Proposition. The Clifford algebra of the configuration form [fdip-000P]
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  • October 17, 2026

Let \(k\) be a commutative ring, \(M\) a \(k\)-module and \(q:M\to k\) a quadratic form. Use the tensor algebra and the square-quotient convention for the Clifford algebra: \[\operatorname {Cl}(q)=T_k(M)/I_q,\] where \(I_q\) is the two-sided ideal generated by \(v\otimes v-q(v)1\) for all \(v\in M\). The canonical linear map \(\iota _q:M\to \operatorname {Cl}(q)\) satisfies \(\iota _q(v)^2=q(v)1\). Scalars in this formula mean their images under the algebra's structure map. No invertibility of \(2\) is required for this construction.

For every unital associative \(k\)-algebra \(A\), a linear map \(f:M\to A\) extends to a unique unital \(k\)-algebra homomorphism \(\widehat f:\operatorname {Cl}(q)\to A\) if and only if \[f(v)^2=q(v)1_A\qquad (v\in M).\] This is the Clifford lift and its universal property.

Proof. 

  • October 17, 2026

Necessity follows by applying an algebra homomorphism to the square relation. For sufficiency, send a pure tensor to its ordered product, \[v_1\otimes \cdots \otimes v_r\longmapsto f(v_1)\cdots f(v_r),\] and send \(c\in k=T_k^0(M)\) to \(c1_A\). The product is multilinear because \(f\) is linear, multiplication in \(A\) is distributive, and the scalar images are central. The tensor-product universal property therefore gives a linear map on each tensor power. Their direct sum defines a linear map \(F:T_k(M)\to A\). Concatenation of pure tensors becomes multiplication of their ordered products. Distributivity then proves multiplicativity on all finite sums, and the degree-zero formula preserves the unit and scalars. Every generator of \(I_q\) maps to \(f(v)^2-q(v)1_A=0\); multiplicativity makes every element of the two-sided ideal map to zero. Thus \(F\) descends to \(\widehat f\). Every tensor is a finite linear combination of pure tensors, whose images in the quotient are products of the \(\iota _q(v)\). An algebra map with the prescribed generator values must have the displayed value on each such product and on every scalar. This proves uniqueness.

Linearity and the square relation give the anticommutator: \[\begin {gathered} \iota _q(u)\iota _q(v)+\iota _q(v)\iota _q(u)\\ =\bigl (q(u+v)-q(u)-q(v)\bigr )1\\ =B_q(u,v)1. \end {gathered}\] Indeed, expand \((\iota _q(u)+\iota _q(v))^2\) and subtract the two squares. This uses no division by \(2\). When \(2\) is invertible, \(b_q=B_q/2\) exists and the relation becomes \(\iota _q(u)\iota _q(v)+\iota _q(v)\iota _q(u)=2b_q(u,v)1\). Conversely that relation with \(u=v\) recovers the square relation by cancellation of \(2\). Thus the two presentations agree in this case. Without that hypothesis, the anticommutator alone need not recover the squares.

Meinrenken gives the bilinear presentation in [meinrenken2013clifford, Definition 2.1, p. 27] and the universal property in [meinrenken2013clifford, Proposition 2.2 and Remark 2.2, p. 29]. His chapter assumes a finite-dimensional vector space over a characteristic-zero field (p. 23); Remark 2.2 includes the ordinary algebra formulation. The direct argument above supplies the stated commutative-ring generality. Mathlib's CliffordAlgebra.ι_sq_scalar, CliffordAlgebra.lift, CliffordAlgebra.lift_unique and CliffordAlgebra.ι_mul_ι_add_swap provide these respective generic square, extension, uniqueness and unhalved-polar statements.

Now take the real configuration form and put \(c_i=\iota _q(\delta _i)\). Write \(\{x,y\}=xy+yx\) for the anticommutator. The matrix entries give \[\begin {aligned} c_i^2&=s1\quad (i=0,1,2,3),\\ \{c_0,c_1\}&=\{c_2,c_3\}=-2\,1,\\ \{c_1,c_2\}&=\{c_3,c_0\}=-2t\,1,\\ \{c_0,c_2\}&=\{c_1,c_3\}=0. \end {aligned}\] These follow from \(q(\delta _i)=Q_{ii}=s\) and \(b_q(\delta _i,\delta _j)=Q_{ij}\). In particular the squares are positive in this convention. The opposite-sign convention would use \(-q\).

For this real form, both the scalar map and \(\iota _q\) are injective. Here is the exterior algebra argument of [meinrenken2013clifford, Proposition 2.1, p. 28], with its operator identities made explicit. On \(\bigwedge V\), let \(\epsilon (v)\omega =v\wedge \omega \), and define contraction by \[d_v(1)=0,\] \[\begin {gathered} d_v(w_1\wedge \cdots \wedge w_r)\\ =\sum _{j=1}^r(-1)^{j-1}b_q(v,w_j)\\ \quad \cdot w_1\wedge \cdots \widehat {w_j}\cdots \wedge w_r. \end {gathered}\] The alternating multilinear expression defines a linear operator on the exterior algebra. Exterior alternation gives \(\epsilon (v)^2=0\). In \(d_v^2\), each pair of removed positions occurs twice with opposite signs and the same scalar coefficient, so \(d_v^2=0\). Separating the first term of \(d_v(v\wedge \omega )\) gives \[d_v\epsilon (v)+\epsilon (v)d_v=b_q(v,v)I.\] Consequently \(C(v)=\epsilon (v)+d_v\) is linear in \(v\) and \(C(v)^2=q(v)I\). The universal property gives a unital algebra map \(\rho :\operatorname {Cl}(q)\to \operatorname {End}_{\mathbb R}(\bigwedge V)\). It satisfies \(\rho (\iota _q(v))1=v\) and \(\rho (r1)1=r1\). Since \(\bigwedge ^1 V=V\) and \(\bigwedge ^0V=\mathbb R\), these equations prove both injections. The contraction identities are also recorded in [meinrenken2013clifford, Section 2.1.3, pp. 25-26]. TauCeti's CliffordAlgebra.ι_injective and CliffordAlgebra.algebraMap_injective give the corresponding injections for an arbitrary quadratic form over a commutative ring with invertible \(2\).

The exterior action just used proves faithfulness on vectors and scalars; it does not assert that \(\rho \) is a faithful algebra representation. Nor does this construction select a physical spinor module, a positive expectation on the Clifford algebra, or a differential operator. The earlier Gaussian expectation acts on polynomial observables of vertex fields; an expectation on this new algebra would require a specified construction.