Sedenion Geometry

1. each element is in 3 triads
2. there are 7 distinct triads
3. any 2 distinct triads intersect in a single unit
4. all such groupings are isomorphic, under permutation of the indices labelling the units

1. each element is in 6 pairs of elements, and each triad containing the element "uses up" 2 of those pairs (since each pair is in a unique triad), yielding 3 triads per unit
2. 3 triads per element, × 7 elements = 21, but each triad is hereby counted thrice, once for each element it contains, so there are 3×7/3 = 7 distinct triads
3. for triads t₁, t₂ let N=o(t₁∩t₂), let e∈(t₁-t₂), and let T={ef:f∈(t₂-t₁)}; then T∩t₁ = T∩t₂ = ∅ and o(T) = o(t₁-t₂) =o (t₂-t₁) = 3-N, so
   o(T∪(t₁-t₂)∪(t₂-t₁)) = 3(3-N) ≤ 7 ⇒ N ≥ 1

Properties for Any Number of Heptads

1. each imaginary unit is in 7 triads
2. there are a total of 35 quaternionic triads
3. each triad intersects 18 others, is disjoint with 16 others
4. if there exist more than one heptad, 2 distinct heptads intersect in a triad
5. any triad can be in at most 3 heptads
6. each triad intersects any heptad in 1 or 3 basis elements

1. each imaginary e_i is in 14 pairs; each pair is in exactly one triad; each triad containing e_i contains 2 of these pairs; hence, there are exactly 7 triads containing e_i
2. each of the 15 elements e_i occurs in 7 triads, giving 105 = 7×15; each triad is thereby counted 3 times, once for each e_i it contains, thus giving 105/3 = 35 distinct triads
3. triad (e_i,e_j,e_k) intersects all other triads containing e_i, e_j, or e_k, and only those; there are 6 such triads for each element, all distinct, hence 18 in total; thus there are 35 - 18 - 1 other triads not containing any of e_i, e_j, or e_k, i.e. 16 triads disjoint with the given triad
4. let heptads h₁, h₂ have intersection h₁∩h₂ containing N elements; let H_i = h_i - (h₁∩h₂), so o(H_i) = 7-N; for E∈H₁, EH₂ is a set of 7-N distinct elements ∉h₁ and ∉h₂; thus o(h₁∪h₂∪EH₂) = N + 3(7-N) = 21-2N ≤ 15 distinct elements, so that N ≥ 3. Moreover, if e_i, e_j∈h₁∩h₂, then e_ie_j∈h₁∩h₂, so if h₁∩h₂ comprises more than a single triad, it must also contain all products of the contained elements, or 7 elements in all, so h₁∩h₂ = h₁ = h₂.
5. if a triad t is in N heptads, each heptad has 4 elements not in any other of the heptads containing t, for a total of 4N + 3 ≤ 15 distinct units
6. let N = o(t∩h) for triad t⊄heptad h; let e∈t-h, and H = {ef: f∈(h-t)}; then o(H) = 7-N, and t∩H = h∩H = ∅; then 15 ≥ o(t)+o(h-t)+o(H) = 3+(7-N)+(7-N) ⇒ N≥1; since h is closed under multiplication, N>1 ⇒ N=3

Systems with 0 Heptads

Systems with 1 Heptad

#1		1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 	I
#16704		1 2 3 4 5 6 7 11 10 9 8 15 14 13 12 	a
#28784		1 2 3 4 5 6 7 13 12 15 14 9 8 11 10 	b
#35153		1 2 3 4 5 6 7 14 15 12 13 10 11 8 9 	c
#26911186439	1 6 7 4 5 2 3 9 8 14 15 13 12 10 11 	d
#26911192746	1 6 7 4 5 2 3 10 11 13 12 14 15 9 8 	e
#26911204250	1 6 7 4 5 2 3 12 13 11 10 8 9 15 14 	f
#26911220407	1 6 7 4 5 2 3 15 14 8 9 11 10 12 13 	g

Systems with 2 Heptads

Systems with 3 Heptads

• e₁, e₂, e₃, e₄, e₅, e₆, e₇,
• e₁, e₂, e₃, e₈, e₉, e₁₀, e₁₁, and
• e₁, e₂, e₃, e₁₂, e₁₃, e₁₄, e₁₅.

#1		1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 	I
#11537		1 2 3 4 5 6 7 10 11 8 9 14 15 12 13 	A
#28784		1 2 3 4 5 6 7 13 12 15 14 9 8 11 10 	B
#40320		1 2 3 4 5 6 7 15 14 13 12 11 10 9 8 	C
#127370897	1 2 3 7 6 5 4 8 9 10 11 14 15 12 13 	D
#127382401	1 2 3 7 6 5 4 10 11 8 9 12 13 14 15 	E
#127399680	1 2 3 7 6 5 4 13 12 15 14 11 10 9 8 	F
#127411184	1 2 3 7 6 5 4 15 14 13 12 9 8 11 10 	G

Systems with 4, 5, or 6 Heptads

Systems with 7 Heptads

• e₁, e₂, e₃, e₄, e₅, e₆, e₇,
• e₁, e₂, e₃, e₈, e₉, e₁₀, e₁₁,
• e₁, e₂, e₃, e₁₂, e₁₃, e₁₄, e₁₅,
• e₁, e₄, e₅, e₈, e₉, e₁₂, e₁₃,
• e₁, e₄, e₅, e₁₀, e₁₁, e₁₄, e₁₅,
• e₁, e₆, e₇, e₈, e₉, e₁₄, e₁₅,
• e₁, e₆, e₇, e₁₀, e₁₁, e₁₂, e₁₃

Systems with 8-14 Heptads

Systems with 15 Heptads

1. any 2 intersecting triads generate a heptad
2. each imaginary unit is in 7 heptads
3. each triad is in 3 heptads

1. each heptad contains 7 triads, hence 7×6/2 pairs of triads; since distinct heptads intersect in a single triad, any pair of triads is thus in only one heptad, and thus there are 15×(7×6/2)=315 distinct pairs in all the 15 heptads; any triad intersects 18 others, so there are a total of 35×18/2=315 pairs of intersecting triads; thus any pair of intersecting triads is contained in exactly one of the heptads
2. each unit e is in 7 triads, so there are 7×6/2 pairs of triads intersecting in e; each such pair generates a heptad, but each such heptad is generated by 3 distinct such pairs, so there are (7×6/2)/3=7 distinct heptads containing e
3. any triad intersects 18 others, and each such intersection generates a heptad; within such a heptad, the given triad intersects 6 others; hence there are 18/6=3 distinct heptads generated by the given triad and its set of intersecting triads, i.e. the given triad is in 3 distinct heptads

Heptad Products