Geometrical algebra's: simple equation

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SUMMARY

The discussion centers on the geometrical algebra equation k F = 0, where k is a vector and F is a bivector, indicating that vector k is perpendicular to plane F. The theorem proposed states that any plane F perpendicular to vector k can be expressed as F = k ∧ A for some vector A. The conversation highlights the importance of understanding the relationship between bivectors and blades in 4-dimensional spacetime, emphasizing that F can be factored as F = k ∧ a. Resources such as the GAViewer program and an interactive tutorial are recommended for further exploration.

PREREQUISITES
  • Understanding of geometrical algebra concepts
  • Familiarity with bivectors and blades in 4-dimensional spacetime
  • Knowledge of vector operations, particularly the wedge product
  • Experience with graphical vector calculators like GAViewer
NEXT STEPS
  • Explore the properties of bivectors and their geometric interpretations
  • Learn how to use GAViewer for practical geometrical algebra calculations
  • Study the implications of the wedge product in higher-dimensional spaces
  • Investigate the relationship between traditional vector algebra and geometrical algebra
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Mathematicians, physicists, game developers, and anyone interested in advanced geometrical algebra concepts and their applications in 4-dimensional spacetime.

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 \mathbf{k} \wedge F = 0, which basically says that the vector k is perpendicular to the plane F.
We are supposed to show (among other things) that F = \mathbf{k} \wedge \mathbf{A} 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 \mathbf{k} \wedge F = 0, 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 F=k\wedge a, where
a is some vector. i.e. F is not only a bivector but is also a blade. Then
k\wedge F = k\wedge k\wedge a = 0. (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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