Twisted Clifford action, reflections, and scalar kernel

For anisotropic $u$, $u^{-1}=u\mathrm{/}Q(u)$ by [F1]. Hence ${\rho }_u(v)=-uvu\mathrm{/}Q(u)$. The polarized relation gives $uv+vu=2B(u,v)1$, so $-uvu\mathrm{/}Q(u)=v-2B(v,u)u\mathrm{/}Q(u)=r_u(v)$.

Let $x\in \mathrm{\Gamma }$ and $y_v={\rho }_x(v)\in V$. Since $\alpha (y_v)=-y_v$, applying $\alpha$ to $y_v=\alpha (x)v{x}^{-1}$ gives $xv\alpha (x{)}^{-1}=\alpha (x)v{x}^{-1}$, that is, ${\rho }_{\alpha (x)}(v)={\rho }_x(v)$. Using this equality in one factor and [F1] gives $2B(y_v,y_w)1=y_v{y}_w+y_w{y}_v=\alpha (x)(vw+wv)\alpha (x{)}^{-1}=2B(v,w)1$. Thus ${\rho }_x$ preserves $B$; it is injective because multiplication by invertible elements is injective, and finite dimensionality makes its image all of $V$.

If $n=0$, [F2] gives $\mathrm{C}\mathrm{l}(V,Q)=\mathbb{R}$, so every kernel element is scalar. If $n>0$ and ${\rho }_x$ is the identity, then $\alpha (x)e_i=e_{i}x$ for every vector of an orthogonal anisotropic basis $e_1,\dots ,e_n$. Write $x=x_0+x_1$ with even and odd parts. The grading gives $x_0{e}_i=e_i{x}_0$ and $-x_1{e}_i=e_i{x}_1$. For a square-free monomial $e_I$ of length $r$, moving $e_i$ past it gives commutation sign $(-1{)}^r$ when $i\mathrm{\notin }I$ and $(-1{)}^{r-1}$ when $i\in I$. For each nonempty even $I$, choose $i\in I$: the required commutation fails; for each odd $I$, choose $i\in I$: the required anticommutation fails. Distinct $I$ containing a fixed $i$ give distinct deleted basis monomials, so no cancellation can remove these coefficients. By [F2], only the empty even monomial remains and $x$ is scalar. Conversely every nonzero scalar acts trivially.

To prove homogeneity, let $A={\rho }_x\in O(V,B)$. By [F4], write $A=r_{u_1}\cdots r_{u_k}$ and put $y=u_1\cdots u_k$. Step 1.1 and homogeneity of each vector give ${\rho }_y=A$. Equality ${\rho }_x={\rho }_y$ implies, after multiplying by $\alpha (y{)}^{-1}$ on the left and $x$ on the right, that $\alpha (y^{-1}x)v=v(y^{-1}x)$ for every $v$. The PBW kernel argument of step 1.3 applies to the even and odd parts of $y^{-1}x$ and makes it a nonzero scalar. Therefore $x$ is a scalar multiple of the homogeneous product $y$.

Products of homogeneous normalizers act by composition: ${\rho }_{xy}={\rho }_x\circ {\rho }_y$. Step 2.1 shows every unrestricted normalizer is homogeneous, and step 1.2 shows its action maps $V$ onto itself. These facts make the set closed under products and inverses, so $\mathrm{\Gamma }$ is a group. Step 1.3 gives its scalar kernel, and step 2.1 proves equality with the homogeneous normalizer.