Norm on a Geometric Algebra

Question

In the literature, for example "New Foundations for Classical Mechanics" by David Hestenes, the author introduces a function on the Geometric Algebra $$||M||^2=\langle M M^\dagger \rangle_0,$$ where dagger is the reverse, and claim that it is a norm. I'm having trouble showing that the triangle inequality is satisfied; I don't know how to estimate such scalars as $$\langle A B^\dagger \rangle_0.$$ I believe the triangle inequality only holds only when the quadratic form on the vector space is positive definite.

Answer 1

Let $C = A + B$. Then $|C|^2 = CC^\dagger = AA^\dagger + BB^\dagger + AB^\dagger + BA^\dagger$, right?

The key isn't to eliminate those cross terms but to bound them. The classic statement of the triangle inequality is

$$|C|^2 \leq (|A| + |B|)^2$$

Expand the right to get

$$|C|^2 \leq |A|^2 + |B|^2 + 2 |A ||B|$$

The classic proof argues that $AB^\dagger+BA^\dagger \leq 2 |A| |B|$. It's not clear to me how exactly one might go about this for the case of a general blade; perhaps you could argue that the $A, B$ must share at least a common plane, and so one can be rotated to the other through a simple rotation, so that $AB^\dagger + BA^\dagger = 2|A||B| \cos \theta$, where $\theta$ is the angle between them. That would be exactly in analogue to the vector case.

Regardless, I don't think you're meant to eliminate these scalars. Rather, you should bound them.