Proposition. The reverse norm descends through the Lipschitz action [fcap-002D]
✍️source

Let $K$ be a field of characteristic different from two, let $V$ be a positive-dimensional finite-dimensional $K$-space, and let $Q$ be nondegenerate. Write $\Gamma (Q)$ for the vector-generated Lipschitz group, and $\tau _Q:\Gamma (Q)\to O(V,Q)$ for its twisted action. For $x\in \Gamma (Q)$, let $N_{\mathrm {rev}}(x)\in K^{\times }$ be the unique scalar unit satisfying $\operatorname {rev}(x)x=N_{\mathrm {rev}}(x)\,1$. Set $\mathcal {S}_K=K^{\times }/(K^{\times })^2$. The top row is exact, and the square commutes:

$\text{[math]}$ The determinant arrow is a separate branch: $SO(V,Q)=\ker (\det )$. It is not part of a longer exact row after $O(V,Q)$. Positive dimension supplies an anisotropic vector, so scalar units form the displayed kernel; the Lipschitz action is onto CliffordAlgebra.scalarUnits_injective CliffordAlgebra.ker_lipschitzToOrthogonal CliffordAlgebra.lipschitzToOrthogonal_surjective. The right vertical map is the orthogonal spinor norm, while $\nu _Q=\nu _Q^O|_{SO(V,Q)}$ is its restriction used in Theorem [fcap-0023]. The square expresses descent through the action, not surjectivity of $\nu _Q^O$ CliffordAlgebra.orthogonalSpinorNorm_lipschitzToOrthogonal.

If $Q(v)$ is nonzero, then \[N_{\mathrm {rev}}(\widehat v)=Q(v),\qquad \nu _Q^O(r_v)=[Q(v)]\] CliffordAlgebra.cliffordNorm_unitι CliffordAlgebra.orthogonalSpinorNorm_reflectionOrthogonal. Consequently, for an even product $x=\widehat v_1\cdots \widehat v_{2k}$ of nonisotropic vector lifts, \[\tau _Q(x)=r_{v_1}\cdots r_{v_{2k}}\in SO(V,Q),\qquad \nu _Q(r_{v_1}\cdots r_{v_{2k}}) =\left [\prod _{i=1}^{2k}Q(v_i)\right ].\] Multiplying a lift by $c\in K^\times $ changes its reverse norm by $c^2$, so the square class is independent of the chosen lift. [chevalley1954algebraic, II.3.5--6, pp. 52--53] gives these vector and paired-product values. The scalar exact sequence, reflection factorization, and the same square-scaling law appear in [meinrenken2013clifford, Theorem 3.1, pp. 50--51]. Chevalley's quotient statement assumes positive dimension; Meinrenken works with a finite-dimensional characteristic-zero nondegenerate bilinear space.

If $e:(V,Q)\simeq (V',Q')$ is an isometry equivalence, then conjugation transport preserves both norms: for $g\in O(V,Q)$ and $h\in SO(V,Q)$, \[\nu _{Q'}^O(ege^{-1})=\nu _Q^O(g),\qquad \nu _{Q'}(ehe^{-1})=\nu _Q(h).\] It therefore carries $\ker \nu _Q$ onto $\ker \nu _{Q'}$ QuadraticMap.IsometryEquiv.orthogonalSpinorNorm_orthogonalGroupCongr QuadraticMap.IsometryEquiv.spinorNorm_specialOrthogonalGroupCongr kernel-transport source. The induced kernel equivalence also commutes with Spin transport Spin-transport source. These are isometry-equivalence statements; they do not assert scalar-extension or orthogonal-product naturality. This is the norm-square companion to the vector-action square in Proposition [fcap-0029].