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.



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


