3D equivalent to complex number plane

Factual Questions
1

I realize one can visualize the imaginary and complex numbers as a rotation of vectors on a plane. Just as one intrinsically thinks that walking forward -5 steps means you actually walked backward 5 steps, one can see walking forward 5i steps as walking left 5 steps. (I think, as if forward is considered the positive x axis, then the positive y-axis would be 90[sup]o[/sup] counterclockwise.)

What I’m wondering is if there is a standard system to take this from two dimensions to three. I came up with the idea of creating a number j which indicates a 90[sup]o[/sup] rotation along the z-axis. But this number has the weird property that, while i[sup]2[/sup] = -1 and j[sup]2[/sup] = -1, ij does not equal -1. In fact, it would seem to have different answers, depending on the order of multiplication. This would make j immune to both the substitution property and the commutative property of multiplication.

All these weird things happening to j seems to make it an unattractive choice.

2

Out of curiosity, what do you mean by “the substitution property”?

Anyway, there are things you can do, but… they depend on exactly what it is you want to do. If you just want some nice number system that corresponds to 3d space or certain linear transformations upon it in a similar way to how complex numbers correspond with both 2d space and combination of rotation and scaling upon it, then, again, there are various possibilities, but it depends on which properties you want to retain in analogy with the complex plane and which you are willing to let go of. [E.g., as you just saw, you might well want to get rid of commutativity]. Ideas like “quaternions” and “Frobenius’s theorem” might be of interest to you.

3

Mathematicians and physicists struggled for years to find a three dimensional analog of the complex numbers. They were looking for a so called “field” that had all the properties of the real numbers, but in three dimensions. In particular, they wanted to be able to divide two such numbers, just as you can do with real numbers and complex numbers, in which there is a well-defined, unique, inverse for every number other than the additive identity (i. e. zero). All attempts failed. Hamilton had a brainstorm and discovered that it could be done in four dimensions. He spent the rest of his life working out the details of this system, called the quaternions. Quaternions indeed have exactly what you are talking about. i, j, and k are all square roots of negative one, and have the following properties:

i^2 = j^2 = k^2 = -1
ij = k, jk = i, ki = j
ij = -ji, jk = -kj, ki = - ik

A general quaternion is a + bi + cj + d*k, where a, b, c, d are real numbers.

The quaternions are closed under mulitplication, addition, multiplication by a scalar, and division (except division by zero). Multiplication and addition are associative and distributive, addition is commutative, but multiplication is not (since ij = -ji, etc.).

Later it was proven that the only dimensions in which it is possible to obey all these axioms are 1 (real numbers), 2 (complex numbers), 4 (quaternions), and 8 (octonions, which are not associative). Thus, you cannot do it in three dimensions. Our system of 3 vectors is not nearly as rich mathematically as the 4 vector quaternions.

In physics, one representation of the the quaternions is the general sum of Pauli spin matrices.

4

Not sure of its relationship to complex numbers. But the 3D equivalent to polar coordinates is spherical coordinates.

5

If a = b, and b = c, then a = c.

It’s very necessary to do anything in algebra. It’s what lets you say, for example, that, if 3x + 5 = 32 and 3(9) + 5 = 32, then x = 9. This wouldn’t work in my attempt at 3D, as ij = j, yet i ≠ 1 (or ji = i, yet j ≠ -1, depending on the order)

I normally just call it substitution, but that’s what it was called on the website I looked up to make sure the commutative property was the one I wanted.

@JWT Kottekoe That’s exactly what I wanted to know. Thanks. I’d heard of ijk, but I didn’t know this was what it was for. Were there proofs that other 2[sup]n[/sup]-dimensional systems won’t work? If the pattern holds, a 16D version would have to lose another multiplicative property; perhaps that just makes it useless for most applications?

6

Not sure about the substitution property, I think what you might be looking for is the cancellation property, namely if ab=ac then b=c. Trouble is, that holds for some systems and not others, so you may well have developed a system in which it doesn’t hold.

It does sound like your more general enterprise is retracing the steps Hamilton took in developing his system of quaternions, so have a look at those.

7

I will answer that question. In 8 dimensions, there is a “number field” sometimes called the Cayley numbers, but more commonly the octonions. Once upon the time I knew the equations they satisfied, but I am sure you could find them on Wiki. There is a basis of 8 elements, one of which is 1 and the remaining seven are all square roots of -1. The problem is that it is no longer associative: (xy)z is not generally the same as x(yz).

It was known long ago that the only dimensions that could support such a structure. Then sometime in 60s or 70s J. Frank Adams showed that no such structure would be possible in any higher dimension. His argument is interesting. I don’t know any details, but I can describe the basic idea. You may have heard that “You can’t comb the hair on a billiard ball.” Leaving aside the fact that billiard ball has no hair, what this is saying is that there is no way of putting a tangent verctor of positive length at each point on a sphere. It is obviously possible on a circle and, using quaternions and octonions, you can comb the hair on a 3-sphere or a 7-sphere. Adams showed that you couldn’t comb the hair on a ball of any dimension other than 1, 3, or 7.

8

30 minutes to a summary answer and 75 minutes to a detailed-to-a-layman answer, and in the middle of the night to boot. Ths place is great. Thanks folks; I learned something today.

9

As Nancarrow points out, “the substitution property” (a = b and b = c implies a = c) continues to hold true for the ring of linear transformations upon 3-dimensional space. It’s just transitivity of equality; there’s nothing much to it. The property you were actually using in your illustrations, though, was cancellability of multiplication (e.g., left-cancellability: if ba = bc, and b is not equal to 0, then a = c), which does indeed disappear in your example.

10

Really? I thought it could be done in any odd dimension.

11

You are right. According to wiki: Vector fields on spheres - Wikipedia, it is only for N = 2,4,8 are there N-1 linearly independent vector fields on an N-1 sphere and not for N = 16. But if there were a 16 dimensional division algebra, there would have to be 15 linearly independent vector fields on the 15-sphere. Adams proved this in 1962, wiki says. Sorry, my memory for these things is increasingly watery. The wiki cite gives the exact number of independent vector fields for any dimension.

It has long been known that division ring structure were possible only for dimensions power of 2. Incidentally, there are many (I think there are 2^{2^{aleph_0}} of them) non-isomorphic 4 dimensional associative division algebras. The trick is that the real mumbers are not central, although the center is always a real closed fleld of index in the complex numbers. To get one of these exotic reals, it suffices to take a transcendence basis of the complex numbers that includes some non-real. Then the rationals with that transcendence basis adjoined is a formally real field that has a real closure and it contains a non-real number, The usual rigidity argument implies it cannot be isomorphic to the reals. Adjoin a square root of -1 and you get the ordinary complex numbers (what else could it be?) and now adjoin j and k and you get exotic quaternions. But I digress.

12

It’s true that Frank Adams solved the vector fields on spheres problem, but that wasn’t the work related to division rings. That was settled when he solved the Hopf invariant one problem. Briefly, if you have a “multiplication” on an n-dimensional euclidean space that is continuous, has a two-sided nonzero unit, and no zero-divisors, then you can construct a map from a (2n-1)-sphere to an n-sphere whose “Hopf invariant” is 1. Frank showed that such a map of spheres exists only when n = 1, 2, 4, or 8.

13

What he said. Also, my “general enterprise” was really just not being able to sleep, and thinking about something else I learned on the Dope (the 2D vector thing mentioned in the OP). I don’t necessarily plan to do any more work in the area.