2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.