Complex numbers in 3 dimensions

Question

There was a question about this topic a while back, see here. It is known that there is no great way to design 3-dim complex numbers: any representation will have a number of drawbacks, such as impossibility to define an inverse, non-commuting products, or a solution that is trivial and of no value. I was first interested in this topic by looking at bicomplex numbers. Though being 4-dim and similar to quaternions, they do commute. But they don't always have an inverse.

The 3-dim complex numbers that I propose here commute, and many have an inverse. I don't know if they are associative and if they accept a matrix representation like complex numbers, quaternions, or bicomplex numbers. I am interested in an answer to these questions.

I define these 3-dim complex numbers as follows. Let $z_1 = a_1 + b_1 i + c_1 j$ and $z_2 = a_2 + b_2 i + c_2 j$. Then the product is defined as

$$z_1 * z_2 = (a_1a_2-b_1b_2+c_1c_2) + (b_1a_2 +a_1b_2) i + (c_1b_2+b_1c_2) j.$$

The first issue is that it is not compatible with the scalar product:

$$(a_1 + b_1 i + c_1 j) * a_2 = a_1 a_2 + b_1 a_2 i \neq (a_1 + b_1 i + c_1 j) \cdot a_2.$$

Note that $i^2=-1, ij = j$ and $j^2=1$. When the inverse exists, it is given by the following formula:

$$(a + bi + cj)^{-1} = \frac{a}{a^2+b^2+c^2} - \frac{bi}{a^2+b^2+c^2}+\frac{cj}{a^2+b^2+c^2}.$$

I used Mathematica to find the inverse, see here.

Update

Another interesting set of 3-dim complex numbers can be found here. The multiplication is associate and commutative, but again, not all of these numbers have an inverse. In fact, fewer than in my own set.

Answer 1

There is another construction of 3D complex numbers that has associative and commutative multiplication: 3dcomplexnumbers.net. This construction, discovered in parallel by Shlomo Jacobi and Reinko Venema (who is behind the website 3dcomplexnumbers), is easier to construct, but hard to find the inverse.

The set constructions, sometimes denominated as $\mathbb R^3$, is $\mathbb J^-_3\circ \mathbb R$, where we have the imaginary unity $j$ where $j^3 = - 1$.

Every 3D number can be represented as $X = (a, b, c) = a + bj + cj^2$.

$$ \begin{align} X Y &= (a, b, c)(d, e, f)\\ &= a(d, e, f) + bj(d, e, f) + cj^2(d, e, f)\\ &= a(d, e, f) + b(- f, d, e) + c(- e, - f, d)\\ &= (ad, ae, af) + (- bf, bd, be) + (- ce, - cf, cd)\\ &= (ad - bf - ce) + (ae + bd - cf)j + (af + be + cd)j^2 \end{align} $$

There's also the matrix representation of these numbers (seeing $X$ as a column vector):

$$ M(X) = \left[\begin{matrix}X & jX & j^2 X\end{matrix}\right] = \left[\begin{matrix} a & - c & - b\\ b & a & - c\\ c & b & a \end{matrix}\right] $$

Also, $XY$ (or $X^n Y^m \;\;\forall n, m\in\mathbb R$) can be calculated by the matrix multiplication (and if you diagonalize the matrix, you can even calc $\exp$ or $\ln$ of any $X$).

Note: the module is not necessarily conservative over multiplication. But the determinant always is.

There are two important numbers to remark:

1. The number $\alpha$, where $n\alpha = \sum_{i = 0}^{n - 1} (j^n)^2$ (or $3\alpha = j^0 + j^2 + j^4 = 1 + j^2 - j = 1 - j + j^2$).
   1. $\alpha^n = \alpha\;\forall n\in\mathbb N^*$.
   2. $\alpha$ is non-invertible.
   3. $X\alpha = (X\cdot \alpha) \alpha =\;<X, \alpha> \alpha$ (multiplying by $\alpha$ projects the number in the $\alpha$ line).
2. The number $\tau := \ln{j}$ (or $j^2$, if $\det M(j) = - 1$)
   1. $\tau$ in the complex numbers ($\mathbb C$) is $\pi i$.
   2. $\tau$ and $\tau^2$ are non-invertible.
   3. $\tau^3 = - k \tau$, in a similar way to $i^3 = - 1 i$.
   4. In 3D, $\tau = {2\pi\over 3\sqrt 3} (j - j^2)$.

The set of non-invertible numbers are the plane $x - y + z = 0$ and the (orthogonal to this plane) line between $0$ and the number $\alpha$. In 4 dimensions constructed this way, the non-invertibles are just two lines.

There's even the exponencial circle $e^{\tau t}$, with center in $\alpha$ and passing by $1$, $- j$, $j^2$...

And finally, as I gave you mostly general rules, it's possible to develop many other nd complex structures. The first site I mention shows it for 3, 4, 5 and 7 dimensions, and the use of $\mathbb C$ as scalars in 3 dimensions.

Final note: in odd dimensions ($\mathbb J^-_n$ where $n = 2m + 1$), there's no solution to $j^n = - 1$.