Problems in understanding exercise 2.2. from Chris Doran "Geometric Algebra for physicists"

Question

I am currently trying to learn geometric algebra with Dorans book. I tried to do excercise 2.2., but I get stuck at the formulation of the exercise:

By expanding the bivector $a \wedge b$ in terms of geometric products, prove that it anticommutes with both a and b, but commutes with any vector perpendicular to the $a \wedge b$ plane.

My thoughts on the excercise were follwing:

a) $$a \wedge b = \frac{1}{2}(ab - ba) = -\frac{1}{2}(ba - ab) = -b \wedge a$$

b) If we want a vector perpendicular to the $a \wedge b$ plane we can write that as $a \wedge (\lambda a \times b)$ with $\lambda \in \mathbb{R} $. Now we have to show that this is equal to $(\lambda a \times b) \wedge a$, right?

Do I interpret his exercise right? So does he mean the geometric product or the outer product when he talks about it, but the wedge product was already defined as anticommutative. I am a bit confused. Thank you for all of your help.

Answer 1

They are asking about the geometric product. They are asking to prove the following three equalities: $$ (a\wedge b)a = -a(a\wedge b), $$$$ (a\wedge b)b = -b(a\wedge b), $$$$ (a\wedge b)c = c(a\wedge b), $$ where $c$ is any vector perpendicular to both $a$ and $b$. You should also not use the cross product for anything, for at least two reasons:

1. One of the points of geometric algebra is to do away with kludges like the cross product which only work in 3D.
2. The notation $\times$ is used in this book for something entirely different, namely the commutator product of multivectors $$ A\times B := \frac12(AB - BA), $$ though I think the authors do also use $\times$ to represent the cross product when they talk about it, which is an unfortunate overloading of notation.

---

A side note: the wedge product is not "defined to be anticommutative". It is anticommutative between two vectors, but it is absolutely not anticommutative in general.