Clifford algebras

We start with $(V, g)$, where $V \isom \bR^n$ and $g$ is a nondegenerate symmetric bilinear form. If $q(v) = g(v,v)$, then $2g(u, v) = q(u + v) - q(u) - q(v)$. Thus $g$ is determined by the corresponding ``quadratic form'' $q$.

The existence of this algebra can be seen in two ways. First of all, let $\sT(V)$ be the tensor algebra on $V$, that is, $\sT(V) := \bigoplus_{k=0}^\infty V^{\ox n}$. Then

$$\Cl(V, g) := \sT(V)/\operatorname{Ideal}\langle u \ox v + v \ox u - 2g(u,v) 1 : u,v \in V\rangle.$$ (1)

Since the relations are not homogeneous, the $\bZ$-grading of $\sT(V)$ is lost, only a $\bZ_2$-grading remains:

$$\Cl(V, g) = \Cl^0(V, g) \oplus \Cl^1(V, g).$$

The second option is to define $\Cl(V, g)$ as a subalgebra of $\End_\bR(\La^\8 V)$ generated by all expressions $c(v) = \eps(v) + \iota(v)$ for $v\in V$, where
$$\begin{align*} \eps(v)\: u_1 \wyw u_k & \mapsto v \w u_1 \wyw u_k \\ \iota(v)\:... ...o \sum_{j=1}^k (-1)^{j-1} g(v,u_j) u_1 \wyw \widehat{u_j} \wyw u_k. \end{align*}$$
Note that $\eps(v)^2 = 0$, $\iota(v)^2 = 0$, and $\eps(v)\iota(u) + \iota(u)\eps(v) = g(v,u) 1$. Thus
$$\begin{align*} c(v)^2 &= g(v, v) 1 \words{for all} v \in V, \\ c(u)c(v) + c(v)c(u) &= 2g(u, v) 1 \words{for all} u,v \in V. \end{align*}$$
Thus these operators on $\La^\8 V$ do provide a representation of the algebra ().

Dimension count: suppose $\{e_1, \dots, e_n\}$ is an orthonormal basis for $(V, g)$, i.e., $g(e_k, e_k) = \pm 1$ and $g(e_j, e_k) = 0$ for $j \neq k$. Then the $c(e_j)$ anticommute and thus a basis for $\Cl(V, g)$ is $\{c(e_{k_1})\dots c(e_{k_r}) : 1 \leq k_1 <\cdots< k_r \leq n\}$, labelled by $K = \{k_1,\dots,k_r\} \subseteq \{1,\dots,n\}$. Indeed,

$$c(e_{k_1})\dots c(e_{k_r})\: 1 \mapsto e_{k_1} \wyw e_{k_r} \equiv e_K \in \La^\8 V$$

and these are linearly independent. Thus the dimension of the subalgebra of $\End_\bR(\La^\8 V)$ generated by all $c(v)$ is just $\dim\La^\8V = 2^n$. Now, a moment's thought shows that in the abstract presentation (), the algebra $\Cl(V, g)$ is generated as a vector space by the $2^n$ products $e_{k_1} e_{k_2}\dots e_{k_r}$, and these are linearly independent since the operators $c(e_{k_1}) \dots c(e_{k_r})$ are linearly independent in $\End_\bR(\La^\8 V)$. Therefore, this representation of $\Cl(V, g)$ is faithful, and $\dim\Cl(V, g) = 2^n$.

The so-called ``symbol map'':

$$\sg : a \mapsto a(1) : \Cl(V, g) \to \La^\8 V$$

is inverted by a ``quantization map'':

$$Q : u_1 \w u_2 \wyw u_r \longmapsto \frac{1}{r!} \sum_{\tau\... ...1)^\tau c(u_{\tau(1)}) c(u_{\tau(2)}) \dots c(u_{\tau(r)}).$$ (2)

To see that it is an inverse to $\sg$, one only needs to check it on the products of elements of an orthonormal basis of $(V, g)$.

From now, we write $uv$ instead of $c(u)c(v)$, etc., in $\Cl(V, g)$.