I am currently self-studying axiomatic geometric algebra from a couple of books(Doran-Lasenby(DL), Hestenes(H), notes on the internet), and I got stuck while checking some calculations.

When defining the outer product of vectors, DL starts by defining it as the completely anti-symmetrized sum, that is

$$\bigwedge_{i=1}^n a_i=\frac1{n!}\sum_{\sigma\in S_n}\text{sgn}(\sigma)\prod_{i=1}^na_{\sigma(i)}$$

and from this goes on to motivate the wedge product $v$ of a vector and a grade-$r$ multivector $A_r$ in terms of the geometric product

$$a\wedge A_r=\frac12(aA_r+(-1)^rA_ra)$$

H also mentions the latter definition but I haven't encountered the former definition yet.

Now, in case $A_r$ is a pure blade, one can go from the first definition to the second definition by an induction process (DL has a proof, albeit with a few typos which are fixable), but I cannot go back from the second to the first. Here's what I attempted.

For me, $r=2$. So let $A_r=A_2=b\wedge c$. Then

\begin{align} a\wedge A_2&=a\wedge (b\wedge c)\\&=\frac12(a(b\wedge c)+(-1)^2(b\wedge c)a)\\ &=\frac12\left(a\left(\frac12(bc-cb)\right)+\left(\frac12(bc-cb)\right)a\right)\\ &=\frac14(abc-acb+bca-cba) \end{align}

but that's not the anti-symmetrized sum.

I can't see where I am going wrong. Any help is greatly appreciated.

  • 1
    $\begingroup$ shouldn't the second sum symbol in your first equation be a product symbol? $\endgroup$ Jun 4, 2022 at 13:57
  • $\begingroup$ Oh definitely, that's a typo right there. Fixed. $\endgroup$ Jun 4, 2022 at 13:57

1 Answer 1


You aren't going wrong, but I'd start in a slightly different way so that you get just two terms in your sum to start with. First expand

$$\begin{aligned}a \wedge \left( { b \wedge c } \right)&=\frac{1}{{2}} \left( { a \left( { b \wedge c } \right)+\left( { b \wedge c } \right) a } \right) \\ &=\frac{1}{{2}} \left( { a \left( { b c - b \cdot c } \right)+\left( { b c - b \cdot c } \right) a } \right) \\ &=\frac{1}{{2}} \left( { a b c + b c a } \right) - a \left( { b \cdot c } \right) \\ &=\frac{1}{{2}} \left( { a b c + b c a } \right) - \frac{1}{{2}} a \left( { b c + c b } \right) \\ &=\frac{1}{{2}} \left( { b c a - a c b } \right).\end{aligned}$$

Now note that we can toggle any pair of vectors using $ y x = 2 x \cdot y - x y $, so $$\begin{aligned}b c a - a c b&=b \left( { 2 c \cdot a - a c } \right) - \left( { 2 a \cdot c - c a } \right) b \\ &=c a b - b a c,\end{aligned}$$ or $$\begin{aligned}b c a - a c b&=\left( { 2 b \cdot c - c b } \right) a - a \left( { 2 c \cdot b - b c } \right) \\ &=a b c - c b a.\end{aligned}$$ That is $$b c a - a c b = c a b - b a c = a b c - c b a.$$

We can plug this into our first expansion of the wedge, by writing

$$\begin{aligned}a \wedge \left( { b \wedge c } \right)&=\frac{1}{{2}} \left( { b c a - a c b } \right) \\ &=\frac{1}{{3 \times 2}} 3 \left( { b c a - a c b } \right) \\ &=\frac{1}{{3!}} \left( { \left( { b c a - a c b } \right) + \left( { c a b - b a c } \right) + \left( { a b c - c b a } \right)} \right) \\ &=\frac{1}{{3!}}\left( { a b c + b c a + c a b - a c b - b a c - c b a } \right).\end{aligned}$$ Observe that we have all the permutations of the products in this sum, each weighted by the sign of the permutation, as desired.


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