Clifford algebra

Let $V$ be a vector space over a field $k$ , and $Q:V\\times V\\to k$ asymmetric bilinear form. Then the Clifford algebra $\\operatorname{Cliff}(Q,V)$ isthe quotient of the tensor algebra $\\mathcal{T}(V)$ by the relations

v\\otimes w+w\\otimes v=-2Q(v,w)\\qquad\\forall v,w\\in V.

Since the above relationship is not homogeneous in the usual $\\mathbb{Z}$ -grading on $\\mathcal{T}(V)$ , $\\operatorname{Cliff}(Q,V)$ does not inherit a $\\mathbb{Z}$ -grading. However, by reducing mod 2, we also have a $\\mathbb{Z}_{2}$ -grading on $\\mathcal{T}(V)$ , and the relations above are homogeneouswith respect to this, so $\\operatorname{Cliff}(Q,V)$ has a natural $\\mathbb{Z}_{2}$ -grading,which makes it into a superalgebra.

In addition, we do have a filtration on $\\operatorname{Cliff}(Q,V)$ (making it afiltered algebra), and the associated graded algebra $\\operatorname{Gr}\\operatorname{Cliff}(Q,V)$ is simply $\\Lambda^{*}V$ , the exterior algebra of $V$ . Inparticular,

\\dim\\operatorname{Cliff}(Q,V)=\\dim\\Lambda^{*}V=2^{\\dim V}.

The most commonly used Clifford algebra is the case $V=\\mathbb{R}^{n}$ , and $Q$ is the standard inner product with orthonormal basis $e_{1},\\ldots,e_{n}$ .In this case, the algebra is generated by $e_{1},\\ldots,e_{n}$ and theidentity of the algebra $1$ , with the relations

	$\\displaystyle e_{i}^{2}$	$\\displaystyle=-1$
	$\\displaystyle e_{i}e_{j}$	$\\displaystyle=-e_{j}e_{i}\\quad(i\eq j)$

Trivially, $\\operatorname{Cliff}(\\mathbb{R}^{0})=\\mathbb{R}$ , and it can be seen from the relationsabove that $\\operatorname{Cliff}(\\mathbb{R})\\cong\\mathbb{C}$ , the complex numbers, and $\\operatorname{Cliff}(\\mathbb{R}^{2})\\cong\\mathbb{H}$ , the quaternions.

On the other hand, for $V=\\mathbb{C}^{n}$ we get the particularly answer of

\\operatorname{Cliff}(\\mathbb{C}^{2k})\\cong\\mathrm{M}_{2^{k}}(\\mathbb{C})\\qquad%\\operatorname{Cliff}(\\mathbb{C}^{2k+1})=\\mathrm{M}_{2^{k}}(\\mathbb{C})\\oplus%\\mathbf{M}_{2^{k}}(\\mathbb{C}).