Vtyjkyuk

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.

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.

$Bbb R^m$ 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 $, 0 le k le n text and 1 le j_1 < dots < j_k le n$. 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 $$| M | := sqrtM * M = sqrtsum_J (M^J)^2$$ 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 $phi : Bbb G^p,q rightarrow Bbb R^p+q$ 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 $$|M|_textEucl := | phi (M) |$$ defines a Euclidean norm . With respect to the selected basis, the Euclidean norm gives $|M|_textEucl = sqrtsum_J (M^J)^2$ 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.