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

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


  • $\begingroup$ $e_i \cdot e_j = 0$ $\endgroup$ – anon Nov 8 '10 at 13:38
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    $\begingroup$ ...and for orthogonal vectors, the Clifford product coincides with the wedge product. $\endgroup$ – Hans Lundmark Nov 8 '10 at 13:48
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    $\begingroup$ What you seem to be calling an "explicit, pedantic, algebraic way" is what most people would call "a proof"... $\endgroup$ – Mariano Suárez-Álvarez Nov 8 '10 at 13:52
  • $\begingroup$ Ok, well I want a proof that for orthogonal vectors the Clifford product coincides with the wedge product. In other words, I certainly realize that $e_i e_j = e_i \cdot e_j + e_i \wedge e_j = 0 + e_{ij}$ when $i \ne j$. But now suppose I have $e_i e_j e_k$ for distinct $i$, $j$, and $k$. Then I get $e_{ij} e_k$ but have no rule for the Clifford product between a bivector and a vector, so I am stuck! Thanks for the help. $\endgroup$ – corsecat Nov 8 '10 at 13:59
  • $\begingroup$ I agree with Hans; there's nothing to say until you tell us what your definition of the Clifford algebra is. $\endgroup$ – Qiaochu Yuan Nov 8 '10 at 16:30

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?


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


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