I started looking into Clifford Algebras, but I think I am conceptually missing some points. Here is how I understand the geometric product now (leaving out coordinate system independence and the generalized definitions of the products for brevity):

Let's say we have a Clifford Algebra $Cl(\mathbb{R^n})$ with $e_1...e_n$ as orthonormal basis, where $e_i^2 = 1$ and $e_ie_j = 0$ for $i \neq j$ for simplicity.

The geometric product is associative and distributive over addition. Also for 1-vectors the geometric product can be represented by $ab = a \cdot b + a \land b$. Where $a \cdot b = b \cdot a$ and $a \land b = -b \land a$, which gives $a \cdot b = 0$ if $a \perp b$ and $a \land b = 0$ if $a \parallel b$.

Now, I have the following problem, which I think gives the most concise way of representing my question:

Given $a = e_1$ and $b = e_2 \land e_1$, what is $ab$?

With what I have above, I can solve this in the following four ways, giving me different answers:

$$ab = e_1e_2 \land e_1 = 0 \land e_2 = 0$$ $$ab = -e_1e_1 \land e_2 = -1 \land e_2 = -e_2$$ $$ab = e_1e_2 \land e_1 = (e_1 \cdot e_2 + e_1 \land e_2) \land e_1 = e_1 \land e_2 \land e_1 = 0$$ $$ab = -e_1e_1 \land e_2 = -(e_1 \cdot e_1 + e_1 \land e_1) \land e_2 = -1 \land e_2 = -e_2$$

What point am I missing here?


The thing you assume in the first step of all of your proposed methods of evaluating $ab$ is that because the geometric product and the wedge product are both associative individually, that they must also be associative together. That is $\require{enclose}\color{red}{\enclose{circle}{\require{cancel}\cancel{\color{black}{u(v\wedge w)=(uv)\wedge w}}}}$. This is NOT TRUE (thus the red strikethrough ;) ).

Now let's go through (one of) the correct method(s) of evaluating this product

$$ab = (e_1)(e_2\wedge e_1) = -(e_1)(e_1\wedge e_2) = -(e_1)(e_1e_2) = -(e_1e_1)e_2 = -(1)e_2 = -e_2$$

  • $\begingroup$ Once again proving that assumption is the mother of all screw-ups. I was trying so hard to decompose the problem into products I understand, that it never occurred to me to reduce the problem to geometric products so I could use the correct associative relations. Thank you for the insight! $\endgroup$ – jvdh Apr 2 '16 at 15:36

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