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The real spin factor satisfies the Jordan identity

Statement

For every real inner-product space $V$ , the product on $S (V)$ in Real spin-factor product is commutative and bilinear, has unit $(0, 1)$ , and satisfies $x \circ (y \circ x^{2}) = (x \circ y) \circ x^{2}$ . Hence it is a unital real Jordan algebra.

Facts & Assumptions

Given: A real inner-product space $V$ , the product in Real spin-factor product, and $x = (v, α)$ , $y = (w, β)$ in $V \oplus R$ .

[F1]

$(v, α) \circ (w, β) = (α w + β v, ⟨ v, w ⟩ + α β)$ (Real spin-factor product).

[F2]

The real inner product is symmetric and bilinear, with $∥ v ∥^{2} : = ⟨ v, v ⟩$ (Real and complex inner product spaces, with the inner product linear in the first argument).

[F3]

The Jordan identity to verify is $x \circ (y \circ x^{2}) = (x \circ y) \circ x^{2}$ (Jordan algebra over a field of characteristic not two).

Proof

1.1givenF1F2algebra

Formula [F1] is symmetric in $(v, α)$ and $(w, β)$ by [F2], and is bilinear. The element $(0, 1)$ is a two-sided unit because $(0, 1) \circ (v, α) = (v, α)$ .

1.2givenF1F2algebra

Put $s : = ∥ v ∥^{2}$ and $r : = ⟨ v, w ⟩$ . Substituting $x$ twice in [F1] gives $x^{2} = (2 α v, s + α^{2})$ .

2.1givenF1F2step 1.2algebra

Using step 1.2 in [F1] first gives $y \circ x^{2} = ((s + α^{2}) w + 2 α β v, 2 α r + β (s + α^{2})) .$ A second substitution yields $x \circ (y \circ x^{2}) = (α (s + α^{2}) w + [2 α r + β (s + 3 α^{2})] v, r (s + 3 α^{2}) + β (3 α s + α^{3})) .$ Each scalar coefficient uses only bilinearity and symmetry in [F2].

3.1givenF1F2step 1.2step 2.1algebra

First $x \circ y = (α w + β v, r + α β)$ . Applying [F1] to $(x \circ y) \circ x^{2}$ and using step 1.2 gives $(x \circ y) \circ x^{2} = (α (s + α^{2}) w + [2 α r + β (s + 3 α^{2})] v, r (s + 3 α^{2}) + β (3 α s + α^{3})) .$ This equals step 2.1 component by component.

4.1step 1.1step 2.1step 3.1F3∎

Steps 1.1, 2.1, and 3.1 verify the unit, commutativity, bilinearity, and [F3]; hence $S (V)$ is a unital real Jordan algebra.

Depends on

Used by

The spin factor on the real plane Example
The spin factor and the plus-quadratic Clifford convention Proposition

Cited to discharge well-definedness by Real spin-factor product.

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John C. Baez, The Octonions, §3.4, PDF p. 27 (standard reference, not scraped)
N. Jacobson, Lectures on Quadratic Jordan Algebras, §1 (standard reference, not scraped)