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The spin factor on the real plane

Statement

For $V = R^{2}$ with its standard dot product, the spin factor has unit $u = (0, 1)$ . For $v = (x, y)$ , $(v, α)^{2} = (2 α v, x^{2} + y^{2} + α^{2}) .$ Its square cone is exactly $α \geq \sqrt{x^{2} + y^{2}}$ , with boundary $α^{2} = x^{2} + y^{2}$ and $α \geq 0$ . The orthonormal basis $e_{1} = ((1, 0), 0)$ , $e_{2} = ((0, 1), 0)$ , $u = ((0, 0), 1)$ satisfies $u \circ e_{i} = e_{i}, e_{1}^{2} = e_{2}^{2} = u, e_{1} \circ e_{2} = 0,$ and for $z = a e_{1} + b e_{2} + c u$ , $z^{2} = 2 a c e_{1} + 2 b c e_{2} + (a^{2} + b^{2} + c^{2}) u .$

Facts & Assumptions

Given: The spin factor $S (R^{2})$ with the standard positive definite inner product.

[F1]

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

[F2]

This product satisfies the Jordan identity (The real spin factor satisfies the Jordan identity).

[F3]

The induced length on a real inner-product space is a norm and satisfies $∥ λ v ∥ = ∣ λ ∣ ∥ v ∥$ (The induced length is a norm).

[F4]

Every nonnegative real number has a unique nonnegative square root (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[F5]

Give $R^{2}$ its standard Euclidean norm topology and $R^{2} \times R$ the product with the usual topology on $R$ .

Proof

1.1givenF1algebra

Substituting $u = (0, 1)$ , $e_{1} = ((1, 0), 0)$ , and $e_{2} = ((0, 1), 0)$ into [F1] gives $u \circ e_{i} = e_{i}$ , $e_{1}^{2} = e_{2}^{2} = u$ , and $e_{1} \circ e_{2} = 0$ , and verifies that $u$ is the unit.

2.1givenF1step 1.1algebra

By [F1] and the standard dot product, $(v, α)^{2} = (2 α v, x^{2} + y^{2} + α^{2})$ for $v = (x, y)$ . Expanding in the basis products of step 1.1 gives $(a e_{1} + b e_{2} + c u)^{2} = 2 a c e_{1} + 2 b c e_{2} + (a^{2} + b^{2} + c^{2}) u$ .

3.1givenF3step 2.1algebra

If $(w, t) = (v, α)^{2}$ , step 2.1 gives $w = 2 α v$ and $t = ∥ v ∥^{2} + α^{2}$ . By norm homogeneity in [F3], $∥ w ∥ = 2 ∣ α ∣ ∥ v ∥$ , and $(∣ α ∣ - ∥ v ∥)^{2} \geq 0$ gives $∥ w ∥ \leq t$ . Thus every square lies in the stated cone.

3.2givenF4step 2.1algebra

Conversely, suppose $t \geq ∥ w ∥$ . If $t = 0$ , norm separation in [F3] gives $w = 0$ and $(w, t) = (0, 0)^{2}$ . If $t > 0$ , then $t^{2} - ∥ w ∥^{2} = (t - ∥ w ∥) (t + ∥ w ∥) \geq 0$ , so both $\sqrt{t^{2} - ∥ w ∥^{2}}$ and, after defining $s = (t + \sqrt{t^{2} - ∥ w ∥^{2}}) / 2 > 0$ , $α = \sqrt{s}$ exist by [F4]. Set $v = w / (2 α)$ . The quadratic equation $s^{2} - t s + ∥ w ∥^{2} / 4 = 0$ gives $s + ∥ w ∥^{2} / (4 s) = t$ and $2 α v = w$ , so step 2.1 gives $(v, α)^{2} = (w, t)$ .

4.1givenstep 3.2algebra

The equality locus of the square cone is $t = ∥ w ∥$ , equivalently $t \geq 0$ and $t^{2} - ∥ w ∥^{2} = 0$ ; for $w = (x, y)$ and $t = α$ , this is $α \geq 0$ and $α^{2} = x^{2} + y^{2}$ .

5.1F2F3F5step 4.1algebra∎

The function $g (w, t) = t - ∥ w ∥$ is continuous in the product topology [F5], since the triangle inequality in [F3] gives $∣ ∥ w ∥ - ∥ w^{'} ∥ ∣ \leq ∥ w - w^{'} ∥$ . Continuity makes ${g > 0}$ and ${g < 0}$ open; the first lies in the cone and the second in its complement. If $g (w, t) = 0$ , then $(w, t + ε)$ is in the cone and $(w, t - ε)$ is outside it for every $ε > 0$ ; both approach $(w, t)$ as $ε ↓ 0$ . Hence exactly the equality locus in step 4.1 is the boundary. The algebra is Jordan by [F2].

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Sources

John C. Baez, The Octonions, §3.3, PDF pp. 27–28 (standard reference, not scraped)
N. Jacobson, Lectures on Quadratic Jordan Algebras, §1 (standard reference, not scraped)