Projections in Geometric Algebra

Question

Given a geometric algebra $G$. Let $I, A,B$ be an $n,r$ and $s$ blade, respectively. Let $P(*)=(*\rfloor I) I^{-1}$ be the projection operator onto $I$. Suppose that $P(A)=A$. The problem is to show that $P(AB)=AP(B)$. I'm able to show this when $r=1$ and $s$ arbitrary but can't generalize to an arbitrary $r$-blade. Any ideas to the proof or counterexamples, would greatly be appreciated.

Answer 1

As a clarification of the question, although $I$ is very commonly used to denote the unit pseudoscalar of the geometric algebra in which all objects in question reside, that isn't the case here. The question does not assume that $P(B)=B$, which means that it also isn't the case in general that simply $P(AB)=AB$. In particular, in the application used in the book section cited in the question's comments, $I$ is the unit pseudoscalar of the tangent space to some point in a manifold that's embedded in some larger ambient space, not the unit pseudoscalar of the entire ambient space.

For readers not familiar with the notation, the left contraction $\rfloor$ is very closely related to the inner product. With $C_p$ and $D_q$ being homogeneous multivectors of grade $p$ and $q$, the left contraction is defined as $$C_p\rfloor D_q\equiv\langle C_p D_q \rangle_{q-p}\ .\tag{1}$$ I.e., $\rfloor$ is the same as the most common definition of $\cdot$ except that the definition uses $q-p$ instead of $|q-p|$.

For a component-free proof of the requested result, see the proof of eq. 2.13e on pages 21-23 of Clifford Algebra to Geometric Calculus. However, in this case I think the result is easier to understand if a basis is chosen.

Since $I$ is a blade of grade $n$, it's possible to write $I$ as $$I=e_1 e_2\cdots e_n\ ,$$ where the vectors $e_i$ are all orthogonal to each other. We will use those $n$ vectors as part of our chosen vector basis. However, we assume that the total vector space we are dealing with may have a dimension greater than $n$, so we arbitrarily choose additional basis vectors $f_1, f_2,\dots$, all of which are orthogonal to all of the $e_i$ and orthogonal to each other, sufficient to span the entire vector space. I.e., the complete set of orthogonal basis vectors we use is $$\{e_1, e_2,\dots, e_n, f_1, f_2,\dots\}\ .$$ The basis blades we use are then formed from the basis vectors as all possible distinct ordered geometric products of the basis vectors, where we can say "geometric product" instead of "outer product" because the basis vectors are all orthogonal.

If $C_p$ is a scalar times a basis blade of grade $p$ and $D_q$ is a scalar times a basis blade of grade $q$, then $$C_p D_q=\langle C_p D_q\rangle_{p+q-2c}\ ,\tag{2}$$ where $c$ is the number of basis vectors that $C_p$ and $D_q$ have in common. It's easy to prove this by repeatedly using $e_i e_j=-e_j e_i$ to reorder the basis vectors in the product to put identical basis vectors together, and then noting that $e_i e_i$ results in two vectors becoming a scalar.

In a restriction which we will remove later, we first consider the case where $A$ and $B$ are basis blades. The question specifies that $A$ and $B$ are blades, but even with that restriction, a blade in general will be a linear sum of the basis blades that we have chosen.

With $B$ a basis blade of grade $s$, formed from the product of $m$ of the $f_i$'s and $s-m$ of the $e_i$'s, from (2) we have that $$\begin{aligned} BI&=\langle BI\rangle_{s+n-2(s-m)}\\ &=\langle BI\rangle_{n-s+2m} \end{aligned}$$ And then from (1) we have that $$\begin{aligned}B\rfloor I&=\langle\langle BI \rangle_{n-s+2m}\rangle_{n-s}\\ &=\begin{cases} \langle BI \rangle_{n-s+2m},&m=0\\ 0,&m>0 \end{cases}\\ &=\begin{cases} BI,&m=0\\ 0,&m>0 \end{cases} \end{aligned}$$

Multiplying both sides of that equation on the right by $I^{-1}$ gives $$P(B)=\begin{cases} B,&m=0\\ 0,&m>0 \end{cases}$$ which is exactly the behavior we're looking for. This equation shows the intuitive meaning of what it means to project onto a blade.

We can then multiply both sides of that equation on the left by any $A$ to give $$A P(B)=\begin{cases} AB,&m=0\\ 0,&m>0 \end{cases}\tag{3}$$

With $A$ a grade $r$ basis blade such that $P(A)=A$, we have from the above that $A$ must not contain any of the $f_i$ as a factor, and hence must be a product of $r$ of the $e_i$'s. The $s-m$ of the $e_i$'s that are factors of $B$ can now be split into two groups, the $t$ of the $e_i$'s that are a factor of both $B$ and $A$, and the $s-m-t$ of the $e_i$'s that are a factor of $B$ but not also a factor of $A$. From (2), we have that $$AB=\langle AB \rangle_{r+s-2t}$$ so from a second application of (2), $$\begin{aligned}ABI&=\langle ABI \rangle_{(r+s-2t)+n-2(r-t+(s-m-t))}\\ &=\langle ABI \rangle_{n-r-s+2t+2m}\ . \end{aligned}$$ Then from (1), we have that $$\begin{aligned}(AB)\rfloor I&=\langle\langle ABI \rangle_{n-r-s+2t+2m}\rangle_{n-(r+s-2t)}\\ &=\begin{cases} \langle ABI \rangle_{n-r-s+2t+2m},&m=0\\ 0,&m>0 \end{cases}\\ &=\begin{cases} ABI,&m=0\\ 0,&m>0 \end{cases} \end{aligned}$$ Multiplying both sides of that equation on the right by $I^{-1}$ gives $$P(AB)=\begin{cases} AB,&m=0\\ 0,&m>0 \end{cases}\tag{4}$$ But the RHS of (3) and (4) are the same, so we conclude that $$P(AB)=AP(B)\ .\tag{5}$$

This looks like the result to be proved, but it isn't quite because we have only proved it for the case of $A$ and $B$ being basis blades. The more general case follows from both $P(AB)$ and $AP(B)$ being linear in both $A$ and $B$.

To spell out the linearity statement:

Let $\{C_i\}$ be the set of all basis blades such that $P(C_i)=C_i$, i.e., the set of all basis blades which don't contain any of the $f_j$ as a factor, and let $\{D_i\}$ be the set of all basis blades. (5) can then be expressed as $$P(C_i D_j)=C_i P(D_j)$$ for any $i$ and $j$.

Then we can split any general $B$ and any appropriate general $A$ into component blades as $$A=\sum_i A^i C_i$$ and $$B=\sum_i B^i D_i\ .$$ We then have $$\begin{aligned} P(AB)&=P\left(\left(\sum_i A^i C_i\right)\left(\sum_j B^j D_j\right)\right)\\ &=\sum_i \sum_j A^i B^j P(C_i D_j)\\ &=\sum_i \sum_j A^i B^j C_i P(D_j)\\ &=\left(\sum_i A^i C_i\right) P\left(\sum_j B^j D_j\right)\\ &=A P(B)\ . \end{aligned}$$ This very general result doesn't depend on $A$ and $B$ being blades, or even being homogeneous.