How does grade projection act on homogeneous multivectors in geometric algebra?

Question

I'm reading Clifford Algebra to Geometric Calculus by Hestenes, and struggling with an early result about reversion inside of a grade-projection operator.

It is noted that $A_r$ and $B_s$ are homogeneous multivectors of grades $r$ and $s$. Just to be sure I understand the basic concepts, a homogeneous multivector is different from a blade in the following way. A homogeneous multivector could be a sum of blades of the same grade, e.g.

$$ A_r = \langle A_r \rangle_3 = abc + abd $$

where $\langle X \rangle_n$ is called the "$n$-vector part of $X$" by Hestenes, and $a,b,c,d$ are linearly independent (anti-commuting) vectors.

Hestenes defines the reversion operator for arbitrary multivectors as follows:

$$ (AB)^\dagger = B^\dagger A^\dagger $$ $$ (A + B)^\dagger = A^\dagger + B^\dagger $$

Then for vectors:

$$ a^\dagger = a $$ $$ (a_1a_2...a_n)^\dagger = a_n...a_2a_1 $$

Good so far. A little farther:

$$ \langle A^\dagger \rangle _r = \langle A \rangle _r^\dagger = (-1)^{r(r-1)/2}\langle A \rangle _r $$

This makes sense because $\langle A \rangle _r$ is homogenous, so the reversion distributes over the r-blades in its sum, the vectors in the blade products anticommute, and you pick up a negative sign for odd permutations.

This is where I'm lost:

$$ \langle A_rB_s \rangle _r = \langle B_s^\dagger A_r \rangle _r = (-1)^{s(s-1)/2}\langle B_sA_r \rangle _r $$

If I tried expanding this in a simple example, I get:

$$ = \langle (a_1a_2)(b_1b_2b_3) \rangle _2 $$ $$ = \langle a_1a_2b_1b_2b_3 \rangle _2 = 0 $$

It seems like any product of $r$-vector and $s$-vector will produce only $(r+s)$-vectors, which don't contribute at all to the r-grade part of the multivector. I feel like I have a fundamental misunderstanding of one (or some) of these concepts.

Answer 1

I think that this is not the same notation as you will find elsewhere. Usually, the Clifford algebra is taken to be $\mathbb Z/2\mathbb Z$-graded, rather than $\mathbb Z$-graded, precisely because an apparently homogeneous multivector of grade $n \ge 2$ can often be traded for the sum of two multivectors, of degrees $n$ and $n - 2$, by switching the order of two factors.

Hestenes seems (I say only 'seems', because I'm confused by the approach of treating the structure $\mathscr G$ and the projections $\langle A\rangle_r$ as existing before the axioms, rather than stating axioms to be satisfied by a later structure) to impose as an axiom the existence of a grading on a geometric algebra, and to require that a piece of grade $r$ be a product of $r$ anticommuting vectors; but not every product of $r$ vectors lives in grade $r$! (For example, the square of any vector lies in grade $0$, not grade $2$.) I think that Hestenes is requiring that every product of $r$ anti-commuting vectors lie in grade $r$. In this case, one has, for example, that $$ (b\cdot b)a = \underbrace{(a\cdot b)b}_{\text{commutes with $b$}} + \underbrace{(b\cdot b)a - (a\cdot b)b}_{\text{anticommutes with $b$}}, $$ so $$ (b\cdot b)a b = \underbrace{(a\cdot b)(b\cdot b)}_{\text{grade $0$}} + \underbrace{((b\cdot b)a - (a\cdot b)b)b}_{\text{grade $2$}}, $$ where $\cdot$ denotes the scalar product, satisfying $2a\cdot b = (a + b)^2 - a^2 - b^2$, introduced in (1.21). (Note that vectors anti-commute if and only if they are orthogonal for this product.) More generally, $\langle A_r B_s\rangle_r$ is $0$ unless $s$ is even. If $b_1$ and $b_2$ are anticommuting vectors, and $a$ is any vector, then $$ (b_1\cdot b_1)(b_2\cdot b_2)a = \underbrace{(b_2\cdot b_2)(a\cdot b_1)b_1}_{\text{commutes with $b_1$}} + \underbrace{(b_1\cdot b_1)(a\cdot b_2)b_2}_{\text{commutes with $b_2$}} + \underbrace\cdots_{\text{anti-commutes with $b_1$ and $b_2$}}, $$ so \begin{multline*} (b_1\cdot b_1)(b_2\cdot b_2)a(b_1 b_2) \\ = \underbrace{(b_1\cdot b_1)(b_2\cdot b_2)(a\cdot b_1)b_2 - (b_1\cdot b_1)(b_2\cdot b_2)(a\cdot b_2)b_1}_{\text{grade $1$}} + \underbrace{(\cdots)(b_1 b_2)}_{\text{grade $3$}} \\ = (b_1\cdot b_1)(b_2\cdot b_2)(b_2 b_1)a \end{multline*} and $$ (b_1\cdot b_1)(b_2\cdot b_2)(b_1 b_2)a = \underbrace{-(b_1\cdot b_1)(b_2\cdot b_2)(a\cdot b_1)b_2 + (b_1\cdot b_1)(b_2\cdot b_2)(a\cdot b_2)b_1}_{\text{grade $1$}} + \underbrace{(b_1 b_2)(\cdots)}_{\text{grade $3$}}. $$ (That unmotivated decomposition of $a$ probably makes more sense if you think of Gram–Schmidt orthogonalisation.) In particular, $$ \langle a(b_1 b_2)\rangle_1 = \langle(b_2 b_1)a\rangle_1 = -\langle(b_1 b_2)a\rangle_1 = (a\cdot b_1)b_2 - (a\cdot b_2)b_1. $$ You can probably see the general picture from this.