Defining i,j,k,l,.. for complex number in a 3D space, in 4D space, etc [duplicate]

Question

Real number is often used to represent a point in a 1-dimensional number line. Real numbers are written as $a$ where $a \in \mathbb{R}$.

Complex number is often used to represent a point in a 2-dimensional plane. Complex numbers are written as $a + bi$ where $a,b \in \mathbb{R}$ and $i^2 = -1$.

What about higher dimensional space?

We can extend the concept of complex number, and write a point in a 3-dimensional space as $a + bi + cj$ where $a,b,c \in \mathbb{R}$, but how do we define i and j?

Certainly, this also applies to 4-dimensional space. We can write a point in 4D as $a + bi + cj + dk$, how do we define $i,j,k$?

How about even higher n-dimensional space? Is there a system to define i,j,k,... that will extend to arbitrary dimensions? Is there a name for it?

Answer 1

It depends upon what structure you want. In $\mathbb{R}^n$ we do not have the operation of multiplying two elements, as we do on $\mathbb{C}$. If you want that, you probably want a normed division algebra, which are available in four and eight dimensions, but no more.

If you just want a vector space they are basis vectors but you can't multiply two elements.

Answer 2

Just for kicks, here's a direct argument that you can't extend $\mathbb{C}$ to something three dimensional where you can still multiply.

Start with your reals, $i$, and $j$ as before. The question is, what should $ij$ be?

Since it will be something in your three dimensional thing, you can express it as $ij = a + bi + cj$ for some real numbers $a$, $b$, and $c$. Multiplying everything on the left by $i$ gives $$-j = ai - b + cij $$ and re-expanding $ij$ then gives \begin{align*} -j &= ai -b +c(a+bi + cj) \\ &=ai-b+ac+cbi+c^2j\\ &= (ac-b) + (a+bc)i +c^2 j\end{align*} and now equation coefficients gives $ -1 = c^2$, contradicting the fact that $c$ was real.

Thus, if you extend to a three dimensional setting, you are not allowed to multiply $i$ and $j$. To me, this makes the quaternions (a 4 dimensional setting mentioned in the other answers) much more surprising!

Answer 3

Over 20 years ago I found the way to generalize the complex numbers to higher dimensions.

Basically it is very simple: on $\mathbb{R}^3$ you define an imaginary unit $j$ via stating $j^3 = -1$.

Your complex numbers now look like $T = A + Bj + Cj^2$.

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During the last 1.5 years I have picked up studying this stuff again, some of the results can be found at a page on my website:

3D Complex Stuff

Compared to the complex plane the 3D complex numbers form a far more complicated structure; so not only university freshmen can learn from it, the same goes for retired math professors...

A nice starter exercise would be the next:

Use the above definition of 3D complex numbers (with $j^3 = -1$) and proof that the equation $X^2 = -1$ has no solution inside $\mathbb{R}^3$.

Answer 4

You should totally look about Quaternions,Octonions and all other sets defined as the same way as them. Quaternions are used to represent numbers in a 4 dimensional space whereas Octonions are used in 8 dimensional space. You could also look to my question and it's answer about numbers defined on a 4 dimensional space.

EDIT : you can also add the Sedenions to the list, numbers on a 16 dimensional space. These kind of numbers, for $n$ dimensions can be written as follows : $$\alpha_1 +\alpha_2e_1+\alpha_3e_2+...+\alpha_{n}e_{n-1}$$ where $\alpha_n\in\Bbb R$ and $e_n^2=-1$ and for all number a and b, $a≠b, e_a≠e_b$. All the $e_n$s are said to be the elementary units of the set. So you can represent these numbers in a $n$ dimensional space using the the coefficients $\alpha_n$. For complex numbers, $n=2$ and $e_1=i$. For Quaternions, $n=4$ and $e_1=i, e_2=j, e_3=k$.

Answer 5

If we take complex number in 3d plan than we found 2 condition, let we take 3 axis x y and z:

1. x is real and yand z are imaginary, that is: $x+i(y+z)$

2. x and z is real but y is imaginary, that is: $x+z+iy$