Norm on a Geometric Algebra 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.  
 A: ${\Bbb R}^m$ has an scalar product (a.k.a. inner product) $ {\bf x} \cdot {\bf y} = \sum_{j=1}^m x^j y^j $ from which one can define the magnitude $\| {\bf x} \| := \sqrt{ {\bf x} \cdot {\bf x} } = \sqrt {\sum_{j=1}^m (x^j)^2}$ of the vector ${\bf x} = \sum_{j=1}^m {\bf e}_j x^j$.  The magnitude is well known to satisfy the triangle inequality and other defining properties of a norm, as in shown in just about any text that discusses ${\Bbb R}^m$.  It is sometimes called the vector norm, a.k.a. $L^2$ norm, on ${\Bbb R}^m$.
Introduce the scalar product on ${\Bbb G}^{p,q}$ by setting $M*N := \langle M^\dagger N \rangle$.  The dagger is multivector reversion, juxtaposition is the geometric product, and the angle brackets indicate the scalar part of the multivector they enclose. Choose an orthonormal basis $\{{\bf e}_1, ..., {\bf e}_n \}$ for the scalar product space $V^{p,q}$ used to generate the geometric algebra ${\Bbb G}^{p,q}$.  This determines a canonical basis for ${\Bbb G}^{p,q}$, namely $\{ {\bf E}_J := {\bf e}_{j_1} \cdots {\bf e}_{j_k} \,|\, 0 \le k \le n \text{ and } 1 \le j_1 < \dots < j_k \le n\}$. Here $n=p+q$ and ${\bf e}_{j_1} \cdots {\bf e}_{j_k} := 1$ for $k=0$.  Any multivector $M$ has an expansion $M = \sum_J {\bf E}_J M^J$, where $J$ runs over the $2^n$ distinct strictly-ordered index values $(j_1, \dots, j_k)$.
The canonical basis is orthonormal with respect to the scalar product, ${\bf E}_I * {\bf E}_J = {\bf e}_{j_1}^2 \cdots {\bf e}_{j_k}^2 \delta_{IJ}$. Consequently the scalar product of two multivectors $M$ and $N$ is 
$$M*N = \sum_J ({\bf e}_I * {\bf e}_J)M^I N^J  = \sum_J {\bf e}_{j_1}^2 \cdots {\bf e}_{j_k}^2 M^J N^J.$$ 
(The product ${\bf e}_{j_1}^2 \cdots {\bf e}_{j_k}^2$ is $\pm 1$.) When the generating scalar product space is Euclidean, i.e. when the signature $(p,q) =(n,0)$, one has ${\bf E}_I * {\bf E}_J = \delta_{IJ}$.  So ${\Bbb G}^{n,0}$ together with the scalar product $M*N$ is isomorphic to the scalar product space ${\Bbb R}^{(2^n)}$.  Consequently $$\| M \| := \sqrt{M * M} = \sqrt{\sum_J (M^J)^2}$$ is a triangle inequality-satisfying norm on ${\Bbb G}^{n,0}$.  
For non-Euclidean cases ($q \gt 0$), there exist multivectors $M$ for which $M * M$ is negative.  Consequently the scalar product is positive definite on neither ${\Bbb R}^{p,q}$ nor ${\Bbb G}^{p,q}$.  In that event one cannot express $M * M$ as the square of some real number $\| M \|$.  However one can still define a norm as follows:  Choose the linear space isomorphism $\phi : {\Bbb G}^{p,q} \rightarrow {\Bbb R}^{(2^{p+q})}$ that sends the $i$th element ($i = 1, 2, ..., 2^{p+q}$) in a selected (ordered) canonical basis $\{ {\bf E}_J \}$ for ${\Bbb G}^{p,q}$ to the $i$th element in the (ordered) standard basis $\{ {\bf e}_i \}$ for ${\Bbb R}^{(2^{p+q})}$.  Then $$\|M\|_\text{Eucl} := \| \phi (M) \|$$ defines a Euclidean norm .  With respect to the selected basis, the Euclidean norm gives $\|M\|_\text{Eucl} = \sqrt{\sum_J (M^J)^2}$ for $M = \sum_J M^J {\bf E}_J $, so we know that the Euclidean norm must satisfy the triangle inequality.  Unfortunately the Euclidean norm  so defined on ${\Bbb G}^{p,q}$ is dependent on the choice of orthonormal basis $\{{\bf e}_1, \dots, {\bf e}_{p+q} \}$ for the generating scalar product space ${\Bbb R}^{p,q}$ of ${\Bbb G}^{p,q}$.  The Euclidean norm does not arise "naturally" from the scalar product structure.
A: 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.
