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 1


$ \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 $\psi^{-1}\circ\pi_\wedge\circ\alt$.


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