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Reference: Page 140 of the book in the title.

I know this question may be difficult to answer on here, but regarding $\mathbb{R}^{n}$ as a subset of the nth Clifford algebra $Cl_{n}$ for each $u \in \Gamma_{n}$, the function $\rho_{u}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ defined by $\rho_u(v)=\alpha(u)vu^{-1}$ is an $\mathbb{R}$-linear isometry, where $\alpha: Cl_{n} \rightarrow Cl_{n}$ is the canonical automorphism and $\Gamma_{n}$ is the nth Clifford group.

I don't comprehend the second equality in his proof given. He says

For any $x \in \mathbb{R}^{n}$, $|\rho_{u}(x)|^{2}=-\rho_{u}(x)^{2}=-((-u)xu^{-1})^{2}$.

How do we get the second equality? If we write out $-\rho_{u}(x)^2=-( \alpha(u)xu^{-1})^{2}$ it appears he's claiming $\alpha(u)=-u$, but I don't think this is true for every $u \in \Gamma_{n}$.

Tuo
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  • Sounds correct, although I'm really only familiar with the author's definition which is simply defined it on the basis elements for $Cl_{n}$, in particular, $\alpha(e_{i_{1}}...e_{i_{k}})=(-1)^{k}e_{i_{1}}...e_{i_{k}}$ – Tuo Apr 10 '17 at 23:04
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    This is true if $u$ is one of the monomials $e_{i_{1}}...e_{i_{k}}$ but the monomials are only a basis, so you have to take linear combinations which this is not true for. For example $u=e_{1}e_{2}+e_{1}e_{2}e_{3}$ you get $\alpha(u)=e_{1}e_{2}-e_{1}e_{2}e_{3} \neq (-1)^{n}u$ for any $n$. – Tuo Apr 10 '17 at 23:17

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