M.E. Irizarry-Gelpí

M.E. Irizarry-Gelpí

Physics impostor. Mathematics interloper. Husband. Father.

Bi-Complex Numbers

Complex numbers are a staple of mathematics and physics. A complex number \(z\) can be written as the sum of two terms:

$$z = a + bi.$$

Here \(a\) and \(b\) are real numbers, and \(i^{2} = -1\). For a long time, I thought of \(i\) as the one and only imaginary unit, the only "number" that squares to -1. But you also have quaternions, which can be written as a sum of four terms:

$$ a + bi + cj + dk. $$

Here \(a\), \(b\), \(c\), and \(d\) are real numbers. The quaternion units \(i\), \(j\), and \(k\) all square to -1:

$$ i^{2} = -1; \qquad j^{2} = -1; \qquad k^{2} = -1. $$

However, multiplication is more complicated:

$$ ij + ji = 0; \qquad jk + kj = 0; \qquad ki + ik = 0; $$
$$ ij - ji = 2k; \qquad jk - kj = 2i; \qquad ki - ik = 2j. $$

Obviously, there are many ways to restrict a quaternion to a complex number:

$$ a +bi; \qquad a + cj; \qquad a + dk. $$

Each of these is a complex number, but with a different unit!

There is another sum of four terms:

$$ a + bJ + CK + dS. $$

Here, again, \(a\), \(b\), \(c\), and \(d\) are real numbers. However, the units \(J\), \(K\), and \(S\) do not all square to -1:

$$ J^{2} = -1; \qquad K^{2} = -1; \qquad S^{2} = +1. $$

The multiplication of these units is different from the quaternion units:

$$ JK + KJ = 2S; \qquad KS + SK = -2J; \qquad SJ + JS = -2K; $$
$$ JK - KJ = 0; \qquad KS - SK = 0; \qquad SJ - JS = 0. $$

That is, this multiplication operation is commutative. Numbers of this form are called bi-complex numbers.

For quaternions, there is only one conjugation operation:

$$ (a+bi+cj+dk)^{*} = a-bi-cj-dk. $$

This conjugation operation is an involution. With this conjugation operation you can define one quadrance operation:

$$ (a+bi+cj+dk)^{*} (a+bi+cj+dk) = a^{2} + b^{2} + c^{2} + d^{2}. $$

For bi-complex numbers it is more useful to define three conjugation operations:

$$ (a+bJ+cK+dS)^{*_{J}} = a-bJ+cK-dS; $$
$$ (a+bJ+cK+dS)^{*_{K}} = a+bJ-cK-dS; $$
$$ (a+bJ+cK+dS)^{*_{S}} = a-bJ-cK+dS. $$

Each of these conjugation operations is an involution. You can define three quadrance operations:

$$ (a+bJ+cK+dS)^{*_{J}} (a+bJ+cK+dS) = a^{2} + b^{2} - c^{2} - d^{2} + 2(ac + bd)K; $$
$$ (a+bJ+cK+dS)^{*_{K}} (a+bJ+cK+dS) = a^{2} - b^{2} + c^{2} - d^{2} + 2(ab + cd)J; $$
$$ (a+bJ+cK+dS)^{*_{S}} (a+bJ+cK+dS) = a^{2} + b^{2} + c^{2} + d^{2} + 2(ad - bc)S. $$

None of these three quadrance operations are real. However, the quadrance of each quadrance gives a real number. This is the bi-quadrance:

$$ (a+bJ+cK+dS)^{*_{S}} (a+bJ+cK+dS)^{*_{K}} (a+bJ+cK+dS)^{*_{J}} (a+bJ+cK+dS). $$

The bi-quadrance is real and can be written in many equivalent forms:

$$ (a^{2} + b^{2} - c^{2} - d^{2})^{2} + 4 (ac + bd)^{2}; $$
$$ (a^{2} - b^{2} + c^{2} - d^{2})^{2} + 4 (ab + cd)^{2}; $$
$$ (a^{2} + b^{2} + c^{2} + d^{2})^{2} - 4 (ad - bc)^{2}. $$

The last form can be factorized:

$$ ((a+d)^{2} + (b-c)^{2}) ((a-d)^{2} + (b+c)^{2}). $$

Two things are clear from this expression. First, the bi-quadrance is always non-negative. Second, the bi-quadrance can be zero for non-trivial bi-complex numbers. A bi-complex number with zero bi-quadrance is a zero divisor. There are two kinds of zero divisors:

$$ a+bJ+bK-aS; \qquad a+bJ-bK+aS. $$

These two kinds of zero divisors are interesting because they are linear combinations of terms that are idempotents or square roots of minus idempotents. For example,

$$ \left( \frac{1 + S}{2} \right)^{2} = \frac{1 + 1 + 2S}{4} = \frac{1 + S}{2}; $$
$$ \left( \frac{1 - S}{2} \right)^{2} = \frac{1 + 1 - 2S}{4} = \frac{1 - S}{2}; $$
$$ -\left( \frac{J + K}{2} \right)^{2} = \frac{1 + 1 - 2S}{4} = \frac{1 - S}{2}; $$
$$ -\left( \frac{J - K}{2} \right)^{2} = \frac{1 + 1 + 2S}{4} = \frac{1 + S}{2}. $$

Note that you also have

$$ \left( \frac{1 + S}{2} \right) \left( \frac{J - K}{2} \right) = \frac{J - K - K + J}{4} = \frac{J - K}{2}; $$
$$ \left( \frac{1 - S}{2} \right) \left( \frac{J + K}{2} \right) = \frac{J + K + K + J}{4} = \frac{J + K}{2}. $$

One kind of zero divisor can be written as

$$ a \left( \frac{1 - S}{2} \right) + b \left( \frac{J + K}{2} \right). $$

It is funny that this follows the pattern of complex numbers in the sense that a complex number is written as

$$ a(1) + b(i), $$

where \(1^{2} = 1\) (1 is the idempotent), \(-i^{2} = 1\) (\(i\) is the square root of minus the idempotent), and \(1 i = i\) (multiplying \(i\) by the idempotent is the identity operation). The same holds for the other kind of zero divisor:

$$ a \left( \frac{1 + S}{2} \right) + b \left( \frac{J - K}{2} \right). $$

Note that \(*_{S}\) does not change the kind of a zero divisor, but \(*_{J}\) and \(*_{K}\) do.

Zero Bi-Quadrance Decomposition

Any bi-complex number can be written in terms of these combinations:

$$ a+bJ+cK+dS = (a + d) \left( \frac{1 + S}{2} \right) + (b - c) \left( \frac{J - K}{2} \right) + (b + c) \left( \frac{J + K}{2} \right) + (a - d) \left( \frac{1 - S}{2} \right). $$

The bi-quadrance of each individual term is zero.

Segre Star Operation

In analogy with the Hodge star operation, you can define a Segre star operation as follows:

$$ (1)^{\star} = S, \qquad (J)^{\star} = -K, \qquad (K)^{\star} = -J, \qquad (S)^{\star} = 1. $$

This is an involution. This means that you can star-decompose any bi-complex number \(z\) into a self-star and an anti-self-star component:

$$ z = \left( \frac{z + z^{\star}}{2} \right) + \left( \frac{z - z^{\star}}{2} \right). $$

If \(z = a+bJ+cK+dS\), then the Segre star is:

$$ z^{\star} = d - cJ - bK + aS. $$

The self-star part of \(z\) is:

$$ \frac{z + z^{\star}}{2} = (a + d)\left( \frac{1 + S}{2} \right) + (b - c)\left( \frac{J - K}{2} \right); $$

and the anti-self-star part of \(z\) is:

$$ \frac{z - z^{\star}}{2} = (a - d)\left( \frac{1 - S}{2} \right) + (b + c)\left( \frac{J + K}{2} \right). $$

Each component of the Segre star decomposition is a zero divisor. The Segre star decomposition is equivalent to the zero bi-quadrance decomposition.

The product of the self-star and anti-self-star components is zero, because

$$ \left( \frac{1 + S}{2} \right) \left( \frac{1 - S}{2} \right) = \frac{1 - 1}{4} = 0; $$
$$ \left( \frac{1 + S}{2} \right) \left( \frac{J + K}{2} \right) = \frac{J + K - K - J}{4} = 0; $$
$$ \left( \frac{1 - S}{2} \right) \left( \frac{J - K}{2} \right) = \frac{J - K + K - J}{4} = 0; $$

and

$$ \left( \frac{J + K}{2} \right) \left( \frac{J - K}{2} \right) = \frac{-1 + 1}{4} = 0; $$

This is a form of orthogonality.

The Segre star operation is equivalent to multiplying by \(S\).

Exponential Function

The exponential of a complex number involves exponential and trigonometric functions of real numbers:

$$ \exp{(a + bi)} = \exp{(a)} \left[ \cos{(b)} + i \sin{(b)} \right]. $$

The exponential of a bi-complex number involves exponential, trigonometric, and hyperbolic functions of real numbers. Since the multiplication is commutative, there is factorization:

$$ \exp{(a + bJ + cK + dS)} = \exp{(a)} \exp{(bJ)} \exp{(cK)} \exp{(dS)}. $$

Using

$$ \exp{(bJ)} = \cos{(b)} + J \sin{(b)} ; $$
$$ \exp{(cK)} = \cos{(c)} + K \sin{(c)}; $$

and

$$ \exp{(dS)} = \cosh{(d)} + S \sinh{(d)}; $$

you can find that

$$ \exp{(a + bJ + cK + dS)} = A + BJ + CK + DS; $$

where

$$ A = \exp{(a)} \left[ \cos{(b)} \cos{(c)} \cosh{(d)} + \sin{(b)} \sin{(c)} \sinh{(d)} \right]; $$
$$ B = \exp{(a)} \left[ \sin{(b)} \cos{(c)} \cosh{(d)} - \cos{(b)} \sin{(c)} \sinh{(d)} \right]; $$
$$ C = \exp{(a)} \left[ \cos{(b)} \sin{(c)} \cosh{(d)} - \sin{(b)} \cos{(c)} \sinh{(d)} \right]; $$

and

$$ D = \exp{(a)} \left[ \sin{(b)} \sin{(c)} \cosh{(d)} + \cos{(b)} \cos{(c)} \sinh{(d)} \right]. $$

Lozenge Invariant

With two complex numbers \(x\) and \(y\) you can construct the expression

$$ x^{*}y. $$

This is invariant under the following transformation:

$$ x \rightarrow \exp(ib) x, \qquad y \rightarrow \exp(ib) y. $$

With four bi-complex numbers \(w\), \(x\), \(y\), and \(z\) you can construct the lozenge expression

$$ w x^{*_{J}} y^{*_{K}} z^{*_{S}}. $$

The lozenge is invariant under the following transformation:

$$ w \rightarrow \exp{(bJ+cK+dS)} w, \qquad x \rightarrow \exp{(bJ+cK+dS)} x, \qquad y \rightarrow \exp{(bJ+cK+dS)} y, \qquad z \rightarrow \exp{(bJ+cK+dS)} z.$$

You can permute the variables in the lozenge to obtain an invariant that is symmetric and an invariant that is anti-symmetric.