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.


1 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.


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