,e₁₁	1
5	e₁,e₁₂,e₁₃	0
6	e₁,e₁₄,e₁₅	0
7	e₂,e₄,e₆	1
8	e₂,e₅,e₇	0
9	e₂,e₈,e₁₀	1
10	e₂,e₉,e₁₁	0
11	e₂,e₁₂,e₁₄	0
12	e₂,e₁₃,e₁₅	1
13	e₃,e₄,e₇	1
14	e₃,e₅,e₆	1
15	e₃,e₈,e₁₁	1
16	e₃,e₉,e₁₀	1
17	e₃,e₁₂,e₁₅	0
18	e₃,e₁₃,e₁₄	0
19	e₄,e₈,e₁₂	1
20	e₄,e₉,e₁₃	1
21	e₄,e₁₀,e₁₄	1
22	e₄,e₁₁,e₁₅	1
23	e₅,e₈,e₁₃	1
24	e₅,e₉,e₁₂	0
25	e₅,e₁₀,e₁₅	0
26	e₅,e₁₁,e₁₄	1
27	e₆,e₈,e₁₄	1
28	e₆,e₉,e₁₅	1
29	e₆,e₁₀,e₁₂	0
30	e₆,e₁₁,e₁₃	0
31	e₇,e₈,e₁₅	1
32	e₇,e₉,e₁₄	0
33	e₇,e₁₀,e₁₃	1
34	e₇,e₁₁,e₁₂	0

Observations on Heptad Structure

To analyze the octonionic nature of the heptads, they must be compared to the known representations of octonions and twisted octonions. Except for H0, the indices of the unit elements must thus be remapped to {1,2,3,4,5,6,7}. The natural map is monotonic; once applied, the renamed triads are used to determine which of the 30 quaternionic groupings corresponds to the heptad, hence which of the signmask values correspond to true octonions. For XOR-based representations, this is easy - the remapped heptad corresponds to the XOR quaternion grouping, and the order of bits in the signmask is unaltered, so direct comparison to the special signmask values 08,0f,11,... determines which triad (if any) is distinguished. Click here for a more complicated (hence more instructive) example.

Thus, H1, H3, H5, H7, H9, H11, H13, and H0 are all true octonions; the distinguished triads in the other heptads are (specified by T#):
H2: 0; H4: 1; H6: 2; H8: 7; H10: 8; H12: 13; H14: 14; these are exactly the triads in H0.