Lemma. Determinant parity detects the even lift [lawson2016spin, I.2, Theorems 2.7 and 2.9, pp. 17--19] [fcap-001A]

Let \(R\) be a commutative ring in which \(2\) is invertible, let \(M\) be a finite free \(R\)-module, and let \(Q\) be a quadratic form on \(M\). A reflection in a vector of invertible quadratic value has determinant \(-1\). Therefore a product of \(r\) such reflections has determinant \[(-1)^r.\] In particular, every product of an even number of reflections belongs to \(SO(M,Q)\).

Lawson--Michelsohn state this parity argument for quadratic vector spaces over a field. The finite-free commutative-ring statement above is the formalized extension: its determinant calculation uses freeness and finiteness, while invertibility of \(2\) identifies the fixed part of the grading involution with the even subalgebra.

The Clifford lift remembers this parity. If \(p\in \operatorname {Pin}(M,Q)\) and its orthogonal action has determinant \(1\), then \(p\) lies in the even Clifford subalgebra. Hence \(p\) is a Spin element. Thus any Pin lift of a special orthogonal transformation is automatically even.