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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!

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$e_i \cdot e_j = 0$ –  anon Nov 8 '10 at 13:38
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...and for orthogonal vectors, the Clifford product coincides with the wedge product. –  Hans Lundmark Nov 8 '10 at 13:48
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What you seem to be calling an "explicit, pedantic, algebraic way" is what most people would call "a proof"... –  Mariano Suárez-Alvarez Nov 8 '10 at 13:52
    
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. –  corsecat Nov 8 '10 at 13:59
    
I agree with Hans; there's nothing to say until you tell us what your definition of the Clifford algebra is. –  Qiaochu Yuan Nov 8 '10 at 16:30
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2 Answers 2

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?

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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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