Pipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 5 results · all verified · 0 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 5 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Jordan Algebras, Spin Factors, and Alternative Examples: Examples and Counterexamples

1 · Prerequisites

2 · Summary

These worked examples check the A-page conventions in explicit low-dimensional models. The symmetrized product on $M_{2} (R)$ is Jordan but not associative, and the self-adjoint part of $M_{2} (C)$ need not be closed under the original multiplication. The standard-plane spin factor gives a three-element multiplication table and a direct square formula. The named octonion units witness a nonzero associator, while an off-diagonal product shows how octonion multiplication enters the Albert algebra.

The symmetrized product need not be associative and The self-adjoint part need not be an associative subalgebra distinguish the Jordan product from the original associative product. The spin factor on the real plane calculates the products and squares for $S (R^{2})$ . A concrete octonion associator witness checks the associator $[i, j, ℓ] = 2 k ℓ$ , and An off-diagonal product in the Albert algebra computes one Hermitian matrix product with $k / 2$ and $- k / 2$ in transposed off-diagonal entries. The induced-norm supplier gives homogeneity and continuity in the spin-factor cone proof; the real square-root theorem supplies roots in the cone construction. Each example is proved from its listed suppliers and coordinate conventions.

Companion A page: jordan-spin-factor-and-alternative-algebras.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

The symmetrized product need not be associative

Statement

The symmetrized product $A^{+}$ of an associative algebra need not be associative. In $M_{2} (R)^{+}$ , put $x = E_{12}$ , $y = E_{21}$ , and $z = E_{11}$ . Then $(x \circ y) \circ z = \frac{1}{2} E_{11}$ while $x \circ (y \circ z) = \frac{1}{4} I_{2}$ .

Facts & Assumptions

Given: $M_{2} (R)$ with usual matrix multiplication and $a \circ b = (a b + b a) / 2$ .

[F1]

The symmetrized product is $a \circ b = (a b + b a) / 2$ (Symmetrized product of an associative algebra).

[F2]

The usual matrix product on $M_{2} (R)$ is bilinear, associative, and unital with identity $I_{2}$ , and $R$ has characteristic zero. For $a, b \in {1, 2}$ , $E_{a b}$ is the matrix with $1$ in position $(a, b)$ and zeros elsewhere; these units satisfy $E_{a b} E_{c d} = δ_{b c} E_{a d}$ . The symmetrized product of an associative algebra is Jordan (The symmetrized product of an associative algebra is Jordan).

Proof

1.1givenF1F2algebra

The matrix-unit rule gives $x \circ y = I_{2} / 2$ and $y \circ z = E_{21} / 2$ . Therefore $(x \circ y) \circ z = (I_{2} / 2) \circ E_{11} = E_{11} / 2$ .

2.1givenF1F2step 1.1algebra∎

It also gives $x \circ (y \circ z) = E_{12} \circ (E_{21} / 2) = \frac{1}{4} (E_{11} + E_{22}) = I_{2} / 4.$ These differ, since their $(1, 1)$ entries are $1 / 2$ and $1 / 4$ . Thus the product is not associative; it is nevertheless a Jordan product by [F2].

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

The self-adjoint part need not be an associative subalgebra

Statement

In $A = M_{2} (C)$ with conjugate-transpose involution, the self-adjoint part is closed under the symmetrized product but not under the original associative multiplication. For $X = E_{11}, Y = E_{12} + E_{21},$ both $X, Y$ are self-adjoint while $X Y = E_{12}$ is not.

Facts & Assumptions

Given: $A = M_{2} (C)$ , with ordinary matrix multiplication and $Z^{*} = {\overline{Z}}^{T}$ .

[F1]

A complex associative algebra with involution has a self-adjoint part $A_{s a} = {Z : Z^{*} = Z}$ , and the involution reverses products (Complex associative algebra with involution).

[F2]

For $a, b \in {1, 2}$ , $E_{a b}$ is the matrix with $1$ in position $(a, b)$ and zeros elsewhere; these units obey $E_{a b} E_{c d} = δ_{b c} E_{a d}$ . The usual product makes $M_{2} (C)$ a unital associative complex algebra, and $C$ has characteristic zero. Conjugate transpose is conjugate-linear, satisfies $(Z^{*})^{*} = Z$ , $(Z W)^{*} = W^{*} Z^{*}$ , and fixes $I_{2}$ , so it is an involution as required by [F1]. It sends each real matrix unit $E_{a b}$ to $E_{b a}$ ; in particular, it fixes $E_{11}$ and $E_{22}$ and interchanges $E_{12}$ with $E_{21}$ . The self-adjoint part is closed under the symmetrized Jordan product by The symmetrized product of an associative algebra is Jordan.

Proof

1.1givenF1F2algebra

Since $E_{11}^{*} = E_{11}$ and $(E_{12} + E_{21})^{*} = E_{21} + E_{12}$ , both $X$ and $Y$ belong to $A_{s a}$ . The matrix-unit rule gives $X Y = E_{12}$ .

2.1givenF1F2step 1.1algebra∎

But $(X Y)^{*} = E_{21} \neq E_{12} = X Y$ , so $X Y \notin A_{s a}$ . Hence the self-adjoint part is not an associative subalgebra, even though it is closed under the symmetrized Jordan product by [F2].

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

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].

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

A concrete octonion associator witness

Statement

In the pair convention for $O$ , the associator $[i, j, ℓ] : = (i j) ℓ - i (j ℓ)$ is $2 k ℓ \neq 0$ . Both alternative laws still hold, so this is a concrete alternative but nonassociative algebra.

Facts & Assumptions

Given: The named basis units and multiplication on $O$ .

[F1]

The pair formula gives $i j = k$ , $(0, k) = k ℓ$ , and $i (j ℓ) = (0, j i)$ with $j i = - k$ (Real octonions by Cayley–Dickson doubling, The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ ).

[F2]

The same algebra satisfies both alternative laws (The Cayley–Dickson octonions are alternative but not associative).

Proof

1.1givenF1algebra

From [F1], $(i j) ℓ = (0, k) = k ℓ$ and $i (j ℓ) = (0, - k) = - k ℓ$ . Thus $[i, j, ℓ] = k ℓ - (- k ℓ) = 2 k ℓ$ . This is nonzero because $k \neq 0$ in $H$ and the real scalar $2$ is nonzero.

2.1step 1.1F2∎

By [F2], the left and right alternative laws hold for every pair of octonions, while step 1.1 gives a nonzero associator. Hence this example is alternative and not associative.

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passOpen item page →

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.

Jordan Algebras, Spin Factors, and Alternative Examples: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The symmetrized product need not be associative

Statement

Facts & Assumptions

Proof

The self-adjoint part need not be an associative subalgebra

Statement

Facts & Assumptions

Proof

The spin factor on the real plane

Statement

Facts & Assumptions

Proof

A concrete octonion associator witness

Statement

Facts & Assumptions

Proof

An off-diagonal product in the Albert algebra

Statement

Facts & Assumptions

Proof

Sources