Geometric algebra

Remarks

Polarizing $v v = v \cdot v$ gives $uv + vu = 2\, u \cdot v$, which is why $u \cdot v = \tfrac{1}{2}(uv + vu)$. As a graded vector space, $\mathcal{G}(V) = \bigoplus_{k=0}^{n} \mathcal{G}^k(V)$ is isomorphic to the exterior algebra $\Lambda V$, of dimension $2^n$ for $n = \dim V$. Example: $V = \R^2$ with the dot product and orthonormal basis $e_1, e_2$. Then $e_1 e_2 = -e_2 e_1$, and $I = e_1 e_2$ satisfies $I^2 = e_1 e_2 e_1 e_2 = -e_1 e_1 e_2 e_2 = -1$, so $\set{a + bI}$ is a copy of $\C$. With $R = \cos(\theta/2) - I \sin(\theta/2)$ and $\tilde R = \cos(\theta/2) + I \sin(\theta/2)$, one checks $R e_1 \tilde R = e_1 (\cos \theta + I \sin \theta) = \cos\theta \, e_1 + \sin\theta \, e_2$, a rotation by $\theta$. Non-example: the exterior algebra $\Lambda V$ with the wedge product alone is not a geometric algebra in this sense, because there $v \wedge v = 0$ for every $v$, while $v v = v \cdot v \ne 0$ for a non-null vector. The notation $\mathcal{G}(p, q)$ or $\mathrm{Cl}_{p,q}(\R)$ is used when the form has $p$ positive and $q$ negative squares.