Learning Geometric/Clifford Algebra

[PatrickPowers]

In the 19th century William Clifford came up with an algebra that generalized to geometry in higher dimensions. Having skimmed through numerous sources, my favorite is a ninety page tutorial by Eric Chisholm (2012).

https://archive.org/details/arxiv-1205.5935

It slices, it dices...

[PatrickPowers]

Another introduction to geometric algebra is at

http://faculty.luther.edu/~macdonal/GA&GC.pdf

It has versions of vector calculus with curl, divergence, Stokes' Theorem, and so forth, all defined in any number of dimensions.

[ICN5D]

Excellent, thanks for sharing this. I strongly feel that somewhere in geometric algebra will exist a method to derive the equation for a 4D Mantis, and related toroidal objects. Perhaps even more can come from it. Something like this literature has been outside my accessibility, since I have no formal math background, other than what I learned how to do here.

[PatrickPowers]

One thing I just learned: don't trust the GA teaching materials!

They keep saying a bivector is a signed area and show you a little diagram. Not true! A bivector is any 2D thing you want it to be. It can be a subspace, that is, it can be infinite in extent. It can be a small circular area centered at a single point if you want. It can be square, pentagonal, any shape you like.

Not only that, the 2Ds don't have to be Euclidian. They can be polar coordinates, cylindrical, anything you can cook up.

All that other meaning comes from context and the particular application.

Another thing that got me was: The sum of vectors is always a vector. Is the sum of bivectors a bivector?

Yes. But it's misleading. A blade is an elementary element. Reasonable basis vectors are (I think) always blades. A vector may always be expressed as the sum of blades. There can always be a change of basis on a vector so that becomes a blade. So the sum of vectors is always a blade, even it it doesn't look like one.

Bivectors can always be expressed as the sum of blades. But if N>3 there can't always be a change of basis on a bivector so that becomes a blade. Consider v = ae₁e₂ + be₃e₄. There is no way to write that as ce'₁e'₂ in some new e' basis. I dunno how to prove it, but Denker claims it and I trust him.

[quickfur]

[...]
Bivectors can always be expressed as the sum of blades. But if N>3 there can't always be a change of basis on a bivector so that becomes a blade. Consider v = ae₁e₂ + be₃e₄. There is no way to write that as ce'₁e'₂ in some new e' basis. I dunno how to prove it, but Denker claims it and I trust him.

It makes sense to me, because a bivector can represent a rotation, and in 4D, it's possible to simultaneously rotate in two orthogonal planes with two independent rates of rotation, and there is no way to rewrite that as a single rotation (which it would be, if it were possible to reduce that sum into a single basis bivector in some basis). The sum is inherently irreducible.

[PatrickPowers]

[...]
Bivectors can always be expressed as the sum of blades. But if N>3 there can't always be a change of basis on a bivector so that becomes a blade. Consider v = ae₁e₂ + be₃e₄. There is no way to write that as ce'₁e'₂ in some new e' basis. I dunno how to prove it, but Denker claims it and I trust him.

It makes sense to me, because a bivector can represent a rotation, and in 4D, it's possible to simultaneously rotate in two orthogonal planes with two independent rates of rotation, and there is no way to rewrite that as a single rotation (which it would be, if it were possible to reduce that sum into a single basis bivector in some basis). The sum is inherently irreducible.

Yes it does seem obvious.

[mr_e_man]

Bivectors can always be expressed as the sum of blades. But if N>3 there can't always be a change of basis on a bivector so that becomes a blade. Consider v = ae₁e₂ + be₃e₄. There is no way to write that as ce'₁e'₂ in some new e' basis. I dunno how to prove it, but Denker claims it and I trust him.

Proof:

Any bivector blade B can be represented as a wedge product B=x^y for some vectors x and y. It follows that

B^B = (x^y)^(x^y) = 0

But, with v=e₁e₂+e₃e₄,

v^v = (e₁e₂)^(e₁e₂) + (e₁e₂)^(e₃e₄) + (e₃e₄)^(e₁e₂) + (e₃e₄)^(e₃e₄)

= 0 + e₁e₂e₃e₄ + e₃e₄e₁e₂ + 0

= 2 e₁e₂e₃e₄ =/= 0

so it cannot be a blade.

By the way, the definition of a blade does not depend on any basis. Any scalar or vector is a blade, by definition. Higher-grade blades are wedge products of vectors.