Bivector = product of two null vectors

Rotation generated by a bivector that is the product of two null vectors. In R(2,2) there are hyperbolic isometries, ae13, circular rotations, ae12, isocline hyperbolic and circular isometries. If a bivector is the product of a null vector such as e1+e3 and a vector, it generates parabolic isometries (rotations). The question involves a bivector that is the product of two orthogonal null vectors such as B=a(e1+e3)(e2+e4). This seems to produce a translation along one of the null vectors. Is this correct? I can’t find literature on this.

I would look at an orthogonal non-null vector basis, and then apply those vectors as versors (so you know what the fundamental reflections are), and then of the product of two of such to find out what the even versors are. You may find some guide from my paper on R(3,3), which is directly associated with the projective transformations in 3-space. Perhaps the 2D versions ofmy versors are precisely what you need (and translations are among them).

By the way. let this be a lesson to all, never to use LaTeX in your paper titles…

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You are incorrect, bivectors in general are not a geometric product. A lot of famous people who write geometric algebra papers are not smart enough to understand this. Every time they try to say that a bivector is a geometric product of two vectors, it is easy for me to destroy their argument with counter examples. I have done this many times:

Proof using the null vectors ni and no using Grassmann.jl:

julia> using Grassmann

julia> @basis S"+-"
(⟨+-⟩, v, v₁, v₂, v₁₂)

julia> ni = v1+v2
1v₁ + 1v₂

julia> no = (v2-v1)/2
-0.5v₁ + 0.5v₂

julia> ni*no
-1.0 + 1.0v₁₂

julia> ni∧no
1.0v₁₂

As you can clearly see, the geometric product of two null vectors is not a bivector, only the Grassmann exterior product of two null vectors is a bivector.

In fact, all bivectors are by definition Grassmann products and not Clifford products. Funnily enough, there have been several prominent members in this community who disagreed with me about this in the past, and I have explained to them several times in the past few years that their misunderstanding leads to logical inconsistencies and incorrectness.

Be careful, you might encounter Eric Wieser or Sudgy when you discuss geometric algebra, both of these are people who vehemently disagreed with me on this and demonstrated their lack of understanding of Grassmann/Clifford geometric algebra. They just could not believe that I proved them wrong in their assumption that a bivectors are not Clifford geometric products, they are Grassmann exterior products.

Thanks Chakravala for pointing this out. I had always though that the product of two orthogonal vectors was a bivector. I’ll have to go back and check the bivector.net software for R(2,2) with the term a(e1+e3)(e2+e4);; when I checked this before it was indeed a bivector. Oh dear,I 'm losing it. And here I thought that I had discovered a new type of rotation! Apologies.

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@wfmcgee is right because the null vectors are actually orthogonal. But in you case aren’t.

My post is intended to be a reminder that the geometric product of two vectors is not always an exterior product, even if there are special cases, in general it’s not.

Yeah. And I’m just pointing out that your example is not a counter example for the thing in question, because of that you ended up confusing @wfmcgee . You also should 've been more explicit and say that it is not in general true but is true for orthogonal vectors.

Thanks for your concern. But my reply was tongue-in-cheek. I’m confused, but not about the product of orthogonal vectors…

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I think the confusion can arise because orthogonality is not exactly the same as linear independence. These two concepts do not entirely overlap.

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