Talk:Octonion

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Please update this rating as the article progresses, or if the rating is inaccurate. Click to show/hide comments. Please add to or update the comments to suggest improvements to the article. Close to B: more history, and some primary references would really lift this. Geometry guy 21:14, 31 May 2007 (UTC)

Could you describe what are they used for ? --Taw

I have a couple of problems with this:

Octonions form the largest normed division algebra. These are algebras in which it is possible to define division and the norm of a number, in the algebra of complex numbers the norm of a number is equal to that number multiplied by its complex conjugate.The norm plays a vital role in the study of abstract algebras.

However, a norm cannot be defined for the elements of any algebra larger than the octonions, a notable example is the case of the sedenions these 16 dimensional numbers have little or no algebraic structure and have very little value.

The second property mentioned above is that of division being defined in an algebra, the octonions form the largest algebra in which division is defined, again take the case of the sedenions, it is possible to choose two non-zero sedenions, S and T such that S*T=0.

This is not possible in any smaller algebra, such as the real numbers, complex numbers, quaternions or octonions.

These facts have brought octonions to the forefront of research in modern physics at present, in superstring theory the existence of norms and division in algebras of dimensions 1, 2, 4 and 8 endows spaces of dimensions 3, 4, 6 and 10 with exceptional properties, the 10 dimensional case being very important to string theory.

First, our article on normed division algebras claims that the quaternions are the largest. This is probably due to the fact that our division algebras have to be associative.

Second, there are smaller algebras where division is not defined; take for instance the dual numbers.

These things need to be cleaned up. We should probably start with cleanly defining the norm of an octonion. AxelBoldt 05:03 Jan 4, 2003 (UTC)

Also the claim that there are five exceptional Lie groups, as opposed to five types or classes of exceptional Lie group. There are five exceptional Dynkin diagrams, but that's not the same thing. (I took out five; that's almost certainly not the best solution. Let who can improve it.) Septentrionalis 03:26, 26 September 2005 (UTC)

If a real number represents the mass of an object, i, j, and k represent the position, and l represents time, than you'd end up with one number for the mass, position, velocity, and time of an object. If you square the position or velocity, you'd always get a negetive real number. Using that, than if mass is positive, than energy (mass times velocy squared) is negitive, and vice versa. — Daniel 22:15, 20 November 2005 (UTC)

[edit] External links

The external link to the fractal site 404s mavhc

[edit] Extension

Shouldn't this link to the extension of Octonions, the 16-nions (don't know the proper name/spelling) just as the Quaternions linked to the Octonions?

I do not know much algebra, as I have always been more of an analysis person, but I think the 16-nions have a lot of special properties. For instance, they do not yield a cross product in 15-space, as the quaternions (resp octonions) do in 3 (resp 7) space. I think this fact was proved by Frobenius, although I have not found a proof of it online. --Jpawloski 14:50, 16 February 2006 (UTC)

It does link to the sedenions (in the see-also section). --Zundark 15:45, 16 February 2006 (UTC)

Missed that. Thanks. --Jpawloski 16:02, 16 February 2006 (UTC)

[edit] Question

How to build that multiply table? --80.178.62.64 12:58, 19 August 2007 (UTC)

• Hello - One way is through the Cayley-Dickson construction. Another would be to begin with quaternions to basis {1,i,j,k} and propose another basis l with l² = − 1 and demanding that the resulting algebra is a vector space over the reals equipped with a multiplicative norm. Jens Koeplinger 12:45, 20 August 2007 (UTC)

It looks backward to me. For example, . What am I doing wrong? Phoenix1304 (talk) 14:58, 6 April 2008 (UTC)

You need to take non-associativity into account: If you look closely at the brackets, you see that i(kl) = jl, however, i(kl) = − (ik)l = − ( − j)l = jl. So, in your equation, the step from the first expression to the second expression misses a minus sign.

But - in case you're interested - not every 3-tuple is non-associative. In general, any three octonion basis elements {a,b,c} are associative if they satisfy one (ore more) of these properties:

1. One (or more) of the {a,b,c} is 1.
2. Two of the {a,b,c} multiply to .
3. The set {a,b,c} is multiplicatively closed (with a factor ), i.e. .

If any of these three conditions are satisfied, then a triplet of octonion basis elements is associative: (ab)c = a(bc). If none are satisfied, as it was in your example, then the triplet is anti-associative, i.e. (ab)c = − a(bc).

Does this answer your question? Thanks, Koeplinger (talk) 16:23, 6 April 2008 (UTC)

I guess at the root of my problem is I don't fully understand where the non-associativity originates. I mean, with quaternions the non-commutivity came about (at least in my education) as an attempt to eliminate the bcij and bcji terms when normalizing. Phoenix1304 (talk) 16:51, 6 April 2008 (UTC)

It goes to the requirement of adding an imaginary basis element to your existing set of basis elements, while maintaining a multiplicative norm, i.e. | | a | | | | b | | = | | ab | | for every element of your number systems. Together with the requirement that every non-zero number has a non-zero norm (to arrive at a "normed division algebra"), you must lose commutativity for quaternions, and you must lose associativity for octonions. Let's see:

For quaternions, beginning with the complex basis {1,i}, you want to add another imaginary basis j with j² = − 1 such that j is different from i, any non-zero element of your space has non-zero norm, and the norm is multiplicative. If you take two numbers A: = i + j and B: = i − j, then their product is AB = − 1 − ij + ji + 1. As you already said, if ij were to be ji, then the norm of the product would be zero, which violates our requirements. With ji: = − ij, however, we are not violating our requirements. After checking all the other basis elements, one can see that the quaternions satisfy are OK.

For octonions, we begin at the quaternion basis {1,i,j,k} and require an additional imaginary basis, l, with l² = − 1. You can show, just as for the quaternions, that l must also be anti-commutative with all your quaternion basis elements. For associativity, we now need three numbers, for example A: = i + j, B: = i − l, and C: = j − l and their product is:

(AB)C = ((i + j)(i − l))(j − l)

= ( − 1 − il + ji − jl)(j − l)

= − j − l − (il)j + (il)l + (ji)j − (ji)l − (jl)j + (jl)l

Assuming that associativity would hold, we would have:

ABC = − j − l + kl − i + i + kl − l − j = − 2j − 2l + 2kl

The norm of this would be , which is wrong. Instead, the two terms − (il)j and − (ji)l are anti-associative and cancel each other out; the correct result is:

(AB)C = − j − l + kl − i + i − kl − l − j = − 2j − 2l

It has norm as it should be. Hope this helps! (And I hope there are no more errors in the calculations!) Thanks, Jens Koeplinger (talk) 21:00, 6 April 2008 (UTC)

(removed the "PS" that was here, because it was wrong; the above example is correct ... unless you can prove me differently? I would have loved to have an example where the product evaluates to zero under the assumption of associativity) -- sorry for that. Thanks, Jens Koeplinger (talk) 04:12, 7 April 2008 (UTC)

Thanks a lot! That really helps me get my head around the article. Phoenix1304 (talk) 14:34, 7 April 2008 (UTC)

Koeplinger's argument can be considerably simplified. Assume we have a real associative algebra with elements i,j,k = ij,l which all anticommute. We make no assumptions as to the squares. Then

kl = ijl = − ilj = + lij = lk = − kl

so kl = 0. This algebra has zero divisors and so the norm cannot be multiplicative. Worse, if k² = − 1 then l = 0, or if l² = − 1 then k = 0.

In other words, if we assume that l anticommutes with both i and j then it must commute with the product ij. One then runs into the same problems as before. The only way to make l anticommute with k and still get a nonzero product is to drop the assumption of associativity. -- Fropuff (talk) 19:29, 7 April 2008 (UTC)

Thanks for that example! It doesn't require a particular definition of norm, either, which is desirable as well. Koeplinger (talk) 22:20, 7 April 2008 (UTC)