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An off-diagonal product in the Albert algebra

Statement

Write $E_{a b} (q)$ for the $3 \times 3$ matrix with $q$ in row $a$ , column $b$ , and zero elsewhere. Let $X = E_{12} (i) + E_{21} (- i)$ and $Y = E_{23} (j) + E_{32} (- j)$ in $H_{3} (O)$ . Their Jordan product has $k / 2$ in the $(1, 3)$ entry and $- k / 2$ in the $(3, 1)$ entry.

Facts & Assumptions

Given: The Hermitian matrix space $H_{3} (O)$ and its product $X \circ Y = (X Y + Y X) / 2$ .

[F1]

The matrix product is $(X Y)_{a b} = \sum_{r} X_{a r} Y_{r b}$ , and $X \circ Y = (X Y + Y X) / 2$ (Hermitian three-by-three octonion matrix data).

[F2]

The embedded quaternion units obey $i j = k$ , $j i = - k$ , and $\overset{ˉ}{i} = - i$ , $\overset{ˉ}{j} = - j$ , $\overset{ˉ}{k} = - k$ (Real octonions by Cayley–Dickson doubling, The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ ).

[F3]

The product maps Hermitian matrices into $H_{3} (O)$ (The real Albert algebra is a Jordan algebra).

Proof

1.1givenF1F2algebra

The identities in [F2] make $X$ and $Y$ Hermitian. In $X Y$ , the only possibly nonzero entry is $(X Y)_{13} = X_{12} Y_{23} = i j = k$ . In $Y X$ , the only possibly nonzero entry is $(Y X)_{31} = Y_{32} X_{21} = (- j) (- i) = j i = - k$ . Thus their half-sum has entries $k / 2$ at $(1, 3)$ and $- k / 2$ at $(3, 1)$ , with all others zero.

2.1givenF3step 1.1∎

The product from step 1.1 is Hermitian, and by [F3] it belongs to the Albert Jordan algebra.

Depends on

Used by

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Sources

N. Jacobson, Lectures on Quadratic Jordan Algebras, §8 (standard reference, not scraped)
John C. Baez, The Octonions, PDF pp. 31–35 (standard reference, not scraped)