Octonions do not form a Clifford Algebra...?

Higher-dimensional geometry (previously "Polyshapes").

Octonions do not form a Clifford Algebra...?

Postby Paul » Thu Dec 02, 2004 4:52 am

Hello Pat and all,

I'm getting the impression that the Octonions do not form a Clifford Algebra... is this correct?

Recall in my post below how I formed both the complex product and the quaternion product from the geometric product... in terms of multiplicative operation tables, all that can be put this way:

Note: I'm not sure I'm using the Cl<sub>p,q</sub> notation correctly... could someone check if I'm using it correctly?

Image

The quaternion product also appears to be a geometric product:

Image
Image

However, I was having allot of trouble with the octonion product...

Image
Image

I just couldn't make it work... and then I consulted Tony Smith's, John Baez's webpages (which is where I got some of the diagrams above), and I believe I've found that I'm not just missing it... but, the Octonions are not a Clifford Algebra. Do the Octonions form what is called an 'exceptional' Lie Algebra?

It appears the octonion product can be put in terms of the geometric product, but only with some rather involved manipulations... see Lounesto's <u>Clifford Algebras and Spinors</u>, Chapter 23.

However, if I'm following Tony Smith correctly, it seems perhaps the easiest way to approach the octonion product through Clifford Algebra is through the formula below, which also applies to quaternions (and complex numbers if the wedge product is substituted for the cross product):

Image

where we want to substitute the 7D cross product to form the octonion product. If I understand correctly, Tony says we're supposed to do this in Cl<sub>0,8</sub>... However, unless I'm misunderstanding the notation, that would seem to imply that we'd be dealing with 8 basis vectors squaring to -1. Don't we only want 7 basis vectors squaring to -1...?

This is where much of my confusion about the notation lies... p+q are supposed to equal n, the dimension of the space which the Clifford Algebra is embedded in... we do want a 8D space here, but we want a space which is a direct sum of a scalar real space and a 7D imaginary space... correct?

Can someone clarify all this? And, show how we form the octonion product from the 7D cross product as indicated above? Which 7D cross product is Tony referring to anyway?
Paul
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Postby pat » Thu Dec 02, 2004 5:11 am

Unfortunately, the octonion multiplication isn't associative whilst all Clifford algebras are associative. So, the octonions won't be a Clifford algebra (or a subalgebra of a Clifford algebra).

I believe all of your notations in your figures was correct.

I don't know if the octonions qualify as an exceptional Lie algebra. I don't know enough about Lie algebras yet. They definitely count as a 'real, normed division algebra'.

And, I'd be confused about Tony saying Cl<sub>0,8</sub>, too. Hmmm. Of course, I'm actually surpised it's not Cl<sup>+</sup><sub>16,0</sub>. But, I have a hard time getting any foot-holds when reading Tony's pages (Baez on the other hand... wonderful writer). I want to understand them.... but I'm not there yet.

Particularly, he claims on one page that there are only real, normed division algebras in dimensions 2<sup>0</sup>, 2<sup>1</sup>, 2<sup>2</sup>, and 2<sup>3</sup> precisely because 0 is prime (run with it), 1 is prime, 2 is prime, 3 is prime, and 4 is composite. I have seen proofs that those are the only real, normed division algebras. But, I don't see how four being composite was the real kicker that made the sequence stop there.
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Postby Paul » Thu Dec 02, 2004 6:38 am

Hello Pat,

Thanks for your response.

I kinda guessing that Tony is referring to the cross product between vectors and simple bivectors which exists in 7D. This is the cross product that we're supposed to use in the above formula... yes?

I believe it's the 7D cross product Lounesto is talking about here... Cross-product in n-dimensions,... which Lounesto also talks about in Chapter 7 of Clifford Algebras and Spinors.

It kinda sounds as though one essentially has to wedge with that nasty-looking bunch of 7D trivectors... e<sub>124</sub>+e<sub>235</sub>+e<sub>346</sub>+e<sub>457</sub>+e<sub>561</sub>+e<sub>672</sub>+e<sub>713</sub>... which Lounesto indicates

It just seems no matter how you attempt to express the octonion product in Clifford terms, things just get nasty-looking pretty quick...

This also kinda scares me when Lounesto says

In R^3 the direction of a x b is unique, up to orientation having two
possibilities, but in R^7 the direction of a x b depends on vectors
defining the cross product; namely (expressed with the contraction "_|"):

a x b = (a ^ b) _| e123 in R^3 when e123 = e1^e2^e3 but

a x b = (a ^ b) _| (e124+e235+e346+e457+e561+e672+e713) in R^7.

Also in R^7 there are other planes than the linear span of a and b
giving the same direction as a x b.


It just kinda sounds like this 7D cross product can't communicate as much information about the nature of it's product as the 3D cross product can...

I'm wondering if there might be some relation here between the loss of commutivity in 3D, and then the additional loss of associativity evident in 7D...? and this resulting loss of information concerning the nature of these two cross products? Do we also lose information concerning the associated wedge products? I wonder if this somewhat accounts for why much order, of various kinds, just seems to breakdown as we progress up through the dimensions?

I'm hoping you'll tell me I'm not understanding correctly, or I've made some silly mistake...
Paul
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