Clifford algebra

Remarks

Universal property: for every associative unital $F$-algebra $A$ and every linear map $f \colon V \to A$ with $f(v)^2 = Q(v) \cdot 1_A$ for all $v \in V$, there is exactly one unital algebra homomorphism $\mathrm{Cl}(V, Q) \to A$ that extends $f$. If $\dim V = n$, then $\dim \mathrm{Cl}(V, Q) = 2^n$, with basis the products $e_{i_1} \cdots e_{i_k}$, $i_1 < \cdots < i_k$, for a basis $e_1, \ldots, e_n$ of $V$. Example: $V = \R^2$ with $Q(x, y) = -(x^2 + y^2)$. The relation $(e_1 + e_2)^2 = Q(e_1 + e_2) = -2$ with $e_1^2 = e_2^2 = -1$ gives $e_1 e_2 = -e_2 e_1$, and then $(e_1 e_2)^2 = -1$; so $\mathrm{Cl}(V, Q)$ is the quaternion algebra with $i = e_1$, $j = e_2$, $k = e_1 e_2$. For $Q = 0$, the Clifford algebra is the exterior algebra of $V$. Non-example: the tensor algebra $T(V)$ for $V \ne 0$ is not a Clifford algebra of $(V, Q)$, because $v \otimes v$ is not a scalar there. Some authors use the opposite sign convention $v^2 = -Q(v)$.