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Clifford Classification, Modules, and Graded Tensor Products: Examples and Counterexamples

1 · Prerequisites

2 · Summary

These examples check the positive-square-first table at low signatures, show how the even subalgebra determines graded-module types, and exhibit the failure of the ordinary tensor product for orthogonal Clifford sums. Each calculation uses the definitions and proved results on real-complex-clifford-classification-and-graded-products.

Worked checks

Low-signature real Clifford algebras as C, H, and matrix algebras compares the first real matrix and division-algebra cases with the classification table.
Even-subalgebra types determine small graded Clifford modules checks split and nonsplit even-algebra rows and the resulting graded-simple counts.
The ordinary tensor product does not model an orthogonal Clifford sum shows that commuting factor generators cannot model the anticommuting generators from orthogonal summands.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

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.

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

Low-signature real Clifford algebras as C, H, and matrix algebras

Example

With $v^{2} = Q (v) 1$ and signatures ordered $(+^{p}, -^{q})$ , the five signatures $(1, 0)$ , $(0, 1)$ , $(2, 0)$ , $(1, 1)$ , and $(0, 2)$ give respectively

R \oplus R, C, M_{2} (R), M_{2} (R), H .

This checks the sign and indexing conventions at the smallest dimensions.

Facts & Assumptions

Given: The five standard real forms listed above.

[F1]

The real classification table depends on $p - q m o d 8$ and the total dimension (Real and complex Clifford algebra classification with explicit signature).

[F2]

The low-dimensional algebra identifications are ${C l}_{1, 0} ≅ R \oplus R$ , ${C l}_{0, 1} ≅ C$ , ${C l}_{2, 0} ≅ M_{2} (R)$ , ${C l}_{1, 1} ≅ M_{2} (R)$ , and ${C l}_{0, 2} ≅ H$ (Low-dimensional real Clifford algebra models).

[F3]

A linear map satisfying every quadratic square relation extends uniquely to a unital Clifford-algebra homomorphism (Universal property of the Clifford algebra).

[F4]

Ordered square-free monomials form a Clifford basis, of dimension $2^{n}$ in dimension $n$ (PBW basis and exterior associated graded for a Clifford algebra).

[F5]

The coefficient algebras have real bases $1, i$ for $C$ and $1, i, j, k$ for $H$ , with $i^{2} = j^{2} = - 1$ and $i j = - j i = k$ in $H$ (Real, complex, and quaternion coefficient algebras and matrix notation).

Verification

1.1givenF1algebra

For $(p, q) = (1, 0)$ , $n = 1$ and $r = 1$ , so [F1] gives $R \oplus R$ . For $(0, 1)$ , $r = 7$ , giving $C$ . For $(2, 0)$ , $n = 2, r = 2$ , giving $M_{2} (R)$ ; for $(1, 1)$ , $r = 0$ , also $M_{2} (R)$ ; and for $(0, 2)$ , $r = 6$ , giving $H$ .

1.2givenconstructalgebra

Set $A = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array})$ , $B_{+} = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ , and $B_{-} = (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ . Direct multiplication gives $A^{2} = B_{+}^{2} = I$ , $B_{-}^{2} = - I$ , $A B_{+} = - B_{+} A = B_{-}$ , and $A B_{-} = - B_{-} A = B_{+}$ . Thus $(x A + y B_{+})^{2} = (x^{2} + y^{2}) I$ and $(x A + y B_{-})^{2} = (x^{2} - y^{2}) I$ . The four matrices $I, A, B_{+}, B_{-}$ are linearly independent: a zero linear combination has diagonal entries $a + b = a - b = 0$ and off-diagonal entries $c + d = c - d = 0$ , forcing all coefficients to vanish.

2.1F2F3F4F5step 1.1step 1.2constructalgebra∎

For the five signatures in order, send the standard generators to $(1, - 1)$ , to $i \in C$ , to $A, B_{+}$ , to $A, B_{-}$ , and to $i, j \in H$ . The one-generator images have the required squares; step 1.2 checks every linear combination in both matrix cases; and [F5] gives $(x i + y j)^{2} = - (x^{2} + y^{2})$ in the quaternion case. Hence [F3] extends each linear map to a unital algebra homomorphism. The images of the ordered monomials are respectively the real bases $(1, 1), (1, - 1)$ ; $1, i$ ; $I, A, B_{+}, B_{-}$ ; $I, A, B_{-}, B_{+}$ ; and $1, i, j, k$ . They span each target and are linearly independent by step 1.2 and [F5], and the source dimensions are $2, 2, 4, 4, 4$ by [F4]. Each homomorphism is therefore an isomorphism, explicitly realizing [F2] and agreeing with step 1.1.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Even-subalgebra types determine small graded Clifford modules

Example

For signatures $(1, 0)$ , $(0, 1)$ , $(2, 0)$ , $(1, 1)$ , and $(0, 2)$ , the even subalgebras are respectively $R$ , $R$ , $C$ , $R \oplus R$ , and $C$ . Hence their graded-simple module counts and coefficient types are respectively one real, one real, one complex, two real, and one complex.

Facts & Assumptions

Given: The positive-square-first convention and the five real Clifford algebras in the Example.

[F1]

The even subalgebra of ${C l}_{p, q}$ is ${C l}_{q, p - 1}$ for $p > 0$ and ${C l}_{0, q - 1}$ for $p = 0, q > 0$ (Even subalgebras of real and complex Clifford algebras).

[F2]

Graded-simple Clifford modules correspond to simple modules over the even subalgebra; matrix algebra factors determine their coefficient types (Irreducible graded Clifford modules and real, complex, and quaternionic types).

[F3]

The Clifford algebra of the zero vector space over $R$ is $R$ (The zero-dimensional Clifford algebra is the base field).

[F4]

${C l}_{0, 1} ≅ C$ and ${C l}_{1, 0} ≅ R \oplus R$ (Low-dimensional real Clifford algebra models).

Verification

1.1givenF1F3F4algebra

For ${C l}_{1, 0}$ and ${C l}_{2, 0}$ , [F1] gives even algebras ${C l}_{0, 0}$ and ${C l}_{0, 1}$ ; [F3]–[F4] identify these with $R$ and $C$ . For ${C l}_{0, 1}$ and ${C l}_{0, 2}$ it gives ${C l}_{0, 0} = R$ and ${C l}_{0, 1} = C$ . For ${C l}_{1, 1}$ it gives ${C l}_{1, 0} = R \oplus R$ by [F4].

2.1F2step 1.1algebra∎

The fields $R$ and $C$ each have one simple module of their own coefficient type, while $R \oplus R$ has two simple modules, one supported on each factor. Applying [F2] therefore gives the five counts and types in the Example. In every positive-dimensional case an invertible odd generator identifies the two parity components, so each resulting graded module has equal dimensions in degrees $0$ and $1$ .