Triality\\\\ A Brief Note

Triality A Brief Note Johar M. Ashfaque A triality among three real vector spaces V1 , V2 and V3 is a tri-linear map F : V1 ⊗ V2 ⊗ V3 → R such that for any non-zero v1 ∈ V1 , v2 ∈ V2 there is a v3 ∈ V3 for which F (v1 , v2 , v3 ) is a non-zero and similarly for the other two cases. Note. If we fix v1 then F (v1 , ◦, ◦) is a duality between V2 and V3 and likewise for the other two cases. If V1 , V2 and V3 are normed, a normed triality amongst them is a triality such that for any v1 ∈ V1 , v2 ∈ V2 and v3 ∈ V3 |F (v1 , v2 , v3 )| ≤ ||v1 ||||v2 ||||v3 || and given any v1 ∈ V1 and v2 ∈ V2 there is some v3 ∈ V3 so that equality is obtained and likewise for other two cases. There is a paucity of trialities when compared to dualities and normed trialities are even harder to find. One way to find a normed triality is to construct one from a product operation on a division algebra. Assuming we have a division algebra D and a product map ·:D⊗D→D which upon dualizing gives · : D ⊗ D ⊗ D∗ → R. We know that such algebras have an inner product so we can dualize to obtain a product ·:D⊗D⊗D

→ R

·(A, B, C)

=

(A · B, C)

To see this as a normed triality note |(A, B, C)| ≤ ||A · B||||C|| = ||A||||B||||C|| and equality is obtained for C = ±A · B. Conversely, given a normed triality, F , we wish to obtain the division algebra. Pick any choice of unit vectors v1 ∈ V1 and v2 ∈ V2 . Then there is a unit vector v3 ∈ V3 such that F (v1 , v2 , v3 ) = 1. Since F (v1 , ◦, ◦) is a duality between V2 and V3 , we have an isometric isomorphism V2 ↔ V3 where v2 ↔ v3 and likewise V1 ↔ V2 and V1 ↔ V3 where v1 ↔ v2 and v1 ↔ v3 respectively. Thus any triality F is in fact F :V ⊗V ⊗V →R which is canonical only after choosing v1 and v2 . We have F : V ⊗ V → V ∗ . Since F (v1 , ◦, ◦) is a duality, we have an isometric isomorphism V → V ∗ . As a result, we obtain a new map ? : V ⊗ V → V. Note. v1 ? b = b and a ? v2 = a, so we can identify v1 , v2 and v3 with the identity element.

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Note. a ? b is an element c ∈ V such that |F (a, b, c)| ≤ ||a||||b||||c|| so ||a ? b|| = ||a||||b||. Hence, we conclude that we have obtained a division algebra.

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