Non-Existence of Three-Dimensional Numbers

Question

I have recently begun my study on Naive Lie Theory by John Stillwell. I am trying to understand his historical argument on the early warning signs that 3-dimensional numbers cannot exist. This is an excerpt from the first chapter of Stillwell.

If we set $z_1 = a_1+ib_1$ and $z_2 = a_2+ib_2$, then the multiplicative property of (squared) absolute value states that $({a_1}^2 + {b_1}^2)({a_2}^2 +{b_2}^2)=(a_1a_2 − b_1b_2)^2 +(a_1b_2+a_2b_1)^2$, as can be checked by working out the product $z_1z_2$ and its squared absolute value. This identity is particularly interesting in the case of integers $a_1,b_1,a_2,b_2,$ because it says that (a sum of two squares)×(a sum of two squares) = (a sum of two squares).

This fact was noticed nearly 2000 years ago by Diophantus, who mentioned an instance of it in Book III, Problem 19, of his Arithmetica. However, Diophantus said nothing about sums of three squares—with good reason, because there is no such three-square identity. For example $(1^2 +1^2+1^2)(0^2+1^2+2^2)=3×5=15$, and $15$ is not a sum of three integer squares. This is an early warning sign that there are no three-dimensional numbers.

What is the basis of this claim? I don’t see the triviality of why the product of two sums of three squares always being a sum of three squares is a requisite for existence of three-dimensional numbers.

Answer 1

Suppose you have a system with numbers of the form $\alpha = a + bi + cj$, and some formula for multiplication with two properties:

1. The squared absolute value $|a + bi + cj|^2 = a^2 + b^2 + c^2$ is multiplicative; i.e. $|\alpha \beta|^2 = |\alpha|^2 |\beta|^2$ for all $\alpha, \beta$; and
2. If $\alpha$ and $\beta$ have integer coordinates, then $\alpha \beta$ does too.

Then let $\alpha = 1 + i + j$ and $\beta = i + 2j$. These both have integer coordinates, so their product does too. The squared absolute value of their product is $$ |\alpha \beta|^2 = |\alpha|^2 |\beta|^2 = (1^2 + 1^2 + 1^2) (0^2 + 1^2 + 2^2) = 3 \cdot 5 = 15. $$ So the product must have the form $a + bi + cj$ where $a, b, c$ are integers and $a^2 + b^2 + c^2 = 15$, which is impossible. So any three-dimensional number system must fail either property (1) or (2).

Answer 2

There can be 3-dimensional commutative hypercomplex number systems. I will give three examples.

3D numbers. Example 1

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 one can see that

$j^2=k^2=1$

$jk=j+k-1$

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

$j^j=j^k=j$

$k^k=k^j=k$

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

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

The division formula would be:

$\frac{a_1+b_1 j+c_1 k}{a_2+b_2 j+c_2 k}=\frac{j}{2} \left(\frac{a_1+b_1+c_1}{a_2+b_2+c_2}-\frac{a_1-b_1+c_1}{a_2-b_2+c_2}\right)+\frac{k}{2} \left(\frac{a_1+b_1+c_1}{a_2+b_2+c_2}-\frac{a_1+b_1-c_1}{a_2+b_2-c_2}\right)+\frac{a_1+b_1-c_1}{2 \left(a_2+b_2-c_2\right)}+\frac{a_1-b_1+c_1}{2 \left(a_2-b_2+c_2\right)}$

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

Particularly, we will see that

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

and

$\log (j k)=i\pi-\frac{i \pi j}{2}-\frac{i \pi k}{2}$

3D numbers. Example 2

This is a realization of triplex numbers, described in this video.

Here,

$1=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$

$j=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{array} \right)$

$k=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right)$

Unprotect[Power]; Power[0, 0] = 1; Protect[Power];
$Pre = (# /. ({j -> {1, E^(2 I \[Pi]/3)}, 
          k -> {1, E^(-2 I \[Pi]/3)}}) /. {x_, y_} -> 
        FullSimplify[(x/3 + Im[y]/Sqrt[3] - Re[y]/3) j + (x/3 - 
             Im[y]/Sqrt[3] - Re[y]/3) k + 
          1/3 (x + y + Conjugate[y])] // FullSimplify // Expand) &;

Particularly, we will see that

$j^2=k$, $k^2=j$, $jk=1$

$j^k=-\frac{1}{3} 2 e^{\frac{\pi }{\sqrt{3}}} j+\frac{j}{3}+\frac{1}{3} e^{\frac{\pi }{\sqrt{3}}} k+\frac{k}{3}+\frac{e^{\frac{\pi }{\sqrt{3}}}}{3}+\frac{1}{3}$

$0^{j + k + 1}=-\frac{j}{3}-\frac{k}{3}+\frac{2}{3}$

$\log j = \frac{2 \pi j}{3 \sqrt{3}}-\frac{2 \pi k}{3 \sqrt{3}}$

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

The division formula is

$\frac{a_1+b_1 j+c_1 k}{a_2+b_2 j+c_2 k}=\frac{a_2^2 \left(a+b j_2+c j_1\right)-a_2 \left(b_2 \left(a j_2+b j_1+c\right)+c_2 \left(a j_1+b+c j_2\right)\right)+c_2^2 \left(a j_2+b j_1+c\right)-b_2 c_2 \left(a+b j_2+c j_1\right)+b_2^2 \left(a j_1+b+c j_2\right)}{a_2^3+b_2^3+c_2^3-3 a_2 b_2 c_2}$

If we add complex unity, we will see that

$i^{j+k}=-\frac{j}{3}-\frac{j}{\sqrt{3}}-\frac{k}{3}+\frac{k}{\sqrt{3}}-\frac{1}{3}$

3D numbers. Example 3.

This system is similar to dual numbers and has nilpotents. It is isomorphic to the matrices of the following form:

$1=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$

$\epsilon=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{array} \right)$

$\varepsilon=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right)$

In this system $\epsilon^3=\varepsilon^2=0$ and $\epsilon^2=\varepsilon$.

The division formula is:

$\frac{a_1+b_1\epsilon+c_1\varepsilon}{a_2+b_2\epsilon+c_2\varepsilon}=\frac{a_1 b_2^2 \varepsilon }{a_2^3}-\frac{b_1 b_2 \varepsilon }{a_2^2}-\frac{a_1 b_2 \epsilon }{a_2^2}+\frac{b_1 \epsilon }{a_2}+\frac{c_1 \varepsilon }{a_2}-\frac{a_1 c_2 \varepsilon }{a_2^2}+\frac{a_1}{a_2}$