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.