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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$2010-11-08
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    ...and for orthogonal vectors, the Clifford product coincides with the wedge product.2010-11-08
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    What you seem to be calling an "explicit, pedantic, algebraic way" is what most people would call "a proof"...2010-11-08
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    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.2010-11-08
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    I agree with Hans; there's nothing to say until you tell us what your definition of the Clifford algebra is.2010-11-08

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