I'm confused by discussion on page 19 of the "Spin Geometry" book by Blaine Lawson and Marie-Louise Michelson. It claims, that when field $k$ is a spin field, $\widetilde P(V,q)/P(V,q)$ is either $0$ or $\mathbb{Z}/2 \mathbb{Z}$ (see definitions below). It looks misleading: under most reasonable assumption, it seems, I can prove (see below), that it is always 0. So either the claim is misleading indeed, or there is a mistake in my understanding.
The question: What does the textbook want to say here? Is it true, that $P(V,q)=\widetilde P(V,q)$, so the second case is nonexistent?
Some clarifications, context and my thoughts about the question.
On page 16 he assumes $\dim V<\infty$ and $q$ is nondegenerate. It looks, that this assumption is made for the rest of the chapter, and in particular, for the claim above. It seems to me, that independently of whether we are making the assumption or not, the claim looks strange.
Case "the assumption is made": Given this assumption, it looks to me, that $\widetilde P(V,q) = P(V,q)$, so $\widetilde P(V,q) / P(V,q) = 0$. Thus, it's not clear, where does the case $\mathbb{Z}/2\mathbb{Z}$ come from. It seems, that the result is even independent on whether $k$ is a spin field or not. See my (probably wrong) proof below. So although statement is true, the dichotomy looks to be misleading. Is it the case?
Case "the assumption is not made": If $V$ and $q$ are arbitrary, one can come up with an example with nontrivial factors: e.g. if $V=k$, $q=0$, then $\widetilde P(V,q) \simeq \text{Inv}(k[\epsilon]/(\epsilon^2))$, where $\epsilon$ is the basis element of $V$. So in this case the statement looks to be wrong.
Definitions and facts from the book: throughout the chapter $V$ is a vector space over a field $k$, $q$ is a quadratic form on $V$. Both $P(V,q)$ and $\widetilde P(V,q)$ are defined to be subgroups of the group of invertible elements $\text{Cl}^*(V,q)$ of the Clifford algebra $\text{Cl}(V,q)$ on that vector space. $P$ is defined (page 13) to be a subgroup, generated by $\{v\in V\colon q(v)\neq 0\} \subset \text{Cl}^*(V,q)$. $\widetilde P(V,q)=\{\varphi \in \text{Cl}^*(V,q)\colon \alpha(\varphi)v\varphi^{-1}\in V\}$. Here $\alpha(v_1\dots v_r) = (-1)^r v_1\dots v_r$ is the sign automorphism of $\text{Cl}(V,q)$. He proves, that under the assumption ($\dim V < \infty$ and $q$ is nondegenerate) we have $P(V,q)\subset \widetilde P(V,q) \to O(V,q)$ and, moreover, the kernel of $\widetilde P(V,q)\to O(V,q)$ is $k^*$ and the composition $P(V,q)\to O(V,q)$ is surjective.
Below is my attempt to prove, that the case $\mathbb{Z}/2\mathbb{Z}$ is nonexistent. I'm not sure if my proof is correct.
Claim: If $\dim V< \infty$, $q$ is nondegenerate, then $P(V,q)=\widetilde P(V,q)$.
Proof: Since we already know, that $P(V,q)\subset \widetilde P(V,q)$, it is enough to prove, that this inclusion is surjective. Let $\varphi\in \widetilde P(V,q)$. Let $\varphi_1\in P(V,q)$ be an element with the same image in $O(V,q)$. Then $t=\varphi \varphi_1^{-1}\in k$. Take $v\in V:q(v)\neq 0$. Then $t=(-(q(v))^{-1}v) (tv)\in P(V,q)$ (by definition $vv=-q(v)$. Thus $\varphi = t \varphi_1 \in P(V,q)$.