## Golden Biquaternions, 3 Generations and Spin

This paper was originally published at Google Knol on December 19, 2009

Google discontinued the Knol platform on May 1, 2012.

Contact: sureshemre at gmail

Abstract

### We show that a particular solution of the golden condition $\displaystyle (g - 1/g = 1)$ for biquaternions and its physical interpretation as the fermion predicts exactly 3 generations and 1/2,  3/2,  2 as the only possibilities for particle spin and hints at the composite nature of bosons. The ‘golden biquaternion as fermion’ idea works if there are 3 fundamental variables that are coupled in a pairwise fashion.

Introduction

J.Baez [1] presented mathematical objects (sets of binary trees) found by R.Houston and J.Propp satisfying the golden condition  $g-1/g=1$.  In this note we show that there are other mathematical objects, namely biquaternions, that satisfy the golden condition and they may be very relevant in modeling elementary particles.

A comprehensive bibliography of quaternions and biquaternions in mathematical physics is maintained by A.Gsponer and J.-P. Hurni [2] [3]. Biquaternionic formulation of relativistic quantum mechanics is well established as demonstrated by the references in the analytical bibliography [3] and a recent review by K.Morita [4]. In the biquaternionic formulation of relativistic quantum mechanics the biquaternion represents a space-time point.

The biquaternions also form the core of the “algebrodynamics over complex space” paradigm discussed by V.V.Kassandrov [5] and the references therein. Another major effort pointing out the “algebraic design” of physics was developed by G.M.Dixon [6].

In this note we mention yet another possibility within the algebraic paradigm. We show that biquaternions satisfying the golden condition  $g-1/g=1$  exist. A particular solution $Q$  involving Steiner’s Roman surface may be relevant in modeling 3 generations of fermions.   We also point to a research direction where $Q,Q^2,Q^3$   represent fermions (spin=1/2), supersymmetric particles (spin=3/2) and graviton (spin=2), respectively. In this approach , $Q,Q^2,Q^3$  represent the internal state of the elementary particle rather than the space-time point.

The $Q,Q^2,Q^3$ framework predicts exactly 3 fermion generations and 1/2, 3/2, and 2 as the only possibilities for spin and hints at the composite nature of bosons. These predictions arise naturally from the mathematical properties of $Q,Q^2,Q^3$.

Complexified Quaternions

Let $\displaystyle {\mathbf{1,e_1,e_2,e_3}}$ be the quaternion basis elements and $z_0,z_1,z_2,z_3$ complex numbers.

$\displaystyle q={z_0} \mathbf{1} + {z_1} \mathbf{e}_1 +{z_2} \mathbf{e}_2 + {z_3} \mathbf{e}_3$                                                                             (1)

is a biquaternion (complexified quaternion). The complex numbers $z_0,z_1,z_2,z_3$ commute with the quaternion basis elements $\displaystyle {\mathbf{1,e_1,e_2,e_3}}$.

The quaternion basis elements have the following properties:

${\mathbf{e}_1}^2 = {\mathbf{e}_2}^2 = {\mathbf{e}_3}^2 = {\mathbf{e}_1}{\mathbf{e}_2}{\mathbf{e}_3}= -\mathbf{1}$
$\mathbf{e}_1 \mathbf{e}_2 = \mathbf{e}_3 \qquad$
$\mathbf{e}_2 \mathbf{e}_3 = \mathbf{e}_1 \qquad$
$\mathbf{e}_3 \mathbf{e}_1 = \mathbf{e}_2$                                                                                                               (2)
$\mathbf{e}_1 \mathbf{e}_3 = -\mathbf{e}_2 \qquad$
$\mathbf{e}_2 \mathbf{e}_1 = -\mathbf{e}_3 \qquad$
$\mathbf{e}_3 \mathbf{e}_2 = -\mathbf{e}_1$

The algebra of quaternions $\mathbb{H}$ is a normed division algebra. The algebra of complexified quaternions $\mathbb{P} = \mathbb{H} \otimes \mathbb{C}$ is not a division algebra since the norm can be zero. For biquaternions with non-zero norm the inverse is defined as

$q^{-1}=\bar{q}/N^2(q)$                                                                                                      (3)

$\bar{q} = {z_0}\mathbf{1} - {z_1}\mathbf{e}_1 - {z_2}\mathbf{e}_2 - {z_3}\mathbf{e}_3$                                                                             (4)

$N^2(q) = {z_0}^2 + {z_1}^2 + {z_2}^2 + {z_3}^2$                                                                            (5)

Golden Biquaternions

Let’s see if golden biquaternions exist

$Q ={z_0}\mathbf{1} +{z_1}\mathbf{e}_1 +$ ${z_2}\mathbf{e}_2 +$ ${z_3}\mathbf{e}_3$

where

$Q - 1/Q = \mathbf{1}$

Let

${z_0}^2 + {z_1}^2 + {z_2}^2 + {z_3}^2=-1$                                                                                  (6)

this is possible because $z_0,z_1,z_2,z_3$ are complex numbers, then

$Q - 1/Q = 2 z_0 \mathbf{1}$

The golden condition will be satisfied if

$z_0=1/2$

${z_1}^2 + {z_2}^2 + {z_3}^2=-5/4$

Let’s transform the variables using

$T(z_1,z_2,z_3)=(iyz, izx, ixy)$

where

$i=\sqrt{-1}$

is the imaginary number of complex algebra and $x,y,z$ are real numbers. Under this transformation

${z_1}^2 + {z_2}^2 + {z_3}^2 = -y^2z^2 - z^2x^2 -x^2y^2 = -\frac{5}{4}$                                                (7)

There is a surface known as the Steiner’s Roman Surface which has the intrinsic equation

$y^2z^2 + z^2x^2 + x^2y^2 - r^2 xyz = 0$                                                                       (8)

$Q$ is a golden biquaternion if

$Q = \left(\frac{1}{2}\right) \mathbf{1}+$ $(iyz)\mathbf{e}_1 +$  $(izx)\mathbf{e}_2 + (ixy)\mathbf{e}_3$

has the property that the $(x,y,z)$ vector span a Steiner’s Roman Surface.

Using Eq.(8), we can write Eq.(7) as

$r^2 xyz = \frac{5}{4}$                                                                                                             (9)

Steiner’s Roman surface is the image of a sphere of radius centered at origin under the projection

$(x,y,z) \rightarrow (yz, zx, xy)$                                                                                       (10)

The radius of the transformed sphere can be set to 1 without any loss of generality. We should mention in passing that the Steiner’s Roman surface also shows up as the momentum locus of a discrete Suslov system on $SO(3)$ [7].

Note on the Other Solutions

If we used the transform $T(z_1,z_2,z_3)=(ix, iy, iz)$ we would have obtained the condition
$x^2 + y^2 + z^2 = \frac{5}{4}$                                                                                                (11)

which corresponds to a sphere of radius $r=\frac{\sqrt{5}}{2}$

This solution is not physically interesting. If we used the transforms

$T(z_1,z_2,z_3)=(ix, y, z)$
$T(z_1,z_2,z_3)=(ix, iy, z)$
$T(z_1,z_2,z_3)=(ix, y, iz)$
$T(z_1,z_2,z_3)=(x, iy, z)$
$T(z_1,z_2,z_3)=(x, iy, iz)$
$T(z_1,z_2,z_3)=(x, y, iz)$

the solution surfaces would be hyperboloids.

Physically interesting solution Q results from the transformation of variables

$T(z_1,z_2,z_3)=(iyz, izx, ixy)$

Properties of Steiner’s Roman Surface

Steiner’s Roman surface is a realization of the real projective plane $\mathbb{R}\mathbf{P}^2$ in $\mathbb{R}^3$. It is non-orientable and closed. It has self-intersections. The Steiner’s Roman surface has 6 Whitney singularities, these are points that have no tangent plane.

To visualize the Steiner’s Roman surface please see the images in Figure(1), Figure(2), Figure(3)  and rotate the 3-D image given in [8].

Figure(1) : one of the 4 “lobes”

Figure(2) : one of the 4 “faces”

Figure(3) : one of the 6 pinch-points

Steiner’s Roman surface has 4 bulbous lobes. The overall symmetry is tetrahedral. Suppose we place this object inside a  tetrahedron with least possible volume, the 4 lobes of the Steiner’s Roman surface would be under the 4 vertices of the tetrahedron. Similarly, the 4 “faces” of the Steiner’s Roman surface would correspond to the 4 faces of the tetrahedron. The 6 edges of the tetrahedron would touch the Steiner’s Roman surface at 6 points known as the pinch-points. Between each pair of lobes there is a line of double-points that end in the 6 pinch points. The double-points are the intersection lines of the projection mentioned in Eq.(10). The $(0,0,0)$ is a triple-point.

The boundary lines (double-points, intersection lines, or the lines connecting the origin to the pinch points) are aligned with the 3 axes: $x,y,z$. On the boundary lines there are points like

$(x,0,0)$
$(0,y,0)$
$(0,0,z)$

where $x,y,z$ can be between $-1$  and $1$. The boundary points belong to 2 different lobes simultaneously. The pinch-points are

$(1,0,0)$
$(-1,0,0)$
$(0,1,0)$
$(0,-1,0)$
$(0,0,1)$
$(0,0,-1)$

Note that there are no

$(0,y,z)$
$(x,0,z)$
$(x,y,0)$

points on the Steiner’s Roman surface.

The Steiner’s Roman surface is non-orientable (one-sided) surface. The 4 bulbous lobes correspond to

$(|x|,|y|,|z|)$
$(|x|,-|y|,-|z|)$
$(-|x|,-|y|,|z|)$
$(-|x|,|y|,-|z|)$

Here  the || notation is used to indicate the absolute value of the variables, the signs are made explicit. In other words $|x|,|y|,|z|$ vary between $0$ and $1$.

Note that, on Steiner’s Roman surface $xyz>0$. Either all three $x,y,z$ are positive or else exactly two are negative. It is impossible for exactly one of $x,y,z$ to be zero. It is possible for exactly two of $x,y,z$ to be zero on the boundary lines.

The $xyz>0$ property can be seen from the parametric expression of Steiner’s Roman Surface. In terms of longitude ( ) and latitude ( ) of the projected sphere

$x= r^2 cos(\theta) cos(\Phi) sin(\Phi)$
$y= r^2 sin(\theta) cos(\Phi) sin(\Phi)$                                                                                     (12)
$z= r^2 cos(\theta) sin(\theta) cos^2(\Phi)$

For other interesting properties of the Steiner’s Roman surface see references [9], [10].

Properties of  $Q,Q^2,Q^3$

The golden number

$\phi = \left(\frac{1}{2} + \frac{\sqrt{5}}{2} \right)$

exhibits this property

${\phi}^n + (-\phi)^{-n} = L_n$       $n=1,2,3,4,5,$                                                             (13)

The $L_n$ are the Lucas numbers which are defined by the recurrence relation

$L_n = L_{n-1} + L_{n-2}$                                                                                               (14)

The first two numbers in the Lucas sequence are 1 and 3, i.e., $L_1=1,L_2=3$. Note the similarity to Fibonacci numbers where the first two numbers in the sequence are 0 and 1.

golden biquaternions

$Q=\left(\frac{1}{2}\right) \mathbf{1}+(iyz)\mathbf{e}_1+(izx)\mathbf{e}_2+(ixy)\mathbf{e}_3$

also exhibit this property with one significant difference.

${Q}^n + (-Q)^{-n} = L_n$        $n=1,2,3$                                                                     (15)

or explicitly,

$Q-1/Q=1$

$Q^2+1/Q^2=3$                                                                                                      (16)

$Q^3-1/Q^3=4$

To see this we need to work out the powers of Q. One can do this by brute force and paying attention to the multiplication table of the quaternion basis elements. There is also a formula

$AB=a_0 b_0-\mathbf{A}\cdot \mathbf{B}+a_0\mathbf{A}+\mathbf{B} b_0+i\mathbf{A}\times\mathbf{B}$                                                   (17)

where A and B are biquaternions, $\mathbf{A}$ and $\mathbf{B}$ are the vector-like part of the biquaternions involving quaternion basis elements $\mathbf{e_1}$, $\mathbf{e_2}$, $\mathbf{e_3}$. The $\cdot$ and $\times$ are the 3-dimensional scalar and vector products, respectively. The product of two biquaternions is a biquaternion.

Explicitly,

${Q} = (1/2)\mathbf{1}+[(iyz)\mathbf{e}_1+(izx)\mathbf{e}_2+(ixy)\mathbf{e}_3]$           $N^2(Q) = -1$
${Q}^2=(3/2)\mathbf{1}+[(iyz)\mathbf{e}_1+(izx)\mathbf{e}_2+(ixy)\mathbf{e}_3]$           $N^2(Q^2) = 1$                     (18)
${Q}^3=(2)\mathbf{1}+2[(iyz)\mathbf{e}_1+(izx)\mathbf{e}_2+(ixy)\mathbf{e}_3]$              $N^2(Q^3) = -1$

Recall that (x,y,z) span a Steiner’s Roman surface

$y^2z^2 + z^2x^2 + x^2y^2 = \frac{5}{4}$                                                                                         (19)

which were used to obtain the results shown in Eq.(18). One can also derive the results

$Q^2 = 1 + Q$                                                                                                            (20)

$Q^3 = Q + Q^2 = 1 + 2Q$                                                                                        (21)

Spin

In a physical interpretation of $Q,Q^2,Q^3$ the spin would be indicated by the $z_0$ component. Fermions (spin=1/2) would be represented by $Q$. The spin=3/2 particles would be represented by $Q^2$ and the graviton (spin=2) would be represented by $Q^3$.

Vector-bosons with spin=1 would be represented by $Q^2-Q$ and $Q^3-2Q$

This is one of the specific predictions of the $Q,Q^2,Q^3$ framework.

Fermion Generations

Another specific prediction of the $Q,Q^2,Q^3$  framework is exactly 3 generations for fermions.

Let’s consider the space spanned by a vector of 3 variables (S,R,T) whose trajectories always remain on a Steiner’s Roman surface excluding the triple-point and the double-points and refer to it as the SRT-manifold.

The SRT-manifold excludes the (0,0,0) point; the pinch-points (1,0,0), (-1,0,0), (0,1,0), (0,-1,0), (0,0,1), (0,0,-1); and the boundary lines (|S|,0,0), (-|S|,0,0), (0,|R|,0), (0,-|R|,0), (0,0,|T|), (0,0,-|T|) of the Steiner Roman surface. The absolute values |S|,|R|,|T| can be between 0 and 1.

We point to a research direction where one represents the elementary particles with $Q,Q^2,Q^3$  whose $\mathbf{e_1},\mathbf{e_2},\mathbf{e_3}$ components are restricted to a specific zone of the SRT-manifold. Each point on the SRT-manifold represents a unique internal state of the particle. The hypothesis is that all quantum numbers except spin can be represented by mathematical constructs defined on the SRT-manifold. We propose SRT-manifold as an alternative to the Calabi-Yau manifold.

We also put forward the hypothesis that the 6 boundary lines of the SRT-manifold are attractors. The state vector (S,R,T) can transition to another lobe of the SRT-manifold through a boundary line but it cannot remain on the boundary line.

Each lobe of the SRT-manifold has 3 distinct zones that are under the influence of 3 of the 6 attractors. The zones are distinguished by which 2 of the 3 attractors is dominant.

Lobe 1: (|S|,|R|,|T|)

Zone 1: dominated by the attractors (|S|,0,0) and (0,|R|,0)
Zone 2: dominated by the attractors (|S|,0,0) and (0,0,|T|)
Zone 3: dominated by the attractors (0,|R|,0) and (0,0,|T|)

Lobe 2: (|S|,-|R|,-|T|

Zone 1: dominated by the attractors (|S|,0,0) and (0,-|R|,0)
Zone 2: dominated by the attractors (|S|,0,0) and (0,0,-|T|)
Zone 3: dominated by the attractors (0,-|R|,0) and (0,0,-|T|)

Lobe 3: (-|S|,-|R|,|T|)

Zone 1: dominated by the attractors (-|S|,0,0) and (0,-|R|,0)
Zone 2: dominated by the attractors (-|S|,0,0) and (0,0,|T|)
Zone 3: dominated by the attractors (0,-|R|,0) and (0,0,|T|)

Lobe 4: (-|S|,|R|,-|T|)

Zone 1: dominated by the attractors (-|S|,0,0) and (0,|R|,0)
Zone 2: dominated by the attractors (-|S|,0,0) and (0,0,-|T|)
Zone 3: dominated by the attractors (0,|R|,0) and (0,0,-|T|)

The 4 lobes of the SRT-manifold can represent the 4 rows of the fermion table shown below. The 3 zones in each lobe represent the 3 fermion generations.

 $\nu_e$ $\nu_\mu$ $\nu_\tau$ $e$ $\mu$ $\tau$ $d$ $s$ $b$ $u$ $c$ $t$
 $Q_{[1,1]}$ $Q_{[1,2]}$ $Q_{[1,3]}$ $Q_{[2,1]}$ $Q_{[2,2]}$ $Q_{[2,3]}$ $Q_{[3,1]}$ $Q_{[3,2]}$ $Q_{[3,3]}$ $Q_{[4,1]}$ $Q_{[4,2]}$ $Q_{[4,3]}$

where Q[ij] is the golden biquaternion representing fermions and i=1,2,3,4 and j=1,2,3  refer to the lobe number and zone number, respectively.

Discussion

Key to the success of the $Q,Q^2,Q^3$  framework is the assumption that there are 3 fundamental variables internal to the elementary particle. The second important assumption is that these 3 fundamental variables are coupled in a pairwise fashion. The third factor is obviously the golden condition which can be thought of as the primordial generator.

Representation of spin and 3 generations in a common framework is a good start but there is much more work to be done in the SRT program. We need to come up with the mathematical constructs defined on the SRT-manifold to explain all the quantum numbers as well as the masses of the elementary particles. Where is time in the $Q,Q^2,Q^3$  framework? Where is dynamics in the $Q,Q^2,Q^3$ framework? These questions remain.

How do we represent the motion of particles in space-time by the internal state variables? What is the relationship between the internal space and the external space? SRT program needs to resolve this fundamental question and explain the 4 fundamental interactions: electromagnetic, weak, strong, and gravitional forces.

References

[1] John Baez, “This Week’s Finds in Mathematical Physics (Week 203) (Feb 28, 2004)“,

[2] Andre Gsponer and Jean-Pierre Hurni, “Quaternions in mathematical physics (1): Alphabetical bibliography”, arXiv:math-ph/0510059.

[3] Andre Gsponer and Jean-Pierre Hurni, “Quaternions in mathematical physics (2): Analytical bibliography”, arXiv:math-ph/0511092.

[4] Katsusada Morita, “Quaternions, Lorentz Group and the Dirac Theory”, Progress of Theoretical Physics, Vol.117, No.3 (2007)

[5] Vladimir V. Kassandrov, “Algebrodynamics over Complex Space and Phase Extension of the Minkowski Geometry”, arXiv:gr-qc/0602088. (2006)

[6] Geoffrey M.Dixon, “Division Algebras: Octonions, Quaternions, Complex Numbers and the Algebraic Design of Physics”, Springer (1994), Kluwer Academic Publishers (2002), ISBN: 0792328906 / 9780792328902

[7] Y.N.Federov and D.V.Zenkov, “Discrete Nonholonomic LL Systems on Lie Groups”, arXiv:math/0409415. (2004)

[9] E.V.Ferapontov, “Integrable Systems in Projective Differential Geometry”, arXiv:math/9903150. (1999)