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?

  • $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.