# I Geometric Algebra: Rejection of one blade in another

#### Hiero

Let the (multi-vector valued) “inner product” between a j-vector U and a k-vector B be defined as the (k-j) grade part of the geometric product UB, (a.k.a. “left contraction”) that is,
$$U\cdot B := <UB>_{k-j}$$
(0 when j > k) as is done in Alan Macdonald’s book “Linear and Geometric Algebra.”

In chapter 7, Alan defines the projection of one blade U in another blade B and then gives a theorem stating it is equal to
$$P_B(U)=(U\cdot B)B^{-1}$$

A natural way to define the rejection of U in B might be as the identity minus the projection, that is,
$$R_B(U):=U-P_B(U)=U-(U\cdot B)B^{-1}$$

This agrees with the previously given (in the book) formula for the rejection of a vector u in a blade B,
$R_B(u)=u-(u\cdot B)B^{-1}=uBB^{-1}-(u\cdot B)B^{-1}=(uB-u\cdot B)B^{-1}=(u∧B)B^{-1}$

This definition of rejection also seems to have the property that it always has zero projection onto B (as we would expect of a rejection) as we can check:

$P_B(R_B(U))=((U-(U\cdot B)B^{-1})\cdot B)B^{-1}=<(U-<UB>_{k-j}B^{-1})B>_{k-j}B^{-1}=<UB-<UB>_{k-j}>_{k-j}B^{-1}=(<UB>_{k-j}-<<UB>_{k-j}>_{k-j})B^{-1}=(<UB>_{k-j}-<UB>_{k-j})B^{-1}=0$

My question is about a comment made by the author in one of the problems. (I can add a picture of the page with the problem if someone confirms it’s allowed.)

He said that this definition of rejection fails in dimensions higher than three. He did not explicitly say the following, but he seems to imply it fails because the projection of the rejection is not always zero... but the above steps seem to at least outline a proof that it is always zero.

So why did he say that this definition is not suitable in higher dimensions??? Have I made a mistake in the above manipulations?

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#### fresh_42

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Yes, you can add 1 or 2 pages without problems, but I wonder if we have members who are fit in the calculus of blades.

#### FactChecker

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2018 Award
I see a statement to that effect in problem 7.1.5c on page 124. Did you look at the counterexample he mentions there?:
Use $A = (e_1+e_2)\wedge(e_3+e_4), B=e_1\wedge e_3$ to disprove $rej_B(A) = (A\wedge B)/B$.

CORRECTION EDIT:
The problem statement is that the A and B above are counterexamples to disprove that $rej_B(A) = A - P_B(A)$ for dimensions greater than 3.

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#### Hiero

I see a statement to that effect in problem 7.1.5c on page 124. Did you look at the counterexample he mentions there?:
Use $A = (e_1+e_2)\wedge(e_3+e_4), B=e_1\wedge e_3$ to disprove $rej_B(A) = (A\wedge B)/B$.
Yes that’s exactly the problem! I don’t have the book with me, I’m at work, but I believe in part b he showed that $rej_B(A) = (A\wedge B)/B$ is an inadequate definition. Then he introduced the definition I agree with, namely $rej_B(A) = A - proj_B(A) = A - (A\cdot B)/B$ but it seemed as though he said this definition fails?

I didn’t mention the “counterexample” because it has zero projection according to this definition. I would type out the details if I had time but you can check yourself.

I suppose it was just a misinterpretation? I’ll reread the wording later.

Thanks!

#### Hiero

Yes, you can add 1 or 2 pages without problems, but I wonder if we have members who are fit in the calculus of blades.
I’ve attached an image of the problem because there seems to be no other interpretation... But I feel like it’s an error? My proof in the OP seems valid, and I even saw this definition on the geometric algebra Wikipedia page today.

#### FactChecker

Gold Member
2018 Award
I stand corrected about the statement of the problem. I corrected my post above.

#### Hiero

I’m in (a boring) math class now so I have time to do the example for anyone curious.

$$A=(e_1+e_2)\wedge (e_3+e_4)= e_1e_3 + e_1e_4 + e_2e_3+e_2e_4$$
$$B=e_1\wedge e_3=e_1e_3$$
$$B^{-1}=e_3e_1$$

The projection of A in B is,

$$P_B(A)=(A\cdot B)B^{-1}=<AB>_0B^{-1}=<(e_1e_3 + e_1e_4 + e_2e_3+e_2e_4)e_1e_3>_0e_3e_1=<-1+ e_3e_4 + e_1e_2-e_1e_2e_3e_4>_0e_3e_1=(-1)e_3e_1=e_1e_3$$

Then the rejection of A onto B is,
$$rej_B(A)=A-P_B(A)=e_1e_4+e_2e_3+e_2e_4$$

Then the projection of the rejection is zero as expected,

$$P_B(rej_B(A))=<rej_B(A)B>_0B^{-1}=<(e_1e_4 + e_2e_3+e_2e_4)e_1e_3>_0e_3e_1=<e_3e_4 + e_1e_2-e_1e_2e_3e_4>_0e_3e_1=(0)e_3e_1=0$$

I’m going to just chalk it up to a small error in the book.

#### fresh_42

Mentor
2018 Award
Shouldn't it be $-B$ instead of $B^{-1}$?

#### Hiero

Shouldn't it be $-B$ instead of $B^{-1}$?
Not sure exactly which step you mean, but in this particular case (where $B=e_1e_3$) we have $B^{-1}=-B$

I again don’t have the book with me so I’ll paraphrase from memory. The inverse of a j-blade B (which can be expressed as an outer product of j vectors (each in $\mathbb R ^n$) $B=x_1\wedge ... \wedge x_j$) is defined by
$$B^{-1}:=\frac{B^†}{|B|^2}$$
Where $B^†$ denotes the “reverse” of the blade ($B^†=x_j\wedge ... \wedge x_1=(-1)^{j(j-1)/2}B$).
(Also the norm is a scalar and so it commutes and so I can write it underneath with no ambiguity about if it’s division on the left or right.)
(I think this applies to j-vectors also, not just j-blades, but not to multi-vectors with mixed grades. I’ll have to reread it later. EDIT: nevermind, it didn’t work for e_1e_2+e_3e_4... I think maybe this definition of inverse works for multi-vectors when they can be expressed purely as a geometric of vectors, but that wasn’t in the book, just a suspicion.)

The reverse is defined for multi-vectors by reversing each term. The norm is also defined for multi-vectors; if a multi-vector M is expressed in a $\mathbb G^n$-basis $\{e_J\}$ as $M = \Sigma_J(M_Je_J)$ then we define $|M|^2:= \Sigma_J(M_J)^2$. Then the book gives a theorem showing the norm is well defined (by giving a coordinate-free way of expressing it) in a theorem which says $|M|^2=<MM^†>_0$ which is the scalar part of the product with the reverse.

I don’t think we can naively extend the definition of inverse from blades to multi-vectors even though reverse and norm are both defined for multivectors. For example, $M=1+e_i$ has no inverse; I remember proving this in a problem, (we basically apply the assumed inverse on the left of $(1+e_i)(1-e_i)=0$ to get the contradiction $1=e_i$) and so the inverse of M is clearly not $M^†/|M|^2=(1+e_i)/2$

So inverses are more subtle for multi-vectors; I’ll have to look when I get home to see if he mentions anything about general inverses in the book.

Nonetheless the problem of multi-vector inverses is immaterial for projections because any subspace B (that we want to project into) is represented with a pure blade (which can be expressed as the outer product of any basis of the subspace that B represents).

Sorry for writing so much, but recalling all of this helps me learn better.

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#### fresh_42

Mentor
2018 Award
I was only wondering if $B=e_1 \wedge e_3 =e_1e_3$ is the anti-commutative multiplication, so $e_3e_1=-e_1e_3=-B$. The notation $B^{-1}$ is simply a bit confusing if we deal with two operations.

#### Hiero

I was only wondering if $B=e_1 \wedge e_3 =e_1e_3$ is the anti-commutative multiplication, so $e_3e_1=-e_1e_3=-B$. The notation $B^{-1}$ is simply a bit confusing if we deal with two operations.
Ohh, I think I understand the confusion now.

$B^{-1}$ the multiplicative inverse not the additive inverse. Multiplication meaning the geometric product not the wedge product.

And yes this wedge product anti-commutes.

Does that clarify?

(For example, if $B=2e_1$, then $-B=-2e_1$, and $B^{-1}=e_1/2$)

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"Geometric Algebra: Rejection of one blade in another"

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