Remark. Exterior symbols preserve vectors, not Clifford products [fdip-008L]

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The Grassmann algebra and the earlier configuration Clifford algebra share an exterior vector-space model. For a quadratic form \(q\) on \(V^*\), the symbol map is a linear isomorphism \[\operatorname {Cl}(V^*,q)\cong \Lambda (V^*).\] This is [meinrenken2013clifford, Proposition 2.5, pp. 32-33] and the equivalence used in the transported exterior-basis coordinates. The formal declaration CliffordAlgebra.equivExterior supplies a module equivalence over any commutative ring in which \(2\) is invertible; it does not identify the products.

For example, an exterior generator satisfies \(\xi ^2=0\), while the corresponding Clifford generator satisfies \(\iota _q(\xi )^2=q(\xi )1\). For \(q(\xi )\ne 0\) these cannot be the same multiplication under a unit-preserving map. The quadratic form remains in Clifford contraction terms. The associated graded algebra forgets those lower terms, but the Clifford algebra itself does not. Therefore the Berezin functional and its determinant formulas belong to the exterior product; the vector-space comparison alone does not transfer them to a Clifford product.