Geometric Algebra: Rejection of one blade in another

In summary, Alan Macdonald defines the inner product between a j-vector U and a k-vector B as the (k-j) grade part of the geometric product UB. He shows that the rejection of U in B is always zero, but fails in dimensions greater than three. He suggests a more adequate definition of rejection which is equal to A - proj_B(A) = A - (A\cdot B)/B.
  • #1
Hiero
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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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  • #2
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.
 
  • #3
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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  • #4
FactChecker said:
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!
 
  • #5
7B3B24D4-0B90-49DF-BCED-C2E1B6DF73ED.jpeg

fresh_42 said:
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.
 
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  • #6
I stand corrected about the statement of the problem. I corrected my post above.
 
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  • #7
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.
 
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  • #8
Shouldn't it be ##-B## instead of ##B^{-1}##?
 
  • #9
fresh_42 said:
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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  • #10
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.
 
  • #11
fresh_42 said:
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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  • #12
old forum post but i'm going to add a reply in case someone (like me) looks here while reading an old print of the book.

in the newer print, the question is rephrased to specify that the rejection is a poor definition because the result you end up with is not a blade, so it doesn't represent a subspace and thus is not very useful as a rejection

edit: in my opinion i think that either: 1) a more suitable definition of the rejection generalizable to any dimension should have been presented (although i don't know if this is possible given that higher dimensions do not have a unique direction orthogonal to a plane), or, 2) that this definition of the rejection should be THE definition regardless of whether or not it represents a subspace by being a blade, just given that it satisfies that the projection of it equals 0 and the projection plus the rejection definition given does equal the original multivector, just for completeness (note: i'm an (amateur) mathematician, not a physicist, but this is the only place discussing the question so it's where i'll give my comments)
 
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Related to Geometric Algebra: Rejection of one blade in another

1. What is geometric algebra?

Geometric algebra is a mathematical framework that extends the concepts of traditional algebra to include geometric operations such as rotations, translations, and reflections. It uses a combination of vectors, scalars, and multivectors to represent geometric objects and operations.

2. What is the "rejection of one blade in another" in geometric algebra?

The rejection of one blade in another is a geometric algebra operation that involves projecting one multivector onto another multivector in order to isolate a specific component or "blade". This operation is useful for breaking down complex multivectors into simpler components for easier manipulation and analysis.

3. How is the rejection of one blade in another performed?

The rejection of one blade in another is performed by taking the inner product of the two multivectors and then dividing the result by the squared magnitude of the second multivector. This results in a new multivector that represents the rejected component of the original multivector.

4. What are some applications of the rejection of one blade in another in geometric algebra?

The rejection of one blade in another has many applications in fields such as computer graphics, robotics, and physics. It can be used for 3D object manipulation, motion planning, and analyzing physical systems with multiple degrees of freedom.

5. Are there any limitations to using the rejection of one blade in another in geometric algebra?

While the rejection of one blade in another is a powerful tool in geometric algebra, it does have some limitations. It can only be performed on multivectors with compatible dimensions, and it may not always produce a meaningful result if the two multivectors are not geometrically related.

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