I stumbled onto to something... unless it turns out I'm making some silly mistake, it appears to be a major problem with the geometric... inner, and outer products.
I think you all know I've been trying to find an easy way to form inner and outer products... some easy to follow, and easy to remember rules. The webpage on using the wedge product (outer product) for tuning musical instruments, wedge-product, is very helpful for learning how to form outer products.
Recently, I came across this Master's Thesis by Mikael Nilsson, Geometric Algebra with Conzilla: Building a Conceptual Web of Mathematics. On pp. 37, 38 he provides an easy way to form outer and inner products:
p. 37:
p. 38:
I believe Mikael is referring to the Hestenes semi-commutative inner product, which is the inner product I've been using, as well.
I made up a couple of Venn diagrams to provide a graphical tool to understanding what Mikael is saying. For the outer product, where A<sub>i</sub> and B<sub>i</sub> are the sets of basis indices for each of the current product operands (all products without explicit operators are geometric products):
![Image](http://paul-mccarthy.us/Misc/out_dia1.jpg)
Then, the inner product:
![Image](http://paul-mccarthy.us/Misc/in_dia1.jpg)
Some of you more visually intuitive than I might wonder immediately... What if...?
![Image](http://paul-mccarthy.us/Misc/N_in_out.jpg)
However, I didn't see this possibility until later...
Now that I had an easy to use way to form both inner and outer products at my disposal, I thought I'd go back and follow up on the work in one of my previous Tetraspace Forum posts, Octonions do not form a Clifford Algebra...?
You'll notice that two of the operation tables for the geometric product that I derived appear to be isomorphic to the complex product... Cl<sub>0,1</sub> and Cl<sub>2</sub><sup>+</sup>... and two appear to be isomorphic to the quaternion product... Cl<sub>0,2</sub> and Cl<sub>3</sub><sup>+</sup>. And, I worked them out to be sure...
![Image](http://paul-mccarthy.us/Misc/CP_prod1.gif)
![Image](http://paul-mccarthy.us/Misc/CP_prod2.gif)
![Image](http://paul-mccarthy.us/Misc/Cl02_q.gif)
However, when I got to Cl<sub>3</sub><sup>+</sup>, I started to encounter some serious problems... the bivector basis...
The Cl<sub>3</sub><sup>+</sup> representation for the quaternions makes the assignments i = -e<sub>23</sub>, j = -e<sub>31</sub>, k = -e<sub>12</sub>.
So, for instance, neither '23' nor '31' are fully contained within the other, nor is their intersection null... that is, we have:
![Image](http://paul-mccarthy.us/Misc/N_in_out.jpg)
This means that all product terms involving the bivectors, excepting those which are the same (that is, '23' and '23', for instance), are neither going to be inner, nor outer products...?
Yet, the definition of the geometric product is: ab = a.b + a^b...? So, where are the product terms formed by the bivectors in the geometric product coming from?
More specifically, let's look at the geometric product table for Cl<sub>3</sub>, as posted on Ian Bell's website, Multivector Methods: The Geometric Product: (I've highlighted the entries of interest here)
Now, if we look at the product tables for the inner and outer products on Ian's website, Multivector Methods: Products of multivectors:
You'll also notice that using Lounesto's (left) contractive inner product won't help with this problem:
So, can someone explain all this? Does anyone have any ideas?