Am I Hallucinating? (confusion in geometric algebra)

[Hornbein]

I'm learning geometric algebra. There is a very simple statement which I think is wrong. But it must be right, because all the experts say so. Arrg!

The only properties used are

1a = a1 = a

aa = 1

if b<>a then ab = -baTheir claim is that abcdabcd = -1.

Let's see:

aa = 1

abab = -abba = -aa = -1

abcabc = -abacbc = ababcc = abab = -1

abcdabcd = -abcadbcd = abcabdcd = -abcabcdd = -abcabc = --1 = 1

But the experts say abcdabcd = -1. See for yourself at
http://geocalc.clas.asu.edu/pdf-preAdobe8/ZBW_I_QM.pdf, eqn 10 and 11

This can't be. To reduce the 4D case to the 3D case there have to be an odd number of swaps. They are of opposite signs and nonzero. They can't possibly be equal.

Elsewhere they say that abcabc and abab equal -1.

?

 

[mnb96]

I think you are ignoring Eqs. (7)-(8) which basically tell you that your 4D space has signature (+,-,-,-). In other words, this means that you are working on ℝ^1,3 and not on ℝ⁴, thus, using your notation, aa=1 and bb=cc=dd=-1.

Another thing worth noticing is that those vectors a,b,c,d are not arbitrary vectors: they are the basis vectors of your 4D space, and b,c,d by definition square to -1. In the context of that paper abcd represents the pseudoscalar of the space. Whether the square of the pseudoscalar is positive or negative (or zero) depends on the metric of the space.

It shall also be pointed out that ab=-ba is not true in general, but only when a,b are orthogonal vectors, which is expressed in Eq. (8).

 

[Hornbein]

I think you are ignoring Eqs. (7)-(8) which basically tell you that your 4D space has signature (+,-,-,-). In other words, this means that you are working on ℝ^1,3 and not on ℝ⁴, thus, using your notation, aa=1 and bb=cc=dd=-1.

Another thing worth noticing is that those vectors a,b,c,d are not arbitrary vectors: they are the basis vectors of your 4D space, and b,c,d by definition square to -1. In the context of that paper abcd represents the pseudoscalar of the space. Whether the square of the pseudoscalar is positive or negative (or zero) depends on the metric of the space.

It shall also be pointed out that ab=-ba is not true in general, but only when a,b are orthogonal vectors, which is expressed in Eq. (8).

Aha, that's it. I had the wrong space.