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Let $C(Q)$ denote the clifford algebra of vector space $Q$ with respect to a quadratic form $q:V \rightarrow \Bbb R$. Hence we have the relation $w^2 = Q(w) \cdot 1$ for $w \in V$.

Let $\alpha:C(Q) \rightarrow C(Q)$ be the canonical automoprhism $\alpha^2=id, \alpha=-x$.

The Clifford group of $Q$ is $$\Gamma (Q) = \{ x \in C(Q)^* \, ; |, \alpha(x) \cdot v \cdot x^{-1} \in V \text{ for all } v \in V \}$$

How is this set closed under inverses?

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  • $\begingroup$ You can rewrite the condition as $\alpha(x)Vx^{-1}=V$. $\endgroup$ – Lord Shark the Unknown Feb 12 at 5:39
  • $\begingroup$ Ok it doesn't seem clear to me how the equality holds: the way is define $f_x :V \rightarrow V$, then as $\alpha(x) \cdot v \cdot x^{-1}$. This map is injective, $V$ finite dimensinoal so bijective. Is there an easier way? $\endgroup$ – CL. Feb 12 at 5:51

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