Chirality

From now on, $n = 2m$ for $n$ even, $n = 2m + 1$ for $n$ odd. We take $\bCl(V) \isom \Cl(V,g) \ox_\bR \bC$ with $g$ always positive definite.

Suppose $\{e_1, \dots, e_n\}$ is an oriented orthonormal basis for $(V, g)$. If $e_k' = \sum_{j=1}^n h_{jk} e_j$ with $H^t H = 1_n$, then $e_1'\dots e_n' = (\det H) e_1\dots e_n$, and $\det H = \pm 1$. We restrict to the oriented case $\det H = +1$, so the expression $e_1 e_2\dots e_n$ is independent of $\{e_1,e_2,\dots,e_n\}$. Thus

$$\ga := (-i)^m e_1 e_2\dots e_n$$

is well-defined in $\bCl(V)$. Now

$$\ga^* = i^m e_n\dots e_2 e_1 = (-i)^m(-1)^m(-1)^{n(n-1)/2}e_1 e_2\dots e_n = (-1)^m(-1)^{n(n-1)/2}\ga,$$

and

$$\frac{n(n-1)}{2} = \begin{cases} m(2m - 1), & \text{$n$ even... ...+ 1)m, & \text{$n$ odd} \end{cases} \Biggr\} \equiv m \bmod 2,$$

so $\ga^* = \ga$. But also $\ga^*\ga = (e_n\dots e_2 e_1)(e_1 e_2\dots e_n) = (+1)^n = 1$, so $\ga$ is ``unitary''. Hence $\ga^2 = 1$, so $\frac{1+\ga}{2}$, $\frac{1-\ga}{2}$ are ``orthogonal projectors'' in $\bCl(V)$.

Since $\ga e_j = (-1)^{n-1} e_j\ga$, we get that if $n$ is odd, then $\ga$ is central in $\bCl(V)$; and for $n$ even, $\ga$ anticommutes with $V$, but is central in the even subalgebra $\Cl^0(V)$. Moreover, when $n$ is even and $v\in V$, then $\ga v \ga = -v$, so that $\ga(\cdot)\ga = \chi \in \Aut(\bCl(V))$.