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

I will assume, as it is usually defined, that $C(Q)^*$ is the set of invertible elements in $C(Q)$. Then, the existence of an element $x^{-1}\in C(Q)$ such that $xx^{-1}=x^{-1}x=1$ is, by definition, guaranteed for each $x\in C(Q)^*$. What we need to show is that $x^{-1}$ is in fact an element in $\Gamma(Q)$.

First, for each $x\in \Gamma(Q)$, the function $\sigma(x):V\rightarrow V$ given by $\sigma(x)(v)=\alpha(x)vx^{-1}$ is a vector space isomorphism. It is clear that it is a linear transformation, since the Clifford product is bilinear. Moreover, $$ \sigma(x)(v)=0 \;\;\Rightarrow\;\; \alpha(x)vx^{-1}=0\;\;\Rightarrow\;\; v=0,$$ where we simply multiplied (using the Clifford product) the right side by $x^{-1}$ and the left side by $\alpha(x)^{-1}$.

Finally, since $\sigma(x)$ is an isomorphism, for each $v\in V$ we have a $w\in V$ such that $\alpha(x)wx^{-1}=v$. Here, we remember that since $\alpha$ is an automorphism, we have $\alpha(x)^{-1}=\alpha(x^{-1})$. It then follows that $$ v=\alpha(x)wx^{-1} \;\;\Rightarrow\;\; vx=\alpha(x)w \;\; \Rightarrow\;\; \alpha(x^{-1})vx=w\in V, $$ that is, $x^{-1}\in \Gamma(Q)$.


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