How do I evaluate the Clifford product in dimensions greater than 3?

Question

The Clifford product of a pair of vectors $a,b$ is an associative operation defined by

$$ ab = a \cdot b + a \wedge b.$$

In sufficiently low dimensions I am used to being able to define the Clifford product on arbitrary $k$-vectors by repeatedly applying the vector definition. For instance, suppose that I build a Clifford algebra over $\mathbb{R}^3$ with the usual (positive) Euclidean inner product. Then I can easily write out the Clifford product of any pair of basis bivectors. For instance,

$$ e_{12}e_{13} = e_1 e_2 e_1 e_3 = -e_2(e_1 e_1)e_3 = -e_2 (1) e_3 = -e_{23}.$$

In four dimensions I get stuck, because it's possible that two of the indices don't "cancel," and then I don't know how to apply the product:

$$e_{12}e_{34} = e_1 e_2 e_3 e_4.$$

Where do I go from here? I strongly suspect that this equals just $e_{1234}$, but I don't know how to show (in an explicit, pedantic, algebraic way) that

$$e_1 e_2 e_3 e_4 = e_1 \wedge e_2 \wedge e_3 \wedge e_4.$$

Thanks!

Answer 1

It's not easy to give a proof without knowing exactly what your definitions of the Clifford and exterior algebras are, so the following argument is still a little handwaving. But let's say that we somehow have defined what the exterior algebra of a vector space $V$ is; we know that its elements are multivectors, and the rule which generates everything is that $x \wedge y + y \wedge x=0$ if $x$ and $y$ are vectors in $V$. The Clifford algebra has the same elements as the exterior algebra, and the same linear space structure, but the multiplication is different: it is generated by the rule $xy+yx=2 \, Q(x,y)$ where $Q$ is the inner product on $V$. Since "orthogonal" means that $Q(e_i,e_j)=0$, it shouldn't be too hard to believe that the Clifford and exterior multiplications agree for orthogonal vectors.

Maybe this section on Wikipedia can be of some help too?

Answer 2

It appears that you are thinking in terms of "geometric algebra". For those unfamliar with this doctrine, it's a way of regarding the Clifford algebra on an inner product space $(V,Q)$ and the exterior algebra on the $V$ as the same set, but with different product operations. Let's stick to a ground field of characteristic zero. Then the Clifford algebra $C$ of $(V,Q)$ is generated by the elements $v\in V$ with relations $vv=Q(v)$. Now for $v_1,\ldots,v_k\in V$ one can define wedge product $$v_1\wedge v_2\wedge\cdots\wedge v_k =\frac1{k!}\sum_{\pi\in S_k}(-1)^{\text{sgn}(\pi)}v_{\pi(1)}v_{\pi(2)}\cdots v_{\pi(k)}\in C.$$ This wedge product identifies $C$ with the exterior algebra $\bigwedge(V)$.

With this definition, Hans's comments are absolutely right. If $v_1,\ldots,v_k$ are pairwise orthogonal then the Clifford product $v_1v_2\cdots v_k$ equals the exterior product $v_1\wedge v_2\wedge\cdots\wedge v_k$, and this is certainly the case for orthogonal basis vectors: $e_1 e_2 e_3 e_4 =e_1\wedge e_2\wedge e_3\wedge e_4$.