Masteries · Jason GlowneyArithmoí Book 1

Masteries

1.8.6 The Theory of Light

This sec­tion re­turns to some of the con­cepts cov­ered in Sec­tion 1.7.2, in­volv­ing as­pects of the di­men­sion­al trans­for­ma­tion map­ping ℂ1:→ℝ2\mathbb{C}^{1}: \rightarrow \mathbb{R}^{2}. This will help us better un­der­stand the nature of light and con­sid­er how the math­e­mat­i­cal frame­work we have been work­ing with is causal to the phys­i­cal phe­nom­e­nol­o­gy per­ceived around us. We start by look­ing at the first cone number in the pos­i­tive ori­en­ta­tion from the iden­ti­ty, (−1)14,( - 1)^{\frac{1}{4}},which has the fol­low­ing equiv­a­lence ex­pres­sion

(−1)14=12+i2,\begin{array}{r} ( - 1)^{\frac{1}{4}} = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}, \end{array}
(1.8.244)

and point on the com­plex unit circle is de­pict­ed in Figure 1.8.10. Based on the two pre­vi­ous sec­tions, this number should stand out as the pos­i­tive min­i­mal el­e­ment of the in­scribed oc­ta­gon, which, in math­e­mat­i­cal no­ta­tion, is

𝓂ℯ[+18]=(e2iπ)(18)=(−1)14,\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{1}{8} \right\rbrack} = \left( e^{2i\pi} \right)^{\left( \frac{1}{8} \right)} = ( - 1)^{\frac{1}{4}}, \end{array}
(1.8.245)

and this cone number be­longs to the set of Lie al­ge­bra gen­er­a­tors Y(ℤ8)Y\left( \mathbb{Z}_{8} \right)

Y(ℤ8)={(−1)0,(−1)14,(−1)12,(−1)34,(−1)1,−(−1)14,−(−1)12,−(−1)34}.\begin{array}{r} Y\left( \mathbb{Z}_{8} \right) = \left\{ ( - 1)^{0},( - 1)^{\frac{1}{4}},( - 1)^{\frac{1}{2}},( - 1)^{\frac{3}{4}},( - 1)^{1},{- ( - 1)}^{\frac{1}{4}},{- ( - 1)}^{\frac{1}{2}},{- ( - 1)}^{\frac{3}{4}} \right\}. \end{array}
(1.8.246)
Figure
Figure 1.8.10 – The cone number (−1)14( - 1)^{\frac{1}{4}} on the com­plex unit circle.

Each el­e­ment in the set above has an 8×88 \times 8 matrix Lie al­ge­bra gen­er­a­tor rep­re­sen­ta­tion, which can be seen using the ex­am­ple for the al­ge­bra Y(ℤ8)2=(−1)14Y\left( \mathbb{Z}_{8} \right)_{2} = ( - 1)^{\frac{1}{4}}

Y(ℤ8)2=(0000000110000000010000000010000000010000000010000000010000000010),\begin{array}{r} Y\left( \mathbb{Z}_{8} \right)_{2} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix}, \end{array}
(1.8.247)

where the set of el­e­ments A(Y(ℤ8))A\left( Y\left( \mathbb{Z}_{8} \right) \right) can be gen­er­at­ed by ((−1)14)k\left( ( - 1)^{\frac{1}{4}} \right)^{k}, given that by de­fault, the pos­i­tive and neg­a­tive-valued min­i­mal el­e­ments are pri­ma­ry gen­er­a­tors, 𝓂ℯ[±18].\mathcal{m}\mathcal{e}_{\left\lbrack \pm \frac{1}{8} \right\rbrack}.The Y(ℤ8)2Y\left( \mathbb{Z}_{8} \right)_{2} matrix has eigen­val­ues rep­re­sent­ing vertex po­si­tions in the com­plex unit circle, com­pris­ing the set

λ(ℤ8)={1,(−1)14,i,(−1)34,−1,−(−1)14,−i,−(−1)34}.\begin{array}{r} \lambda\left( \mathbb{Z}_{8} \right) = \left\{ 1,( - 1)^{\frac{1}{4}},i,( - 1)^{\frac{3}{4}}, - 1, - ( - 1)^{\frac{1}{4}}, - i, - ( - 1)^{\frac{3}{4}} \right\}. \end{array}
(1.8.248)

The pri­ma­ry gen­er­a­tor el­e­ment subset 𝓅ℊ(Y(ℤ8))\mathcal{pg}\left( Y\left( \mathbb{Z}_{8} \right) \right) rep­re­sents the cone num­bers ∈Y(ℤ8)\in Y\left( \mathbb{Z}_{8} \right)

𝓅ℊ(A(Y(ℤ8)))={(−1)14,(−1)34,−(−1)14,−(−1)34},\begin{array}{r} \mathcal{pg}\left( A\left( Y\left( \mathbb{Z}_{8} \right) \right) \right) = \left\{ ( - 1)^{\frac{1}{4}},( - 1)^{\frac{3}{4}}, - ( - 1)^{\frac{1}{4}},{- ( - 1)}^{\frac{3}{4}} \right\}, \end{array}
(1.8.249)

and the re­main­ing el­e­ments of A(Y(ℤ8))A\left( Y\left( \mathbb{Z}_{8} \right) \right) are the axis num­bers, which define the in­scribed square geo­met­ric in­te­ger Y(ℤ4)Y\left( \mathbb{Z}_{4} \right),

A(Y(ℤ8))−𝓅ℊ(A(Y(ℤ8)))=Y(ℤ4)={i,−1,−i,1}.\begin{array}{r} A\left( Y\left( \mathbb{Z}_{8} \right) \right)- \mathcal{pg}\left( A\left( Y\left( \mathbb{Z}_{8} \right) \right) \right) = Y\left( \mathbb{Z}_{4} \right) = \left\{ i, - 1, - i,1 \right\}. \end{array}
(1.8.250)

The 8×88 \times 8 matrix rep­re­sen­ta­tion of the (−1)14( - 1)^{\frac{1}{4}} pri­ma­ry gen­er­a­tor has a total of 82=648^{2} = 64 en­tries, which can take on values of either 11 or 00, sig­ni­fy­ing bits of in­for­ma­tion. As a pri­ma­ry gen­er­a­tor, the spa­tial arrange­ment of the non-zero en­tries also en­codes the entire Y(ℤ8)Y\left( \mathbb{Z}_{8} \right) group by way of the mul­ti­pli­ca­tion op­er­a­tion ((−1)14)k\left( ( - 1)^{\frac{1}{4}} \right)^{k}. As a group, Y(ℤ8)Y\left( \mathbb{Z}_{8} \right) is en­cod­ed in 6464 bits of in­for­ma­tion, which can be seen when we con­sid­er the eight cyclic shifts of the Lie al­ge­bra gen­er­a­tors in binary

Y(ℤ8)1=000,   Y(ℤ8)2=001,   Y(ℤ8)3=010,   Y(ℤ8)4=011,Y\left( \mathbb{Z}_{8} \right)_{1} = 000,\ \ \ Y\left( \mathbb{Z}_{8} \right)_{2} = 001,\ \ \ Y\left( \mathbb{Z}_{8} \right)_{3} = 010,\ \ \ Y\left( \mathbb{Z}_{8} \right)_{4} = 011,
Y(ℤ8)5=100,   Y(ℤ8)6=101,   Y(ℤ8)7=110,Y(ℤ8)8=111,\begin{array}{r} Y\left( \mathbb{Z}_{8} \right)_{5} = 100,\ \ \ Y\left( \mathbb{Z}_{8} \right)_{6} = 101,\ \ \ Y\left( \mathbb{Z}_{8} \right)_{7} = 110,Y\left( \mathbb{Z}_{8} \right)_{8} = 111, \end{array}
(1.8.251)

and given that there are eight non-zero el­e­ments in each cyclic shift matrix, there are 82=648^{2} = 64 bits of in­for­ma­tion.

Pre­vi­ous­ly in the chap­ter, we con­sid­ered di­men­sion­al trans­forms of num­bers, re­call­ing the fol­low­ing ex­am­ple for the di­men­sion­al map­ping from a com­plex number to a real matrix

ℂ1:→ℝ2=i:→(0−110).\begin{array}{r} \mathbb{C}^{1}: \rightarrow \mathbb{R}^{2} = i: \rightarrow \begin{pmatrix} 0 & - 1 \\ 1 & 0 \end{pmatrix}. \end{array}
(1.8.252)

We apply the same di­men­sion­al trans­form map­ping in the op­po­site di­rec­tion for our min­i­mal el­e­ment Y(ℤ8)2=(−1)14Y\left( \mathbb{Z}_{8} \right)_{2} = ( - 1)^{\frac{1}{4}}, by using the map­ping no­ta­tion Y(ℤ[+8])2:→Y(ℤ[±4])2,Y\left( \mathbb{Z}_{\lbrack + 8\rbrack} \right)_{2}: \rightarrow Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{2},

Y(ℤ[+8])2:→Y(ℤ[±4])2=(0000000110000000010000000010000000010000000010000000010000000010):→(000−1100001000010).\begin{array}{r} Y\left( \mathbb{Z}_{\lbrack + 8\rbrack} \right)_{2}: \rightarrow Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{2} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix}: \rightarrow \begin{pmatrix} 0 & 0 & 0 & - 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}. \end{array}
(1.8.253)

We have gone from an 8×88 \times 8 sized matrix to a 4×44 \times 4 matrix and in­tro­duced a neg­a­tive-valued in­te­ger in the a4,1a_{4,1} po­si­tion of the mapped result. As we know, the Y(ℤ4)2Y\left( \mathbb{Z}_{4} \right)_{2} matrix rep­re­sents the in­scribed square, and a look at the eigen­val­ues of the Y(ℤ[±4])2Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{2} matrix, re­veals the cone num­bers

λ(Y(ℤ[±4])2)={(−1)14,(−1)34,−(−1)14,−(−1)34}.\begin{array}{r} \lambda\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{2} \right) = \left\{ ( - 1)^{\frac{1}{4}},( - 1)^{\frac{3}{4}}, - ( - 1)^{\frac{1}{4}},{- ( - 1)}^{\frac{3}{4}} \right\}. \end{array}
(1.8.254)

The eigen­val­ues sig­ni­fy that the in­scribed square has been ro­tat­ed by iπ4\frac{i\pi}{4} ra­di­ans when com­pared to Y(ℤ4)2Y\left( \mathbb{Z}_{4} \right)_{2}, where Y(ℤ[±4])2Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{2} is seen in the fol­low­ing figure as the red square, and Y(ℤ4)2Y\left( \mathbb{Z}_{4} \right)_{2} as the blue square.

Figure
Figure 1.8.11 –The in­scribed po­si­tion of Y(ℤ4)2Y\left( \mathbb{Z}_{4} \right)_{2} in blue, is ro­tat­ed to the Y(ℤ[±4])2Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{2} po­si­tion in red.

The Y(ℤ[±4])2Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{2} gen­er­a­tor forms the oc­ta­gon by way of its suc­ces­sive powers, (Y(ℤ[±4])2)k,\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{2} \right)^{k},which gen­er­ates the fol­low­ing set of eight unique el­e­ments

A(Y(ℤ[±4])2)={(1000010000100001)(000−1100001000010),(00−10000−110000100),(0−10000−10000−11000),(−10000−10000−10000−1),(0001−10000−10000−10),(00100001−10000−100),(010000100001−1000)}.\begin{array}{r} A\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{2} \right) = \left\{ \begin{array}{r} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & 0 & 0 & - 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & - 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \\ 1 & 0 & 0 & 0 \end{pmatrix}, \\ \begin{pmatrix} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \end{pmatrix},\begin{pmatrix} 0 & 0 & 0 & 1 \\ - 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ - 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ - 1 & 0 & 0 & 0 \end{pmatrix} \end{array} \right\}. \end{array}
(1.8.255)

In the set above the eigen­val­ues of the Y(ℤ[±4])1Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{1}, Y(ℤ[±4])3,Y(ℤ[±4])5Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{3},Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{5}el­e­ments are

λ(Y(ℤ[±4])1)={1},  λ(Y(ℤ[±4])3)={i,−i} ,  λ(Y(ℤ[±4])5)={1},\begin{array}{r} \lambda\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{1} \right) = \left\{ 1 \right\},\ \ \lambda\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{3} \right) = \left\{ i, - i \right\}\ ,\ \ \lambda\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{5} \right) = \left\{ 1 \right\}, \end{array}
(1.8.256)

which, to­geth­er with the eigen­val­ues of the pri­ma­ry gen­er­a­tor

λ(Y(ℤ[±4])2)={(−1)14,(−1)34,−(−1)14,−(−1)34},\begin{array}{r} \lambda\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{2} \right) = \left\{ ( - 1)^{\frac{1}{4}},( - 1)^{\frac{3}{4}}, - ( - 1)^{\frac{1}{4}},{- ( - 1)}^{\frac{3}{4}} \right\}, \end{array}
(1.8.257)

en­com­pass the λ(ℤ8)\lambda\left( \mathbb{Z}_{8} \right) eigen­val­ue set.

Figure
Figure 1.8.12 –The vertex values as­so­ci­at­ed with the A(Y(ℤ[±4])2)A\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{2} \right) set of el­e­ments.

We show the in­scribed geom­e­try of the reg­u­lar oc­ta­gon, the result of the Y(ℤ[±4])2Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{2} pri­ma­ry gen­er­a­tor's ro­ta­tions and rep­re­sent the el­e­ments in the A(Y(ℤ[±4])2)A\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{2} \right) set. It may be coun­ter­in­tu­itive at first, but if we first con­sid­er the Y(ℤ[±4])3Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{3} and Y(ℤ[±4])4Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{4} el­e­ments

Y(ℤ[±4])3=(00−10000−110000100),  Y(ℤ[±4])4=(0−10000−10000−11000),\begin{array}{r} Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{3} = \begin{pmatrix} 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix},\ \ Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{4} = \begin{pmatrix} 0 & - 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \\ 1 & 0 & 0 & 0 \end{pmatrix}, \end{array}
(1.8.258)

to rep­re­sent the vertex lo­ca­tions at ii and (−1)34( - 1)^{\frac{3}{4}} on the com­plex unit circle, the con­sis­ten­cy as vertex lo­ca­tions is ap­pre­ci­at­ed by squar­ing each

(Y(ℤ[±4])3)2=(−10000−10000−10000−1),  (Y(ℤ[±4])4)2=(00100001−10000−100).\begin{array}{r} \left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{3} \right)^{2} = \begin{pmatrix} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \end{pmatrix},\ \ \left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{4} \right)^{2} = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ - 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \end{pmatrix}. \end{array}
(1.8.259)

So, (Y(ℤ[±4])3)2\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{3} \right)^{2} is found to equal the neg­a­tive iden­ti­ty, which is the ex­pect­ed result given that i2=−1,i^{2} = - 1,and then we find (Y(ℤ[±4])4)2=−Y(ℤ[±4])3\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{4} \right)^{2} = - Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{3}, rep­re­sent­ing −i.- i.The A(Y(ℤ[±4]))A\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right) \right) set cycles twice in terms of shifts before re­turn­ing to the iden­ti­ty el­e­ment. The eight Lie al­ge­bra gen­er­a­tors of the set A(Y(ℤ[±4]))A\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right) \right) can also be rep­re­sent­ed as the fol­low­ing binary set

A⇀(Y(ℤ[±4])j)={000,   001,   010,   011,100,   101,   110,   111},\begin{array}{r} \overset{\rightharpoonup}{A}\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{j} \right) = \left\{ \begin{array}{r} 000,\ \ \ 001,\ \ \ 010,\ \ \ 011, \\ 100,\ \ \ 101,\ \ \ 110,\ \ \ 111 \end{array} \right\}, \end{array}
(1.8.260)

where the di­rec­tion­al arrow for A⇀\overset{\rightharpoonup}{A} sig­ni­fies the above is a shift­ing set. Given that non-zero en­tries take on two values, (−1,1)( - 1,1), there are eight bits of in­for­ma­tion in each Lie al­ge­bra gen­er­a­tor ∈A(Y(ℤ[±4]))\in A\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right) \right), which, when mul­ti­plied by the number of el­e­ments in A⇀(Y(ℤ[±4])j)\overset{\rightharpoonup}{A}\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{j} \right), we find 82=648^{2} = 64 bits of in­for­ma­tion are present in the group, re­veal­ing that a con­ser­va­tion of in­for­ma­tion exists in the map­ping Y(ℤ[+8])2:→Y(ℤ[±4])2.Y\left( \mathbb{Z}_{\lbrack + 8\rbrack} \right)_{2}: \rightarrow Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{2}.As ex­pect­ed, in the A(Y(ℤ[±4]))A\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right) \right) set there are also four pri­ma­ry gen­er­a­tors

𝓅ℊ(A(Y(ℤ[±4])))={(000−1100001000010),(0−10000−10000−11000),(0001−10000−10000−10)(010000100001−1000)},\begin{array}{r} \mathcal{pg}\left( A\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right) \right) \right) = \left\{ \begin{array}{r} \begin{pmatrix} 0 & 0 & 0 & - 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix},\begin{pmatrix} 0 & - 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \\ 1 & 0 & 0 & 0 \end{pmatrix}, \\ \begin{pmatrix} 0 & 0 & 0 & 1 \\ - 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ - 1 & 0 & 0 & 0 \end{pmatrix} \end{array} \right\}, \end{array}
(1.8.261)

which differ from the non-gen­er­a­tors based on their lack of sym­me­try for non-zero en­tries about the di­ag­o­nal and the odd number dis­tri­b­u­tion of pos­i­tive to neg­a­tive values. These pri­ma­ry gen­er­a­tors rep­re­sent cone num­bers given that when squared, all the re­sult­ing el­e­ments are imag­i­nary-valued Lie al­ge­bras

𝓅ℊ(A(Y(ℤ[±4])2))={(00−10000−110000100),(00100001−10000−100)(00−10000−110000100),(00100001−10000−100)},\begin{array}{r} \mathcal{pg}\left( A\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)^{2} \right) \right) = \left\{ \begin{array}{r} \begin{pmatrix} 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ - 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \end{pmatrix} \\ \begin{pmatrix} 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ - 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \end{pmatrix} \end{array} \right\}, \end{array}
(1.8.262)

ev­i­denced by the fact that when squared, the el­e­ments equal the neg­a­tive iden­ti­ty

(00−10000−110000100)2=(00100001−10000−100)2=(−10000−10000−10000−1).\begin{array}{r} \begin{pmatrix} 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}^{2} = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ - 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \end{pmatrix}^{2} = \begin{pmatrix} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \end{pmatrix}. \end{array}
(1.8.263)

The Y(ℤ4)Y\left( \mathbb{Z}_{4} \right) geo­met­ric in­te­ger set is con­tained in the A(Y(ℤ[±4]))A\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right) \right) set, A(Y(ℤ4))∈A(Y(ℤ[±4]))A\left( Y\left( \mathbb{Z}_{4} \right) \right) \in A\left( Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right) \right), where A(Y(ℤ4))A\left( Y\left( \mathbb{Z}_{4} \right) \right) is the subset

A(Y(ℤ4))={(1000010000100001),(00−10000−110000100),(−10000−10000−10000−1),(00100001−10000−100)}.\begin{array}{r} A\left( Y\left( \mathbb{Z}_{4} \right) \right) = \left\{ \begin{array}{r} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix},\begin{pmatrix} 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}, \\ \begin{pmatrix} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & - 1 \end{pmatrix},\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ - 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \end{pmatrix} \end{array} \right\}. \end{array}
(1.8.264)

Geo­met­ri­cal­ly A(Y(ℤ4))A\left( Y\left( \mathbb{Z}_{4} \right) \right) is a subset since the ver­tices of the in­scribed square, Y(ℤ4)Y\left( \mathbb{Z}_{4} \right), are aligned on the four axes points of the com­plex unit circle and are lim­it­ed to vertex ro­ta­tions of iπ2\frac{i\pi}{2} ra­di­ans, there­fore, Y(ℤ4)Y\left( \mathbb{Z}_{4} \right) can be ro­tat­ed in four dif­fer­ent ways to align its ver­tices with the in­scribed oc­ta­gon. Next, we reduce the size of the Y(ℤ[±4])2Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{2} di­men­sion while con­serv­ing its in­for­ma­tion con­tent with the map­ping

Y(ℤ[±4])2:→Y(ℤ[±i2])2=(000−1100001000010):→(01i0),\begin{array}{r} Y\left( \mathbb{Z}_{\lbrack \pm 4\rbrack} \right)_{2}: \rightarrow Y\left( \mathbb{Z}_{\lbrack \pm i2\rbrack} \right)_{2} = \begin{pmatrix} 0 & 0 & 0 & - 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}: \rightarrow \begin{pmatrix} 0 & 1 \\ i & 0 \end{pmatrix}, \end{array}
(1.8.265)

where it is easy to verify that (01i0)\begin{pmatrix} 0 & 1 \\ i & 0 \end{pmatrix} matrix rep­re­sents a cone number by its square

(01i0)2=(i00i),\begin{array}{r} \begin{pmatrix} 0 & 1 \\ i & 0 \end{pmatrix}^{2} = \begin{pmatrix} i & 0 \\ 0 & i \end{pmatrix}, \end{array}
(1.8.266)

and more specif­i­cal­ly, that

(01i0)=(−1)14,\begin{array}{r} \begin{pmatrix} 0 & 1 \\ i & 0 \end{pmatrix} = ( - 1)^{\frac{1}{4}}, \end{array}
(1.8.267)

since the pos­i­tive ro­ta­tion ((−1)14)2=i.\left( ( - 1)^{\frac{1}{4}} \right)^{2} = i.The A(Y(ℤ[±i2])j)A\left( Y\left( \mathbb{Z}_{\lbrack \pm i2\rbrack} \right)_{j} \right) set of Lie al­ge­bra gen­er­a­tors is the fol­low­ing

A(Y(ℤ[±i2])j)={(1001),   (01i0),   (i00i),   (0i−10),(−100−1),   (0−1−i0),   (−i00−i),   (0−i10)},\begin{array}{r} A\left( Y\left( \mathbb{Z}_{\lbrack \pm i2\rbrack} \right)_{j} \right) = \left\{ \begin{array}{r} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 1 \\ i & 0 \end{pmatrix},\ \ \ \begin{pmatrix} i & 0 \\ 0 & i \end{pmatrix},\ \ \ \begin{pmatrix} 0 & i \\ - 1 & 0 \end{pmatrix}, \\ \begin{pmatrix} - 1 & 0 \\ 0 & - 1 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & - 1 \\ - i & 0 \end{pmatrix},\ \ \ \begin{pmatrix} - i & 0 \\ 0 & - i \end{pmatrix},\ \ \ \begin{pmatrix} 0 & - i \\ 1 & 0 \end{pmatrix} \end{array} \right\}, \end{array}
(1.8.268)

which, like our other sets, con­tains a total of eight el­e­ments rep­re­sent­ing the oc­ta­gon. The pri­ma­ry gen­er­a­tors have the fol­low­ing cone number values for their eigen­val­ues

λ(Y(ℤ[±2i])2,6)={(−1)14,−(−1)14},\begin{array}{r} \lambda\left( Y\left( \mathbb{Z}_{\lbrack \pm 2i\rbrack} \right)_{2,6} \right) = \left\{ ( - 1)^{\frac{1}{4}},{- ( - 1)}^{\frac{1}{4}} \right\}, \end{array}
(1.8.269)
λ(Y(ℤ[±2i])4,8)={(−1)34,−(−1)34},\begin{array}{r} \lambda\left( Y\left( \mathbb{Z}_{\lbrack \pm 2i\rbrack} \right)_{4,8} \right) = \left\{ ( - 1)^{\frac{3}{4}},{- ( - 1)}^{\frac{3}{4}} \right\}, \end{array}
(1.8.270)

while the re­main­ing el­e­ments have eigen­val­ues that are equal to the axis values of the in­scribed square

λ(Y(ℤ[±2i])1)={1,1},  λ(Y(ℤ[±2i])3)={i,i},\lambda\left( Y\left( \mathbb{Z}_{\lbrack \pm 2i\rbrack} \right)_{1} \right) = \left\{ 1,1 \right\},\ \ \lambda\left( Y\left( \mathbb{Z}_{\lbrack \pm 2i\rbrack} \right)_{3} \right) = \left\{ i,i \right\},
λ(Y(ℤ[±2i])5)={−1,−1},  λ(Y(ℤ[±2i])7)={−i,−i},\begin{array}{r} \lambda\left( Y\left( \mathbb{Z}_{\lbrack \pm 2i\rbrack} \right)_{5} \right) = \left\{ - 1, - 1 \right\},\ \ \lambda\left( Y\left( \mathbb{Z}_{\lbrack \pm 2i\rbrack} \right)_{7} \right) = \left\{ - i, - i \right\}, \end{array}
(1.8.271)

where these axis num­bers have only one unique eigen­val­ue. This lack of unique­ness of eigen­val­ues is a math­e­mat­i­cal in­di­ca­tor that these are not pri­ma­ry gen­er­a­tors of the A(Y(ℤ[±2i])j)A\left( Y\left( \mathbb{Z}_{\lbrack \pm 2i\rbrack} \right)_{j} \right) set. As the reader may have guessed, we can also fur­ther reduce the di­men­sions of the matrix rep­re­sen­ta­tion for the A(Y(ℤ[±i2])j)A\left( Y\left( \mathbb{Z}_{\lbrack \pm i2\rbrack} \right)_{j} \right) set to the 1×11 \times 1 form of the set A(Y(ℤ[±(−1)(14)ϵ])j)A\left( Y\left( \mathbb{Z}_{\left\lbrack \pm ( - 1)^{\left( \frac{1}{4} \right)}\epsilon \right\rbrack} \right)_{j} \right). This rep­re­sents the com­plete trans­for­ma­tion­al map­ping of the ℤ8\mathbb{Z}_{8} geo­met­ric in­te­ger to the com­plex plane and the set is com­prised of the fol­low­ing Lie al­ge­bra gen­er­a­tors

A(Y(ℤ[±(−1)(14)ϵ])j)={(1),   ((−1)14),   ((−1)12),   ((−1)34),(−1),   (−(−1)14),   (−(−1)12),   (−(−1)34)},\begin{array}{r} A\left( Y\left( \mathbb{Z}_{\left\lbrack \pm ( - 1)^{\left( \frac{1}{4} \right)}\epsilon \right\rbrack} \right)_{j} \right) = \left\{ \begin{array}{r} (1),\ \ \ \left( ( - 1)^{\frac{1}{4}} \right),\ \ \ \left( ( - 1)^{\frac{1}{2}} \right),\ \ \ \left( ( - 1)^{\frac{3}{4}} \right), \\ ( - 1),\ \ \ \left( {- ( - 1)}^{\frac{1}{4}} \right),\ \ \ \left( - ( - 1)^{\frac{1}{2}} \right),\ \ \ \left( {- ( - 1)}^{\frac{3}{4}} \right) \end{array} \right\}, \end{array}
(1.8.272)

for which, each el­e­ment in the set is a point on the com­plex unit circle, rep­re­sent­ing the eigen­val­ues of the ℤ8\mathbb{Z}_{8} geo­met­ric in­te­ger. The 8×88 \times 8 matrix rep­re­sen­ta­tion of the min­i­mal el­e­ment 𝓂ℯ[+18]\mathcal{m}\mathcal{e}_{\left\lbrack + \frac{1}{8} \right\rbrack} we began with, is a com­plex number, and the di­men­sion-re­duc­ing trans­form

𝓂ℯ[+18]:(Y(ℤ8)1:→Y(ℤ[±(−1)(14)ϵ])1)=(0000000110000000010000000010000000010000000010000000010000000010):→((−1)14),\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{1}{8} \right\rbrack}:\left( Y\left( \mathbb{Z}_{8} \right)_{1}: \rightarrow Y\left( \mathbb{Z}_{\left\lbrack \pm ( - 1)^{\left( \frac{1}{4} \right)}\epsilon \right\rbrack} \right)_{1} \right) = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix}: \rightarrow \left( ( - 1)^{\frac{1}{4}} \right), \end{array}
(1.8.273)

makes this math­e­mat­i­cal­ly clear.

Pre­vi­ous­ly men­tioned was that the min­i­mal el­e­ment 𝓂ℯ[+18]=(e2iπ)18\mathcal{m}\mathcal{e}_{\left\lbrack + \frac{1}{8} \right\rbrack} = \left( e^{2i\pi} \right)^{\frac{1}{8}} is phys­i­cal­ly as­so­ci­at­ed with light. Ex­pand­ing on this con­nec­tion is ap­pro­pri­ate with our back­ground from the pre­vi­ous sec­tion on the geo­met­ric in­te­gers and primes. In physics, light, specif­i­cal­ly elec­tro-mag­net­ism, is rep­re­sent­ed in the form of the U(1)U(1) group, which is con­sis­tent with the result we ar­rived at using our di­men­sion­al trans­form Y(ℤ8)1:→Y(ℤ[±(−1)(14)ϵ])1.Y\left( \mathbb{Z}_{8} \right)_{1}: \rightarrow Y\left( \mathbb{Z}_{\left\lbrack \pm ( - 1)^{\left( \frac{1}{4} \right)}\epsilon \right\rbrack} \right)_{1}.We have seen that there is a fun­da­men­tal geom­e­try that is cen­tral to the char­ac­ter­is­tics of groups, and the case for A(Y(ℤ[±(−1)(14)ϵ])j)A\left( Y\left( \mathbb{Z}_{\left\lbrack \pm ( - 1)^{\left( \frac{1}{4} \right)}\epsilon \right\rbrack} \right)_{j} \right) is no ex­cep­tion. Light is based on the first prime number p1=2p_{1} = 2, and the in­scribed geom­e­try of the gp1gp_{1} geo­met­ric prime. Early in the book, we of­fered a sim­i­lar rep­re­sen­ta­tion to Figure 1.8.13 below, where the geo­met­ric forms of the com­plex circle, the hy­per­bo­la, and the hy­per­bo­la's as­ymp­totes are pre­sent­ed. In the figure, we have la­beled points on the unit circle using the no­ta­tion cjc_{j}, rep­re­sent­ing the or­dered A(cj)A\left( c_{j} \right) set

A(cj)={(1),   ((−1)14),   ((−1)12),   ((−1)34),(−1),   (−(−1)14),   (−(−1)12),   (−(−1)34)},\begin{array}{r} A\left( c_{j} \right) = \left\{ \begin{array}{r} (1),\ \ \ \left( ( - 1)^{\frac{1}{4}} \right),\ \ \ \left( ( - 1)^{\frac{1}{2}} \right),\ \ \ \left( ( - 1)^{\frac{3}{4}} \right), \\ ( - 1),\ \ \ \left( {- ( - 1)}^{\frac{1}{4}} \right),\ \ \ \left( - ( - 1)^{\frac{1}{2}} \right),\ \ \ \left( {- ( - 1)}^{\frac{3}{4}} \right) \end{array} \right\}, \end{array}
(1.8.274)

which can be seen to in­clude points on the as­ymp­totes and the axes. Also present in the figure is cur­va­ture, in the form of pos­i­tive cur­va­ture for the unit circle, neg­a­tive cur­va­ture for the hy­per­bo­la, and light having flat cur­va­ture as the per­fect bal­ance be­tween the other two.

Figure
Figure 1.8.13 –Points in the com­plex plane from the light set A(cj)A\left( c_{j} \right) ra­di­al­ly pro­ject­ing from the origin.

In 1905, Albert Ein­stein pub­lished a paper named "On the Elec­tro­dy­nam­ics of Moving Bodies"1, in which he pro­posed his theory of spe­cial rel­a­tiv­i­ty. The theory is based on the pos­tu­lates that the laws of physics are in­vari­ant, and the speed of light is in­vari­ant for all ob­servers. The im­pe­tus behind the theory stemmed from the com­pat­i­bil­i­ty issues that New­ton­ian me­chan­ics had with Maxwell's equa­tions, and the theory cor­rects me­chan­ics for bodies moving at close to the speed of light, often re­ferred to as rel­a­tivis­tic ve­loc­i­ties.

Spe­cial rel­a­tiv­i­ty is a con­se­quence of hy­per­bol­ic geom­e­try. We will touch upon the sub­ject briefly and will ul­ti­mate­ly go more in-depth into the sub­ject matter in the en­su­ing books on physics in the Arith­moí series. The most common form of the Lorentz trans­for­ma­tion rep­re­sent­ing ve­loc­i­ty unique to the xx-di­rec­tion is

t′=γ(t−vxc2),   x′=γ(x−vt),t^{'} = \gamma\left( t - \frac{vx}{c^{2}} \right),\ \ \ x' = \gamma(x - vt),
y′=y,   z′=z,\begin{array}{r} y^{'} = y,\ \ \ z^{'} = z, \end{array}
(1.8.275)

where F=(t,x,y,z)F = (t,x,y,z) and F′=(t′,x′,y′,z′)F^{'} = \left( t^{'},x^{'},y^{'},z' \right), rep­re­sent the two co­or­di­nate frames of an ob­served event, for the sta­tion­ary frame FF, the moving frame F′F', the speed of light cc, and the Lorentz factor γ\gamma

γ=11−v2c2.\begin{array}{r} \gamma = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}. \end{array}
(1.8.276)

Space and time are one and the same, and by choos­ing to mea­sure them with the same units, ctct, we re­place the tt time vari­able in the Lorentz trans­for­ma­tion to mea­sure the co­or­di­nate frames of our ob­served result as (ct,x)(ct,x) and (ct′,x′)\left( {ct}^{'},x' \right), where the trans­for­ma­tions become 2

ct′=γ(ct−vcx),ct' = \gamma\left( ct - \frac{v}{c}x \right),
x′=γ(x−vcct),\begin{array}{r} x^{'} = \gamma\left( x - \frac{v}{c}ct \right), \end{array}
(1.8.277)

The sym­me­try be­tween the space­time and ve­loc­i­ty trans­forms be­comes more ev­i­dent in the ex­pres­sions above. There are also in­verse Lorentz trans­for­ma­tions, given that no priv­i­leged ref­er­ence frame exists in the spe­cial rel­a­tiv­i­ty theory. These in­verse trans­for­ma­tions are found simply by solv­ing the trans­for­ma­tions above for ctct and xx

ct=γ(ct′+vcx′),ct = \gamma\left( ct' + \frac{v}{c}x' \right),
x=γ(x′+vcct′),\begin{array}{r} x = \gamma\left( x^{'} + \frac{v}{c}ct^{'} \right), \end{array}
(1.8.278)

and these rep­re­sent the frame FF moving with a neg­a­tive ve­loc­i­ty −v- v that is equal in mag­ni­tude rel­a­tive to the F′F'frame. We sub­sti­tute the term, tanh⁡(β)\tanh(\beta), for vc\ \frac{v}{c}, where β\beta is the ra­pid­i­ty term, and for the unit hy­per­bol­ic equa­tion

cosh2⁡(β)−sinh2⁡(β)=1,\begin{array}{r} \cosh^{2}{(\beta) - \sinh^{2}{(\beta) = 1,}} \end{array}
(1.8.279)

we solve for cosh⁡(β)\cosh(\beta) as fol­lows

cosh2⁡(β)cosh2⁡(β)−sinh2⁡(β)cosh2⁡(β)=1cosh2⁡(β)\frac{\cosh^{2}(\beta)}{\cosh^{2}(\beta)} - \frac{\sinh^{2}(\beta)}{\cosh^{2}(\beta)} = \frac{1}{\cosh^{2}(\beta)}
=1−tanh2⁡(β)=1cosh2⁡(β)= 1 - \tanh^{2}(\beta) = \frac{1}{\cosh^{2}(\beta)}
=cosh⁡(β)=11−tanh2⁡(β),\begin{array}{r} = \cosh{(\beta) = \frac{1}{\sqrt{1 - \tanh^{2}(\beta)}},} \end{array}
(1.8.280)

for which cosh⁡(β)\cosh(\beta) is the hy­per­bol­ic Lorentz factor, γ\gamma, and we set the equiv­a­lence

γ=cosh⁡(β)=11−tanh2⁡(β).\begin{array}{r} \gamma = \cosh{(\beta) = \frac{1}{\sqrt{1 - \tanh^{2}(\beta)}}.} \end{array}
(1.8.281)

The fol­low­ing equiv­a­lence is then the case for the sinh⁡(β)\sinh(\beta) func­tion

sinh⁡(β)=cosh⁡(β)tanh⁡(β)=γ(vc),\begin{array}{r} \sinh(\beta) = \cosh{(\beta)\tanh(\beta)} = \gamma\left( \frac{v}{c} \right), \end{array}
(1.8.282)

and now, we can sub­sti­tute our hy­per­bol­ic equiv­a­lents back into the trans­for­ma­tions

ct′=cosh⁡(β)(ct−tanh⁡(β)x)=cosh(β)ct−sinh⁡(β)x,\begin{array}{r} ct^{'} = \cosh(\beta)\left( ct - \tanh(\beta)x \right) = cosh(\beta)ct - \sinh(\beta)x, \end{array}
(1.8.283)
x′=cosh⁡(β)(x−tanh⁡(β)ct)=cosh(β)x−sinh⁡(β)ct,\begin{array}{r} x^{'} = \cosh(\beta)\left( x - \tanh(\beta)ct \right) = cosh(\beta)x - \sinh{(\beta)ct}, \end{array}
(1.8.284)

then put the hy­per­bol­ic trans­for­ma­tions into matrix form, giving us

(x′ct′)=(cosh⁡(β)−sinh⁡(β)−sinh⁡(β)cosh⁡(β))(xct).\begin{array}{r} \begin{pmatrix} x^{'} \\ ct^{'} \end{pmatrix} = \begin{pmatrix} \cosh(\beta) & {- \sinh}(\beta) \\ {- \sinh}(\beta) & \cosh(\beta) \end{pmatrix}\begin{pmatrix} x \\ ct \end{pmatrix}. \end{array}
(1.8.285)

We take the ex­am­ple for a ra­pid­i­ty value of β=π6,\beta = \frac{\pi}{6},for tanh⁡(β)\tanh(\beta), then being ap­prox­i­mate­ly equal to

tanh⁡(π6)≈0.4805,\begin{array}{r} \tanh{\left( \frac{\pi}{6} \right) \approx 0.4805}, \end{array}
(1.8.286)

and given that tanh⁡(β)=vc\tanh(\beta) = \frac{v}{c} rep­re­sents a frame F′F'that is moving at ≈0.4805\approx 0.4805 the speed of light rel­a­tive to the ob­serv­er's frame. We insert the value β=π6\beta = \frac{\pi}{6} into the matrix form, to find

(x′ct′)=(cosh⁡(π6)−sinh⁡(π6)−sinh⁡(π6)cosh⁡(π6))(xct)\begin{pmatrix} x' \\ ct' \end{pmatrix} = \begin{pmatrix} \cosh\left( \frac{\pi}{6} \right) & {- \sinh}\left( \frac{\pi}{6} \right) \\ {- \sinh}\left( \frac{\pi}{6} \right) & \cosh\left( \frac{\pi}{6} \right) \end{pmatrix}\begin{pmatrix} x \\ ct \end{pmatrix}
→(x′ct′)≈(1.1402−0.5479−0.54791.1402)(xct)≈(1.1402x−0.5479ct1.1403ct−0.5479x),\begin{array}{r} \rightarrow \begin{pmatrix} x^{'} \\ ct^{'} \end{pmatrix} \approx \begin{pmatrix} 1.1402 & - 0.5479 \\ - 0.5479 & 1.1402 \end{pmatrix}\begin{pmatrix} x \\ ct \end{pmatrix} \approx \begin{pmatrix} 1.1402x - 0.5479ct \\ 1.1403ct - 0.5479x \end{pmatrix}, \end{array}
(1.8.287)

which sig­ni­fies the values of the space­time co­or­di­nates, dis­tance, and elapsed time that the moving ob­serv­er would record com­pared to the sta­tion­ary ob­serv­er. The in­verse trans­for­ma­tion can be found by left mul­ti­ply­ing by the in­verse of the trans­for­ma­tion matrix

(cosh⁡(π6)−sinh⁡(π6)−sinh⁡(π6)cosh⁡(π6))−1(x′ct′)=(cosh⁡(π6)−sinh⁡(π6)−sinh⁡(π6)cosh⁡(π6))−1(cosh⁡(π6)−sinh⁡(π6)−sinh⁡(π6)cosh⁡(π6))(xct)\begin{pmatrix} \cosh\left( \frac{\pi}{6} \right) & {- \sinh}\left( \frac{\pi}{6} \right) \\ {- \sinh}\left( \frac{\pi}{6} \right) & \cosh\left( \frac{\pi}{6} \right) \end{pmatrix}^{- 1}\begin{pmatrix} x' \\ ct' \end{pmatrix} = \begin{pmatrix} \cosh\left( \frac{\pi}{6} \right) & {- \sinh}\left( \frac{\pi}{6} \right) \\ {- \sinh}\left( \frac{\pi}{6} \right) & \cosh\left( \frac{\pi}{6} \right) \end{pmatrix}^{- 1}\begin{pmatrix} \cosh\left( \frac{\pi}{6} \right) & {- \sinh}\left( \frac{\pi}{6} \right) \\ {- \sinh}\left( \frac{\pi}{6} \right) & \cosh\left( \frac{\pi}{6} \right) \end{pmatrix}\begin{pmatrix} x \\ ct \end{pmatrix}
→(cosh⁡(π6)sinh⁡(π6)sinh⁡(π6)cosh⁡(π6))(x′ct′)=(xct).\begin{array}{r} \rightarrow \begin{pmatrix} \cosh\left( \frac{\pi}{6} \right) & \sinh\left( \frac{\pi}{6} \right) \\ \sinh\left( \frac{\pi}{6} \right) & \cosh\left( \frac{\pi}{6} \right) \end{pmatrix}\begin{pmatrix} x^{'} \\ ct^{'} \end{pmatrix} = \begin{pmatrix} x \\ ct \end{pmatrix}. \end{array}
(1.8.288)

Fig­ures 1.8.14,15 are plots of hy­per­bol­ic and Eu­clid­ean ro­ta­tions, which are both or­thog­o­nal ro­ta­tions in their re­spec­tive geome­tries. In the hy­per­bol­ic case, two lines are or­thog­o­nal when they are re­flec­tions across the as­ymp­tote of the spe­cif­ic hy­per­bo­la. The as­ymp­tote found in Figure 1.8.14 is la­beled c1c_{1}, and rep­re­sents the cone number (−1)14( - 1)^{\frac{1}{4}} as the speed of light. Hy­per­bol­ic or­thog­o­nal­i­ty is cen­tral to spe­cial rel­a­tiv­i­ty for defin­ing si­mul­ta­ne­ous events, where two events will be si­mul­ta­ne­ous when or­thog­o­nal to a spe­cif­ic time­line.

Figure
Figure 1.8.14 –Pre­served hy­per­bol­ic or­thog­o­nal­i­ty for the ro­ta­tion with β=π6.\beta = \frac{\pi}{6}.

The hy­per­bol­ic ro­ta­tion in Figure 1.8.14 rep­re­sents the in­verse trans­for­ma­tion for values of ct′ct'and x′x', which are of equal mag­ni­tude. By choos­ing ct′,x′=1ct^{'},x^{'} = 1, this im­plies that elapsed time and dis­tance are equal to 11, and cal­cu­lat­ing its matrix-vector, we find

(xct)=(cosh⁡(π6)sinh⁡(π6)sinh⁡(π6)cosh⁡(π6))(11)\begin{pmatrix} x \\ ct \end{pmatrix} = \begin{pmatrix} \cosh\left( \frac{\pi}{6} \right) & \sinh\left( \frac{\pi}{6} \right) \\ \sinh\left( \frac{\pi}{6} \right) & \cosh\left( \frac{\pi}{6} \right) \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix}
→(xct)≈(1.14020.54790.54791.1402)(11)≈(1.68811.6881),\begin{array}{r} \rightarrow \begin{pmatrix} x \\ ct \end{pmatrix} \approx \begin{pmatrix} 1.1402 & 0.5479 \\ 0.5479 & 1.1402 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix} \approx \begin{pmatrix} 1.6881 \\ 1.6881 \end{pmatrix}, \end{array}
(1.8.289)

which sig­ni­fies the space­time co­or­di­nates for an event seen by an ob­serv­er in the frame F′=(1,1,y′=y,z′=z)F^{'} = \left( 1,1,y^{'} = y,z^{'} = z \right) was seen for the same by the ob­serv­er in frame FF for co­or­di­nates F≈(1.6881,1.6881,y=y′,z=z′).F \approx \left( 1.6881,1.6881,{y = y}^{'},{z = z}^{'} \right).

Figure
Figure 1.8.15 –Pre­served Eu­clid­ean or­thog­o­nal­i­ty for the ro­ta­tion with β=π6\beta = \frac{\pi}{6}.

More ev­i­dence that hy­per­bol­ic geom­e­try truly un­der­lies spe­cial rel­a­tiv­i­ty is found by look­ing at the Lorentz factor

γ=11−v2c2,\begin{array}{r} \gamma = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}, \end{array}
(1.8.290)

for which, values of vv are ex­pect­ed to be en­tered for v<c.v < c.If the value for the ve­loc­i­ty is en­tered as equal to the speed of light, v=cv = c, the Lorentz factor is in­fi­nite

γ=11−(v=c)2c2=10=∞,\begin{array}{r} \gamma = \frac{1}{\sqrt{1 - \frac{(v = c)^{2}}{c^{2}}}} = \frac{1}{0} = \infty, \end{array}
(1.8.291)

and for su­per­lu­mi­nal values where v>cv > c, the Lorentz factor be­comes imag­i­nary valued

γ=11−(v>c)2c2∈Im.\begin{array}{r} \gamma = \frac{1}{\sqrt{1 - \frac{(v > c)^{2}}{c^{2}}}} \in Im. \end{array}
(1.8.292)

The hy­per­bol­ic ro­ta­tion in Figure 1.8.15 rep­re­sents the in­verse trans­for­ma­tion for values of ct′ct'and x′x', which are of equal mag­ni­tude. By choos­ing ct′,x′=1ct^{'},x^{'} = 1, this im­plies that elapsed time and dis­tance are equal to 11, and cal­cu­lat­ing its matrix-vector, we find

(xct)=(cosh⁡(π6)sinh⁡(π6)sinh⁡(π6)cosh⁡(π6))(11)\begin{pmatrix} x \\ ct \end{pmatrix} = \begin{pmatrix} \cosh\left( \frac{\pi}{6} \right) & \sinh\left( \frac{\pi}{6} \right) \\ \sinh\left( \frac{\pi}{6} \right) & \cosh\left( \frac{\pi}{6} \right) \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix}
→(xct)≈(1.14020.54790.54791.1402)(11)≈(1.68811.6881),\begin{array}{r} \rightarrow \begin{pmatrix} x \\ ct \end{pmatrix} \approx \begin{pmatrix} 1.1402 & 0.5479 \\ 0.5479 & 1.1402 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix} \approx \begin{pmatrix} 1.6881 \\ 1.6881 \end{pmatrix}, \end{array}
(1.8.293)

which sig­ni­fies the space­time co­or­di­nates for an event seen by an ob­serv­er in the frame F′=(1,1,y′=y,z′=z)F^{'} = \left( 1,1,y^{'} = y,z^{'} = z \right) was seen for the same by the ob­serv­er in frame FF for co­or­di­nates F≈(1.6881,1.6881,y=y′,z=z′).F \approx \left( 1.6881,1.6881,{y = y}^{'},{z = z}^{'} \right).Ev­i­dence that the hy­per­bol­ic geom­e­try is truly what un­der­lies spe­cial rel­a­tiv­i­ty is found by look­ing at the Lorentz factor

γ=11−v2c2,\begin{array}{r} \gamma = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}, \end{array}
(1.8.294)

into which values of vv are ex­pect­ed to be en­tered for v<c.v < c.If the value for the ve­loc­i­ty is en­tered as equal to the speed of light, v=cv = c, the Lorentz factor is in­fi­nite

γ=11−(v=c)2c2=10=∞,\begin{array}{r} \gamma = \frac{1}{\sqrt{1 - \frac{(v = c)^{2}}{c^{2}}}} = \frac{1}{0} = \infty, \end{array}
(1.8.295)

and for su­per­lu­mi­nal values where v>cv > c, the Lorentz factor be­comes imag­i­nary valued

γ=11−(v>c)2c2∈Im.\begin{array}{r} \gamma = \frac{1}{\sqrt{1 - \frac{(v > c)^{2}}{c^{2}}}} \in Im. \end{array}
(1.8.296)

The tanh⁡(β)\tanh(\beta) func­tion rep­re­sents the ratio of the hy­per­bol­ic func­tions sinh⁡(β)cosh⁡(β)\frac{\sinh(\beta)}{\cosh(\beta)}, and in­her­ent in its math­e­mat­i­cal form, it only ap­proach­es the value of one as β→+∞\beta \rightarrow + \infty

limβ→+∞⁡(tanh⁡(β))=1,\begin{array}{r} \lim_{\beta \rightarrow + \infty}\left( \tanh(\beta) \right) = 1, \end{array}
(1.8.297)

and neg­a­tive one as β→−∞\beta \rightarrow - \infty

limβ→−∞⁡(tanh⁡(β))=−1,\begin{array}{r} \lim_{\beta \rightarrow - \infty}\left( \tanh(\beta) \right) = - 1, \end{array}
(1.8.298)

where these limits can be seen clear­ly in the graph of the tanh⁡(β)\tanh(\beta) func­tion in the fol­low­ing figure.

Figure
Figure 1.8.16 –The plot of the tanh⁡(β)\tanh(\beta) func­tion.

So, the hy­per­bol­ic form of the Lorentz factor

γ=11−tanh2⁡(β),\begin{array}{r} \gamma = \frac{1}{\sqrt{1 - \tanh^{2}(\beta)}}, \end{array}
(1.8.299)

is not pred­i­cat­ed upon en­ter­ing a ve­loc­i­ty less than the speed of light, and in­stead, we can see that γ\gamma is the result of hy­per­bol­ic geom­e­try, and more pre­cise­ly, it is the cosh⁡(β)\cosh(\beta) func­tion. Geo­met­ri­cal­ly, as the ra­pid­i­ty β\beta be­comes larger, the degree of neg­a­tive cur­va­ture de­creas­es, and as β→±∞\beta \rightarrow \pm \infty, the cur­va­ture be­comes flat, like that of the light as­ymp­tote.

Re­turn­ing to the A(cj)A\left( c_{j} \right) set of light el­e­ments

A(cj)={(1),   ((−1)14),   ((−1)12),   ((−1)34),(−1),   (−(−1)14),   (−(−1)12),   (−(−1)34)},\begin{array}{r} A\left( c_{j} \right) = \left\{ \begin{array}{r} (1),\ \ \ \left( ( - 1)^{\frac{1}{4}} \right),\ \ \ \left( ( - 1)^{\frac{1}{2}} \right),\ \ \ \left( ( - 1)^{\frac{3}{4}} \right), \\ ( - 1),\ \ \ \left( {- ( - 1)}^{\frac{1}{4}} \right),\ \ \ \left( - ( - 1)^{\frac{1}{2}} \right),\ \ \ \left( {- ( - 1)}^{\frac{3}{4}} \right) \end{array} \right\}, \end{array}
(1.8.300)

we con­sid­er the pos­i­tive­ly ori­ent­ed min­i­mal el­e­ment 𝓂ℯ[+14]=(−1)14,\mathcal{m}\mathcal{e}_{\left\lbrack + \frac{1}{4} \right\rbrack} = ( - 1)^{\frac{1}{4}},where its suc­ces­sive powers can be seen to move along the axes and light as­ymp­totes, seen by its abil­i­ty to gen­er­ate the A(cj)A\left( c_{j} \right) set

{((−1)14)k}={(−1)14,   i,   (−1)34,   (−1),−(−1)14,   −i,   −(−1)34,   1}.\begin{array}{r} \left\{ \left( ( - 1)^{\frac{1}{4}} \right)^{k} \right\} = \left\{ \begin{array}{r} ( - 1)^{\frac{1}{4}},\ \ \ i,\ \ \ ( - 1)^{\frac{3}{4}},\ \ \ ( - 1), \\ {- ( - 1)}^{\frac{1}{4}},\ \ \ - i,\ \ \ {- ( - 1)}^{\frac{3}{4}},\ \ \ 1 \end{array} \right\}. \end{array}
(1.8.301)

This axis-as­ymp­tote prop­a­ga­tion char­ac­ter­is­tic is how light moves math­e­mat­i­cal­ly and phys­i­cal­ly, which dif­fers sig­nif­i­cant­ly from the phe­nom­e­nol­o­gy of mas­sive ob­jects bound by the rel­a­tiv­i­ty in­her­ent to hy­per­bol­ic geom­e­try. Light is out­side of the geo­met­ric con­fines of space­time, but, at the same time, it moves freely through the space­time geom­e­try.

Light is cen­tral to gain­ing a rea­son­able un­der­stand­ing of the uni­verse, and any theory that does not con­sid­er its im­por­tance will be woe­ful­ly in­com­plete. In Chap­ter 4, we con­sid­er conic sec­tions, no­tably the light cone, which was sig­nif­i­cant be­cause the light cone gen­er­at­ed the SU(2)SU(2) sub­group. This sub­group stands out, given its dis­tance metric is Eu­clid­ean, and in real space at non-rel­a­tivis­tic ve­loc­i­ties the world around us ap­pears dis­tinct­ly Eu­clid­ean. Light does not just rep­re­sent a ve­loc­i­ty limit for mas­sive ob­jects in space or pho­tons/elec­tro­mag­net­ic waves; it is math­e­mat­i­cal­ly re­spon­si­ble for cre­at­ing the phys­i­cal spaces in which we exist. This not only in­cludes the real space we move around in, but also the imag­i­nary space in which we move through time.

To un­der­stand the math­e­mat­ics behind light, we return to the geo­met­ric in­te­gers to con­sid­er the do­decagon formed by cyclic ro­ta­tions of the equi­lat­er­al tri­an­gle. The gp2gp_{2} geo­met­ric prime set A(Y(gp2)j)A\left( Y\left( gp_{2} \right)_{j} \right) had three unique el­e­ments, which co­or­di­nat­ed with the three cyclic shifts of its ver­tices in­scribed in the com­plex unit sphere

A(Y(gp2)j)={(100010001),(001100010),(010001100)}.\begin{array}{r} A\left( Y\left( gp_{2} \right)_{j} \right) = \left\{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix},\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} \right\}. \end{array}
(1.8.302)

The in­scribed do­dec­a­he­dron is rep­re­sent­ed by ro­tat­ing the equi­lat­er­al tri­an­gle by iπ2\frac{i\pi}{2} ra­di­ans, seen as the trans­for­ma­tion in the image below.

Figure
Figure 1.8.17 – The Y(gp2)2×e(iπ2)Y\left( gp_{2} \right)_{2} \times e^{\left( \frac{i\pi}{2} \right)}ro­ta­tion in the com­plex unit circle.

The first matrix rep­re­sen­ta­tion of the ro­tat­ed min­i­mal el­e­ment 𝓂ℯ[+igp2]\mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack} from the A(Y(gp2)j)A\left( Y\left( gp_{2} \right)_{j} \right) set, we denote as fol­lows

𝓂ℯ[+igp2]=Y(gp2)2×e(iπ2)=(i000e(3iπ4)000e(iπ4))(001100010)=(00i(−1)(34)000(−1)(14)0),\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack} = Y\left( gp_{2} \right)_{2} \times e^{\left( \frac{i\pi}{2} \right)} = \begin{pmatrix} i & 0 & 0 \\ 0 & e^{\left( \frac{3i\pi}{4} \right)} & 0 \\ 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & i \\ ( - 1)^{\left( \frac{3}{4} \right)} & 0 & 0 \\ 0 & ( - 1)^{\left( \frac{1}{4} \right)} & 0 \end{pmatrix}, \end{array}
(1.8.303)

which has one imag­i­nary entry, and the other two are cone num­bers sit­u­at­ed ±iπ4\pm \frac{i\pi}{4} ra­di­ans away. By find­ing the eigen­val­ues of the matrix, we find con­sis­ten­cy with the vertex po­si­tion­ing de­pict­ed in the figure above

λ(𝓂ℯ[+igp2])=eig(00i(−1)(34)000(−1)(14)0)={e(iπ2),e(−5iπ6),e(−iπ6)},\begin{array}{r} \lambda\left( \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack} \right) = eig\begin{pmatrix} 0 & 0 & i \\ ( - 1)^{\left( \frac{3}{4} \right)} & 0 & 0 \\ 0 & ( - 1)^{\left( \frac{1}{4} \right)} & 0 \end{pmatrix} = \left\{ e^{\left( \frac{i\pi}{2} \right)},e^{\left( - \frac{5i\pi}{6} \right)},e^{\left( - \frac{i\pi}{6} \right)} \right\}, \end{array}
(1.8.304)

which are all so­lu­tions to its char­ac­ter­is­tic poly­no­mi­al

𝒸𝓅(𝓂ℯ[+112])=λ3+i=0.\begin{array}{r} \mathcal{cp}\left( \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{1}{12} \right\rbrack} \right) = \lambda^{3} + i = 0. \end{array}
(1.8.305)

To see the in­tu­ition behind the result for the 𝒸𝓅(𝓂ℯ[+igp2])\mathcal{cp}\left( \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack} \right) eigen­val­ues, each of these are so­lu­tions to the equa­tion |𝓂ℯ[+igp2]−Iλ|=0\left| \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack} - I\lambda \right| = 0. The eigen­val­ues are cal­cu­lat­ed as fol­lows

|−λ0i(−1)(34)−λ00(−1)(14)−λ|=0,\begin{array}{r} \left| \begin{matrix} - \lambda & 0 & i \\ ( - 1)^{\left( \frac{3}{4} \right)} & - \lambda & 0 \\ 0 & ( - 1)^{\left( \frac{1}{4} \right)} & - \lambda \end{matrix} \right| = 0, \end{array}
(1.8.306)

for which, taking the de­ter­mi­nant of the matrix yields

→−λ(λ2−0)−0+i(((−1)(34)×(−1)(14))−0)=0\rightarrow - \lambda\left( \lambda^{2} - 0 \right) - 0 + i\left( \left( ( - 1)^{\left( \frac{3}{4} \right)} \times ( - 1)^{\left( \frac{1}{4} \right)} \right) - 0 \right) = 0
→−λ3−i=0\rightarrow - \lambda^{3} - i = 0
→λ3+i=0,\begin{array}{r} \rightarrow \lambda^{3} + i = 0, \end{array}
(1.8.307)

and the eigen­val­ues are the said so­lu­tions to this char­ac­ter­is­tic poly­no­mi­al in the com­plex plane. As we have seen pre­vi­ous­ly, the matrix 𝓂ℯ[+igp2]\mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack} is also a so­lu­tion to the char­ac­ter­is­tic poly­no­mi­al, given that

(00i(−1)(34)000(−1)(14)0)3=(−i000−i000−i).\begin{array}{r} \begin{pmatrix} 0 & 0 & i \\ ( - 1)^{\left( \frac{3}{4} \right)} & 0 & 0 \\ 0 & ( - 1)^{\left( \frac{1}{4} \right)} & 0 \end{pmatrix}^{3} = \begin{pmatrix} - i & 0 & 0 \\ 0 & - i & 0 \\ 0 & 0 & - i \end{pmatrix}. \end{array}
(1.8.308)

We gen­er­ate the sub­group by taking suc­ces­sive powers of the min­i­mal el­e­ment, (𝓂ℯ[+igp2])k,\left( \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack} \right)^{k},to find the fol­low­ing set of twelve el­e­ments, where (↖)( \nwarrow ) presents a pos­i­tive­ly ori­ent­ed ro­ta­tion,

A(Y(gp[i2](↖)))={(00i(−1)34000(−1)140),   (0(−1)34000−(−1)14−100),   (−i000−i000−i),(001(−1)14000−(−1)340),   (0(−1)14000(−1)34i00),   (−1000−1000−1),(00−i−(−1)34000−(−1)140),   (0−(−1)34000(−1)14100),   (i000i000i),(00−1−(−1)14000(−1)340),   (0−(−1)14000−(−1)34−i00),   (100010001),}.\begin{array}{r} A\left( Y\left( {gp}_{\lbrack i2\rbrack}( \nwarrow ) \right) \right) = \left\{ \begin{array}{r} \begin{pmatrix} 0 & 0 & i \\ ( - 1)^{\frac{3}{4}} & 0 & 0 \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & ( - 1)^{\frac{3}{4}} & 0 \\ 0 & 0 & - ( - 1)^{\frac{1}{4}} \\ - 1 & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} - i & 0 & 0 \\ 0 & - i & 0 \\ 0 & 0 & - i \end{pmatrix}, \\ \begin{pmatrix} 0 & 0 & 1 \\ ( - 1)^{\frac{1}{4}} & 0 & 0 \\ 0 & {- ( - 1)}^{\frac{3}{4}} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{3}{4}} \\ i & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix}, \\ \begin{pmatrix} 0 & 0 & - i \\ {- ( - 1)}^{\frac{3}{4}} & 0 & 0 \\ 0 & {- ( - 1)}^{\frac{1}{4}} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & - ( - 1)^{\frac{3}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \\ 1 & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} i & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & i \end{pmatrix}, \\ \begin{pmatrix} 0 & 0 & - 1 \\ {- ( - 1)}^{\frac{1}{4}} & 0 & 0 \\ 0 & ( - 1)^{\frac{3}{4}} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & - ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & - ( - 1)^{\frac{3}{4}} \\ - i & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \end{array} \right\}. \end{array}
(1.8.309)

Geo­met­ri­cal­ly, these are rep­re­sent­ed in the figure below, where each vertex may orient in one of twelve pos­si­ble po­si­tions on the com­plex unit circle.

Figure
Figure 1.8.18 –The in­scribed geome­tries of the A(Y(gp[2i](↖)))A\left( Y\left( {gp}_{\lbrack 2i\rbrack}( \nwarrow ) \right) \right) set as

equi­lat­er­al tri­an­gles in­scribed in the do­decagon.

The group char­ac­ter­is­tic poly­no­mi­al has re­mark­able sym­me­try, seen by the fol­low­ing

𝒞𝒫(Y(gp[2i](↖)))=((λ−1)3+(λ−i)3+(λ+1)3+(λ+i)3+2(λ3−1)+2(λ3−i)+2(λ3+1)+2(λ3+i)),\begin{array}{r} \mathcal{CP}\left( Y\left( {gp}_{\lbrack 2i\rbrack}( \nwarrow ) \right) \right) = \left( \begin{array}{r} (\lambda - 1)^{3} + (\lambda - i)^{3} + (\lambda + 1)^{3} + (\lambda + i)^{3} \\ + 2\left( \lambda^{3} - 1 \right) + 2\left( \lambda^{3} - i \right) + 2\left( \lambda^{3} + 1 \right) + 2\left( \lambda^{3} + i \right) \end{array} \right), \end{array}
(1.8.310)

which rep­re­sents the com­po­si­tion of the fol­low­ing sub­groups of the gp2gp_{2} geo­met­ric prime

A(igp2)=2(λ3−i)+(λ+i)3:{(00−i−(−1)34000−(−1)140),(0−(−1)14000−(−1)34−i00),(−i000−i000−i)},\begin{array}{r} A\left( igp_{2} \right) = {2\left( \lambda^{3} - i \right) + (\lambda + i)}^{3}:\left\{ \begin{pmatrix} 0 & 0 & - i \\ {- ( - 1)}^{\frac{3}{4}} & 0 & 0 \\ 0 & {- ( - 1)}^{\frac{1}{4}} & 0 \end{pmatrix},\begin{pmatrix} 0 & - ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & - ( - 1)^{\frac{3}{4}} \\ - i & 0 & 0 \end{pmatrix},\begin{pmatrix} - i & 0 & 0 \\ 0 & - i & 0 \\ 0 & 0 & - i \end{pmatrix} \right\}, \end{array}
(1.8.311)
A(−gp2)=2(λ3+1)+(λ+1)3:{(00−1−(−1)14000(−1)340),(0(−1)34000−(−1)14−100),(−1000−1000−1)},\begin{array}{r} A\left( - gp_{2} \right){= 2\left( \lambda^{3} + 1 \right) + (\lambda + 1)}^{3}:\left\{ \begin{pmatrix} 0 & 0 & - 1 \\ {- ( - 1)}^{\frac{1}{4}} & 0 & 0 \\ 0 & ( - 1)^{\frac{3}{4}} & 0 \end{pmatrix},\begin{pmatrix} 0 & ( - 1)^{\frac{3}{4}} & 0 \\ 0 & 0 & - ( - 1)^{\frac{1}{4}} \\ - 1 & 0 & 0 \end{pmatrix},\begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix} \right\}, \end{array}
(1.8.312)
A(−igp2)=2(λ3+i)+(λ−i)3:{(00i(−1)34000(−1)140),(0(−1)14000(−1)34i00),(i000i000i)},\begin{array}{r} {A\left( - igp_{2} \right) = 2\left( \lambda^{3} + i \right) + (\lambda - i)}^{3}:\left\{ \begin{pmatrix} 0 & 0 & i \\ ( - 1)^{\frac{3}{4}} & 0 & 0 \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \end{pmatrix},\begin{pmatrix} 0 & ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{3}{4}} \\ i & 0 & 0 \end{pmatrix},\begin{pmatrix} i & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & i \end{pmatrix} \right\}, \end{array}
(1.8.313)
A(gp2)=2(λ3−1)+(λ−1)3:{(001(−1)14000−(−1)340),(0−(−1)34000(−1)14100),(100010001)},\begin{array}{r} {A\left( gp_{2} \right) = 2\left( \lambda^{3} - 1 \right) + (\lambda - 1)}^{3}:\left\{ \begin{pmatrix} 0 & 0 & 1 \\ ( - 1)^{\frac{1}{4}} & 0 & 0 \\ 0 & {- ( - 1)}^{\frac{3}{4}} & 0 \end{pmatrix},\begin{pmatrix} 0 & - ( - 1)^{\frac{3}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \\ 1 & 0 & 0 \end{pmatrix},\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right\}, \end{array}
(1.8.314)

The el­e­ments in the dis­tinct sub­sets have de­ter­mi­nants equal to kk in the ex­pres­sion A(kgp2)A\left( kgp_{2} \right). By its sym­me­try, the 𝒞𝒫(Y(gp[2i])(↖))\mathcal{CP}\left( Y\left( {gp}_{\lbrack 2i\rbrack} \right)( \nwarrow ) \right) group char­ac­ter­is­tic poly­no­mi­al re­duces to the sim­pli­fied form

𝒞𝒫(Y(gp[2i])(↖))=12λ3,\begin{array}{r} \mathcal{CP}\left( Y\left( {gp}_{\lbrack 2i\rbrack} \right)( \nwarrow ) \right) = 12\lambda^{3}, \end{array}
(1.8.315)

show­ing its do­dec­a­he­dron char­ac­ter­is­tic form. Only the triv­ial so­lu­tion exists for the char­ac­ter­is­tic poly­no­mi­al equa­tion 12λ3=0,12\lambda^{3} = 0,sig­ni­fy­ing that the func­tion cross­es the real-axis only at the origin, as seen in the figure below.

Figure
Figure 1.8.19 –The plot of the group char­ac­ter­is­tic func­tion 𝒞𝒫(Y(gp[2i])(↖)).\mathcal{CP}\left( Y\left( {gp}_{\lbrack 2i\rbrack} \right)( \nwarrow ) \right).

The net-zero char­ac­ter­is­tic is present in the graph above, mean­ing that for any point on the pos­i­tive por­tion of the curve, there is an­oth­er point on the neg­a­tive por­tion that, when added to it, will equal zero. In other words, the char­ac­ter­is­tic poly­no­mi­al is sym­met­ric about the hor­i­zon­tal axis. Net-zero is es­sen­tial for any theory of every­thing to be con­sis­tent math­e­mat­i­cal­ly and philo­soph­i­cal­ly. The­o­ries pred­i­cat­ed upon a uni­verse spring­ing from the zero-point with­out a net-zero bal­ance of its parts equates to the free-lunch model, for which any claim of causal­i­ty is void. The cyclic group nature is in­her­ent in the el­e­ments of the A(Y(gp[i2])(↖))A\left( Y\left( {gp}_{\lbrack i2\rbrack} \right)( \nwarrow ) \right) set, ap­pre­ci­at­ed by first defin­ing its sub­sets of el­e­ments

Aj[i]={(00−i−(−1)34000−(−1)140),(0−(−1)14000−(−1)34−i00),(−i000−i000−i)},\begin{array}{r} A_{j}^{\lbrack i\rbrack} = \left\{ \begin{pmatrix} 0 & 0 & - i \\ {- ( - 1)}^{\frac{3}{4}} & 0 & 0 \\ 0 & {- ( - 1)}^{\frac{1}{4}} & 0 \end{pmatrix},\begin{pmatrix} 0 & - ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & - ( - 1)^{\frac{3}{4}} \\ - i & 0 & 0 \end{pmatrix},\begin{pmatrix} - i & 0 & 0 \\ 0 & - i & 0 \\ 0 & 0 & - i \end{pmatrix} \right\}, \end{array}
(1.8.316)
Aj[−1]={(00−1−(−1)14000(−1)340),(0(−1)34000−(−1)14−100),(−1000−1000−1)},\begin{array}{r} A_{j}^{\lbrack - 1\rbrack} = \left\{ \begin{pmatrix} 0 & 0 & - 1 \\ {- ( - 1)}^{\frac{1}{4}} & 0 & 0 \\ 0 & ( - 1)^{\frac{3}{4}} & 0 \end{pmatrix},\begin{pmatrix} 0 & ( - 1)^{\frac{3}{4}} & 0 \\ 0 & 0 & - ( - 1)^{\frac{1}{4}} \\ - 1 & 0 & 0 \end{pmatrix},\begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix} \right\}, \end{array}
(1.8.317)
Aj[−i]={(00i(−1)34000(−1)140),(0(−1)14000(−1)34i00),(i000i000i)},\begin{array}{r} A_{j}^{\lbrack - i\rbrack} = \left\{ \begin{pmatrix} 0 & 0 & i \\ ( - 1)^{\frac{3}{4}} & 0 & 0 \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \end{pmatrix},\begin{pmatrix} 0 & ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{3}{4}} \\ i & 0 & 0 \end{pmatrix},\begin{pmatrix} i & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & i \end{pmatrix} \right\}, \end{array}
(1.8.318)
Aj[1]={(001(−1)14000−(−1)340),(0−(−1)34000(−1)14100),(100010001)}.\begin{array}{r} A_{j}^{\lbrack 1\rbrack} = \left\{ \begin{pmatrix} 0 & 0 & 1 \\ ( - 1)^{\frac{1}{4}} & 0 & 0 \\ 0 & {- ( - 1)}^{\frac{3}{4}} & 0 \end{pmatrix},\begin{pmatrix} 0 & - ( - 1)^{\frac{3}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \\ 1 & 0 & 0 \end{pmatrix},\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right\}. \end{array}
(1.8.319)

These can be sit­u­at­ed into an array as

A(Y(gp[i2])(↖))=(A1[i]A2[−1]A3[−i]A1[1]A2[i]A3[−1]A1[−i]A2[1]A3[i]A1[−1]A2[−i]A3[1]),\begin{array}{r} A\left( Y\left( {gp}_{\lbrack i2\rbrack} \right)( \nwarrow ) \right) = \begin{pmatrix} A_{1}^{\lbrack i\rbrack} & A_{2}^{\lbrack - 1\rbrack} & A_{3}^{\lbrack - i\rbrack} \\ A_{1}^{\lbrack 1\rbrack} & A_{2}^{\lbrack i\rbrack} & A_{3}^{\lbrack - 1\rbrack} \\ A_{1}^{\lbrack - i\rbrack} & A_{2}^{\lbrack 1\rbrack} & A_{3}^{\lbrack i\rbrack} \\ A_{1}^{\lbrack - 1\rbrack} & A_{2}^{\lbrack - i\rbrack} & A_{3}^{\lbrack 1\rbrack} \end{pmatrix}, \end{array}
(1.8.320)

where the el­e­ments of each sub­group are arranged in three-el­e­ment cyclic di­ag­o­nals and the Ak(1)A_{k}^{(1)} define the min­i­mal el­e­ments, or, in other words, the pri­ma­ry gen­er­a­tors. For clar­i­ty, we note as an ex­am­ple that Ak(2)=(Ak(1))2.A_{k}^{(2)} = {{(A}_{k}^{(1)})}^{2}.Now, we take the A(igp2)A\left( igp_{2} \right) sub­group el­e­ments found in the 3×33 \times 3 upper-left sub-matrix and define the min­i­mal el­e­ment sub­group, 𝒜(igp2),\mathcal{A}\left( igp_{2} \right),as

𝒜(igp2)=(A1[i]000A2[i]000A2[i]).\begin{array}{r} \mathcal{A}\left( igp_{2} \right) = \begin{pmatrix} A_{1}^{\lbrack i\rbrack} & 0 & 0 \\ 0 & A_{2}^{\lbrack i\rbrack} & 0 \\ 0 & 0 & A_{2}^{\lbrack i\rbrack} \end{pmatrix}. \end{array}
(1.8.321)

Suc­ces­sive powers of the min­i­mal el­e­ment sub­group 𝒜(igp2)\mathcal{A}\left( igp_{2} \right) will form a co­he­sive twelve el­e­ment su­per­group, 𝒜(Y(gp[i2])(↖)),\mathcal{A}\left( Y\left( {gp}_{\lbrack i2\rbrack} \right)( \nwarrow ) \right),com­posed of the fol­low­ing el­e­ments

𝒜(Y(gp[i2])(↖))={(A1[i]000A2[i]000A2[i]),(A2[−1]000A1[−1]000A3[−1]),(A3[−i]000A3[−i]000A3[−i]),(A1[1]000A2[1]000A2[1]),(A2[i]000A1[i]000A3[i]),(A3[−1]000A3[−1]000A3[−1]),(A1[−i]000A2[−i]000A2[−i]),(A2[1]000A1[1]000A3[1]),(A3[i]000A3[i]000A3[i]),(A1[−1]000A2[−1]000A2[−1]),(A2[−i]000A1[−i]000A3[−i]),(A3[1]000A3[1]000A3[1])}.\begin{array}{r} \mathcal{A}\left( Y\left( {gp}_{\lbrack i2\rbrack} \right)( \nwarrow ) \right) = \left\{ \begin{array}{r} \begin{pmatrix} A_{1}^{\lbrack i\rbrack} & 0 & 0 \\ 0 & A_{2}^{\lbrack i\rbrack} & 0 \\ 0 & 0 & A_{2}^{\lbrack i\rbrack} \end{pmatrix},\begin{pmatrix} A_{2}^{\lbrack - 1\rbrack} & 0 & 0 \\ 0 & A_{1}^{\lbrack - 1\rbrack} & 0 \\ 0 & 0 & A_{3}^{\lbrack - 1\rbrack} \end{pmatrix},\begin{pmatrix} A_{3}^{\lbrack - i\rbrack} & 0 & 0 \\ 0 & A_{3}^{\lbrack - i\rbrack} & 0 \\ 0 & 0 & A_{3}^{\lbrack - i\rbrack} \end{pmatrix}, \\ \begin{pmatrix} A_{1}^{\lbrack 1\rbrack} & 0 & 0 \\ 0 & A_{2}^{\lbrack 1\rbrack} & 0 \\ 0 & 0 & A_{2}^{\lbrack 1\rbrack} \end{pmatrix},\begin{pmatrix} A_{2}^{\lbrack i\rbrack} & 0 & 0 \\ 0 & A_{1}^{\lbrack i\rbrack} & 0 \\ 0 & 0 & A_{3}^{\lbrack i\rbrack} \end{pmatrix},\begin{pmatrix} A_{3}^{\lbrack - 1\rbrack} & 0 & 0 \\ 0 & A_{3}^{\lbrack - 1\rbrack} & 0 \\ 0 & 0 & A_{3}^{\lbrack - 1\rbrack} \end{pmatrix}, \\ \begin{array}{r} \begin{pmatrix} A_{1}^{\lbrack - i\rbrack} & 0 & 0 \\ 0 & A_{2}^{\lbrack - i\rbrack} & 0 \\ 0 & 0 & A_{2}^{\lbrack - i\rbrack} \end{pmatrix},\begin{pmatrix} A_{2}^{\lbrack 1\rbrack} & 0 & 0 \\ 0 & A_{1}^{\lbrack 1\rbrack} & 0 \\ 0 & 0 & A_{3}^{\lbrack 1\rbrack} \end{pmatrix},\begin{pmatrix} A_{3}^{\lbrack i\rbrack} & 0 & 0 \\ 0 & A_{3}^{\lbrack i\rbrack} & 0 \\ 0 & 0 & A_{3}^{\lbrack i\rbrack} \end{pmatrix}, \\ \begin{pmatrix} A_{1}^{\lbrack - 1\rbrack} & 0 & 0 \\ 0 & A_{2}^{\lbrack - 1\rbrack} & 0 \\ 0 & 0 & A_{2}^{\lbrack - 1\rbrack} \end{pmatrix},\begin{pmatrix} A_{2}^{\lbrack - i\rbrack} & 0 & 0 \\ 0 & A_{1}^{\lbrack - i\rbrack} & 0 \\ 0 & 0 & A_{3}^{\lbrack - i\rbrack} \end{pmatrix},\begin{pmatrix} A_{3}^{\lbrack 1\rbrack} & 0 & 0 \\ 0 & A_{3}^{\lbrack 1\rbrack} & 0 \\ 0 & 0 & A_{3}^{\lbrack 1\rbrack} \end{pmatrix} \end{array} \end{array} \right\}. \end{array}
(1.8.322)

This has a very sim­i­lar struc­ture to the A(Y(gp[i2])(↖))A\left( Y\left( {gp}_{\lbrack i2\rbrack} \right)( \nwarrow ) \right) set it was de­rived from, ev­i­denced by the re­peat­ing three-el­e­ment cyclic di­ag­o­nals, the pres­ence in the bottom half of the neg­a­tive-valued el­e­ments co­in­cid­ing with re­spec­tive el­e­ments in the top half, and the pres­ence of axes iden­ti­ties in the right-most column.

The su­per­group process can be re­peat­ed in­def­i­nite­ly, con­sid­er­ing the 𝒜(Y(gp[i2])(↖))\mathcal{A}\left( Y\left( {gp}_{\lbrack i2\rbrack} \right)( \nwarrow ) \right) set can be arranged into the array

𝒜(Y(gp[i2])(↖))=(𝒜1[i]𝒜2[−1]𝒜3[−i]𝒜1[1]𝒜2[i]𝒜3[−1]𝒜1[−i]𝒜2[1]𝒜3[i]𝒜1[−1]𝒜2[−i]𝒜3[1]),\begin{array}{r} \mathcal{A}\left( Y\left( {gp}_{\lbrack i2\rbrack} \right)( \nwarrow ) \right) = \begin{pmatrix} \mathcal{A}_{1}^{\lbrack i\rbrack} & \mathcal{A}_{2}^{\lbrack - 1\rbrack} & \mathcal{A}_{3}^{\lbrack - i\rbrack} \\ \mathcal{A}_{1}^{\lbrack 1\rbrack} & \mathcal{A}_{2}^{\lbrack i\rbrack} & \mathcal{A}_{3}^{\lbrack - 1\rbrack} \\ \mathcal{A}_{1}^{\lbrack - i\rbrack} & \mathcal{A}_{2}^{\lbrack 1\rbrack} & \mathcal{A}_{3}^{\lbrack i\rbrack} \\ \mathcal{A}_{1}^{\lbrack - 1\rbrack} & \mathcal{A}_{2}^{\lbrack - i\rbrack} & \mathcal{A}_{3}^{\lbrack 1\rbrack} \end{pmatrix}, \end{array}
(1.8.323)

where the min­i­mal el­e­ment sub­group 𝒜[2](igp2)\mathcal{A}_{\lbrack 2\rbrack}\left( igp_{2} \right) can also be de­fined as

𝒜(2)(igp2)=(𝒜1[i]000𝒜2[i]000𝒜3[i]),\begin{array}{r} \mathcal{A}_{(2)}\left( igp_{2} \right) = \begin{pmatrix} \mathcal{A}_{1}^{\lbrack i\rbrack} & 0 & 0 \\ 0 & \mathcal{A}_{2}^{\lbrack i\rbrack} & 0 \\ 0 & 0 & \mathcal{A}_{3}^{\lbrack i\rbrack} \end{pmatrix}, \end{array}
(1.8.324)

whose suc­ces­sive powers 𝒜[2](igp2)k\mathcal{A}_{\lbrack 2\rbrack}\left( igp_{2} \right)^{k} also form the twelve el­e­ment 𝒜[2](Y(gp[i2])(↖))\mathcal{A}_{\lbrack 2\rbrack}\left( Y\left( {gp}_{\lbrack i2\rbrack} \right)( \nwarrow ) \right) set

𝒜(2)(Y(gp[i2])(↖))=(𝒜(2,1)[i]𝒜(2,2)[−1]𝒜(2,3)[−i]𝒜(2,1)[1]𝒜(2,2)[i]𝒜(2,3)[−1]𝒜(2,1)[−i]𝒜(2,2)[1]𝒜(2,3)[i]𝒜(2,1)[−1]𝒜(2,2)[−i]𝒜(2,3)[1]).\begin{array}{r} \mathcal{A}_{(2)}\left( Y\left( {gp}_{\lbrack i2\rbrack} \right)( \nwarrow ) \right) = \begin{pmatrix} \mathcal{A}_{(2,1)}^{\lbrack i\rbrack} & \mathcal{A}_{(2,2)}^{\lbrack - 1\rbrack} & \mathcal{A}_{(2,3)}^{\lbrack - i\rbrack} \\ \mathcal{A}_{(2,1)}^{\lbrack 1\rbrack} & \mathcal{A}_{(2,2)}^{\lbrack i\rbrack} & \mathcal{A}_{(2,3)}^{\lbrack - 1\rbrack} \\ \mathcal{A}_{(2,1)}^{\lbrack - i\rbrack} & \mathcal{A}_{(2,2)}^{\lbrack 1\rbrack} & \mathcal{A}_{(2,3)}^{\lbrack i\rbrack} \\ \mathcal{A}_{(2,1)}^{\lbrack - 1\rbrack} & \mathcal{A}_{(2,2)}^{\lbrack - i\rbrack} & \mathcal{A}_{(2,3)}^{\lbrack 1\rbrack} \end{pmatrix}. \end{array}
(1.8.325)

The (𝓂ℯ[+igp2])k\left( \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack} \right)^{k} min­i­mal el­e­ment is ori­entable in the other di­men­sions of ℂ3\mathbb{C}^{3}, and our ini­tial min­i­mal el­e­ment can be dis­tin­guished as

𝓂ℯ[+igp2](↖)=(00i(−1)34000(−1)140),\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nwarrow ) = \begin{pmatrix} 0 & 0 & i \\ ( - 1)^{\frac{3}{4}} & 0 & 0 \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \end{pmatrix}, \end{array}
(1.8.326)

where (↖)( \nwarrow ) presents a pos­i­tive­ly ori­ent­ed ro­ta­tion. The min­i­mal el­e­ment 𝓂ℯ[+igp2](↗)\mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nearrow ) is the trans­pose of 𝓂ℯ[+igp2](↖)\mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nwarrow )

𝓂ℯ[+igp2](↗)=(𝓂ℯ[+igp2](↖))T=(0(−1)34000(−1)14i00),\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nearrow ) = \left( \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nwarrow ) \right)^{T} = \begin{pmatrix} 0 & ( - 1)^{\frac{3}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \\ i & 0 & 0 \end{pmatrix}, \end{array}
(1.8.327)

has the same eigen­val­ues as 𝓂ℯ[+igp2](↖)\mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nwarrow )

λ(𝓂ℯ[+igp2](↗))=eig(0(−1)34000(−1)14i00)={−e(iπ6),e(iπ2),−e(5iπ6)},\begin{array}{r} \lambda\left( \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nearrow ) \right) = eig\begin{pmatrix} 0 & ( - 1)^{\frac{3}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \\ i & 0 & 0 \end{pmatrix} = \left\{ {- e}^{\left( \frac{i\pi}{6} \right)},e^{\left( \frac{i\pi}{2} \right)},{- e}^{\left( \frac{5i\pi}{6} \right)} \right\}, \end{array}
(1.8.328)

and the set of el­e­ments 𝓂ℯ[+igp2](↗)\mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nearrow ) gen­er­at­ing A(Y(gp[2i])(↗))A\left( Y\left( {gp}_{\lbrack 2i\rbrack} \right)( \nearrow ) \right), is the fol­low­ing

A(Y(gp[i2])(↗))={(0(−1)34000(−1)14i00),(00−1(−1)34000−(−1)140),(−i000−i000−i),(0(−1)14000−(−1)34100),(00i(−1)14000(−1)340),(−1000−1000−1),(0−(−1)34000−(−1)14−i00),(001−(−1)34000(−1)140),(i000i000i),(0−(−1)14000(−1)34−100),(00−i−(−1)14000−(−1)340),(100010001)}.\begin{array}{r} A\left( Y\left( {gp}_{\lbrack i2\rbrack} \right)( \nearrow ) \right) = \left\{ \begin{array}{r} \begin{pmatrix} 0 & ( - 1)^{\frac{3}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \\ i & 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 & - 1 \\ ( - 1)^{\frac{3}{4}} & 0 & 0 \\ 0 & {- ( - 1)}^{\frac{1}{4}} & 0 \end{pmatrix},\begin{pmatrix} - i & 0 & 0 \\ 0 & - i & 0 \\ 0 & 0 & - i \end{pmatrix}, \\ \begin{pmatrix} 0 & ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & {- ( - 1)}^{\frac{3}{4}} \\ 1 & 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 & i \\ ( - 1)^{\frac{1}{4}} & 0 & 0 \\ 0 & ( - 1)^{\frac{3}{4}} & 0 \end{pmatrix},\begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix}, \\ \begin{pmatrix} 0 & {- ( - 1)}^{\frac{3}{4}} & 0 \\ 0 & 0 & {- ( - 1)}^{\frac{1}{4}} \\ - i & 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 & 1 \\ {- ( - 1)}^{\frac{3}{4}} & 0 & 0 \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \end{pmatrix},\begin{pmatrix} i & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & i \end{pmatrix}, \\ \begin{pmatrix} 0 & - ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{3}{4}} \\ - 1 & 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 & - i \\ {- ( - 1)}^{\frac{1}{4}} & 0 & 0 \\ 0 & - ( - 1)^{\frac{3}{4}} & 0 \end{pmatrix},\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{array} \right\}. \end{array}
(1.8.329)

The re­main­ing com­plex di­men­sion has the fol­low­ing min­i­mal el­e­ment

𝓂ℯ[+igp2](→)=(−(−1)(16)000i000−(−1)(56)),\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \rightarrow ) = \begin{pmatrix} - ( - 1)^{\left( \frac{1}{6} \right)} & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & - ( - 1)^{\left( \frac{5}{6} \right)} \end{pmatrix}, \end{array}
(1.8.330)

which has the fol­low­ing re­la­tion­ships with the other com­plex di­men­sion­al min­i­mal el­e­ments

𝓂ℯ[+igp2](→)=(0(−1)112000(−1)14−(−1)1300)×𝓂ℯ[+igp2](↖),\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \rightarrow ) = \begin{pmatrix} 0 & ( - 1)^{\frac{1}{12}} & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \\ {- ( - 1)}^{\frac{1}{3}} & 0 & 0 \end{pmatrix}\times \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nwarrow ), \end{array}
(1.8.331)
𝓂ℯ[+igp2](→)=𝓂ℯ[+igp2](↗)×(00−(−1)13(−1)112000(−1)140),\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \rightarrow )= \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nearrow ) \times \begin{pmatrix} 0 & 0 & {- ( - 1)}^{\frac{1}{3}} \\ ( - 1)^{\frac{1}{12}} & 0 & 0 \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \end{pmatrix}, \end{array}
(1.8.332)

and the ro­ta­tion ma­tri­ces, R(↖),R(↗)R( \nwarrow ),R( \nearrow ) rep­re­sent ori­ent­ed tri­an­gles in ℂ3\mathbb{C}^{3}, given the eigen­val­ues for both are λ={e(2iπ3),e(4iπ3),1}\lambda = \left\{ e^{\left( \frac{2i\pi}{3} \right)},e^{\left( \frac{4i\pi}{3} \right)},1 \right\}.

Figure
Figure 1.8.20 –The in­scribed tri­an­gu­lar ro­ta­tions ma­tri­ces R(↖)R( \nwarrow ) and R(↗).R( \nearrow ).

Al­though 𝓂ℯ[+igp2](→)\mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \rightarrow ) may appear dif­fer­ent from the other min­i­mal el­e­ments, 𝓂ℯ[+igp2](↖)\mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nwarrow ) and 𝓂ℯ[+igp2](↗)\mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nearrow ), its eigen­val­ues are the same

λ(𝓂ℯ[+igp2](→))=eig(−(−1)(16)000i000−(−1)(56))={−e(iπ6),e(iπ2),−e(5iπ6)},\begin{array}{r} \lambda\left( \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \rightarrow ) \right) = eig\begin{pmatrix} - ( - 1)^{\left( \frac{1}{6} \right)} & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & - ( - 1)^{\left( \frac{5}{6} \right)} \end{pmatrix} = \left\{ {- e}^{\left( \frac{i\pi}{6} \right)},e^{\left( \frac{i\pi}{2} \right)},{- e}^{\left( \frac{5i\pi}{6} \right)} \right\}, \end{array}
(1.8.333)

which is the case given that a di­ag­o­nal­ized matrix has its el­e­ments as eigen­val­ues. Graph­i­cal­ly we can rep­re­sent the two ma­tri­ces, R(↖),R(↗)R( \nwarrow ),R( \nearrow ), which rotate 𝓂ℯ[+igp2](↖)\mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nwarrow ) and 𝓂ℯ[+igp2](↗)\mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nearrow ) to 𝓂ℯ[+igp2](→)\mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \rightarrow ), as points on the com­plex unit circle (Figure 1.8.20).

The set of el­e­ments 𝓂ℯ[+igp2](→)\mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \rightarrow ) gen­er­at­ing A(Y(gp[2i])(→))A\left( Y\left( {gp}_{\lbrack 2i\rbrack} \right)( \rightarrow ) \right), is the fol­low­ing

A(Y(gp[2i])(→))={(−(−1)(16)000i000−(−1)(56)),((−1)(13)000−1000−(−1)(23)),(−i000−i000−i),((−1)(23)0001000−(−1)(13)),(−(−1)(56)000i000−(−1)(16)),(−1000−1000−1),((−1)(16)000−i000(−1)(56)),(−(−1)(13)0001000(−1)(23)),(i000i000i),(−(−1)(23)000−1000(−1)(13)),((−1)(56)000−i000(−1)(16)),(100010001)},\begin{array}{r} A\left( Y\left( {gp}_{\lbrack 2i\rbrack} \right)( \rightarrow ) \right) = \left\{ \begin{array}{r} \begin{pmatrix} - ( - 1)^{\left( \frac{1}{6} \right)} & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & - ( - 1)^{\left( \frac{5}{6} \right)} \end{pmatrix},\begin{pmatrix} ( - 1)^{\left( \frac{1}{3} \right)} & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - ( - 1)^{\left( \frac{2}{3} \right)} \end{pmatrix},\begin{pmatrix} - i & 0 & 0 \\ 0 & - i & 0 \\ 0 & 0 & - i \end{pmatrix}, \\ \begin{pmatrix} ( - 1)^{\left( \frac{2}{3} \right)} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - ( - 1)^{\left( \frac{1}{3} \right)} \end{pmatrix},\begin{pmatrix} {- ( - 1)}^{\left( \frac{5}{6} \right)} & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & - ( - 1)^{\left( \frac{1}{6} \right)} \end{pmatrix},\begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix}, \\ \begin{pmatrix} ( - 1)^{\left( \frac{1}{6} \right)} & 0 & 0 \\ 0 & - i & 0 \\ 0 & 0 & ( - 1)^{\left( \frac{5}{6} \right)} \end{pmatrix},\begin{pmatrix} {- ( - 1)}^{\left( \frac{1}{3} \right)} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & ( - 1)^{\left( \frac{2}{3} \right)} \end{pmatrix},\begin{pmatrix} i & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & i \end{pmatrix}, \\ \begin{pmatrix} {- ( - 1)}^{\left( \frac{2}{3} \right)} & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & ( - 1)^{\left( \frac{1}{3} \right)} \end{pmatrix},\begin{pmatrix} ( - 1)^{\left( \frac{5}{6} \right)} & 0 & 0 \\ 0 & - i & 0 \\ 0 & 0 & ( - 1)^{\left( \frac{1}{6} \right)} \end{pmatrix},\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{array} \right\}, \end{array}
(1.8.334)

and the set of all sets A(Y(gp[2i](↖,↗,→)))A\left( Y\left( {gp}_{\lbrack 2i\rbrack}( \nwarrow , \nearrow , \rightarrow ) \right) \right), all have the same group char­ac­ter­is­tic poly­no­mi­al

𝒞𝒫(Y(gp[2i](↖,↗,→)))=((λ−1)3+(λ−i)3+(λ+1)3+(λ+i)3+2(λ3−1)+2(λ3−i)+2(λ3+1)+2(λ3+i)),\begin{array}{r} \mathcal{CP}\left( Y\left( {gp}_{\lbrack 2i\rbrack}( \nwarrow , \nearrow , \rightarrow ) \right) \right) = \left( \begin{array}{r} (\lambda - 1)^{3} + (\lambda - i)^{3} + (\lambda + 1)^{3} + (\lambda + i)^{3} \\ + 2\left( \lambda^{3} - 1 \right) + 2\left( \lambda^{3} - i \right) + 2\left( \lambda^{3} + 1 \right) + 2\left( \lambda^{3} + i \right) \end{array} \right), \end{array}
(1.8.335)

which, as we are aware, re­duces to

𝒞𝒫(Y(gp[2i](↖,↗,→)))=12λ3.\begin{array}{r} \mathcal{CP}\left( Y\left( {gp}_{\lbrack 2i\rbrack}( \nwarrow , \nearrow , \rightarrow ) \right) \right) = 12\lambda^{3}. \end{array}
(1.8.336)

The di­men­sion­al trans­form ma­tri­ces are de­fined as

Y(→,↖,↗)={Y(→),Y(↖),Y(↗)}={(00e(− 5iπ12)010e(5iπ12)00),   (001010100),   (001010100)},\begin{array}{r} Y( \rightarrow , \nwarrow , \nearrow ) = \left\{ Y( \rightarrow ),Y( \nwarrow ),Y( \nearrow ) \right\} = \left\{ \begin{pmatrix} 0 & 0 & e^{\left( - \ \frac{5i\pi}{12} \right)} \\ 0 & 1 & 0 \\ e^{\left( \frac{5i\pi}{12} \right)} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} \right\}, \end{array}
(1.8.337)

which, when map­ping sim­i­lar point­ing arrows, rotate the min­i­mal el­e­ments. The prod­uct of A(Y(gp[2i](↖,↗,→)))A\left( Y\left( {gp}_{\lbrack 2i\rbrack}( \nwarrow , \nearrow , \rightarrow ) \right) \right) with matched ma­tri­ces Y(↖,↗,→)Y( \nwarrow , \nearrow , \rightarrow )

A(Y(gp[2i](↖,↗,→)))×Y(↖,↗,→)={A(Y(gp[i2](↖)))×Y(↖),A(Y(gp[i2](↗)))×Y(↗),A(Y(gp[i2](→)))×Y(→)},\begin{array}{r} A\left( Y\left( {gp}_{\lbrack 2i\rbrack}( \nwarrow , \nearrow , \rightarrow ) \right) \right) \times Y( \nwarrow , \nearrow , \rightarrow ) = \left\{ \begin{array}{r} A\left( Y\left( {gp}_{\lbrack i2\rbrack}( \nwarrow ) \right) \right) \times Y( \nwarrow ), \\ A\left( Y\left( {gp}_{\lbrack i2\rbrack}( \nearrow ) \right) \right) \times Y( \nearrow ), \\ A\left( Y\left( {gp}_{\lbrack i2\rbrack}( \rightarrow ) \right) \right) \times Y( \rightarrow ) \end{array} \right\}, \end{array}
(1.8.338)

will yield a set that in­cludes all the cone gen­er­a­tor Lie al­ge­bras, seen by pre­sent­ing the com­po­si­tion of el­e­ments for each subset

A(Y(gp[2i](↖)))×Y(↖)={(i0000e(3iπ4)0e(iπ4)0),   (0e(3iπ4)0e(−3iπ4)0000−1),   (00−i0−i0−i00),(10000e(iπ4)0e(− iπ4)0),   (0e(iπ4)0e(3iπ4)0000i),   (00−10−10−100),(−i0000e(− iπ4)0e(− 3iπ4)0),   (0e(− iπ4)0e(iπ4)00001),   (00i0i0i00),(−10000e(− 3iπ4)0e(3iπ4)0),   (0e(− 3iπ4)0e(− iπ4)0000−i),   (001010100),},\begin{array}{r} A\left( Y\left( {gp}_{\lbrack 2i\rbrack}( \nwarrow ) \right) \right) \times Y( \nwarrow ) = \left\{ \begin{array}{r} \begin{pmatrix} i & 0 & 0 \\ 0 & 0 & e^{\left( \frac{3i\pi}{4} \right)} \\ 0 & e^{\left( \frac{i\pi}{4} \right)} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & e^{\left( \frac{3i\pi}{4} \right)} & 0 \\ e^{\left( - \frac{3i\pi}{4} \right)} & 0 & 0 \\ 0 & 0 & - 1 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & - i \\ 0 & - i & 0 \\ - i & 0 & 0 \end{pmatrix}, \\ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & e^{\left( \frac{i\pi}{4} \right)} & 0 \\ e^{\left( \frac{3i\pi}{4} \right)} & 0 & 0 \\ 0 & 0 & i \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & - 1 \\ 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{pmatrix}, \\ \begin{pmatrix} - i & 0 & 0 \\ 0 & 0 & e^{\left( - \ \frac{i\pi}{4} \right)} \\ 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \\ e^{\left( \frac{i\pi}{4} \right)} & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & i \\ 0 & i & 0 \\ i & 0 & 0 \end{pmatrix}, \\ \begin{pmatrix} - 1 & 0 & 0 \\ 0 & 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} \\ 0 & e^{\left( \frac{3i\pi}{4} \right)} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} & 0 \\ e^{\left( - \ \frac{i\pi}{4} \right)} & 0 & 0 \\ 0 & 0 & - i \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}, \end{array} \right\}, \end{array}
(1.8.339)
A(Y(gp[2i](↗)))×Y(↗)={(0e(3iπ4)0e(iπ4)0000i),   (−10000e(3iπ4)0e(− 3iπ4)0),   (00−i0−i0−i00),(0e(iπ4)0e(− iπ4)00001),   (i0000e(iπ4)0e(3iπ4)0),   (00−10−10−100),(0e(− iπ4)0e(− 3iπ4)0000−i),   (10000e(− iπ4)0e(iπ4)0),   (00i0i0i00),(0e(− 3iπ4)0e(3iπ4)0000−1),   (−i0000e(− 3iπ4)0e(− iπ4)0),   (001010100),},\begin{array}{r} A\left( Y\left( {gp}_{\lbrack 2i\rbrack}( \nearrow ) \right) \right) \times Y( \nearrow ) = \left\{ \begin{array}{r} \begin{pmatrix} 0 & e^{\left( \frac{3i\pi}{4} \right)} & 0 \\ e^{\left( \frac{i\pi}{4} \right)} & 0 & 0 \\ 0 & 0 & i \end{pmatrix},\ \ \ \begin{pmatrix} - 1 & 0 & 0 \\ 0 & 0 & e^{\left( \frac{3i\pi}{4} \right)} \\ 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & - i \\ 0 & - i & 0 \\ - i & 0 & 0 \end{pmatrix}, \\ \begin{pmatrix} 0 & e^{\left( \frac{i\pi}{4} \right)} & 0 \\ e^{\left( - \ \frac{i\pi}{4} \right)} & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix},\ \ \ \begin{pmatrix} i & 0 & 0 \\ 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ 0 & e^{\left( \frac{3i\pi}{4} \right)} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & - 1 \\ 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{pmatrix}, \\ \begin{pmatrix} 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \\ e^{\left( - \ \frac{3i\pi}{4} \right)} & 0 & 0 \\ 0 & 0 & - i \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & e^{\left( - \ \frac{i\pi}{4} \right)} \\ 0 & e^{\left( \frac{i\pi}{4} \right)} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & i \\ 0 & i & 0 \\ i & 0 & 0 \end{pmatrix}, \\ \begin{pmatrix} 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} & 0 \\ e^{\left( \frac{3i\pi}{4} \right)} & 0 & 0 \\ 0 & 0 & - 1 \end{pmatrix},\ \ \ \begin{pmatrix} - i & 0 & 0 \\ 0 & 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} \\ 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}, \end{array} \right\}, \end{array}
(1.8.340)
A(Y(gp[2i](→)))×Y(→)={(00e(3iπ4)0i0e(iπ4)00),   (00e(− iπ12)0−10e(iπ12)00),   (00e(− 11iπ12)0−i0e(− iπ12)00),(00e(iπ4)010e(− iπ4)00),   (00e(− 7iπ12)0i0e(− 5iπ12)00),   (00e(7iπ12)0−10e(− 7iπ12)00),(00e(− iπ4)0−i0e(− 3iπ4)00),   (00e(11iπ12)010e(− 11iπ12)00),   (00e(iπ12)0i0e(11iπ12)00),(00e(− 3iπ4)0−10e(3iπ4)00),   (00e(5iπ12)0−i0e(7iπ12)00),   (00e(− 5iπ12)010e(5iπ12)00)}.\begin{array}{r} A\left( Y\left( {gp}_{\lbrack 2i\rbrack}( \rightarrow ) \right) \right) \times Y( \rightarrow ) = \left\{ \begin{array}{r} \begin{pmatrix} 0 & 0 & e^{\left( \frac{3i\pi}{4} \right)} \\ 0 & i & 0 \\ e^{\left( \frac{i\pi}{4} \right)} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & e^{\left( - \ \frac{i\pi}{12} \right)} \\ 0 & - 1 & 0 \\ e^{\left( \frac{i\pi}{12} \right)} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & e^{\left( - \ \frac{11i\pi}{12} \right)} \\ 0 & - i & 0 \\ e^{\left( - \ \frac{i\pi}{12} \right)} & 0 & 0 \end{pmatrix}, \\ \begin{pmatrix} 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ 0 & 1 & 0 \\ e^{\left( - \ \frac{i\pi}{4} \right)} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & e^{\left( - \ \frac{7i\pi}{12} \right)} \\ 0 & i & 0 \\ e^{\left( - \ \frac{5i\pi}{12} \right)} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & e^{\left( \frac{7i\pi}{12} \right)} \\ 0 & - 1 & 0 \\ e^{\left( - \ \frac{7i\pi}{12} \right)} & 0 & 0 \end{pmatrix}, \\ \begin{pmatrix} 0 & 0 & e^{\left( - \ \frac{i\pi}{4} \right)} \\ 0 & - i & 0 \\ e^{\left( - \ \frac{3i\pi}{4} \right)} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & e^{\left( \frac{11i\pi}{12} \right)} \\ 0 & 1 & 0 \\ e^{\left( - \ \frac{11i\pi}{12} \right)} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & e^{\left( \frac{i\pi}{12} \right)} \\ 0 & i & 0 \\ e^{\left( \frac{11i\pi}{12} \right)} & 0 & 0 \end{pmatrix}, \\ \begin{pmatrix} 0 & 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} \\ 0 & - 1 & 0 \\ e^{\left( \frac{3i\pi}{4} \right)} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & e^{\left( \frac{5i\pi}{12} \right)} \\ 0 & - i & 0 \\ e^{\left( \frac{7i\pi}{12} \right)} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & e^{\left( - \ \frac{5i\pi}{12} \right)} \\ 0 & 1 & 0 \\ e^{\left( \frac{5i\pi}{12} \right)} & 0 & 0 \end{pmatrix} \end{array} \right\}. \end{array}
(1.8.341)

In the A(Y(gp[2i](→)))×Y(→)A\left( Y\left( {gp}_{\lbrack 2i\rbrack}( \rightarrow ) \right) \right) \times Y( \rightarrow ) subset, the square of the last el­e­ment in each row is as fol­lows:

(00e(− 11iπ12)0−i0e(− iπ12)00)2=(−1000−1000−1),  (00e(7iπ12)0−10e(− 7iπ12)00)2=(100010001),\begin{pmatrix} 0 & 0 & e^{\left( - \ \frac{11i\pi}{12} \right)} \\ 0 & - i & 0 \\ e^{\left( - \ \frac{i\pi}{12} \right)} & 0 & 0 \end{pmatrix}^{2} = \begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix},\ \ \begin{pmatrix} 0 & 0 & e^{\left( \frac{7i\pi}{12} \right)} \\ 0 & - 1 & 0 \\ e^{\left( - \ \frac{7i\pi}{12} \right)} & 0 & 0 \end{pmatrix}^{2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},
(00e(iπ12)0i0e(11iπ12)00)2=(−1000−1000−1),  (00e(− 5iπ12)010e(5iπ12)00)2=(100010001),\begin{array}{r} \begin{pmatrix} 0 & 0 & e^{\left( \frac{i\pi}{12} \right)} \\ 0 & i & 0 \\ e^{\left( \frac{11i\pi}{12} \right)} & 0 & 0 \end{pmatrix}^{2} = \begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix},\ \ \begin{pmatrix} 0 & 0 & e^{\left( - \ \frac{5i\pi}{12} \right)} \\ 0 & 1 & 0 \\ e^{\left( \frac{5i\pi}{12} \right)} & 0 & 0 \end{pmatrix}^{2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \end{array}
(1.8.342)

where all are con­sis­tent with 3×33 \times 3 matrix rep­re­sen­ta­tions of the com­plex num­bers {−i,−1,i,1}.\left\{ - i, - 1,i,1 \right\}.With the subset re­sults after the di­men­sion­al trans­for­ma­tion, a pri­ma­ry gen­er­a­tor, that is alone ca­pa­ble of gen­er­at­ing all the other el­e­ments, does not exist. Using the ex­am­ple of the trans­formed min­i­mal el­e­ment

𝓂ℯ[+igp2](↖)×Y(↖)=(00ie(3iπ4)000e(iπ4)0)(001010100)=(i0000e(3iπ4)0e(iπ4)0),\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nwarrow ) \times Y( \nwarrow ) = \begin{pmatrix} 0 & 0 & i \\ e^{\left( \frac{3i\pi}{4} \right)} & 0 & 0 \\ 0 & e^{\left( \frac{i\pi}{4} \right)} & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} i & 0 & 0 \\ 0 & 0 & e^{\left( \frac{3i\pi}{4} \right)} \\ 0 & e^{\left( \frac{i\pi}{4} \right)} & 0 \end{pmatrix}, \end{array}
(1.8.343)

the gen­er­at­ed set from the result A(𝓂ℯ[+igp2](↖)×Y(↖))A\left( \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nwarrow ) \times Y( \nwarrow ) \right), has the subset rep­re­sen­ta­tion

A(𝓂ℯ[+igp2](↖)×Y(↖))={(i0000e(3iπ4)0e(iπ4)0),(−1000−1000−1),(−i0000e(− iπ4)0e(− 3iπ4)0),(100010001)},\begin{array}{r} A\left( \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nwarrow ) \times Y( \nwarrow ) \right) = \left\{ \begin{array}{r} \begin{pmatrix} i & 0 & 0 \\ 0 & 0 & e^{\left( \frac{3i\pi}{4} \right)} \\ 0 & e^{\left( \frac{i\pi}{4} \right)} & 0 \end{pmatrix},\begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix}, \\ \begin{pmatrix} - i & 0 & 0 \\ 0 & 0 & e^{\left( - \ \frac{i\pi}{4} \right)} \\ 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} & 0 \end{pmatrix},\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{array} \right\}, \end{array}
(1.8.344)

which con­tains the two cone por­tions ∈ℂ3\in \mathbb{C}^{3} ori­ent­ed with pos­i­tive and neg­a­tive imag­i­nary axes. Also present in the subset is the neg­a­tive min­i­mal el­e­ment

𝓂ℯ[−igp2](↖)×Y(↖)=(−i0000e(− iπ4)0e(− 3iπ4)0),\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack - \frac{i}{gp_{2}} \right\rbrack}( \nwarrow ) \times Y( \nwarrow ) = \begin{pmatrix} - i & 0 & 0 \\ 0 & 0 & e^{\left( - \ \frac{i\pi}{4} \right)} \\ 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} & 0 \end{pmatrix}, \end{array}
(1.8.345)

which is bound to the same subset, given that its gen­er­at­ed set is the same

A(𝓂ℯ[+igp2](↖)×Y(↖))=A(𝓂ℯ[−igp2](↖)×Y(↖)).\begin{array}{r} A\left( \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{i}{gp_{2}} \right\rbrack}( \nwarrow ) \times Y( \nwarrow ) \right) = A\left( \mathcal{m}\mathcal{e}_{\left\lbrack - \frac{i}{gp_{2}} \right\rbrack}( \nwarrow ) \times Y( \nwarrow ) \right). \end{array}
(1.8.346)

Next, we take the trans­for­ma­tion of the 𝓂ℯ[−1gp2](↖)\mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack}( \nwarrow ) min­i­mal el­e­ment

𝓂ℯ[−1gp2](↖)×Y(↖)=(00−1e(− 3iπ4)000e(i3π4)0)(001010100)=(−10000e(− 3iπ4)0e(3iπ4)0),\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack}( \nwarrow ) \times Y( \nwarrow ) = \begin{pmatrix} 0 & 0 & - 1 \\ e^{\left( - \ \frac{3i\pi}{4} \right)} & 0 & 0 \\ 0 & e^{\left( \frac{i3\pi}{4} \right)} & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} - 1 & 0 & 0 \\ 0 & 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} \\ 0 & e^{\left( \frac{3i\pi}{4} \right)} & 0 \end{pmatrix}, \end{array}
(1.8.347)

which is equal to the cone por­tion ∈ℂ3\in \mathbb{C}^{3} ori­ent­ed with the neg­a­tive real-axis, and its gen­er­at­ed set, A(𝓂ℯ[−1gp2](↖)×Y(↖)),A\left( \mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack}( \nwarrow ) \times Y( \nwarrow ) \right),con­tains only two el­e­ments, itself and the iden­ti­ty

A(𝓂ℯ[−1gp2](↖)×Y(↖))={(−10000e(−3iπ4)0e(3iπ4)0),(100010001)}.\begin{array}{r} A\left( \mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack}( \nwarrow ) \times Y( \nwarrow ) \right) = \left\{ \begin{pmatrix} - 1 & 0 & 0 \\ 0 & 0 & e^{\left( - \frac{3i\pi}{4} \right)} \\ 0 & e^{\left( \frac{3i\pi}{4} \right)} & 0 \end{pmatrix},\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right\}. \end{array}
(1.8.348)

We find the same holds for trans­form­ing the 𝓂ℯ[1gp2](↖)\mathcal{m}\mathcal{e}_{\left\lbrack \frac{1}{gp_{2}} \right\rbrack}( \nwarrow ) min­i­mal el­e­ment

𝓂ℯ[1gp2](↖)×Y(↖)=(001e(iπ4)000e(− iπ4)0)(001010100)=(10000e(iπ4)0e(− iπ4)0),\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack \frac{1}{gp_{2}} \right\rbrack}( \nwarrow ) \times Y( \nwarrow ) = \begin{pmatrix} 0 & 0 & 1 \\ e^{\left( \frac{i\pi}{4} \right)} & 0 & 0 \\ 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix}, \end{array}
(1.8.349)

which also has itself and the iden­ti­ty in its gen­er­at­ed subset

A(𝓂ℯ[1gp2](↖)×Y(↖))={(10000e(iπ4)0e(−iπ4)0),(100010001)}.\begin{array}{r} A\left( \mathcal{m}\mathcal{e}_{\left\lbrack \frac{1}{gp_{2}} \right\rbrack}( \nwarrow ) \times Y( \nwarrow ) \right) = \left\{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ 0 & e^{\left( - \frac{i\pi}{4} \right)} & 0 \end{pmatrix},\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right\}. \end{array}
(1.8.350)

So, a beau­ti­ful thing hap­pens in the di­men­sion­al trans­for­ma­tion process, where sub­spaces are formed with el­e­ments bound in the sub­space.

The light num­bers can take on many forms as matrix rep­re­sen­ta­tions, and we use the ex­am­ple for the min­i­mal el­e­ment 𝓂ℯ[−1gp2]\mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack}, having the matrix rep­re­sen­ta­tion

𝓂ℯ[−1gp2]=(001−100010),\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack} = \begin{pmatrix} 0 & 0 & 1 \\ - 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}, \end{array}
(1.8.351)

which has eigen­val­ues λ(𝓂ℯ[−1gp2])\lambda\left( \mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack} \right) equal to {−1,e(2iπ3),e(4iπ3)},\left\{ - 1,e^{\left( \frac{2i\pi}{3} \right)},e^{\left( \frac{4i\pi}{3} \right)} \right\},and its char­ac­ter­is­tic poly­no­mi­al is 𝒸𝓅(𝓂ℯ[−1gp2])=λ3+1\mathcal{cp}\left( \mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack} \right) = \lambda^{3} + 1, all of which are con­sis­tent with its nomen­cla­ture 𝓂ℯ[−1gp2]\mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack}. Its gen­er­at­ed set is com­posed of the fol­low­ing six el­e­ments

A(𝓂ℯ[−1gp2])={(001−100010),   (01000−1−100),   (−1000−1000−1),(00−11000−10),   (0−10001100),   (100010001)},\begin{array}{r} A\left( \mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack} \right) = \left\{ \begin{array}{r} \begin{pmatrix} 0 & 0 & 1 \\ - 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & - 1 \\ - 1 & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix}, \\ \begin{pmatrix} 0 & 0 & - 1 \\ 1 & 0 & 0 \\ 0 & - 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & - 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{array} \right\}, \end{array}
(1.8.352)

and we take the di­men­sion­al trans­for­ma­tion of the set A(𝓂ℯ[−1gp2]×Y(↖)),A\left( \mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack} \times Y( \nwarrow ) \right),to find

A(𝓂ℯ[−1gp2]×Y(↖))={(10000−1010),   (010−10000−1),   (00−10−10−100),(−1000010−10),   (0−10100001),  (001010100)}.\begin{array}{r} A\left( \mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack} \times Y( \nwarrow ) \right) = \left\{ \begin{array}{r} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 1 & 0 \\ - 1 & 0 & 0 \\ 0 & 0 & - 1 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & - 1 \\ 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{pmatrix}, \\ \begin{pmatrix} - 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & - 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix},\ \ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} \end{array} \right\}. \end{array}
(1.8.353)

The eigen­val­ues of four of the el­e­ments in the set above in­di­cate that these are ro­ta­tion ma­tri­ces,

λ({(10000−1010),   (0−10100001)})={1,i,−i},\begin{array}{r} \lambda\left( \left\{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right\} \right) = \left\{ 1,i, - i \right\}, \end{array}
(1.8.354)
λ({(010−10000−1),   (−1000010−10)})={−1,i,−i},\begin{array}{r} \lambda\left( \left\{ \begin{pmatrix} 0 & 1 & 0 \\ - 1 & 0 & 0 \\ 0 & 0 & - 1 \end{pmatrix},\ \ \ \begin{pmatrix} - 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & - 1 & 0 \end{pmatrix} \right\} \right) = \left\{ - 1,i, - i \right\}, \end{array}
(1.8.355)

where each of these ma­tri­ces gen­er­ates

A[(10000−1010)]={(10000−1010),   (1000−1000−1),   (1000010−10),   (100010001)},\begin{array}{r} A\left\lbrack \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{pmatrix} \right\rbrack = \left\{ \begin{array}{r} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & - 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{array} \right\}, \end{array}
(1.8.356)
A[(0−10100001)]={(0−10100001),   (−1000−10001),   (010−100001),   (100010001)},\begin{array}{r} A\left\lbrack \begin{pmatrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right\rbrack = \left\{ \begin{array}{r} \begin{pmatrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix},\ \ \ \begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 1 & 0 \\ - 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{array} \right\}, \end{array}
(1.8.357)
A[(−1000010−10)]={(−1000010−10),   (1000−1000−1),   (−10000−1010),   (100010001)},\begin{array}{r} A\left\lbrack \begin{pmatrix} - 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & - 1 & 0 \end{pmatrix} \right\rbrack = \left\{ \begin{array}{r} \begin{pmatrix} - 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & - 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix},\ \ \ \begin{pmatrix} - 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{array} \right\}, \end{array}
(1.8.358)
A[(010−10000−1)]={(010−10000−1),   (−1000−10001),   (0−1010000−1),   (100010001)}.\begin{array}{r} A\left\lbrack \begin{pmatrix} 0 & 1 & 0 \\ - 1 & 0 & 0 \\ 0 & 0 & - 1 \end{pmatrix} \right\rbrack = \left\{ \begin{array}{r} \begin{pmatrix} 0 & 1 & 0 \\ - 1 & 0 & 0 \\ 0 & 0 & - 1 \end{pmatrix},\ \ \ \begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & - 1 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{array} \right\}. \end{array}
(1.8.359)

Except for the iden­ti­ty ma­tri­ces, all of these gen­er­a­tors act to rotate/re­flect in the X2/X3X_{2}/X_{3} and X1/X2X_{1}/X_{2}-planes by ±i\pm i and −1- 1. We select the two non-di­ag­o­nal ma­tri­ces from these sets with their axis ori­ent­ed along the real X1X_{1}-plane,

(10000−1010), (1000010−10),\begin{array}{r} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{pmatrix},\ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & - 1 & 0 \end{pmatrix}, \end{array}
(1.8.360)

which are in­verse­ly re­lat­ed to one an­oth­er, given that

(10000−1010)−1=(1000010−10),\begin{array}{r} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{pmatrix}^{- 1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & - 1 & 0 \end{pmatrix}, \end{array}
(1.8.361)

and the eigen­val­ues of the ma­tri­ces are

λ[(10000−1010)]={1,i,−i},   λ[(1000010−10)]={1,−i,i}.\begin{array}{r} \lambda\left\lbrack \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{pmatrix} \right\rbrack = \left\{ 1,i, - i \right\},\ \ \ \lambda\left\lbrack \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & - 1 & 0 \end{pmatrix} \right\rbrack = \left\{ 1, - i,i \right\}. \end{array}
(1.8.362)

The light cone we first en­coun­tered in the chap­ter on conic sec­tions ori­ent­ed along the real X1X_{1}-axis, is the result of the prod­uct of these two ma­tri­ces, with the real-axis ori­ent­ed gen­er­a­tor from the A(Y(gp[2](↖)))×Y(↖)A\left( Y\left( {gp}_{\lbrack 2\rbrack}( \nwarrow ) \right) \right) \times Y( \nwarrow ) set

(10000e(iπ4)0e(− iπ4)0).\begin{array}{r} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix}. \end{array}
(1.8.363)

We rotate this cone sec­tion gen­er­a­tor with the fol­low­ing matrix-matrix prod­uct

(10000e(iπ4)0e(− iπ4)0)(10000−1010)=(1000e(iπ4)000e(3iπ4)),\begin{array}{r} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & e^{\left( \frac{i\pi}{4} \right)} & 0 \\ 0 & 0 & e^{\left( \frac{3i\pi}{4} \right)} \end{pmatrix}, \end{array}
(1.8.364)

which re­sults in the other light cone Lie al­ge­bra gen­er­a­tor de­fined in the chap­ter on conic sec­tions. We can now define the pri­ma­ry el­e­ments which form the light cone ori­ent­ed in the X1X_{1}-axis ∈ℝ3+\in \mathbb{R}^{3 +}, as

A(ℒ𝒞[+X1]j)={(10000e(iπ4)0e(− iπ4)0),(1000e(iπ4)000e(3iπ4))},\begin{array}{r} A\left( \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{j} \right) = \left\{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix},\begin{pmatrix} 1 & 0 & 0 \\ 0 & e^{\left( \frac{i\pi}{4} \right)} & 0 \\ 0 & 0 & e^{\left( \frac{3i\pi}{4} \right)} \end{pmatrix} \right\}, \end{array}
(1.8.365)

which to­geth­er form the fol­low­ing prod­ucts

ℒ𝒞[+X1]1×ℒ𝒞[+X1]2=(10000e(iπ4)0e(− iπ4)0)(1000e(iπ4)000e(3iπ4))=(10000−1010),\begin{array}{r} \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{1}\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ 0 & e^{\left( \frac{i\pi}{4} \right)} & 0 \\ 0 & 0 & e^{\left( \frac{3i\pi}{4} \right)} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{pmatrix}, \end{array}
(1.8.366)
ℒ𝒞[+X1]2×ℒ𝒞[+X1]1=(1000e(iπ4)000e(3iπ4))(10000e(iπ4)0e(− iπ4)0)=(10000i0i0),\begin{array}{r} \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{2}\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & e^{\left( \frac{i\pi}{4} \right)} & 0 \\ 0 & 0 & e^{\left( \frac{3i\pi}{4} \right)} \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & i \\ 0 & i & 0 \end{pmatrix}, \end{array}
(1.8.367)
(ℒ𝒞[+X1]1)2=(10000e(iπ4)0e(− iπ4)0)2=(100010001),\begin{array}{r} \left( \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{1} \right)^{2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix}^{2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \end{array}
(1.8.368)
(ℒ𝒞[+X1]2)2=(1000e(iπ4)000e(3iπ4))2=(1000i000−i).\begin{array}{r} \left( \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{2} \right)^{2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & e^{\left( \frac{i\pi}{4} \right)} & 0 \\ 0 & 0 & e^{\left( \frac{3i\pi}{4} \right)} \end{pmatrix}^{2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & - i \end{pmatrix}. \end{array}
(1.8.369)

The Lie al­ge­bra re­sults above are the 3×33 \times 3 ver­sions of the SU(2)SU(2) gen­er­a­tors seen pre­vi­ous­ly in the book. Keep­ing the axis entry equal to the iden­ti­ty, we can rep­re­sent this group of el­e­ments with vari­ables a,b,c,da,b,c,d, as

ℒ𝒞[+X1](SU(2))=(1000a+ib−c+id0c+ida−ib),\begin{array}{r} \mathcal{L}\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}\left( SU(2) \right) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & a + ib & - c + id \\ 0 & c + id & a - ib \end{pmatrix}, \end{array}
(1.8.370)

whose de­ter­mi­nant rep­re­sents the dis­tance metric in the space

det⁡(ℒ𝒞[+X1][SU(2)])=a2+b2+c2+d2,\begin{array}{r} \det{\left( \mathcal{L}\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}\left\lbrack SU(2) \right\rbrack \right) = a^{2} + b^{2} + c^{2} + d^{2},} \end{array}
(1.8.371)

find­ing the Eu­clid­ean dis­tance metric of the space we per­ceive phe­nom­e­nal­ly. With the re­sult­ing ma­tri­ces formed by the light cone gen­er­a­tor prod­ucts, we define the ℒ𝒞[+X1][SU(2)]\mathcal{L}\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}\left\lbrack SU(2) \right\rbrack set, as

ℒ𝒞[+X1][SU(2)]={(100010001),(1000i000−i),(10000−1010),(10000i0i0)}.\begin{array}{r} \mathcal{L}\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}\left\lbrack SU(2) \right\rbrack = \left\{ \begin{array}{r} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\begin{pmatrix} 1 & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & - i \end{pmatrix}, \\ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{pmatrix},\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & i \\ 0 & i & 0 \end{pmatrix} \end{array} \right\}. \end{array}
(1.8.372)

The light cone along the neg­a­tive real X1X_{1}-axis is formed sim­i­lar­ly, where we rotate the A(ℒ𝒞[+X1]j)A\left( \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{j} \right) gen­er­a­tors by eiπe^{i\pi}

eiπ×(10000e(iπ4)0e(− iπ4)0)=(−10000e(− 3iπ4)0e(3iπ4)0),\begin{array}{r} e^{i\pi} \times \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix} = \begin{pmatrix} - 1 & 0 & 0 \\ 0 & 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} \\ 0 & e^{\left( \frac{3i\pi}{4} \right)} & 0 \end{pmatrix}, \end{array}
(1.8.373)
eiπ×(1000e(iπ4)000e(3iπ4))=(−1000e(− 3iπ4)000e(− iπ4)).\begin{array}{r} e^{i\pi} \times \begin{pmatrix} 1 & 0 & 0 \\ 0 & e^{\left( \frac{i\pi}{4} \right)} & 0 \\ 0 & 0 & e^{\left( \frac{3i\pi}{4} \right)} \end{pmatrix} = \begin{pmatrix} - 1 & 0 & 0 \\ 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} & 0 \\ 0 & 0 & e^{\left( - \ \frac{i\pi}{4} \right)} \end{pmatrix}. \end{array}
(1.8.374)

We define the pri­ma­ry el­e­ments which form the light cone ori­ent­ed in the X1X_{1}-axis ∈ℝ3−\in \mathbb{R}^{3 -}, as

A(ℒ𝒞[−X1]j)={(−10000e(− 3iπ4)0e(3iπ4)0),(−1000e(− 3iπ4)000e(− iπ4))},\begin{array}{r} A\left( \mathcal{L}{\mathcal{C}_{\left\lbrack - X_{1} \right\rbrack}}_{j} \right) = \left\{ \begin{pmatrix} - 1 & 0 & 0 \\ 0 & 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} \\ 0 & e^{\left( \frac{3i\pi}{4} \right)} & 0 \end{pmatrix},\begin{pmatrix} - 1 & 0 & 0 \\ 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} & 0 \\ 0 & 0 & e^{\left( - \ \frac{i\pi}{4} \right)} \end{pmatrix} \right\}, \end{array}
(1.8.375)

which to­geth­er form the fol­low­ing prod­ucts

ℒ𝒞[−X1]1×ℒ𝒞[−X1]2=(−10000e(− 3iπ4)0e(3iπ4)0)(−1000e(− 3iπ4)000e(− iπ4))=(10000−1010),\begin{array}{r} \mathcal{L}{\mathcal{C}_{\left\lbrack - X_{1} \right\rbrack}}_{1}\times \mathcal{L}{\mathcal{C}_{\left\lbrack - X_{1} \right\rbrack}}_{2} = \begin{pmatrix} - 1 & 0 & 0 \\ 0 & 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} \\ 0 & e^{\left( \frac{3i\pi}{4} \right)} & 0 \end{pmatrix}\begin{pmatrix} - 1 & 0 & 0 \\ 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} & 0 \\ 0 & 0 & e^{\left( - \ \frac{i\pi}{4} \right)} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{pmatrix}, \end{array}
(1.8.376)
ℒ𝒞[−X1]2×ℒ𝒞[−X1]1=(−1000e(− 3iπ4)000e(− iπ4))(−10000e(− 3iπ4)0e(3iπ4)0)=(10000i0i0),\begin{array}{r} \mathcal{L}{\mathcal{C}_{\left\lbrack - X_{1} \right\rbrack}}_{2}\times \mathcal{L}{\mathcal{C}_{\left\lbrack - X_{1} \right\rbrack}}_{1} = \begin{pmatrix} - 1 & 0 & 0 \\ 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} & 0 \\ 0 & 0 & e^{\left( - \ \frac{i\pi}{4} \right)} \end{pmatrix}\begin{pmatrix} - 1 & 0 & 0 \\ 0 & 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} \\ 0 & e^{\left( \frac{3i\pi}{4} \right)} & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & i \\ 0 & i & 0 \end{pmatrix}, \end{array}
(1.8.377)
(ℒ𝒞[−X1]1)2=(−10000e(− 3iπ4)0e(3iπ4)0)2=(100010001),\begin{array}{r} \left( \mathcal{L}{\mathcal{C}_{\left\lbrack - X_{1} \right\rbrack}}_{1} \right)^{2} = \begin{pmatrix} - 1 & 0 & 0 \\ 0 & 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} \\ 0 & e^{\left( \frac{3i\pi}{4} \right)} & 0 \end{pmatrix}^{2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \end{array}
(1.8.378)
(ℒ𝒞[−X1]2)2=(−1000e(− 3iπ4)000e(− iπ4))=(1000i000−i).\begin{array}{r} \left( \mathcal{L}{\mathcal{C}_{\left\lbrack - X_{1} \right\rbrack}}_{2} \right)^{2} = \begin{pmatrix} - 1 & 0 & 0 \\ 0 & e^{\left( - \ \frac{3i\pi}{4} \right)} & 0 \\ 0 & 0 & e^{\left( - \ \frac{i\pi}{4} \right)} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & - i \end{pmatrix}. \end{array}
(1.8.379)

We take these Lie al­ge­bra re­sults to define the ℒ𝒞[−X1][SU(2)]\mathcal{L}\mathcal{C}_{\left\lbrack - X_{1} \right\rbrack}\left\lbrack SU(2) \right\rbrack set as

ℒ𝒞[−X1][SU(2)]={(100010001),   (1000i000−i),(10000−1010),   (10000i0i0)},\begin{array}{r} \mathcal{L}\mathcal{C}_{\left\lbrack - X_{1} \right\rbrack}\left\lbrack SU(2) \right\rbrack = \left\{ \begin{array}{r} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & - i \end{pmatrix}, \\ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & i \\ 0 & i & 0 \end{pmatrix} \end{array} \right\}, \end{array}
(1.8.380)

which we find is equiv­a­lent to the ℒ𝒞[X1][SU(2)]\mathcal{L}\mathcal{C}_{\left\lbrack X_{1} \right\rbrack}\left\lbrack SU(2) \right\rbrack set. It is im­por­tant to note el­e­ments from the ℒ𝒞[+X1][SU(2)]\mathcal{L}\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}\left\lbrack SU(2) \right\rbrack and the ℒ𝒞[−X1][SU(2)]\mathcal{L}\mathcal{C}_{\left\lbrack - X_{1} \right\rbrack}\left\lbrack SU(2) \right\rbrack can gen­er­ate other el­e­ments which depend on the order of the matrix prod­uct. For ex­am­ple, we con­sid­er the prod­uct of the fol­low­ing two el­e­ments ∈ℒ𝒞[+X1][SU(2)]\in \mathcal{L}\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}\left\lbrack SU(2) \right\rbrack

(1000i000−i)(10000−1010)=(10000−i0−i0),\begin{array}{r} \begin{pmatrix} 1 & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & - i \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - i \\ 0 & - i & 0 \end{pmatrix}, \end{array}
(1.8.381)

whose result has neg­a­tive values in the bottom-right 2×22 \times 2 sub­ma­trix, com­pared with the pos­i­tive el­e­ment found

∈ℒ𝒞[−X1][SU(2)], ℒ𝒞[X1][SU(2)]\in \mathcal{L}\mathcal{C}_{\left\lbrack - X_{1} \right\rbrack}\left\lbrack SU(2) \right\rbrack,\ \mathcal{L}\mathcal{C}_{\left\lbrack X_{1} \right\rbrack}\left\lbrack SU(2) \right\rbrack. These neg­a­tive-valued el­e­ments rep­re­sent ro­ta­tions in the op­pos­ing di­rec­tions, and by taking these into ac­count, the light cone set ℒ𝒞[X1][SU(2)]\mathcal{L}\mathcal{C}_{\left\lbrack X_{1} \right\rbrack}\left\lbrack SU(2) \right\rbrack is de­fined as the fol­low­ing set of Lie al­ge­bra gen­er­a­tors

ℒ𝒞[X1][SU(2)]={(100010001),   (1000i000−i),   (10000−1010),   (10000i0i0),(1000−1000−1),   (1000−i000i),   (1000010−10),   (10000−i0−i0)}.\begin{array}{r} \mathcal{L}\mathcal{C}_{\left\lbrack X_{1} \right\rbrack}\left\lbrack SU(2) \right\rbrack = \left\{ \begin{array}{r} \begin{array}{r} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & - i \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & i \\ 0 & i & 0 \end{pmatrix} \end{array}, \\ \begin{pmatrix} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & - i & 0 \\ 0 & 0 & i \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & - 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - i \\ 0 & - i & 0 \end{pmatrix} \end{array} \right\}. \end{array}
(1.8.382)

This is a beau­ti­ful math­e­mat­i­cal result, given that the spaces cre­at­ed by two light cones, which are an­tipo­dal to one an­oth­er, create seam­less Eu­clid­ean space. So, moving in one space will feel like moving in the other space, and the two spaces will be in­dis­cernible to any­thing bound in the ℒ𝒞[X1][SU(2)]\mathcal{L}\mathcal{C}_{\left\lbrack X_{1} \right\rbrack}\left\lbrack SU(2) \right\rbrack subset. Light is re­spon­si­ble for this, par­tic­u­lar­ly the cone num­bers of light. The other light cones that are ori­ent­ed in the X2,X3X_{2},X_{3}-axes will yield sim­i­lar sets, and we will ded­i­cate a whole physics book on the sub­ject in Arith­moí -Volume II, Part 1: "Phe­nom­e­nol­o­gy and the Geom­e­try of Light".

To vi­su­al­ize the space along the X1X_{1}-axis that the min­i­mal el­e­ments

𝓂ℯ[±1gp2](↖)=(001e(iπ4)000e(− iπ4)0),(00−1e(− 3iπ4)000e(3iπ4)0)\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack \pm \frac{1}{gp_{2}} \right\rbrack}( \nwarrow ) = \begin{pmatrix} 0 & 0 & 1 \\ e^{\left( \frac{i\pi}{4} \right)} & 0 & 0 \\ 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 & - 1 \\ e^{\left( - \ \frac{3i\pi}{4} \right)} & 0 & 0 \\ 0 & e^{\left( \frac{3i\pi}{4} \right)} & 0 \end{pmatrix} \end{array}
(1.8.383)

create, we plot in Figure 1.8.21 the eigen­val­ues of 𝓂ℯ[±1gp2](↖)\mathcal{m}\mathcal{e}_{\left\lbrack \pm \frac{1}{gp_{2}} \right\rbrack}( \nwarrow ), as the ver­tices of the in­scribed equi­lat­er­al tri­an­gle. Em­a­nat­ing from the com­plex unit sphere in the pos­i­tive and neg­a­tive di­rec­tions are the light cones of the X1X_{1}-axis.

For com­plete­ness, we con­sid­er ro­tat­ing to the A(ℒ𝒞[+X2]j)A\left( \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{2} \right\rbrack}}_{j} \right) and A(ℒ𝒞[+X3]j)A\left( \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{3} \right\rbrack}}_{j} \right) light cone sets, ori­ent­ed along the X2X_{2} and X3X_{3}- axes, using the start­ing point of the A(ℒ𝒞[+X1]j)A\left( \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{j} \right) light cone set

A(ℒ𝒞[+X1]j)={(10000e(iπ4)0e(− iπ4)0),(1000e(iπ4)000e(3iπ4))}.\begin{array}{r} A\left( \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{j} \right) = \left\{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix},\begin{pmatrix} 1 & 0 & 0 \\ 0 & e^{\left( \frac{i\pi}{4} \right)} & 0 \\ 0 & 0 & e^{\left( \frac{3i\pi}{4} \right)} \end{pmatrix} \right\}. \end{array}
(1.8.384)
Figure
Figure 1.8.21 – In­scribed min­i­mal el­e­ments and the X1X_{1}-axis light cone. The pos­i­tive­ly ori­ent­ed min­i­mal el­e­ment, 𝓂ℯ[1gp2](↖)\mathcal{m}\mathcal{e}_{\left\lbrack \frac{1}{gp_{2}} \right\rbrack}( \nwarrow ), is pic­tured as the red tri­an­gle, and the neg­a­tive­ly ori­ent­ed min­i­mal el­e­ment, 𝓂ℯ[−1gp2](↖)\mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack}( \nwarrow ), is in blue.

Using the simple case of an X1X_{1}-type Lie al­ge­bra

X1=(100001010),\begin{array}{r} X_{1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}, \end{array}
(1.8.385)

the Y2Y_{2} and Y3Y_{3}-types of Lie al­ge­bra will rotate X1X_{1} as fol­lows

X1×Y3=(100001010)(001100010)=(001010100),\begin{array}{r} X_{1} \times Y_{3} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}, \end{array}
(1.8.386)
X1×Y2=(100001010)(010001100)=(010100001),\begin{array}{r} X_{1} \times Y_{2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \end{array}
(1.8.387)

where we see that X1×Y3X_{1} \times Y_{3} ro­tates to the X2X_{2}-type al­ge­bra, and X1×Y2X_{1} \times Y_{2} ro­tates to the X3X_{3}-type al­ge­bra. It should now be clear that the Y2,Y3Y_{2},Y_{3} gen­er­a­tors we have worked with through­out the book are the gp2gp_{2} geo­met­ric prime in its sim­plest in­scribed ori­en­ta­tions, having eigen­val­ues equal to λ(Y2,Y3)={e(2iπ3),e(4iπ3),1}.\lambda\left( Y_{2},Y_{3} \right) = \left\{ e^{\left( \frac{2i\pi}{3} \right)},e^{\left( \frac{4i\pi}{3} \right)},1 \right\}.

We have seen before that chang­ing the order of each prod­uct will yield ro­ta­tions in op­pos­ing di­rec­tions

Y3×X1=X3:(001100010)(100001010)=(010100001),\begin{array}{r} Y_{3} \times X_{1} = X_{3}:\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \end{array}
(1.8.388)
Y2×X1=X2:(010001100)(100001010)=(001010100),\begin{array}{r} Y_{2} \times X_{1} = X_{2}:\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}, \end{array}
(1.8.389)

Given these ro­ta­tions, we know the right mul­ti­pli­ca­tion of the ℒ𝒞[+X1]1\mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{1} gen­er­a­tor by some Y3Y_{3}-type arrange­ment of el­e­ments will yield ℒ𝒞[+X2]1\mathcal{L}{\mathcal{C}_{\left\lbrack + X_{2} \right\rbrack}}_{1}, and right mul­ti­pli­ca­tion by a Y2Y_{2}-type arrange­ment will result in the ℒ𝒞[+X3]1\mathcal{L}{\mathcal{C}_{\left\lbrack + X_{3} \right\rbrack}}_{1} gen­er­a­tor.

The Y3Y_{3}-type gen­er­a­tor is found by taking the min­i­mal el­e­ment from the A(Y(gp[2i](↖)))A\left( Y\left( {gp}_{\lbrack 2i\rbrack}( \nwarrow ) \right) \right) set, 𝓂ℯ[igp2](↖),\mathcal{m}\mathcal{e}_{\left\lbrack \frac{i}{gp_{2}} \right\rbrack}( \nwarrow ),which has the matrix rep­re­sen­ta­tion

𝓂ℯ[igp2](↖)=(00i(−1)34000(−1)140),\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack \frac{i}{gp_{2}} \right\rbrack}( \nwarrow ) = \begin{pmatrix} 0 & 0 & i \\ ( - 1)^{\frac{3}{4}} & 0 & 0 \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \end{pmatrix}, \end{array}
(1.8.390)

then switch the values for the a2,1a_{2,1} and a1,3a_{1,3} en­tries, which gives us

𝓂ℯ[igp2](↖)(a2,1:a1,3↔)=(00(−1)34i000(−1)140).\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack \frac{i}{gp_{2}} \right\rbrack}( \nwarrow )\left( \overset{\leftrightarrow}{a_{2,1}:a_{1,3}} \right) = \begin{pmatrix} 0 & 0 & ( - 1)^{\frac{3}{4}} \\ i & 0 & 0 \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \end{pmatrix}. \end{array}
(1.8.391)

As a pri­ma­ry gen­er­a­tor, 𝓂ℯ[igp2](↖)(a2,1:a1,3↔)\mathcal{m}\mathcal{e}_{\left\lbrack \frac{i}{gp_{2}} \right\rbrack}( \nwarrow )\left( \overset{\leftrightarrow}{a_{2,1}:a_{1,3}} \right) yields the fol­low­ing set

A(Y(gp[2i](↖)(a2,1:a1,3↔)))={(00(−1)34i000(−1)140),   (0−1000−(−1)14(−1)3400),   (−i000−i000−i),(00(−1)141000−(−1)340),   (0i000(−1)34(−1)1400),   (−1000−1000−1),(00−(−1)34−i000−(−1)140),   (01000(−1)14−(−1)3400),   (i000i000i),(00−(−1)14−1000(−1)340),   (0−i000−(−1)34−(−1)1400),   (100010001),},\begin{array}{r} A\left( Y\left( {gp}_{\lbrack 2i\rbrack}( \nwarrow )\left( \overset{\leftrightarrow}{a_{2,1}:a_{1,3}} \right) \right) \right) = \left\{ \begin{array}{r} \begin{pmatrix} 0 & 0 & ( - 1)^{\frac{3}{4}} \\ i & 0 & 0 \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & - 1 & 0 \\ 0 & 0 & - ( - 1)^{\frac{1}{4}} \\ ( - 1)^{\frac{3}{4}} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} - i & 0 & 0 \\ 0 & - i & 0 \\ 0 & 0 & - i \end{pmatrix}, \\ \begin{pmatrix} 0 & 0 & ( - 1)^{\frac{1}{4}} \\ 1 & 0 & 0 \\ 0 & {- ( - 1)}^{\frac{3}{4}} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & i & 0 \\ 0 & 0 & ( - 1)^{\frac{3}{4}} \\ ( - 1)^{\frac{1}{4}} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix}, \\ \begin{pmatrix} 0 & 0 & {- ( - 1)}^{\frac{3}{4}} \\ - i & 0 & 0 \\ 0 & {- ( - 1)}^{\frac{1}{4}} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \\ - ( - 1)^{\frac{3}{4}} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} i & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & i \end{pmatrix}, \\ \begin{pmatrix} 0 & 0 & {- ( - 1)}^{\frac{1}{4}} \\ - 1 & 0 & 0 \\ 0 & ( - 1)^{\frac{3}{4}} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & - i & 0 \\ 0 & 0 & - ( - 1)^{\frac{3}{4}} \\ - ( - 1)^{\frac{1}{4}} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \end{array} \right\}, \end{array}
(1.8.392)

which in­ter­changes a2,1a_{2,1},a1,3a_{1,3} en­tries for the ma­tri­ces in the left column (Y3Y_{3}-type), and their trans­pose en­tries a1,2a_{1,2},a3,1a_{3,1} in the ma­tri­ces of the middle column (Y2Y_{2}-type).

We define the fol­low­ing geo­met­ric prime ro­ta­tion ma­tri­ces using the no­ta­tion R[j,k]R_{\lbrack j,k\rbrack}

R[1,2]=(00e(iπ4)1000e(− iπ4)0),  R[2,1]=(01000e(iπ4)e(− iπ4)00),\begin{array}{r} R_{\lbrack 1,2\rbrack} = \begin{pmatrix} 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ 1 & 0 & 0 \\ 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix},\ \ R_{\lbrack 2,1\rbrack} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ e^{\left( - \ \frac{i\pi}{4} \right)} & 0 & 0 \end{pmatrix}, \end{array}
(1.8.393)
R[2,3]=(00e(iπ4)e(− iπ4)00010),  R[3,2]=(0e(iπ4)0001e(− iπ4)00),\begin{array}{r} R_{\lbrack 2,3\rbrack} = \begin{pmatrix} 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ e^{\left( - \ \frac{i\pi}{4} \right)} & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix},\ \ R_{\lbrack 3,2\rbrack} = \begin{pmatrix} 0 & e^{\left( \frac{i\pi}{4} \right)} & 0 \\ 0 & 0 & 1 \\ e^{\left( - \ \frac{i\pi}{4} \right)} & 0 & 0 \end{pmatrix}, \end{array}
(1.8.394)
R[3,1]=(00ie(− iπ4)000e(− iπ4)0),  R[1,3]=(0e(iπ4)000e(iπ4)−i00),\begin{array}{r} R_{\lbrack 3,1\rbrack} = \begin{pmatrix} 0 & 0 & i \\ e^{\left( - \ \frac{i\pi}{4} \right)} & 0 & 0 \\ 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix},\ \ R_{\lbrack 1,3\rbrack} = \begin{pmatrix} 0 & e^{\left( \frac{i\pi}{4} \right)} & 0 \\ 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ - i & 0 & 0 \end{pmatrix}, \end{array}
(1.8.395)

for which the j,kj,k values in the sub­script in­di­cate an ori­ent­ed ro­ta­tion from the jj-axis to the kk-axis. Each pair­ing of the ro­ta­tion ma­tri­ces R[j,k],R[k,j]R_{\lbrack j,k\rbrack},R_{\lbrack k,j\rbrack} are in­verse­ly re­lat­ed to one an­oth­er, and these ro­ta­tion ma­tri­ces are de­rived from the prod­ucts of the ℒ𝒞[+Xj]1\mathcal{L}{\mathcal{C}_{\left\lbrack + X_{j} \right\rbrack}}_{1} light cone seg­ments as fol­lows

ℒ𝒞[+X1]1×ℒ𝒞[+X2]1=R[1,2],  ℒ𝒞[+X2]1×ℒ𝒞[+X1]1=R[2,1],\begin{array}{r} \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{1}\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{2} \right\rbrack}}_{1} = R_{\lbrack 1,2\rbrack},\ \ \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{2} \right\rbrack}}_{1}\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{1} = R_{\lbrack 2,1\rbrack}, \end{array}
(1.8.396)
ℒ𝒞[+X2]1×ℒ𝒞[+X3]1=R[2,3],  ℒ𝒞[+X3]1×ℒ𝒞[+X2]1=R[3,2],\begin{array}{r} \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{2} \right\rbrack}}_{1}\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{3} \right\rbrack}}_{1} = R_{\lbrack 2,3\rbrack},\ \ \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{3} \right\rbrack}}_{1}\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{2} \right\rbrack}}_{1} = R_{\lbrack 3,2\rbrack}, \end{array}
(1.8.397)
ℒ𝒞[+X3]1×ℒ𝒞[+X1]1=R[3,1],  ℒ𝒞[+X1]1×ℒ𝒞[+X3]1=R[1,3].\begin{array}{r} \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{3} \right\rbrack}}_{1}\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{1} = R_{\lbrack 3,1\rbrack},\ \ \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{1}\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{3} \right\rbrack}}_{1} = R_{\lbrack 1,3\rbrack}. \end{array}
(1.8.398)

The R[j,k]R_{\lbrack j,k\rbrack} ma­tri­ces rep­re­sent gp2gp_{2}-type geo­met­ric primes, which result from the prod­ucts of the ℒ𝒞[+Xj]1\mathcal{L}{\mathcal{C}_{\left\lbrack + X_{j} \right\rbrack}}_{1} light cone seg­ments. All these ma­tri­ces have de­ter­mi­nants equal to one, which is con­sis­tent with the value of light being set to unity. Given these ma­tri­ces, we can define the fol­low­ing light cone seg­ment ro­ta­tions and re­spec­tive in­verse ro­ta­tions as

ℒ𝒞[+X1]1×R[1,2]=ℒ𝒞[+X2]1,  ℒ𝒞[+X2]1×R[2,1]=ℒ𝒞[+X1]1,\begin{array}{r} \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{1} \times R_{\lbrack 1,2\rbrack}= \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{2} \right\rbrack}}_{1},\ \ \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{2} \right\rbrack}}_{1} \times R_{\lbrack 2,1\rbrack}= \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{1}, \end{array}
(1.8.399)
ℒ𝒞[+X2]1×R[2,3]=ℒ𝒞[+X3]1,  ℒ𝒞[+X3]1×R[3,2]=ℒ𝒞[+X2]1,\begin{array}{r} \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{2} \right\rbrack}}_{1} \times R_{\lbrack 2,3\rbrack}= \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{3} \right\rbrack}}_{1},\ \ \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{3} \right\rbrack}}_{1} \times R_{\lbrack 3,2\rbrack}= \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{2} \right\rbrack}}_{1}, \end{array}
(1.8.400)
ℒ𝒞[+X3]1×R[3,1]=ℒ𝒞[+X1]1,  ℒ𝒞[+X1]1×R[1,3]=ℒ𝒞[+X3]1.\begin{array}{r} \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{3} \right\rbrack}}_{1} \times R_{\lbrack 3,1\rbrack}= \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{1},\ \ \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{1} \times R_{\lbrack 1,3\rbrack}= \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{3} \right\rbrack}}_{1}. \end{array}
(1.8.401)

The math­e­mat­i­cal rea­son­ing as to how the ro­ta­tions work, is based upon the fact that each ℒ𝒞[Xj]1\mathcal{L}{\mathcal{C}_{\left\lbrack X_{j} \right\rbrack}}_{1} light cone seg­ment will square to the iden­ti­ty, and the ro­ta­tor is de­fined as ℒ𝒞[+Xj]1×ℒ𝒞[+Xk]1=R[j,k]\mathcal{L}{\mathcal{C}_{\left\lbrack + X_{j} \right\rbrack}}_{1}\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{k} \right\rbrack}}_{1} = R_{\lbrack j,k\rbrack}. So, given that (ℒ𝒞[+Xj]1)2=ϵ\left( \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{j} \right\rbrack}}_{1} \right)^{2} = \epsilon, the de­fined ro­ta­tions will all gen­er­al­ize as fol­lows

ℒ𝒞[+Xj]1×R[j,k]→ℒ𝒞[+Xj]1×(ℒ𝒞[+Xj]1×ℒ𝒞[+Xk]1)\mathcal{L}{\mathcal{C}_{\left\lbrack + X_{j} \right\rbrack}}_{1} \times R_{\lbrack j,k\rbrack}\rightarrow \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{j} \right\rbrack}}_{1} \times \left( \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{j} \right\rbrack}}_{1}\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{k} \right\rbrack}}_{1} \right)
→(ℒ𝒞[+Xj]1)2×ℒ𝒞[+Xk]1=ϵ×ℒ𝒞[+Xk]1=ℒ𝒞[+Xk]1.\begin{array}{r} \rightarrow \left( \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{j} \right\rbrack}}_{1} \right)^{2}\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{k} \right\rbrack}}_{1} = \epsilon\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{k} \right\rbrack}}_{1}= \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{k} \right\rbrack}}_{1}. \end{array}
(1.8.402)

The straight­for­ward R[j,]R_{\lbrack j,\rbrack} matrix prod­ucts

R[2,1]×R[3,2]=(001100010)=𝓂ℯ[+1gp2],\begin{array}{r} R_{\lbrack 2,1\rbrack} \times R_{\lbrack 3,2\rbrack} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}= \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{1}{gp_{2}} \right\rbrack}, \end{array}
(1.8.403)
R[2,3]×R[1,2]=(010001100)=𝓂ℯ[−1gp2],\begin{array}{r} R_{\lbrack 2,3\rbrack} \times R_{\lbrack 1,2\rbrack} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}= \mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack}, \end{array}
(1.8.404)

yield the pos­i­tive and neg­a­tive gp2gp_{2} min­i­mal el­e­ments. There are sev­er­al ways to rotate the ℒ𝒞[+Xj]2\mathcal{L}{\mathcal{C}_{\left\lbrack + X_{j} \right\rbrack}}_{2} light cone sec­tions, and we can choose to use the gp2gp_{2} min­i­mal el­e­ments, along with the R[2,3],R[3,2]R_{\lbrack 2,3\rbrack},R_{\lbrack 3,2\rbrack} ro­ta­tors to define the fol­low­ing bi­jec­tive ro­ta­tion prod­ucts

𝓂ℯ[−1gp2]×ℒ𝒞[+X1]2×𝓂ℯ[+1gp2]=ℒ𝒞[+X3]2,  𝓂ℯ[+1gp2]×ℒ𝒞[+X3]2×𝓂ℯ[−1gp2]=ℒ𝒞[+X1]2,\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack}\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{2}\times \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{1}{gp_{2}} \right\rbrack}= \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{3} \right\rbrack}}_{2},\ \ \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{1}{gp_{2}} \right\rbrack}\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{3} \right\rbrack}}_{2}\times \mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack}= \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{2}, \end{array}
(1.8.405)
R[2,3]×ℒ𝒞[+X3]2×𝓂ℯ[−1gp2]=ℒ𝒞[+X2]2,  R[3,2]×ℒ𝒞[+X2]2×𝓂ℯ[+1gp2]=ℒ𝒞[+X3]2,\begin{array}{r} R_{\lbrack 2,3\rbrack}\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{3} \right\rbrack}}_{2}\times \mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack}= \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{2} \right\rbrack}}_{2},\ \ R_{\lbrack 3,2\rbrack}\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{2} \right\rbrack}}_{2}\times \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{1}{gp_{2}} \right\rbrack}= \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{3} \right\rbrack}}_{2}, \end{array}
(1.8.406)
R[2,3]×𝓂ℯ[−1gp2]×ℒ𝒞[+X1]2=ℒ𝒞[+X2]2,  𝓂ℯ[+1gp2]×R[3,2]×ℒ𝒞[+X2]2=ℒ𝒞[+X1]2.\begin{array}{r} R_{\lbrack 2,3\rbrack}\times \mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{gp_{2}} \right\rbrack}\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{2}= \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{2} \right\rbrack}}_{2},\ \ \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{1}{gp_{2}} \right\rbrack} \times R_{\lbrack 3,2\rbrack}\times \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{2} \right\rbrack}}_{2}= \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{2}. \end{array}
(1.8.407)

Note, as de­fined in the conic sec­tions chap­ter, ℒ𝒞[+X2]2\mathcal{L}{\mathcal{C}_{\left\lbrack + X_{2} \right\rbrack}}_{2}, ℒ𝒞[+X3]2\mathcal{L}{\mathcal{C}_{\left\lbrack + X_{3} \right\rbrack}}_{2} have the same matrix rep­re­sen­ta­tions

ℒ𝒞[+X2]2=(e(iπ4)0001000e(3iπ4)),\begin{array}{r} \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{2} \right\rbrack}}_{2} = \begin{pmatrix} e^{\left( \frac{i\pi}{4} \right)} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & e^{\left( \frac{3i\pi}{4} \right)} \end{pmatrix}, \end{array}
(1.8.408)
ℒ𝒞[+X3]2=(e(iπ4)000e(3iπ4)0001).\begin{array}{r} \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{3} \right\rbrack}}_{2} = \begin{pmatrix} e^{\left( \frac{i\pi}{4} \right)} & 0 & 0 \\ 0 & e^{\left( \frac{3i\pi}{4} \right)} & 0 \\ 0 & 0 & 1 \end{pmatrix}. \end{array}
(1.8.409)

With light cone sec­tion ro­ta­tions de­fined, we see that there is a very suc­cinct in­ter­play be­tween the in­scribed geo­met­ric primes and the light cone sec­tions (ℒ𝒞[+Xj]1,ℒ𝒞[+Xj]2),\left( \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{j} \right\rbrack}}_{1},\mathcal{L}{\mathcal{C}_{\left\lbrack + X_{j} \right\rbrack}}_{2} \right),where the prod­ucts of light cone sec­tions yield in­scribed geo­met­ric prime re­sults, and these geo­met­ric primes also act to rotate the light cone sec­tions them­selves.

Up to this point, we have been con­sid­er­ing light number so­lu­tions to the char­ac­ter­is­tic poly­no­mi­al equal to axis num­bers. We let the YℒkY_{\mathcal{L}k} no­ta­tion define the fol­low­ing YY-type ma­tri­ces

Yℒ1=(a000b000c),  Yℒ2=(0a000bc00),  Yℒ3=(00ab000c0),\begin{array}{r} Y_{\mathcal{L}1} = \begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix},\ \ Y_{\mathcal{L}2} = \begin{pmatrix} 0 & a & 0 \\ 0 & 0 & b \\ c & 0 & 0 \end{pmatrix},\ \ Y_{\mathcal{L}3} = \begin{pmatrix} 0 & 0 & a \\ b & 0 & 0 \\ 0 & c & 0 \end{pmatrix}, \end{array}
(1.8.410)

for which all eigen­val­ue and eigen­vec­tor so­lu­tions will be as fol­lows

Yℒ1[V,D]=[(100010001),(a000b000c) ],\begin{array}{r} Y_{\mathcal{L}1}\lbrack V,D\rbrack = \left\lbrack \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix}\ \right\rbrack, \end{array}
(1.8.411)
Yℒ2[V,D]=Yℒ2[V,D]=[(−(−1)13((abc)(13))000(−1)23(abc)(13)000(abc)(13)),(−(−1)13((abc)(13)c)(−1)23((abc)(13)c)(abc)(13)c(−1)23((abc)(23)ac)−(−1)13((abc)(23)ac)(abc)(23)ac111)],\begin{array}{r} Y_{\mathcal{L}2}\lbrack V,D\rbrack = Y_{\mathcal{L}2}\lbrack V,D\rbrack = \left\lbrack \begin{array}{r} \begin{pmatrix} - ( - 1)^{\frac{1}{3}}\left( (abc)^{\left( \frac{1}{3} \right)} \right) & 0 & 0 \\ 0 & ( - 1)^{\frac{2}{3}}(abc)^{\left( \frac{1}{3} \right)} & 0 \\ 0 & 0 & (abc)^{\left( \frac{1}{3} \right)} \end{pmatrix}, \\ \begin{pmatrix} - ( - 1)^{\frac{1}{3}}\left( \frac{(abc)^{\left( \frac{1}{3} \right)}}{c} \right) & ( - 1)^{\frac{2}{3}}\left( \frac{(abc)^{\left( \frac{1}{3} \right)}}{c} \right) & \frac{(abc)^{\left( \frac{1}{3} \right)}}{c} \\ ( - 1)^{\frac{2}{3}}\left( \frac{(abc)^{\left( \frac{2}{3} \right)}}{ac} \right) & - ( - 1)^{\frac{1}{3}}\left( \frac{(abc)^{\left( \frac{2}{3} \right)}}{ac} \right) & \frac{(abc)^{\left( \frac{2}{3} \right)}}{ac} \\ 1 & 1 & 1 \end{pmatrix} \end{array} \right\rbrack, \end{array}
(1.8.412)
Yℒ3[V,D]=[(−(−1)13((abc)(13))000(−1)23(abc)(13)000(abc)(13)),((−1)23((abc)(23)bc)−(−1)13((abc)(23)bc)(abc)(23)c−(−1)13((abc)(13)c)(−1)23((abc)(13)c)(abc)(13)ac111)].\begin{array}{r} Y_{\mathcal{L}3}\lbrack V,D\rbrack = \left\lbrack \begin{array}{r} \begin{pmatrix} - ( - 1)^{\frac{1}{3}}\left( (abc)^{\left( \frac{1}{3} \right)} \right) & 0 & 0 \\ 0 & ( - 1)^{\frac{2}{3}}(abc)^{\left( \frac{1}{3} \right)} & 0 \\ 0 & 0 & (abc)^{\left( \frac{1}{3} \right)} \end{pmatrix}, \\ \begin{pmatrix} ( - 1)^{\frac{2}{3}}\left( \frac{(abc)^{\left( \frac{2}{3} \right)}}{bc} \right) & - ( - 1)^{\frac{1}{3}}\left( \frac{(abc)^{\left( \frac{2}{3} \right)}}{bc} \right) & \frac{(abc)^{\left( \frac{2}{3} \right)}}{c} \\ - ( - 1)^{\frac{1}{3}}\left( \frac{(abc)^{\left( \frac{1}{3} \right)}}{c} \right) & ( - 1)^{\frac{2}{3}}\left( \frac{(abc)^{\left( \frac{1}{3} \right)}}{c} \right) & \frac{(abc)^{\left( \frac{1}{3} \right)}}{ac} \\ 1 & 1 & 1 \end{pmatrix} \end{array} \right\rbrack. \end{array}
(1.8.413)

The YℒkY_{\mathcal{L}k} char­ac­ter­is­tic poly­no­mi­als are found to be the fol­low­ing

𝒸𝓅(Yℒ1)=(λ−a)(λ−b)(λ−c)=0,\begin{array}{r} \mathcal{cp}\left( Y_{\mathcal{L}1} \right) = (\lambda - a)(\lambda - b)(\lambda - c) = 0, \end{array}
(1.8.414)
𝒸𝓅(Yℒ2)=λ3−abc=0,\begin{array}{r} \mathcal{cp}\left( Y_{\mathcal{L}2} \right) = \lambda^{3} - abc = 0, \end{array}
(1.8.415)
𝒸𝓅(Yℒ3)=λ3−abc=0,\begin{array}{r} \mathcal{cp}\left( Y_{\mathcal{L}3} \right) = \lambda^{3} - abc = 0, \end{array}
(1.8.416)

which sum to the group char­ac­ter­is­tic poly­no­mi­al

𝒞𝒫(Yℒk)=∑𝒸𝓅(Yℒk)=(λ−a)(λ−b)(λ−c)+2λ3−2abc=0.\begin{array}{r} \mathcal{CP}\left( Y_{\mathcal{L}k} \right) = \sum_{}^{}{\mathcal{cp}\left( Y_{\mathcal{L}k} \right) = (\lambda - a)(\lambda - b)(\lambda - c) + 2\lambda^{3} - 2abc} = 0. \end{array}
(1.8.417)

As stated ear­li­er in the sec­tion, the num­bers re­lat­ed to light will travel via as­ymp­totes and axes. More specif­i­cal­ly, the di­men­sion­al­ly trans­formed light num­bers move in this manner. The in­scribed geom­e­try of the as­ymp­tot­ic light num­bers will have Lie gen­er­a­tors of the form

Yℒ1=(ca000cb000cc),  Yℒ2=(0ca000cbcc00),  Yℒ3=(00cacb000cc0),\begin{array}{r} Y_{\mathcal{L}1} = \begin{pmatrix} c_{a} & 0 & 0 \\ 0 & c_{b} & 0 \\ 0 & 0 & c_{c} \end{pmatrix},\ \ Y_{\mathcal{L}2} = \begin{pmatrix} 0 & c_{a} & 0 \\ 0 & 0 & c_{b} \\ c_{c} & 0 & 0 \end{pmatrix},\ \ Y_{\mathcal{L}3} = \begin{pmatrix} 0 & 0 & c_{a} \\ c_{b} & 0 & 0 \\ 0 & c_{c} & 0 \end{pmatrix}, \end{array}
(1.8.418)

where the ca,b,cc_{a,b,c} vari­ables are the as­ymp­tote subset el­e­ments from the A(cj)A\left( c_{j} \right) light set, which was de­fined ear­li­er in the sec­tion, as

A(cj)={(1),   ((−1)14),   ((−1)12),   ((−1)34),(−1),   (−(−1)14),   (−(−1)12),   (−(−1)34)},\begin{array}{r} A\left( c_{j} \right) = \left\{ \begin{array}{r} (1),\ \ \ \left( ( - 1)^{\frac{1}{4}} \right),\ \ \ \left( ( - 1)^{\frac{1}{2}} \right),\ \ \ \left( ( - 1)^{\frac{3}{4}} \right), \\ ( - 1),\ \ \ \left( {- ( - 1)}^{\frac{1}{4}} \right),\ \ \ \left( - ( - 1)^{\frac{1}{2}} \right),\ \ \ \left( {- ( - 1)}^{\frac{3}{4}} \right) \end{array} \right\}, \end{array}
(1.8.419)

and the subset A(c⊗)A\left( c_{\otimes} \right) is

A(c⊗)={((−1)14),   ((−1)34),   (−(−1)14),   (−(−1)34)},\begin{array}{r} A\left( c_{\otimes} \right) = \left\{ \begin{array}{r} \left( ( - 1)^{\frac{1}{4}} \right),\ \ \ \left( ( - 1)^{\frac{3}{4}} \right),\ \ \ \left( {- ( - 1)}^{\frac{1}{4}} \right),\ \ \ \left( {- ( - 1)}^{\frac{3}{4}} \right) \end{array} \right\}, \end{array}
(1.8.420)

where the symbol ⊗\otimes de­notes as­ymp­tot­ic and ⊕︎\oplus rep­re­sents axes for the axis subset A(c⊕︎)A\left( c_{\oplus} \right)

A(c⊕︎)={(1),   ((−1)12),   (−1),   (−(−1)12)},\begin{array}{r} A\left( c_{\oplus} \right) = \left\{ \begin{array}{r} (1),\ \ \ \left( ( - 1)^{\frac{1}{2}} \right),\ \ \ ( - 1),\ \ \ \left( - ( - 1)^{\frac{1}{2}} \right) \end{array} \right\}, \end{array}
(1.8.421)

which were pic­tured as the cjc_{j} values in Figure 1.8.13.

We define the fol­low­ing min­i­mal el­e­ments 𝓂ℯ[(−1)14(1gp1)](↖)\mathcal{m}\mathcal{e}_{\left\lbrack ( - 1)^{\frac{1}{4}}\left( \frac{1}{gp_{1}} \right) \right\rbrack}( \nwarrow ), 𝓂ℯ[(−1)14(1gp1)](↗)\mathcal{m}\mathcal{e}_{\left\lbrack ( - 1)^{\frac{1}{4}}\left( \frac{1}{gp_{1}} \right) \right\rbrack}( \nearrow )

𝓂ℯ[(−1)14(1gp1)](↖)=(00(−1)14(−1)14000(−1)140),   𝓂ℯ[(−1)14(1gp1)](↗)=(0(−1)14000(−1)14(−1)1400)\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack ( - 1)^{\frac{1}{4}}\left( \frac{1}{gp_{1}} \right) \right\rbrack}( \nwarrow ) = \begin{pmatrix} 0 & 0 & ( - 1)^{\frac{1}{4}} \\ ( - 1)^{\frac{1}{4}} & 0 & 0 \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \end{pmatrix},\ \ \ \mathcal{m}\mathcal{e}_{\left\lbrack ( - 1)^{\frac{1}{4}}\left( \frac{1}{gp_{1}} \right) \right\rbrack}( \nearrow ) = \begin{pmatrix} 0 & ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \\ ( - 1)^{\frac{1}{4}} & 0 & 0 \end{pmatrix} \end{array}
(1.8.422)

which are both pri­ma­ry gen­er­a­tors. We cal­cu­late the set the 𝓂ℯ[(−1)14(1gp1)](↖)\mathcal{m}\mathcal{e}_{\left\lbrack ( - 1)^{\frac{1}{4}}\left( \frac{1}{gp_{1}} \right) \right\rbrack}( \nwarrow ) el­e­ment, find­ing twenty-four el­e­ments in the set

A(𝓂ℯ[(−1)14(1gp1)](↖))=\begin{array}{r} A\left( \mathcal{m}\mathcal{e}_{\left\lbrack ( - 1)^{\frac{1}{4}}\left( \frac{1}{gp_{1}} \right) \right\rbrack}( \nwarrow ) \right) = \end{array}
(1.8.423)
{(00(−1)14(−1)14000(−1)140),   (0i000ii00),   ((−1)34000(−1)34000(−1)34),   (00−1−1000−10),   (0−(−1)14000−(−1)14−(−1)1400),   (−i000−i000−i),(00−(−1)34−(−1)34000−(−1)340),   (010001100),   ((−1)14000(−1)14000(−1)14),   (00ii000i0),   (0(−1)34000(−1)34(−1)3400),   (−1000−1000−1),(00−(−1)14−(−1)14000−(−1)140),   (0−i000−i−i00),   (−(−1)34000−(−1)34000−(−1)34),   (001100010),   (0(−1)14000(−1)14(−1)1400),   (i000i000i),(00(−1)34(−1)34000(−1)340),   (0−1000−1−100),   (−(−1)14000−(−1)14000−(−1)14),   (00−i−i000−i0),   (0−(−1)34000−(−1)34−(−1)3400),   (100010001)}.\left\{ \begin{array}{r} \begin{pmatrix} 0 & 0 & ( - 1)^{\frac{1}{4}} \\ ( - 1)^{\frac{1}{4}} & 0 & 0 \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & i & 0 \\ 0 & 0 & i \\ i & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} ( - 1)^{\frac{3}{4}} & 0 & 0 \\ 0 & ( - 1)^{\frac{3}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{3}{4}} \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & - 1 \\ - 1 & 0 & 0 \\ 0 & - 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & - ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & - ( - 1)^{\frac{1}{4}} \\ - ( - 1)^{\frac{1}{4}} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} - i & 0 & 0 \\ 0 & - i & 0 \\ 0 & 0 & - i \end{pmatrix}, \\ \begin{pmatrix} 0 & 0 & - ( - 1)^{\frac{3}{4}} \\ - ( - 1)^{\frac{3}{4}} & 0 & 0 \\ 0 & - ( - 1)^{\frac{3}{4}} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} ( - 1)^{\frac{1}{4}} & 0 & 0 \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & i \\ i & 0 & 0 \\ 0 & i & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & ( - 1)^{\frac{3}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{3}{4}} \\ ( - 1)^{\frac{3}{4}} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix}, \\ \begin{pmatrix} 0 & 0 & - ( - 1)^{\frac{1}{4}} \\ - ( - 1)^{\frac{1}{4}} & 0 & 0 \\ 0 & - ( - 1)^{\frac{1}{4}} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & - i & 0 \\ 0 & 0 & - i \\ - i & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} {- ( - 1)}^{\frac{3}{4}} & 0 & 0 \\ 0 & {- ( - 1)}^{\frac{3}{4}} & 0 \\ 0 & 0 & {- ( - 1)}^{\frac{3}{4}} \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \\ ( - 1)^{\frac{1}{4}} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} i & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & i \end{pmatrix}, \\ \begin{pmatrix} 0 & 0 & ( - 1)^{\frac{3}{4}} \\ ( - 1)^{\frac{3}{4}} & 0 & 0 \\ 0 & ( - 1)^{\frac{3}{4}} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & - 1 & 0 \\ 0 & 0 & - 1 \\ - 1 & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} {- ( - 1)}^{\frac{1}{4}} & 0 & 0 \\ 0 & - ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & {- ( - 1)}^{\frac{1}{4}} \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & - i \\ - i & 0 & 0 \\ 0 & - i & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & {- ( - 1)}^{\frac{3}{4}} & 0 \\ 0 & 0 & {- ( - 1)}^{\frac{3}{4}} \\ {- ( - 1)}^{\frac{3}{4}} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{array} \right\}.

The twenty-four el­e­ments are a due to the fact that the com­pos­ite number ℤ24\mathbb{Z}_{24} is the result of the prod­uct of the prime num­bers gp1×gp1×gp1×gp2gp_{1} \times gp_{1} \times gp_{1} \times gp_{2}, and there is a sym­met­ric dis­tri­b­u­tion of the eight unique el­e­ments into sub­sets of the form

Yℒ1(cj)=(cj000cj000cj),  Yℒ2(cj)=(0cj000cjcj00),  Yℒ3(cj)=(00cjcj000cj0).\begin{array}{r} Y_{\mathcal{L}1}\left( c_{j} \right) = \begin{pmatrix} c_{j} & 0 & 0 \\ 0 & c_{j} & 0 \\ 0 & 0 & c_{j} \end{pmatrix},\ \ Y_{\mathcal{L}2}\left( c_{j} \right) = \begin{pmatrix} 0 & c_{j} & 0 \\ 0 & 0 & c_{j} \\ c_{j} & 0 & 0 \end{pmatrix},\ \ Y_{\mathcal{L}3}\left( c_{j} \right) = \begin{pmatrix} 0 & 0 & c_{j} \\ c_{j} & 0 & 0 \\ 0 & c_{j} & 0 \end{pmatrix}. \end{array}
(1.8.424)

We con­sid­er the group char­ac­ter­is­tic poly­no­mi­al

𝒞𝒫(Yℒk)=∑𝒸𝓅(Yℒk)=(λ−a)(λ−b)(λ−c)+2λ3−2abc=0,\begin{array}{r} \mathcal{CP}\left( Y_{\mathcal{L}k} \right) = \sum_{}^{}{\mathcal{cp}\left( Y_{\mathcal{L}k} \right) = (\lambda - a)(\lambda - b)(\lambda - c) + 2\lambda^{3} - 2abc} = 0, \end{array}
(1.8.425)

and given that for each Yℒk(cj)Y_{\mathcal{L}k}\left( c_{j} \right) subset, we have a=b=c=cja = b = c = c_{j}, the subset group char­ac­ter­is­tic poly­no­mi­al 𝒞𝒫(Yℒk)\mathcal{CP}\left( Y_{\mathcal{L}k} \right) is then equal to

𝒞𝒫(Yℒk(cj))=(λ−cj)3+2(λ3−cj3)=0.\begin{array}{r} \mathcal{CP}\left( Y_{\mathcal{L}k}\left( c_{j} \right) \right) = \left( \lambda - c_{j} \right)^{3} + 2\left( \lambda^{3} - c_{j}^{3} \right) = 0. \end{array}
(1.8.426)

Each value for cjc_{j} is the result of ro­ta­tion on the com­plex unit circle, and we can ex­press each cjc_{j} value as equal­ing

cj=((−1)14)j,\begin{array}{r} c_{j} = \left( ( - 1)^{\frac{1}{4}} \right)^{j}, \end{array}
(1.8.427)

and the group char­ac­ter­is­tic poly­no­mi­al can be found by the sum­ma­tion

𝒞𝒫(A(𝓂ℯ[(−1)14(1gp1)](↖)))=∑𝒞𝒫(Yℒk(cj))=0.\begin{array}{r} \mathcal{CP}\left( A\left( \mathcal{m}\mathcal{e}_{\left\lbrack ( - 1)^{\frac{1}{4}}\left( \frac{1}{gp_{1}} \right) \right\rbrack}( \nwarrow ) \right) \right) = \sum_{}^{}{\mathcal{CP}\left( Y_{\mathcal{L}k}\left( c_{j} \right) \right) = 0}. \end{array}
(1.8.428)

We sub­sti­tute cj=((−1)14)jc_{j} = \left( ( - 1)^{\frac{1}{4}} \right)^{j}, and cal­cu­late the sum to find the group char­ac­ter­is­tic poly­no­mi­al re­duces to

∑j=18(λ−((−1)14)j)3+2(λ3−(((−1)14)j)3)=(24λ3=0),\begin{array}{r} \sum_{j = 1}^{8}{\left( \lambda - \left( ( - 1)^{\frac{1}{4}} \right)^{j} \right)^{3} + 2\left( \lambda^{3} - \left( \left( ( - 1)^{\frac{1}{4}} \right)^{j} \right)^{3} \right)} = \left( 24\lambda^{3} = 0 \right), \end{array}
(1.8.429)

which is very sim­i­lar to what we found for the group char­ac­ter­is­tic poly­no­mi­al for our light cone num­bers

𝒞𝒫(Y(gp[2i])(↖))=(12λ3=0).\begin{array}{r} \mathcal{CP}\left( Y\left( {gp}_{\lbrack 2i\rbrack} \right)( \nwarrow ) \right) = \left( 12\lambda^{3} = 0 \right). \end{array}
(1.8.430)

This sim­i­lar­i­ty is based upon the same sym­me­try in the el­e­ments of the group, and the two func­tions have the same plots except for a scaled dif­fer­ence of two. So, once again, we find the pres­ence of the net-zero char­ac­ter­is­tic in the as­ymp­tot­ic light num­bers.

The hy­per­bol­ic geom­e­try of spe­cial rel­a­tiv­i­ty is also due to light num­bers. Each hy­per­bo­la along the X1,X2,X3X_{1},X_{2},X_{3}-axes is the di­men­sion­al trans­form of the gp2gp_{2} set el­e­ments con­sid­ered ear­li­er

A(gp2)={(001100010),   (010001100),   (100010001)},\begin{array}{r} A\left( gp_{2} \right) = \left\{ \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right\}, \end{array}
(1.8.431)

which also play an im­por­tant role in ℒ𝒞[+Xj]2\mathcal{L}{\mathcal{C}_{\left\lbrack + X_{j} \right\rbrack}}_{2} light cone ro­ta­tions. Using our ro­ta­tion set no­ta­tion, we can define the gp2gp_{2} set as

A(𝓂ℯ[1gp2]×Y(↖))={(100010001),   (001100010),   (010001100)},\begin{array}{r} A\left( \mathcal{m}\mathcal{e}_{\left\lbrack \frac{1}{gp_{2}} \right\rbrack} \times Y( \nwarrow ) \right) = \left\{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} \right\}, \end{array}
(1.8.432)

given that this is a pos­i­tive­ly ori­ent­ed ro­ta­tion of the in­scribed equi­lat­er­al tri­an­gle. The di­men­sion­al trans­form will yield the XX-type Lie al­ge­bra gen­er­a­tors

A(𝓂ℯ[1gp2]×Y(↖))×Y(↖)={(001010100),(100001010),(010100001)},\begin{array}{r} A\left( \mathcal{m}\mathcal{e}_{\left\lbrack \frac{1}{gp_{2}} \right\rbrack} \times Y( \nwarrow ) \right) \times Y( \nwarrow ) = \left\{ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix},\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix},\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right\}, \end{array}
(1.8.433)

all of which are hy­per­bol­ic gen­er­a­tors, seen by their vari­able-re­lat­ed matrix ex­po­nen­tials

exp⁡((x10000x10x10),   (0x30x30000x3),   (00x20x20x200))=\exp{\left( \begin{pmatrix} x_{1} & 0 & 0 \\ 0 & 0 & x_{1} \\ 0 & x_{1} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & x_{3} & 0 \\ x_{3} & 0 & 0 \\ 0 & 0 & x_{3} \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & x_{2} \\ 0 & x_{2} & 0 \\ x_{2} & 0 & 0 \end{pmatrix} \right) =}
={(ex1000cosh⁡(x1)sinh⁡(x1)0sinh⁡(x1)cosh⁡(x1)),   (cosh⁡(x3)sinh⁡(x3)0sinh⁡(x3)cosh⁡(x3)000ex3),   (cosh⁡(x2)0sinh⁡(x2)0ex20sinh⁡(x2)0cosh⁡(x2))}.\begin{array}{r} = \left\{ \begin{pmatrix} e^{x_{1}} & 0 & 0 \\ 0 & \cosh\left( x_{1} \right) & \sinh\left( x_{1} \right) \\ 0 & \sinh\left( x_{1} \right) & \cosh\left( x_{1} \right) \end{pmatrix},\ \ \ \begin{pmatrix} \cosh\left( x_{3} \right) & \sinh\left( x_{3} \right) & 0 \\ \sinh\left( x_{3} \right) & \cosh\left( x_{3} \right) & 0 \\ 0 & 0 & e^{x_{3}} \end{pmatrix},\ \ \ \begin{pmatrix} \cosh\left( x_{2} \right) & 0 & \sinh\left( x_{2} \right) \\ 0 & e^{x_{2}} & 0 \\ \sinh\left( x_{2} \right) & 0 & \cosh\left( x_{2} \right) \end{pmatrix} \right\}. \end{array}
(1.8.434)

So, the hy­per­bol­ic geom­e­try that un­der­scores spe­cial rel­a­tiv­i­ty is also a man­i­fes­ta­tion of light due to the gp2gp_{2} geo­met­ric prime and its re­sult­ing dy­nam­ics. We also saw the role that these play in cone sec­tion ro­ta­tions, and one way to think about this re­la­tion­ship is to con­sid­er the R[j,k]R_{\lbrack j,k\rbrack} ro­ta­tions ma­tri­ces as a split­ting of the gp2gp_{2} geo­met­ric prime, which im­plies they are the result of light split­ting. This notion can be ex­pand­ed upon by con­sid­er­ing that each R[j,k]R_{\lbrack j,k\rbrack} matrix is the in­verse of its R[k,j]R_{\lbrack k,j\rbrack} pair, and the prod­uct

R[j,k]×R[k,j]=(100010001),\begin{array}{r} R_{\lbrack j,k\rbrack} \times R_{\lbrack k,j\rbrack} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \end{array}
(1.8.435)

we see yields the iden­ti­ty of the A(gp2)A\left( gp_{2} \right) set. The square for the ℒ𝒞[+Xj]1\mathcal{L}{\mathcal{C}_{\left\lbrack + X_{j} \right\rbrack}}_{1} seg­ments is also equal to the iden­ti­ty

(ℒ𝒞[+Xj]1)2=(100010001),\begin{array}{r} \left( \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{j} \right\rbrack}}_{1} \right)^{2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \end{array}
(1.8.436)

rep­re­sent­ing a sym­met­ric split­ting of the iden­ti­ty from the A(gp2)A\left( gp_{2} \right) set, and this should remind the reader of the split com­plex number jj, whose square is equal to the iden­ti­ty as well. A closer look at the ℒ𝒞[+Xj]1\mathcal{L}{\mathcal{C}_{\left\lbrack + X_{j} \right\rbrack}}_{1} num­bers using the ℒ𝒞[+X1]1\mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{1} ex­am­ple

ℒ𝒞[+X1]1=(10000e(iπ4)0e(− iπ4)0),\begin{array}{r} \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & e^{\left( \frac{i\pi}{4} \right)} \\ 0 & e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix}, \end{array}
(1.8.437)

has its axis rep­re­sent­ed by the a1,1a_{1,1} entry, along with the 2×22 \times 2 sub­ma­trix in the bottom right corner, rep­re­sent­ing the sym­met­ric iden­ti­ty split, man­i­fest­ed by ro­tat­ing equal dis­tances in ra­di­ans away from ϵ\epsilon in pos­i­tive and neg­a­tive di­rec­tions. This sub­ma­trix is also an­oth­er form of a split com­plex number, given that

(ℒ𝒞[+X1]1(a2:3,2:3))2=(0e(iπ4)e(−iπ4)0)2=(1001),\begin{array}{r} \left( \mathcal{L}{\mathcal{C}_{\left\lbrack + X_{1} \right\rbrack}}_{1}\left( a_{2:3,2:3} \right) \right)^{2} = \begin{pmatrix} 0 & e^{\left( \frac{i\pi}{4} \right)} \\ e^{\left( - \frac{i\pi}{4} \right)} & 0 \end{pmatrix}^{2} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \end{array}
(1.8.438)

which is con­sis­tent with our de­f­i­n­i­tion that j=eiπj = e^{i\pi}, nor­mal­ly taking the 2×22 \times 2 matrix rep­re­sen­ta­tion form

j∈ℝ2=(0−1−10).\begin{array}{r} j \in \mathbb{R}^{2} = \begin{pmatrix} 0 & - 1 \\ - 1 & 0 \end{pmatrix}. \end{array}
(1.8.439)

This rep­re­sen­ta­tion ob­scures that it is also a sym­met­ric split­ting of the iden­ti­ty in pos­i­tive and neg­a­tive di­rec­tions and right­ful­ly sits in the ℂ2\mathbb{C}^{2} space as

j∈ℂ2=(0eiπe−iπ0).\begin{array}{r} j \in \mathbb{C}^{2} = \begin{pmatrix} 0 & e^{i\pi} \\ e^{- i\pi} & 0 \end{pmatrix}. \end{array}
(1.8.440)

Fur­ther ev­i­dence that both split com­plex num­bers are close­ly re­lat­ed is that they share the same de­ter­mi­nants and eigen­val­ues

det⁡{(0e(iπ4)e(− iπ4)0),(0eiπe−iπ0)}=−1,\begin{array}{r} \det\left\{ \begin{pmatrix} 0 & e^{\left( \frac{i\pi}{4} \right)} \\ e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix},\begin{pmatrix} 0 & e^{i\pi} \\ e^{- i\pi} & 0 \end{pmatrix} \right\} = - 1, \end{array}
(1.8.441)
λ{(0e(iπ4)e(− iπ4)0),(0eiπe−iπ0)}={1,−1},\begin{array}{r} \lambda\left\{ \begin{pmatrix} 0 & e^{\left( \frac{i\pi}{4} \right)} \\ e^{\left( - \ \frac{i\pi}{4} \right)} & 0 \end{pmatrix},\begin{pmatrix} 0 & e^{i\pi} \\ e^{- i\pi} & 0 \end{pmatrix} \right\} = \left\{ 1, - 1 \right\}, \end{array}
(1.8.442)

and sub­se­quent­ly have the same char­ac­ter­is­tic poly­no­mi­al λ2=1.\lambda^{2} = 1.In the A(gp2)A\left( gp_{2} \right) set, the non-iden­ti­ty el­e­ments should right­ful­ly be seen as sym­met­ric cyclic split­tings of the iden­ti­ty, given that

(001100010)(010001100)=(100010001),\begin{array}{r} \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \end{array}
(1.8.443)

and more pre­cise­ly, both el­e­ments are pos­i­tive and neg­a­tive di­rec­tion­al splits of the iden­ti­ty into thirds, which is the math­e­mat­i­cal un­der­pin­nings for the fact that

(001100010)3=(010001100)3=(100010001).\begin{array}{r} \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}^{3} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}^{3} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}. \end{array}
(1.8.444)

The R[j,k]R_{\lbrack j,k\rbrack} ro­ta­tion ma­tri­ces are close­ly re­lat­ed to the cyclic split­ting of the iden­ti­ty above, given that each will also equate to the iden­ti­ty when cubed

(R[j,k])3=(100010001),\begin{array}{r} \left( R_{\lbrack j,k\rbrack} \right)^{3} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \end{array}
(1.8.445)

and by as­so­cia­tiv­i­ty, all R[j,k]×R[l,m]R_{\lbrack j,k\rbrack} \times R_{\lbrack l,m\rbrack} prod­ucts will as well when j≠l and k≠mj \neq l\ \text{and}\ k \neq m,

(R[j,k]×R[l,m])3=(100010001).\begin{array}{r} \left( R_{\lbrack j,k\rbrack} \times R_{\lbrack l,m\rbrack} \right)^{3} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}. \end{array}
(1.8.446)

This ap­plies to the prod­ucts we en­coun­tered

R[2,1]×R[3,2]=(001100010),  R[2,3]×R[1,2]=(010001100),\begin{array}{r} R_{\lbrack 2,1\rbrack} \times R_{\lbrack 3,2\rbrack} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix},\ \ R_{\lbrack 2,3\rbrack} \times R_{\lbrack 1,2\rbrack} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}, \end{array}
(1.8.447)

where a three-way-split­ting of the iden­ti­ty is causal to both

(R[2,1]×R[3,2])3×(R[2,3]×R[1,2])3=ϵ.\begin{array}{r} \left( R_{\lbrack 2,1\rbrack} \times R_{\lbrack 3,2\rbrack} \right)^{3} \times \left( R_{\lbrack 2,3\rbrack} \times R_{\lbrack 1,2\rbrack} \right)^{3} = \epsilon. \end{array}
(1.8.448)

The log­i­cal con­clu­sion is that all on­to­log­i­cal­ly co­he­sive al­ge­bras are a split of the iden­ti­ty, known as the number one. This brings us full circle and back to the de­f­i­n­i­tion of the min­i­mal el­e­ment

𝓂ℯ[+1k]=(e2iπ)(1k),\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{1}{k} \right\rbrack} = \left( e^{2i\pi} \right)^{\left( \frac{1}{k} \right)}, \end{array}
(1.8.449)

which is ex­act­ly this split­ting of the iden­ti­ty, given that e2iπ=1e^{2i\pi} = 1, and the net-zero re­quire­ment for co­he­sive­ness. For each 𝓂ℯ[+1k]\mathcal{m}\mathcal{e}_{\left\lbrack + \frac{1}{k} \right\rbrack}, when k≥1k \geq 1, there exists its split analog, 𝓂ℯ[−1k]\mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{k} \right\rbrack}, for which the result of the fol­low­ing prod­uct always holds

𝓂ℯ[+1k]×𝓂ℯ[−1k]=(e2iπk)(e−2iπk)=e0=ϵ.\begin{array}{r} \mathcal{m}\mathcal{e}_{\left\lbrack + \frac{1}{k} \right\rbrack}\times \mathcal{m}\mathcal{e}_{\left\lbrack - \frac{1}{k} \right\rbrack} = \left( e^{\frac{2i\pi}{k}} \right)\left( e^{\frac{- 2i\pi}{k}} \right) = e^{0} = \epsilon. \end{array}
(1.8.450)

How and why does the iden­ti­ty exist? The answer lies in the fact that it is the volume of the zero-di­men­sion­al point, which is indeed equal to 11, and is based on the volume de­f­i­n­i­tion VnV_{n} of the nn-di­men­sion­al Eu­clid­ean ball

Vn=πn2Γ(n2+1)Rn,\begin{array}{r} V_{n} = \frac{\pi^{\frac{n}{2}}}{\Gamma\left( \frac{n}{2} + 1 \right)}R^{n}, \end{array}
(1.8.451)

where RR is the radius, and Γ\Gamma is the gamma func­tion. By solv­ing for n=0n = 0, we find the stated result

V0=π02Γ(02+1)R0=1Γ(1)=1.\begin{array}{r} V_{0} = \frac{\pi^{\frac{0}{2}}}{\Gamma\left( \frac{0}{2} + 1 \right)}R^{0} = \frac{1}{\Gamma(1)} = 1. \end{array}
(1.8.452)

The net-zero split­ting of the zero-di­men­sion­al point is causal to all the Lie al­ge­bras and groups we have con­sid­ered in this book, and it is the origin of light in its raw and pri­mor­dial form. The zero-di­men­sion­al point is eter­nal and un­caused, and with the birth and death of it­er­a­tions of uni­vers­es, it per­sists intact.

Light orig­i­nates from inside of the sin­gu­lar­i­ty, where it is based on the dy­nam­ics of geo­met­ric primes and their polyg­o­nal forms, ro­tat­ing in the com­plex unit sphere. Through the di­men­sion­al trans­for­ma­tion, light man­i­fests itself as space­time, rel­a­tiv­i­ty, matter, forces, and all the other phe­nom­e­na per­ceived lo­cal­ly. Inside the sin­gu­lar­i­ty, as the geo­met­ric prime gains more angles of free­dom for its ver­tices, it takes on the com­pos­ite geom­e­try of the geo­met­ric in­te­ger and its poly­gon with that number of ver­tices.

This sec­tion con­sid­ers the first and second geo­met­ric primes and their com­pos­ite arrange­ments. The gp2gp_{2} geo­met­ric prime will di­men­sion­al­ly trans­form to the gp1gp_{1} prime and its com­pos­ites, with an axis point that varies from fixed to ge­o­des­ic on the com­plex unit sphere once ex­po­nen­ti­at­ed.

  1. Albert Ein­stein (1905) "Zur Elek­tro­dy­namik be­wegter Körper", An­nalen der Physik 17: 891
  2. Dray, Tevian, "The Geom­e­try of Spe­cial Rel­a­tiv­i­ty", 2012, CRC Press