# Motivation behind the definition of the outer product in geometric algebra in terms of sums of geometric products

In chapter 4 of Doran and Lasenby's Geometric Algebra for Physicists, the authors define the outer product of an arbitrary number of 1-vectors as the totally antisymmetrized sum of those vectors, so that

$$a_1 \wedge a_2 \wedge \dots \wedge a_r = \frac{1}{r!} \sum{(-1)^\epsilon a_{k_1} a_{k_2} \dots a_{k_r}} \quad,$$

where the sum runs over all permutations, and $$(-1)^\epsilon$$ is $$+1$$ for an even permutation , and $$-1$$ for an odd permutation.

The authors point out that the antisymmetry ensures that if any vector is a linear combination of the others, the whole product is zero, which is certainly a desirable property. But, apart from this, what motivates such a definition? Plugging in $$r=2$$ immediately gives the desired result, and, for $$r=3$$, after a bunch of algebra, it can be shown that it is exactly equal to the definition given in chapter 2:

$$a \wedge b \wedge c = \frac{1}{2} \left( a(b \wedge c) + (b \wedge c)a \right) = \frac{1}{4} \left( a(bc-cb) + (bc-cb)a \right) = \frac{1}{4} \left( abc + bca - acb - cba \right) \quad.$$

It is not at all obvious to me that this "should" be the case, nor that it in fact is for $$r>3$$. Why does such a definition encapsulate what is desired of a geometric algebra?

## 1 Answer

$$\newcommand\Ext{\mathop{\textstyle\bigwedge}} \newcommand\Cl{\mathrm{Cl}} \newcommand\Alt{\mathrm{Alt}} \newcommand\alt{\mathrm{alt}} \newcommand\tensor\otimes$$Let $$V$$ be the finite dimensional (real) vector space in question.

Your definition of $$\wedge$$ ultimately gives the canonical linear isomorphism between a Clifford algebra and the exterior algebra; the exterior algebra has the property that $$a_1\wedge\dotsb\wedge a_r = 0 \iff \text{these vectors are linearly dependent} \tag{*}$$ and it is exactly this property that gives all the geometry of wedge products. Indeed, the exterior algebra is characterized by the following universal property:

• Let $$A$$ be any associative algebra. Then every linear $$f : V \to A$$ such that $$f(v)^2 = 0$$ for all $$v \in V$$ extends uniquely to an algebra homomorphism $$f' : \Ext V \to A$$ with $$f'(v) = f(v)$$ for all $$v \in V$$.

Since any $$v \in V$$ is linearly dependent to itself, any algebra $$A \supseteq V$$ with the the property ($$*$$) has a homomorphism $$\phi : \Ext V \to A$$ such that $$\phi|_V(v) = v$$. If $$A$$ is built only from sums of products of vectors (like $$\Ext V$$) then $$\phi$$ is surjective. Now suppose $$\phi(x) = 0$$ for some $$x \in \Ext V$$; then $$\phi(y\wedge x) = \phi(y)\phi(x) = 0$$ for any $$y \in \Ext V$$. In particular, we can choose $$y$$ such that $$y\wedge x$$ is a pseudoscalar $$I = v_1\wedge\dotsb\wedge v_n$$ with these vectors linearly independent, whence $$v_1\dotsb v_n = \phi(v_1)\dotsb\phi(v_n) = \phi(I) = 0.$$ But this is impossible by ($$*$$) so $$\phi$$ is an isomorphism. It is in this sense that ($$*$$) forces us to consider the exterior algebra.

The particular definition you've given $$a_1\wedge\dotsb\wedge a_r = \frac1{r!}\sum_{\sigma\in S_r}\mathrm{sgn}(\sigma)a_{\sigma(1)}\dotsb a_{\sigma(r)} \tag{**}$$ in a Clifford algebra $$\Cl(V)$$ comes from the identification of $$\Ext V$$ with alternating tensors $$\Alt(V) \subseteq T(V)$$. $$T(V)$$ is the tensor algebra, and we define alternating tensor via the antisymmetrization map $$\alt : T(V) \to T(V)$$ $$\alt(a_1\tensor\dotsb\tensor a_r) = \sum_{\sigma\in S_r}\mathrm{sgn}(\sigma)a_{\sigma(1)}\tensor\dotsb\tensor a_{\sigma(r)}$$ whence $$\Alt(V)$$ is the image of $$\alt$$. Both $$\Ext V$$ and $$\Cl(V)$$ can be defined as quotients of $$T(V)$$ by appropriate two-sided ideals, yielding projections $$\pi_\wedge : T(V) \to \Ext(V),\quad \pi_\Cl : T(V) \to \Cl(V).$$ The restriction of $$\pi_\wedge$$ to $$\Alt(V)$$ turns out to be an isomorphism when $$\Alt(V)$$ is given the product $$X\wedge Y = \frac{r!s!}{(r+s)!}\alt(X\tensor Y)$$ where $$X, Y \in \Alt(V)$$ are tensors of degree $$r, s$$ respectively. (Notice that this product is undefined over a field of nonzero characteristic; there is no isomorphism between $$\Alt(V)$$ and $$\Ext V$$ in this case, but there is still an intimate relationship between $$\Ext V$$ and $$\Cl(V)$$.) All together we have the following diagram $$\require{AMScd}\begin{CD} T(V) @>{\alt}>> \Alt(V) \\ @V{\pi_\Cl}VV @V{\pi_\wedge}VV \\ \Cl(V) @>\psi>> \Ext V \end{CD}.$$ $$\psi$$ is what makes the diagram commute and is necessarily the canonical linear isomorphism between $$\Cl(V)$$ and $$\Ext V$$. The definition ($$**$$) in $$\Cl(V)$$ comes from the composition $$\psi^{-1}\circ\pi_\wedge\circ\alt$$.