How do I evaluate the Clifford product in dimensions greater than 3? 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!
 A: 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?
A: 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: $e_1 e_2 e_3 e_4
=e_1\wedge e_2\wedge e_3\wedge e_4$.
