Can I keep adding more dimensions to complex numbers?

Question

I know about the concept of the complex plane, but is it possible to move to the third dimension? What about arbitrary many dimensions?

Edit: could you please give me some examples of 3D numbers?

Answer 1

This was a popular pursuit in the 19th century, called hypercomplex numbers, based on the successful example of quaternions and the more esoteric case of octonions. Hypercomplex numbers are now called "finite dimensional associative algebras" but of such algebras, only quaternions and octonions retain features that resemble complex numbers. Matrices form another algebra similar to complex numbers in important ways, and have their own geometric interpretation different from that of complex numbers and quaternions.

Quaternions are related to 3-dimensional rotations and this aspect is generalized to all higher dimensions by Clifford algebras and the representation theory of the $n$-dimensional rotation group.

To answer the EDIT, there are no nontrivial 3-dimensional examples but you can write down trivial examples by listing the multiplication table. Take $1$ and $i$ with the multiplication law the same as in the complex numbers, and add a new element $N$ with some rule for how to multiply it by $1$ and by $i$, such as $N 1 = 1 N = iN = Ni = N$. This gives an associative commutative multiplication law on "3-d numbers" of the form $a.1 + b.i + c.N$, which can be thought of as a rule for combining triples $(a,b,c)$ and $(a',b',c')$ into a third triple. Examples like this are easy to write down, but there is no rule of combination for triples that goes beyond a minor variations on real or complex numbers. Another example of a trivial variation is to multiply the first two components as complex numbers and the third as real numbers. This is called the "direct product" or "direct sum" of the real and complex number system but there is no new structure there.

In dimension 4 there appear new and interesting examples, the quaternions and the algebra of 2x2 matrices.

Answer 2

You might want to look at the Wikipedia article on quaternions. This gets us to dimension $4$, and is useful (and used) for computer graphics. Beyond dimension $4$, there are the octonions, which are not too badly behaved, though we lose associativity of multiplication. Beyond octonions, we lose too many of the algebraic properties of "numbers."

Answer 3

The Cayley–Dickson construction is less than the whole truth, but it's there. It's an infinite sequence beginning with the reals; the second step is the complex field; the third is the quaternions; the fourth is the octonions.

Quaternion multiplication is not commutative; octonion multiplication is not associative.

Answer 4

Historically, people have accepted as "numbers" only a limited number of systems that "break" the properties we are familiar with for real numbers. But this is really only a matter of terminology. If you focus on generalizations of the transformation rules that rotations and dilations of the plane satisfy, you start to glimpse a vast part of the landscape of advanced mathematics. Linear algebra, Lie group theory, operator theory and dynamical systems, just for a start, are aimed at describing useful kinds of transformations on complicated kinds of data, and the rules that govern them.

A simple example in 3D is Euler's rotation theorem: the composition of two rotations in 3D, about possibly different axes, is another rotation, whose axis can be computed. Note that the order matters: 3D rotations do not necessarily commute!

So yes, there are many ways to generalize 2D rotations and get objects with fantastic and useful rules. It's just that we don't call them "numbers."

Answer 5

Example of 3D numbers.

Take $\mathbb{R}^3$ with Hadamard product. In other words, triplets of numbers with element-wise multiplication.

Now assign $(1,1,1)=1,(-1,1,1)=j, (1,1,-1)=k$.

A number would be written in the form $a+bj+ck$. Algebraically it will be a commutative ring with zero divisors (hence, not a field, but that's OK). For instance $(j-1)(k-1)=0$.

Here is a Mathematica code to experiment with:

Unprotect[Power]; Power[0, 0] = 1; Protect[Power];
$Pre = (# /. {j -> {-1, 1, 1}, k -> {1, 1, -1}}) /. {x_, y_, z_} -> 
     x/2 + z/2 + (j (y - x))/2 + (k (y - z))/2 &;

Using this code you can see that

$j^2=k^2=1$

$\log (j+k+1)=\frac{1}{2} j \log (3)+\frac{1}{2} k \log (3)$

$j^j=j^k=j$

$k^k=k^j=k$

$\sqrt{j+k}=\frac{j}{\sqrt{2}}+\frac{k}{\sqrt{2}}$

$0^{j+k}=1-\frac{j}{2}-\frac{k}{2}$

If you add a complex unity $i$, you will get a 6-dimensional number system.

Particularly, you will see that

$i^j=ij$

$j^i=\frac{1}{2} \left(1-(-1)^i\right) j+\frac{(-1)^i}{2}+\frac{1}{2}$

$i^{j+k}=1-j-k$

and

$\sqrt{j}=\left(\frac{1}{2}-\frac{i}{2}\right) j+\left(\frac{1}{2}+\frac{i}{2}\right)$

Answer 6

It is very simple to go to three dimensions, recall that the complex plane is more or less crafted from the fact that $i^2 = -1$. For $\mathbb{R}^3$ you simply define a number $j$ that has the property $j^3 = -1$.

You get a perfect number system (even a perfect complex analysis thing) but it is not a field because there are so called divisors of zero to be found.

Analog to the complex plane $\mathbb{C}$ where it is usual to identify every point $(x, y)$ with a complex number $z = x + iy$, in $\mathbb{R}^3$ you indentify $(x, y, z)$ with $X = x + jy + j^2z$.

An amazing fact is that the equation $X^2 = -1$ has no solution inside this $\mathbb{R}^3$ number system. But if you think about it, that is good because otherwise going to 3 dimensions adds nothing new...

(Yeah yeah, think about that.)