Re: Imaginary numbers, complex plane and beyond.

From: Ryan Reich (ryanr_at_uchicago.edu)
Date: 06/10/04 

Date: Thu, 10 Jun 2004 09:27:47 -0500

On Wed, 09 Jun 2004 23:30:12 -0700, Clifford J. Nelson wrote:

> On 2004-06-09 22:01:22 -0700, Clifford J. Nelson <cnelson9@adelphia.net>
> said:
>
>> On 2004-06-09 19:59:21 -0700, Ryan Reich <ryanr@uchicago.edu> said:
>>
>>> On Wed, 09 Jun 2004 22:40:56 -0400, Sgt. Sausage wrote:
>>>>
>>>> Is there a "higher order" complex number:
>>>>
>>>> a + bi + cj + dk ....
>>>>
>>>> and is it useful for anything?
>>>>
>>>>
>>>> Was just wondering. Nothing serious. Pointers to any reading on this
>>>> are welcome -- if it's not a totally worthless and uninteresting
>>>> question.
>>>
>>> In short: sort of. Look up quaternions and octonions, or the "Cayley
>>> algebra". Quaternions are what you are thinking of (in fact, you just
>>> wrote out what they are above), and octonions are a relative in that
>>> they give a multiplication of sorts on 8-dimensional space. Other than
>>> that...nothing. Not that this is easy to prove or anything.
>>
>> Just stick to an exact rational number for each dimension and there are
>> plenty; look up "number fields".
>>
>> Cliff Nelson
>
> Sorry, I didn't look up "number fields" on Google before I recommended it.
> There aren't any good examples. I meant something like:
>
> http://www.mathpages.com/home/kmath531/kmath531.htm
>
> or:
>
> http://users.adelphia.net/~cnelson9/

These look like examples of algebraic extensions of the rationals. These
are of course finite-dimensional vector spaces over Q so yeah, if you
throw away most of the irrational coordinates in R^n you get a field
structure on the remainder. But you can't use this to make all of R^n
into a field, and you can't try this with finite extensions of R since
there's only one. It's sort of amazing, from this perspective, that there
are ANY higher analogues of the complex numbers.

Thinking about it, the quaternions are actually constructed from the
complex numbers the same way the complexes are constructed from the real
numbers. You adjoin another square root of -1, call it j, and write
everything in the form (a + bi) + (c + di)j = a + bi + cj + d(ij), and
then you automatically get ij = k. Then you require that k^2 = -1 also
and derive that ij = -ji, from which ijk = -1 is pretty easy. Taking k^2
= -1 is necessary for the formula (a + bi + cj + dk)(a - bi - cj - ck) =
a^2 + b^2 + c^2 + d^2 to work, but you ALSO need ij = -ji for that, so
it's just luck that the two come together. And you need the
formula for the quaternions to be a division algebra. Without doing the
algebra, it looks to me like trying this trick again will require even
more luck, which accounts for the octonions not being associative, and
again might fail entirely because there are too many relations and some of
them are mutually exclusive. Not a proof of anything, but interesting
anyway. 

-- 
Ryan Reich
ryanr@uchicago.edu
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