# Why is clifford group a group?

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?

• You can rewrite the condition as $\alpha(x)Vx^{-1}=V$. – Lord Shark the Unknown Feb 12 at 5:39
• 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? – 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, $$x^{-1}\in \Gamma(Q)$$.