CliffordAlgebra2 2002-12-21 18:19:28 2005-10-19 10:04:35 Definition % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} \newcommand{\Cliff}{\operatorname{Cliff}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Gr}{\operatorname{Gr}} Let $V$ be a vector space over a field $k$, and $Q:V\times V\to k$ a symmetric bilinear form. Then the Clifford algebra $\Cliff(Q,V)$ is the quotient of the tensor algebra $\mc{T}(V)$ 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 $\Z$-grading on $\mc{T}(V)$, $\Cliff(Q,V)$ does not inherit a $\Z$-grading. However, by reducing mod 2, we also have a $\Z_2$-grading on $\mc{T}(V)$, and the relations above are homogeneous with respect to this, so $\Cliff(Q,V)$ has a natural $\Z_2$-grading, which makes it into a superalgebra. In addition, we do have a filtration on $\Cliff(Q,V)$ (making it a filtered algebra), and the associated graded algebra $\Gr\Cliff(Q,V)$ is simply $\Lambda^*V$, the exterior algebra of $V$. In particular, $$\dim\Cliff(Q,V)=\dim\Lambda^*V=2^{\dim V}.$$ The most commonly used Clifford algebra is the case $V=\R^n$, 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 the identity of the algebra $1$, with the relations \begin{align*} e_i^2&=-1\\ e_ie_j&=-e_je_i \quad (i\neq j) \end{align*} Trivially, $\Cliff(\R^0)=\R$, and it can be seen from the relations above that $\Cliff(\R)\cong\C$, the complex numbers, and $\Cliff(\R^2)\cong\mathbb{H}$, the quaternions. On the other ha nd, for $V=\C^n$ we get the particularly \PMlinkescapetext{simple} answer of $$\Cliff(\C^{2k}) \cong \mathrm{M}_{2^k}(\C) \qquad \Cliff(\C^{2k+1}) = \mathrm{M}_{2^k}(\C) \oplus \mathbf{M}_{2^k}(\C).$$