Understanding bicomplex numbers

Question

I found by chance, the set of Bicomplex numbers. These numbers took particularly my attention because of their similarity to my previous personal research and question. I should say that I can't really understand the fact that $j^2=+1$ (and must of other abstract algebra) without using matricial interpretations. When I look a bit on the Bicomplex numbers, the think that surprised me a lot was the fact that $ij=ji=k$ and $k^2=-1$. Because using matricial representations, we get : $$ij=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}=\begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}$$ $$ji=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}$$ So, $ij$ should be different then $ji$. Then if we define $k=ij$ or $k=ji$, we get both way $$k^2= I$$ Looked at the Wikipedia article and did some internet search but couldn't find a matricial representation for the bicomplex $k$. So I wanna ask some clarifications about where from the relations $ij=ji$ and $k^2=-1$ come from.

Answer 1

I think the confusion here comes from the the fact that The Bicomplex numbers and the Quaternions are completely different things that use the same letters.

So $i$,$j$ and $k$ mean different things when they represent Bicomplex numbers and when they represent Quaternions.

Quaternions may be represented as a subset of the real $2\times 2$ matrices, but Bicomplex numbers cannot. They may however be represented as $2\times 2$ matrices over $\mathbb C$. So I can set

$$ I = \begin{pmatrix} \mathbf i & 0\\ 0 & \mathbf i \end{pmatrix}$$ $$ J = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$$ $$ K = \begin{pmatrix} 0 & \mathbf i\\ \mathbf i & 0 \end{pmatrix}$$

Where $\mathbf i\in\mathbb C$ is the square root of $-1$.

You can check that all the calculations work.

As for where they come from, inventing crazy algebras was considered a very gentlemanly pursuit in the 1840s and someone just wrote down the rules and noticed it "worked" as an algebra.

Answer 2

You can generate algebras suiting any such set of constraints, but the catch is you might wind up collapsing more than you wanted to :) In the following, I'm glossing over a few details in the interest of getting broader points across.

As a toy example, think of the polynomial ring $\Bbb R[x]$ in one unknown. We can pass to a ring quotient $\Bbb R[x]/(x^2+1)$. The $(x^2+1)$ bit denotes the ideal generated by the polynomial $x^2+1$. Quotients rings can only be formed by using an ideal underneath, so this is necessary. The idea behind a quotient ring is that everything in the ideal is "the same as zero now". So, in particular, $x^2+1=0$, and thus $x^2=-1$. (Actually I can't say that $x^2=-1$, it is really that they are part of the same equivalence class, which we might have to deal with later... ). So, using the map $x\mapsto i$, it turns out $\Bbb R[x]/(x^2+1)\cong \Bbb C$!

In the case of your ring, think of the polynomial ring $\Bbb R[x,y]$ in two unknowns $x$. We want to force $ij=ji$ and $i^2=-1$ and $j^2=1$, so we can try to do something simliar in this polynomial ring by looking at the quotient ring $\Bbb R[x,y]/( x^2+1, y^2-1)$. In this ring, we know that all three of those polynomials in the bottom are "the same as zero", so $x^2=-1$ and $y^2=1$ look a lot like $i,j$ and their commutation relations! Maybe we should map $x\to i$, $y\to j$.

What about $k$? Well, we know we want $xy$ to be what represents $k$. Does $(xy)^2=1$? Well sure, since $(xy)^2=x^2y^2=-1*1=-1$. It turns out that indeed, under this map $\Bbb R[x,y]/(x^2+1,y^2-1)\cong \Bbb B$ the bicomplex numbers.

This method allows you to construct lots of commutative rings satisfying relations you have in mind. Noncommutative ones can be constructed too, it's just that rather than the polynomial ring you use $\Bbb F\langle x,y\dots \rangle$ to denote a polynomial ring in which you don't assume the unknowns can commute. (That's assumed in normal polynomial rings.)

Remember we were looking at Clifford algebras before? They can be constructed using these noncommuting polynomial rings. Suppose you have your multiplication table of $e_i$'s set up so that $e_ie_j=-e_je_i$ for every $i\neq j$, and $e_i^2=c_i\in \Bbb R$.

Then the quotient ring $\Bbb R \langle e_1,\dots,e_n \rangle/(e_1^2-c_1,\dots e_n^2-c_n,e_1e_2+e_2e_1,e_1e_3+e_3e_1,\dots, )$ is isomorphic to that algebra.

Answer 3

The algebra you discribed in the question is tessarines.

So, there are two matrical representations of tessarines.

1. For a tessarine $z=w_1+w_2i+w_3j+w_4ij$ the matrix representation is as follows:

$$\left( \begin{array}{cccc} {w_0} & -{w_1} & {w_2} & -{w_3} \\ {w_1} & {w_0} & {w_3} & {w_2} \\ {w_2} & -{w_3} & {w_0} & -{w_1} \\ {w_3} & {w_2} & {w_1} & {w_0} \\ \end{array} \right)$$

One has to use the 4x4 matrix. After performing the operations, the coefficients of the resulting tessarine can be extracted from the first column of the resulting matrix.

2. Alternatively, one can use a 2x2 matrix of the form $z=a+bj=\left( \begin{array}{cc} a & b \\ b & a \\ \end{array} \right)$

where both $a$ and $b$ are complex numbers. Since complex numbers are embedded in most computer algebra systems, I prefer the second method.

Answer 4

In Mathematica, the following code assd support for the split-complex unity J along with complex unity I. This allows to calculate any expressions with tessarines:

 $Pre = If[FreeQ[#, J], #, Module[{tmp},
tmp = Evaluate[
    MatrixFunction[Function[J, #], {{0, 1}, {1, 0}}]] // 
   FullSimplify;
       tmp /. {{a_, b_}, {b_, a_}} -> a + J b]] &;
J^I
Out= 1/2 (1 + (-1)^I) + 1/2 J (1 - Cosh[Pi] + Sinh[Pi])