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