Normalized Pin products and Clifford norm

For a vector product, [F1] makes $N(x)$ a real scalar. By [F3], every $x\in \mathrm{\Gamma }$ is $\lambda u_1\cdots u_k$ for $\lambda \in {\mathbb{R}}^{\times }$ and anisotropic $u_j$; $N(\lambda )={\lambda }^2$, so $N(x)$ is scalar. For scalar-valued norms, $N(xy)=xy\bar{y}\bar{x}=xN(y)\bar{x}=N(y)N(x)$, which proves multiplicativity.

The empty product has norm $N(1)=1$. Every normalized vector belongs to $\mathrm{\Gamma }$ by [F3], has inverse $u\mathrm{/}Q(u)=\pm u$, and has norm $N(u)=-Q(u)\in \{1,-1\}$. Thus subgroup words are normalized-vector products, and multiplicativity in step 1.1 shows that they have norm $\pm 1$. In dimension zero Pin is $\{1,-1\}$, and both scalars belong to $\mathrm{\Gamma }$ and have norm $1$. Hence Pin is contained in the asserted norm locus in every dimension.

Conversely, take $x\in \mathrm{\Gamma }$ with $N(x)=\pm 1$. By [F4], factor ${\rho }_x$ into reflections and lift them by normalized vectors to a product $y$ with ${\rho }_y={\rho }_x$. By [F3], $x{y}^{-1}=\lambda$ is a nonzero scalar, and step 1.1 gives $N(x)={\lambda }^{2}N(y)$. Since both norms are $\pm 1$ and ${\lambda }^2>0$, ${\lambda }^2=1$, so $\lambda =\pm 1$. If $\dim V>0$, $-1$ is itself a product of normalized vectors: for $Q(u)=1$, $u(-u)=-1$, and for $Q(u)=-1$, $u^2=-1$. If $V=0$, it is included by definition. Thus every norm-$\pm 1$ element lies in Pin.

The grading parity of a vector product is its number of factors modulo two, and nonzero scalars are even. In positive dimension, the even elements of Pin are therefore exactly the even normalized products, which are ${\mathrm{Spin}}_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}$ by [F2]. In dimension zero every Clifford element is scalar by [F3], so both signs are even and have norm $1$, giving all three groups $\{1,-1\}$. In every dimension the definition of ${\mathrm{Spin}}_{N=1}$ selects the norm-one subgroup; multiplicativity in step 1.1 ensures that it is a subgroup.