CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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The ordinary tensor product does not model an orthogonal Clifford sum

Statement refuted

For all quadratic spaces, the ordinary algebra tensor product $C l (V, Q) \otimes_{R} C l (W, R)$ , with multiplication $(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'}$ , is the Clifford algebra of the orthogonal sum with its canonical factor maps.

Facts & Assumptions

Given: The real Clifford algebras use $v^{2} = Q (v) 1$ and the ordinary tensor-product multiplication is the one displayed in the refuted statement.

[F1]

${C l}_{1, 0} ≅ R \oplus R$ and ${C l}_{2, 0} ≅ M_{2} (R)$ , with the specified positive-square-first convention (Low-dimensional real Clifford algebra models).

Counterexample

1.1givenF1choosealgebra

Choose $V = W = R$ with positive quadratic form $x \mapsto x^{2}$ . By [F1], each Clifford factor is $R \oplus R$ , hence commutative. For simple tensors, $(a \otimes b) (c \otimes d) = a c \otimes b d = c a \otimes d b = (c \otimes d) (a \otimes b)$ , and bilinearity extends commutativity to the whole ordinary tensor product.

2.1F1step 1.1algebra∎

By [F1], the orthogonal-sum Clifford algebra is $M_{2} (R)$ , which is noncommutative because $E_{11} E_{12} = E_{12}$ while $E_{12} E_{11} = 0$ . Thus it cannot be isomorphic as an algebra to the commutative ordinary tensor product in step 1.1, refuting the claim.

Remarks

The graded tensor product The graded tensor product for orthogonal sums of quadratic spaces inserts the Koszul sign when the two odd generators are multiplied in the reverse order; that sign produces the required anticommutation.

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