Geometrical algebra's: simple equation

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Discussion Overview

The discussion revolves around the properties of geometrical algebra, specifically focusing on the equation k F = 0, where k is a vector and F is a bivector. Participants explore whether any plane F that is perpendicular to a vector k can be expressed as F = k ∧ A for some vector A, within the context of 4-dimensional spacetime.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states that if k F = 0, then k is perpendicular to the plane F, leading to the assertion that F can be expressed as k wedged with another vector A.
  • Another participant argues that k lies in the subspace F, suggesting that F can be factored as F = k ∧ a, where a is some vector, indicating that F is not only a bivector but also a blade.
  • The same participant notes that in 4 dimensions, a bivector does not necessarily have to represent a plane.
  • A participant shares a resource for learning geometrical algebra, highlighting its usefulness for those familiar with traditional vector algebra/calculus.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between the vector k and the bivector F, with some asserting that F can be expressed in terms of k while others challenge this notion. The discussion remains unresolved regarding the generality of the theorem proposed.

Contextual Notes

The discussion includes assumptions about the dimensionality of the space and the nature of bivectors, which may affect the conclusions drawn. There are also references to specific mathematical properties that may not be universally applicable.

CompuChip
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Hello,

In a geometrical algebra exercise I have to solve the equation
k F = 0,​
where k is a vector and F is a bivector (plane).
From this it follows that also [itex]\mathbf{k} \wedge F = 0[/itex], which basically says that the vector k is perpendicular to the plane F.
We are supposed to show (among other things) that [itex]F = \mathbf{k} \wedge \mathbf{A}[/itex] for some vector A.
My question is: can you prove that this is always possible (or give a counterexample)? So
(Theorem) - any plane F perpendicular to a vector k can be written as k wedged with another vector A
(I'm working in 4 spacetime dimensions. Probably the statement can be (dis)proven by choosing a basis - and then separating the cases where F and k are timelike, spacelike, etc - but apparently the power of geometrical algebras is that one can do without...).
 
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CompuChip said:
From this it follows that also [itex]\mathbf{k} \wedge F = 0[/itex], which basically says that the vector k is perpendicular to the plane F.

On the contrary, it means that k lies in the subspace F, and this implies that F can be factored as [tex]F=k\wedge a[/tex], where
[itex]a[/itex] is some vector. i.e. F is not only a bivector but is also a blade. Then
[itex]k\wedge F = k\wedge k\wedge a = 0[/itex]. (Recall that, in 4 dimensions, a bivector does not in general have to be a plane).
 
OK, that was really stupid.
Thanks a lot.
 
I stumbled on the following the other day:

http://www.science.uva.nl/ga/tutorials/

It's a interactive GA tutorial/presentation for a game programmers conference that provides a really good intro and has a lot of examples that I found helpful to get an intuitive feel for all the various product operations and object types.

Even if you weren't trying to learn GA, if you have done any traditonal vector algebra/calculus, IMO its worthwhile to download this just to just to see the animation of how the old cross product varies with changes to the vectors.

You have to download the GAViewer program (graphical vector calculator) to run the presentation. Once you do that you can use it for other calculation examples. See:

http://www.geometricalgebra.net/downloads/gaviewerexercises.pdf

for some examples of how to use this as a standalone tool (note that the book the drills are from use a different notation for dot product (with a slightly different meaning and uses an oriented L symbol dependent on the grades of the blades).
 
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