I have a problem with the geometric product:

In my book the unit trivector is defined like this: $(e_{1}e_{2})e_{3}=e_{1}e_{2}e_{3}$ But that would mean $(e_{1}e_{2})e_{3}= (e_{1} \wedge e_{2})\cdot e_{3}+(e_{1} \wedge e_{2} \wedge e_{3})$ But I thougt it is just $e_{1} \wedge e_{2} \wedge e_{3}$? I could somehow imagine in my head that the plane spanned by $e_{1} \wedge e_{2}$ is perpendicular to the line $e_{3}$ but I'm not sure that it works like that. Is that right? Would make sense. Ok but why is this true $(e_{1}\wedge e_{2})e_{1}=(-e_{2}e_{1})e_{1}=-e_{2}e_{1}e_{1}=-e_{2}$ because $(e_{1}\wedge e_{2})e_{1}=(e_{1} \wedge e_{2})\cdot e_{1}+ e_{1}\wedge e_{2} \wedge e_{1}$ where the last term is zero. I don't know how the term $e_{1} \wedge e_{2}$ interacts as dot product with $e_{1}$. Similar is $(e_{1}\wedge e_{2})(e_{2}\wedge e_{3})=e_{1}e_{3}$ why isn't this just zero? Because $(e_{1}\wedge e_{2})(e_{2} \wedge e_{3})=(e_{1}\wedge e_{2}) \cdot (e_{2} \wedge e_{3})+e_{1}\wedge e_{2} \wedge e_{2} \wedge e_{3}$ any help?


1 Answer 1


I think it's often easier to treat the dot and wedge products as shorthands for grade projection of geometric products.

For example, for vectors $a, b, c$:

$$a \cdot b = \langle ab \rangle_1, \quad a \wedge b = \langle ab \rangle_2$$


$$a \wedge b \wedge c = \langle abc \rangle_3$$

With that in mind, your book defines the unit trivector as $(e_1 e_2) e_3 = e_1 e_2 e_3$. The geometric product is associative, so that clearly must be true. Breaking it down into grades, however, gives this:

$$\begin{align*}e_1 e_2 e_3 &= \langle e_1 e_2 e_3 \rangle_1 + \langle e_1 e_2 e_3 \rangle_3 \\ &= (e_1 \wedge e_2) \cdot e_3 + (e_1 \cdot e_2) e_3 + e_1 \wedge e_2 \wedge e_3\end{align*}$$

You already realized that the first term is zero. The second term is clearly zero. So $e_1 e_2 e_3 = e_1 \wedge e_2 \wedge e_3$.

Because $e_1, e_2, e_3$ are all orthogonal, most geometric products between them reduce to wedge products and vice versa. Consider:

$$e_1 e_2 = \langle e_1 e_2 \rangle_0 + \langle e_1 e_2 \rangle_2 = e_1 \cdot e_2 + e_1 \wedge e_2 = e_1 \wedge e_2$$

Since the dot product orthogonal vectors is zero.

Hence, when you try to evaluate $(e_1 \wedge e_2) \cdot e_1$, you get

$$(e_1 \wedge e_2) \cdot e_1 = (e_1 e_2) \cdot e_1 = \langle e_1 e_2 e_1 \rangle_1 = -\langle e_2 e_1 e_1 \rangle_1$$

Group the $(e_1 e_1)$ together as $+1$, and you're done. The result is $-e_2$.

Part of your problem is that you're using $ab = a \cdot b + a \wedge b$ for everything, even when $a, b$ are not vectors. That's not correct practice.

Two general multivectors $A, B$ of grades $p, q$ have their geometric product decompose as follows:

$$AB = \langle AB \rangle_{p+q} + \langle AB \rangle_{p+q-2} + \ldots + \langle AB \rangle_{|p-q|}$$

For any term where the grade is greater than the dimension of the space, the grade projection is zero.

Hence, for two bivectors $p=q=2$, the terms are

$$AB = \langle AB \rangle_4 + \langle AB \rangle_2 + \langle AB \rangle_0$$

The grade-4 term is zero in 3d space, but it would be denoted with the wedge product. The grade-2 term is typically denoted $A \times B$ and called the commutator product. The grade-0 term is typically denoted with a dot product.

So when you multiply $(e_1 \wedge e_2)(e_2 \wedge e_3)$, the easy thing to do is to eliminate the wedges in favor of geometric products, and then compute the terms.

$$(e_1 \wedge e_2)(e_2 \wedge e_3) = e_1e_2 e_2 e_3$$

Group those $e_2 e_2$ together as $+1$, and you're done. The result is $+e_1 e_3$. This is the commutator product term that you didn't realize was there.

The main convenience of using an orthogonal basis is that you can do a lot of arithmetic directly with the geometric product acting on basis vectors. Basis vectors can be swapped at the cost of just a minus sign, and you can reduce the overall product until there are no more contractions of vectors to perform.

  • $\begingroup$ Hi, thank you for your answer ! What about the last equation $(e_{1}\wedge e_{2})(e_{2} \wedge e_{3}) = (e_{1} \wedge e_{2}) \cdot (e_{2} \wedge e_{3})+ (e_{1} \wedge e_{2} \wedge e_{2} \wedge e_{3})$ But this is not zero. What's wrong with it? I know you can rewrite it as gp and forget what's going in reality but I want to know what's happening. $\endgroup$ Aug 27, 2014 at 20:08
  • 1
    $\begingroup$ I addressed this in the section starting with "Part of your problem...". For two bivectors $A, B$, the general form of the geometric product is $AB = A \wedge B + A\times B + A \cdot B$. I defined what all those terms mean, including their grades, in the answer already. It's correct to observe that the dot and wedge product are zero here. The commutator product, however, is nonzero. Geometric products are no less real than these other products; I think you should consider it more fundamental, personally, and consider the other products as derived from it using grade projection. $\endgroup$
    – Muphrid
    Aug 27, 2014 at 20:11
  • $\begingroup$ Nothing in my book is referring to a commutator product at this moment, so maybe it is explained later or something. Thx $\endgroup$ Aug 27, 2014 at 20:22
  • $\begingroup$ What book are you using? I might be able to point to some chapters that would be helpful. $\endgroup$
    – Muphrid
    Aug 27, 2014 at 20:23
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    $\begingroup$ I see; you won't run into that stuff until chapter 4 then. What I wrote above is heavily inspired by Doran and Lasenby chapter 4. They introduce grade projection there, for instance, and the whole chapter is more rigorous than the quick and dirty introduction in chapter 2. $\endgroup$
    – Muphrid
    Aug 27, 2014 at 20:39

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