Coordinate representations [fcap-0002]
✍️sourceAGENTDRAFTED
Coordinate representations [fcap-0002]
✍️sourceAGENTDRAFTED
A basis of the generating module gives coordinates on the exterior algebra, and the exterior equivalence transports those coordinates to the Clifford algebra. The empty and singleton coordinates fix the scalar and generator conventions. Orthogonality then controls how certain ordered products pass between the two algebras.
Definition 1. Transported exterior-basis coordinates [wieser2024formalizing, 9.3.5] [fcap-0003]AGENTDRAFTED
Definition 1. Transported exterior-basis coordinates [wieser2024formalizing, 9.3.5] [fcap-0003]AGENTDRAFTED
Let \(R\) be a commutative ring in which \(2\) is invertible, let \(M\) be an \(R\)-module, and choose a basis \(b\) indexed by a linearly ordered type. Write \(E_b(s)\) for the exterior-basis vector indexed by a finite subset \(s\). For a quadratic form \(Q\), the Mathlib linear equivalence CliffordAlgebra.equivExterior from \(\mathcal {C}\kern -2pt\ell (Q)\) to \(\bigwedge _R M\) gives a coordinate basis
\[B_{Q,b}(s) = \operatorname {equivExterior}(Q)^{-1}(E_b(s)).\]
This basis is transported through a linear equivalence. It does not say that the equivalence preserves multiplication or that Clifford multiplication becomes exterior multiplication. The basis-free construction is developed in [wieser2022formalizing].
The exterior equivalence is treated in [wieser2024formalizing, 9.3.5].
Two calculations anchor the choice. The empty subset maps to the scalar unit, \[B_{Q,b}(\varnothing ) = 1,\] and a singleton maps to the corresponding Clifford generator, \[B_{Q,b}(\{i\}) = \iota _Q(b(i)).\] These are module-level identities: they determine the first two coordinate types but do not determine multiplication of arbitrary coordinates.
For subsets with several elements, a coordinate formula additionally needs an ordering convention and its attendant signs. No such convention is implicit in the transported basis.
Lemma 2. Orthogonal pair transport [wieser2024formalizing, 8.4] [fcap-0004]AGENTDRAFTED
Lemma 2. Orthogonal pair transport [wieser2024formalizing, 8.4] [fcap-0004]AGENTDRAFTED
Let \(R\) be a commutative ring in which \(2\) is invertible, let \(M\) be an \(R\)-module, and let \(Q\) be a quadratic form on \(M\). Write \(T_Q\) for the linear equivalence from \(\mathcal {C}\kern -2pt\ell (Q)\) to the exterior algebra supplied by CliffordAlgebra.equivExterior. It is a module equivalence, not an algebra equivalence.
For vectors \(m,n \in M\), the change-of-form calculation for \(T_Q(\iota _Q(m)\iota _Q(n))\) has one extra scalar contribution, evaluated by the bilinear form associated to \(-Q\). If \(m\) and \(n\) are \(Q\)-orthogonal, that contribution vanishes. The resulting two-generator calculation is \[T_Q\bigl (\iota _Q(m)\iota _Q(n)\bigr )=\iota _0(m)\iota _0(n).\] This is the first point at which an orthogonality hypothesis changes the transport calculation. It does not turn \(T_Q\) into a multiplicative map.
The change-of-form construction is the relevant mathematical background in [wieser2024formalizing, 8.4], and the Lean construction of \(\operatorname {equivExterior}\) is described in [wieser2024formalizing, 9.3.5]. The concrete geometric-product discussion in [wieser2024formalizing, 2.1.2] motivates the role of orthogonality, but does not provide this note's formal interface.
Thus orthogonality removes precisely the scalar correction in the two-generator change-of-form identity. No basis or reordering convention is needed.