# What Are the Relations for the Polar Hypercomplex form $a + bi + cj + dk$?

Olariu in "Complex Numbers in $N$ Dimensions" has polar hypercomplex numbers described by its generators as \begin{gather} \alpha^2 = \beta, \\ \beta^2 = 1, \\ \gamma^2 = \beta, \\ \alpha\beta =\beta\alpha = \gamma, \\ \alpha\gamma =\gamma\alpha = 1,\\ \beta\gamma = \gamma\beta = \alpha. \end{gather} A commutative hypercomplex number can be expressed as a linear combination of real and non-real roots of $1$ and $-1$.

My goal is to find an expression for these polar hypercomplex numbers as $a + bi + cj + dk$.

Any help, reference would be greatly appreciated.

• @Paul Sundheim I'm not certain what your question is at this point. How are $i,j,k$ defined? Do you intend a quaternionic representation? Commented Jul 29, 2013 at 6:24
• @Paul Sundheim, the paper you mention is 243 pages long. Is there a particular part you are working on? I will say this, quaternions are a skew-field hence have no zero-divisors. On the other hand, hyperbolic numbers do have zero divisors. The system you post is built over hyperbolic numbers so it will have zero divisors. So... Commented Jul 29, 2013 at 6:31
• @James S. Cook, The relations written can be found on page 53 although more can be found using the index. I have referred to commutative hypercomplex numbers, so I am aware that there will be zero divisors. In fact from the relations above (1 + \beta)(1 - \beta) = 0. These are not hyperbolic since there, the squares of, what I called, i, j, k are all one in that system. Commented Jul 30, 2013 at 4:38
• @YACP, The system is commutative and the question is algebraic and thus the Commutative Algebra tag. Please forgive my ignorance in proper tagging; this is my first post. Commented Jul 30, 2013 at 4:49
• @PaulSundheim I took a stab at your question, I will admit, I'm still not sure what you intend by $i,j,k$ and almost certainly my eventual use of $j$ differs from your intent. That said, I'm pleased with my result :) Commented Jul 30, 2013 at 6:11

I will attempt an answer at this point. First, I think you would like us to read section 3.4 which is found at pages 113-137. I make no claim to understand all those results, clearly the author has spent some time developing exponentials and trigonometric functions as well as studying the structure of zero-divisors and so forth. That said, the basic structure here is simply an algebra $\mathcal{A} = \mathbb{R} \oplus \alpha \mathbb{R} \oplus \beta \mathbb{R} \oplus \gamma \mathbb{R}$ where the multiplication is given by the relations in your post. \begin{gather} \alpha^2 = \beta, \\ \beta^2 = 1, \\ \gamma^2 = \beta, \\ \alpha\beta =\beta\alpha = \gamma, \\ \alpha\gamma =\gamma\alpha = 1,\\ \beta\gamma = \gamma\beta = \alpha. \end{gather} You ask for an expression. I would start with $X=t+x\alpha+y\beta+ z\gamma$ this is a typical number in $\mathcal{A}$. We can multiply them as follows, suppose $A=a+b\alpha+c\beta+ d\gamma$ is another number in $\mathcal{A}$ then \begin{align} AX &= (a+b\alpha+c\beta+ d\gamma)( t+x\alpha+y\beta+ z\gamma) \\ &= a( t+x\alpha+y\beta+ z\gamma)+b\alpha( t+x\alpha+y\beta+ z\gamma)+ \\ &\qquad +c\beta ( t+x\alpha+y\beta+ z\gamma)+ d\gamma ( t+x\alpha+y\beta+ z\gamma) \\ &= 1(at+dx+cy+bz)+\alpha(bt+ax+dy+cz) \\ &\qquad + \beta( ct+bx+ay+dz)+ \gamma( dt+cx+by+az) \end{align} Our notation here is that $e_1=1$ and $e_2 = \alpha$, $e_3=\beta$ and $e_4=\gamma$. It follows that I can read off a matrix representative of the number $A$ as follows: $$M_A = \left[ \begin{array}{cccc} a & d & c & b \\ b & a & d & c \\ c & b & a & d \\ d & c & b & a \end{array} \right]$$ The matrix above represents the linear map $L_A: \mathcal{A} \rightarrow \mathcal{A}$ defined by $L_A(X)=AX$ with respect to the basis of generators. Notice in particular, $$M_1 = \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] \ M_{\alpha} = \left[ \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right] \ M_{\beta}=\left[ \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right] \ M_{\gamma}=\left[ \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{array} \right]$$ You can verify that $M_AM_B = M_{AB}$ hence $M: \mathcal{A} \rightarrow \mathbb{R}^{ 4 \times 4}$ provides an algebra homomorphism.

Interesting, this reminds me of $\mathcal{A}_2 = \mathbb{R} \oplus j\mathbb{R} \oplus j^2 \mathbb{R}\oplus j^3\mathbb{R}$ where $j^4=1$ which has matrix representatives of the following form: $$M_{a+bj+cj^2+dj^3} = \left[ \begin{array}{cccc} a & d & c & b \\ b & a & d & c \\ c & b & a & d \\ d & c & b & a \end{array} \right]$$ Apparently, your algebra is isomorphic to the cyclotomic numbers of order $4$. Identify that $\alpha = j$, $\beta=j^2$ and $\gamma=j^3$. You can recover all your relations by imposing $j^4=1$.

Woke up this morning and it occurred to me you might be after the isomorphism of $\mathbb{R} \oplus j\mathbb{R} \oplus j^2 \mathbb{R}\oplus j^3\mathbb{R}$ to $\mathbb{C} \oplus \mathcal{H}$ where $\mathbb{C}$ is the usual complex numbers and $\mathcal{H}$ are the hyperbolic numbers. The space $\mathcal{A}_3=\mathbb{C} \oplus \mathcal{H}$. I'll use $\mathbb{C} = \mathbb{R} \oplus i \mathbb{R}$ and $\mathcal{H} = \mathbb{R} \oplus \eta \mathbb{R}$ where $i^2=-1$ and $\eta^2=1$. Be carefull though, the identity for $\mathcal{A}_3$ in our current context is $(1,1)$ $$\mathcal{A}_3 = \{ (z_1,z_2) \ | \ z_1 \in \mathbb{C}, z_2 \in \mathcal{H} \}$$ In this notation $(0,\eta)^2 = (0, \eta^2)= (0,1)$ and $(i,0)^2 = (i^2, 0)= (-1,0)$. The isomorphism $\Psi$ from $\mathcal{A}_3$ to $\mathcal{A}_2$ will be fixed by its image on generator $j$. Clearly we need: $$\Psi(j) = (x_1+iy_1,x_2+\eta y_2)$$ such that $(x_1+iy_1,x_2+\eta y_2)^4=(1,1)$ but this is to ask: $$(x_1+iy_1)^4=1, \qquad (x_2+\eta y_2)^4=1$$ One solution is given by: $$x_1 = 0, \ \ y_1 = 1, \ \ x_2=0, y_2 = 1$$ note $(i,\eta)^2 = (i^2,\eta^2) = (-1,1)$ and $(i,\eta)^4 = (i^4,\eta^4) = (1,1)$. Therefore, $\Psi(j) = (i,\eta)$ and you can cipher that: omitting the $\Psi$, $$j = (i,\eta) = \alpha$$ $$j^2 = (-1,1) = \beta$$ $$j^3 = (-i,\eta) = \gamma$$ so, you can use a polar representation in terms of sine and cosine for the copy of the complex numbers, however, as I suspected from the outset, there is a copy of the hyperbolic numbers implicit withing your algebra. Now you can use the relations above to make that explicit.

Incidentally, anytime you have a semi-simple real associative algebra which is commutative it will allow an isomorphism to a direct sum of copies of $\mathbb{C}$ and $\mathcal{H}$. If we allow noncommutativity then the algebra is isomorphic to a direct sum of matrix algebras over $\mathbb{R}$, $\mathbb{C}$ or the quaternions $\mathbb{H}$.

• +1 Extremely thorough answer and further comment regarding the isomorphism that occurred the following morning. Bravo! Commented Jul 30, 2013 at 14:45
• @James S. Cook, First and foremost, your reply completely impresses me with it's depth and clarity. Second, if I read you correctly, the relations I need are that the squares of the generators 1,i,j,k are all one and so the system is hyperbolic. (So you were right from the beginning). And the last relation needed is that the product of the generators: 1ijk = -1. This gives ij = -k, ik = -j, and jk = -i. My copy of Olariu is missing those pages :( Am I right about these generators and relations? Sorry I wasn't clearer about the format of the generators. I needed {i^2, j^2, k^2} = {1, -1} Commented Jul 30, 2013 at 14:58
• @PaulSundheim glad to help, I'm not quite sure what you mean by $i,j,k$ as it relates to your posted relations. If $i=\alpha, j = \beta, k = \gamma$ then surely $1ijk = \alpha \beta \gamma = \gamma^2 = \beta \neq -1$. By the way, my use of $j$ need not conform to your intent, so beware the notation... I found that paper on the Archive at arxiv.org/abs/math/0011044 it has all the pages, maybe with some errors, but hey, it's free! Commented Jul 30, 2013 at 17:59
• @James S. Cook, The i, j, k, are non-real square roots of 1 and/or -1. I was hoping to see how they related to the \alpha, \beta, \gamma, but if I know that an element of the ring can be expressed as a linear combination of square roots of 1 (the 1, i, j and k) and ijk = -1 then I am answered. Am I reading your previous comment correctly? BTW, on page 113 of the paper, equations 3.463 contain an error, I think. You wrote \alpha\beta\gamma = \beta, but the pg 113 relations have \alpha\gamma = -1 so that \alpha\gamma\beta = -\beta. Maybe i = \beta but what j and k would be, I don't know. Commented Jul 30, 2013 at 21:10
• @PaulSundheim I think that's just a typo. If you look at the calculation just below he has $ty'$ as a term arising from multiplying $t\alpha$ and $y'\gamma$ and also $yt'$ again with a $+$ both of these terms in $uu'$ arise from $\alpha \gamma =1$. So, I think that's just a typo. Otherwise, I'd have to revise my whole post... ouch. I think you should think more about what you really mean by $i,j,k$. Commented Jul 30, 2013 at 23:18

The answer is that it can't be done. Because the ring is principal (as is described in answer 2) i.e. $a + bi + cj + dk = A+ B \alpha +C \alpha^2 + D \alpha^3$ there are no relations that would allow for $i^2 = \pm 1$, $j^2 = \pm 1$ and $k^2 = \pm 1$.