Action of New Cayley-Dickson Variants on Quaternions

Using the new Cayley-Dickson formula (a,b)(c,d) = (ac - b*d, da* + bc) 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_je_i*) and (0, e_j)(e_i, 0) = (0, e_je_i)
(0, e_i)(0, e_j) = (-e_i*e_j, 0)

Applying these to the right-handed quaternions extends their multiplication table thus (where the "canonical" mapping is assumed,
i.e. e_i ↔ (e_i, 0) for i = 1,2,3; e₄ ↔ (0, 1); e_i+4 ↔ (0, e_i)):

	e₀	e₁	e₂	e₃
e₀	e₀	e₁	e₂	e₃
e₁	e₁	-e₀	e₃	-e₂
e₂	e₂	-e₃	-e₀	e₁
e₃	e₃	e₂	-e₁	-e₀

→

	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

The triads and their handedness (0=right, 1=left):

Triad	R/L
e₁,e₂,e₃	0
e₁,e₄,e₅	1
e₁,e₆,e₇	0
e₂,e₄,e₆	1
e₂,e₅,e₇	1
e₃,e₄,e₇	1
e₃,e₅,e₆	0

The signmask is seen to be 0111010 = 3a = 3b XOR 01, showing that the distinguished triad is (e₁,e₂,e₃).

The 2nd Variant

Likewise, using the new Cayley-Dickson formula (a,b)(c,d) = (ac - bd*, ad + c*b) 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_ie_j) and (0, e_j)(e_i, 0) = (0, e_i*e_j)
(0, e_i)(0, e_j) = (-e_ie_j*, 0)

	e₀	e₁	e₂	e₃
e₀	e₀	e₁	e₂	e₃
e₁	e₁	-e₀	e₃	-e₂
e₂	e₂	-e₃	-e₀	e₁
e₃	e₃	e₂	-e₁	-e₀

→

	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

The triads and their handedness (0=right, 1=left):

Triad	R/L
e₁,e₂,e₃	0
e₁,e₄,e₅	0
e₁,e₆,e₇	0
e₂,e₄,e₆	0
e₂,e₅,e₇	1
e₃,e₄,e₇	0
e₃,e₅,e₆	0

The signmask is seen to be 0010000 = 10 = 11 XOR 01, showing that the distinguished triad is (e₁,e₂,e₃).