Masteries · Jason GlowneyArithmoí Book 1

Masteries

1.4.9 The SU(2) Subgroup of the Symmetric Conic Group

Now that we have de­scribed the sym­met­ric conic group, the cone sub­group and the axis sub­group, we can turn our at­ten­tion to fur­ther ex­plor­ing the sub­groups that lay within. It should be ob­vi­ous that all the di­he­dral groups ex­plored in pre­vi­ous chap­ters are sub­groups of the sym­met­ric conic group, in­clud­ing S3/D3, D3ℂ,D3𝕫S_{3}{/D}_{3},\ D_{3\mathbb{C}},D_{3\mathbb{z}}. The reader should be think­ing about Cayley's The­o­rem, dis­cussed in Chap­ter 3, and its im­pli­ca­tion that every group is a sub­group of some sym­met­ric group. What might not be ob­vi­ous, while at the same time, re­mark­able is the SU(2)SU(2) group can be seen as a sub­group of the sym­met­ric conic group SU(2)⊂S3ℂΔSU(2) \subset S_{3\mathbb{C}}^{\Delta}. To remind the reader, the SU(2)SU(2) group's Lie al­ge­bra 𝔰𝔲(2)\mathfrak{su}(2) is rep­re­sent­ed by

𝔰𝔲(2)=(ia−zz−ia)={M:a∈ℝ, z∈ℂ},\begin{array}{r} \mathfrak{su}(2) = \begin{pmatrix} ia & - \overline{z} \\ z & - ia \end{pmatrix} = \left\{ M:a\in \mathbb{R},\ z\in \mathbb{C} \right\}, \end{array}
(1.4.135)

gen­er­at­ed by the fol­low­ing ma­tri­ces,

u1=(0ii0),  u2=(0−110),  u3=(i00−i).\begin{array}{r} u_{1} = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix},\ \ u_{2} = \begin{pmatrix} 0 & - 1 \\ 1 & 0 \end{pmatrix},\ \ u_{3} = \begin{pmatrix} i & 0 \\ 0 & - i \end{pmatrix}. \end{array}
(1.4.136)

We define two 3×33 \times 3 ma­tri­ces, using the no­ta­tion ΔY,ΔX\ \Delta Y,\Delta X

 ΔY=(1000(−1)14000(−1)34),    ΔX=(10000(−1)140−(−1)340),\begin{array}{r} \ \Delta Y = \begin{pmatrix} 1 & 0 & 0 \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{3}{4}} \end{pmatrix},\ \ \ \ \Delta X = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \\ 0 & - ( - 1)^{\frac{3}{4}} & 0 \end{pmatrix}, \end{array}
(1.4.137)

which are two per­mu­ta­tions ∈S3ℂΔ\in S_{3\mathbb{C}}^{\Delta}, with de­ter­mi­nants equal to −1- 1. These ma­tri­ces form the SU(2)SU(2) sub­group by having sim­i­lar prod­ucts re­la­tions stud­ied for the tan­gent spaces of the cone. The re­la­tions are stated by the prod­ucts

 ΔX2=(100010001)=u0=ϵ,   ΔYΔX=(10000i0i0)=𝓊1,{\ \Delta X}^{2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = u_{0} = \epsilon,\ \ \ \Delta Y\Delta X = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & i \\ 0 & i & 0 \end{pmatrix} = \mathcal{u}_{1},
 ΔX ΔY=(10000−1010)=𝓊2,   ΔY2=(1000i000−i)=𝓊3.\begin{array}{r} \ \Delta X\ \Delta Y = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{pmatrix} = \mathcal{u}_{2},\ \ \ \Delta Y^{2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & - i \end{pmatrix} = \mathcal{u}_{3}. \end{array}
(1.4.138)

Al­though the ma­tri­ces are 3×33 \times 3, the en­tries are fixed for a1,1=1a_{1,1} = 1, re­main­ing the same value for all prod­ucts in the sub­group. Like the Lie al­ge­bra ma­tri­ces de­scribed using 2×22 \times 2 ma­tri­ces, the 3×33 \times 3 forms above also have de­ter­mi­nants of 11. We could have also just cre­at­ed each matrix of the Lie al­ge­bra in­di­vid­u­al­ly but seeing how the prod­ucts of ΔY,ΔX\Delta Y,\Delta X create the sub­group SU(2)SU(2), this lends itself to a better geo­met­ric per­spec­tive about how the cone num­bers and axis num­bers relate. By as­so­ci­at­ing the vari­ables a,b,c∈ℝa,b,c\in \mathbb{R} to the 𝓊j\mathcal{u}_{j} ma­tri­ces, as well as defin­ing kk to be a scalar con­stant, we rewrite the ma­tri­ces as the set 𝓊j′\mathcal{u}_{j}^{'}:

𝓊j′={(k0000ic0ic0),   (k0000−b0b0),   (k000ia000−ia)},\begin{array}{r} \mathcal{u}_{j}^{'} = \left\{ \begin{pmatrix} k & 0 & 0 \\ 0 & 0 & ic \\ 0 & ic & 0 \end{pmatrix},\ \ \ \begin{pmatrix} k & 0 & 0 \\ 0 & 0 & - b \\ 0 & b & 0 \end{pmatrix},\ \ \ \begin{pmatrix} k & 0 & 0 \\ 0 & ia & 0 \\ 0 & 0 & - ia \end{pmatrix} \right\}, \end{array}
(1.4.139)

and define 𝔰𝔲′(2)\mathfrak{su}'(2) as the group

𝔰𝔲′(2)=(3k000ia−z0z−ia)={M:a,k∈ℝ, z∈ℂ},\begin{array}{r} \mathfrak{s}\mathfrak{u}^{'}(2) = \begin{pmatrix} 3k & 0 & 0 \\ 0 & ia & - \overline{z} \\ 0 & z & - ia \end{pmatrix} = \left\{ M:a,k\in \mathbb{R},\ z\in \mathbb{C} \right\}, \end{array}
(1.4.140)

with de­ter­mi­nants

det⁡(𝔰𝔲′(2))=3k(a2+b2+c2),\begin{array}{r} \det\left( \mathfrak{s}\mathfrak{u}^{'}(2) \right) = 3k\left( a^{2} + b^{2} + c^{2} \right), \end{array}
(1.4.141)

where these ma­tri­ces have a Eu­clid­ean dis­tance metric, scaled by the con­stant 3k3k. By set­ting k=13k = \frac{1}{3}, the de­ter­mi­nant re­duces to a2+b2+c2a^{2} + b^{2} + c^{2}, and is the fa­mil­iar de­ter­mi­nant of the 𝔰𝔲(2)\mathfrak{su}(2) group with its 2×22 \times 2 ma­tri­ces. As seen in Sec­tion 1.4.2, the chord-re­lat­ed tan­gent spaces of the cone yield­ed the SU(2)SU(2) group, and these chords can be en­vi­sioned as the en­tries in our ΔY,ΔX\Delta Y,\Delta X ma­tri­ces, high­light­ed in bold, in the fol­low­ing

ΔY=(1000(−𝟏)𝟏𝟒000(−𝟏)𝟑𝟒),   ΔX=(10000(−𝟏)𝟏𝟒0−(−𝟏)𝟑𝟒0),\begin{array}{r} \Delta Y = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \left( - \mathbf{1} \right)^{\frac{\mathbf{1}}{\mathbf{4}}} & 0 \\ 0 & 0 & \left( - \mathbf{1} \right)^{\frac{\mathbf{3}}{\mathbf{4}}} \end{pmatrix},\ \ \ \Delta X = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & \left( - \mathbf{1} \right)^{\frac{\mathbf{1}}{\mathbf{4}}} \\ 0 & \mathbf{-}\left( - \mathbf{1} \right)^{\frac{\mathbf{3}}{\mathbf{4}}} & 0 \end{pmatrix}, \end{array}
(1.4.142)

and as per­mu­ta­tion el­e­ments of the sym­met­ric conic group, they are re­spon­si­ble for these chord re­la­tion­ships. Taking the matrix ex­po­nen­tials of the 𝓊j′\mathcal{u}_{j}^{'} ma­tri­ces, we find

exp⁡(𝓊1′)=(ek000cos⁡(c)isin(c)0isin(c)cos⁡(c)),   exp⁡(𝓊2′)=(ek000cos⁡(b)−sin(b)0sin⁡(b)cos⁡(b)),   exp⁡(𝓊3′)=(ek000eia000e−ia),\begin{array}{r} \exp\left( \mathcal{u}_{1}^{'} \right) = \begin{pmatrix} e^{k} & 0 & 0 \\ 0 & \cos(c) & isin(c) \\ 0 & isin(c) & \cos(c) \end{pmatrix},\ \ \ \exp\left( \mathcal{u}_{2}^{'} \right) = \begin{pmatrix} e^{k} & 0 & 0 \\ 0 & \cos(b) & - sin(b) \\ 0 & \sin(b) & \cos(b) \end{pmatrix},\ \ \ \exp\left( \mathcal{u}_{3}^{'} \right) = \begin{pmatrix} e^{k} & 0 & 0 \\ 0 & e^{ia} & 0 \\ 0 & 0 & e^{- ia} \end{pmatrix}, \end{array}
(1.4.143)

which all have de­ter­mi­nants equal to eke^{k}, rep­re­sent­ing a sym­met­ric scal­ing by way of the ro­ta­tion­al axes. In Figure 1.4.18, we rep­re­sent the ex­am­ple of the ΔX\Delta X-el­e­ment in re­la­tion to the circle, rep­re­sent­ing a slice of the in­scribed SU(2)SU(2) sphere group, for the axis of ro­ta­tion of the ΔX\ \Delta X-el­e­ment, rep­re­sent­ing the min­i­mal slice vector point­ing toward FF

 ΔX=(F0000E0G0)=(10000(−1)140−(−1)340).\begin{array}{r} \ \Delta X = \begin{pmatrix} F & 0 & 0 \\ 0 & 0 & E \\ 0 & G & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \\ 0 & - ( - 1)^{\frac{3}{4}} & 0 \end{pmatrix}. \end{array}
(1.4.144)
Figure
Figure 1.4.18 – The  ΔX\ \Delta X-el­e­ment for E=(−1)(14), G=−(−1)(34)E = ( - 1)^{\left( \frac{1}{4} \right)},\ G = - ( - 1)^{\left( \frac{3}{4} \right)}and F=1\ F = 1, rep­re­sents the axis of ro­ta­tion for X\mathrm{X}, a slice of the SU(2)SU(2) sphere, as a disc through the X1X_{1}-plane of radius =1= 1, B=2B = \sqrt{2} is the origin of the SU(2)SU(2) sphere.

As we saw with the in­scribed spheres of the tan­gent cone, the other it­er­a­tions of the SU(2)SU(2) group are found the same way. The SU(2)SU(2) group with op­po­site chi­ral­i­ty can be de­rived from the fol­low­ing el­e­ments of the sym­met­ric conic group

ΔX2,1=(1000−(−1)34000(−1)14),   ΔX1,2=(10000(−1)340(−1)140),\begin{array}{r} \overline{\Delta X_{2,1}} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & {- ( - 1)}^{\frac{3}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \end{pmatrix},\ \ \ \Delta X_{1,2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & ( - 1)^{\frac{3}{4}} \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \end{pmatrix}, \end{array}
(1.4.145)

for the matrix ΔX2,1\overline{\Delta X_{2,1}}, taking the di­ag­o­nal form of a ΔY\Delta Y-type matrix, de­not­ing the el­e­ment with a bar to show its sub­group re­la­tion­ship. The matrix prod­ucts ΔX2,1\overline{\Delta X_{2,1}}, ΔX1,2\Delta X_{1,2} are found to rep­re­sent the fol­low­ing matrix el­e­ments,

ΔX1,22=(1000−1000−1)=𝓊0,  ΔX2,1×ΔX1,2=(10000i0i0)=𝓊1,\Delta X_{1,2}^{2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix} = \mathcal{u}_{0},\ \ \overline{\Delta X_{2,1}} \times \Delta X_{1,2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & i \\ 0 & i & 0 \end{pmatrix} = \mathcal{u}_{1},
ΔX1,2×ΔX2,1=(10000−1010)=𝓊2,  (ΔX2,1)2=(1000−i000i)=𝓊3.\begin{array}{r} \Delta X_{1,2} \times \overline{\Delta X_{2,1}} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 0 & 1 & 0 \end{pmatrix} = \mathcal{u}_{2},\ \ \left( \overline{\Delta X_{2,1}} \right)^{2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & - i & 0 \\ 0 & 0 & i \end{pmatrix} = \mathcal{u}_{3}. \end{array}
(1.4.146)

The above ma­tri­ces differ from the prior 𝓊j\mathcal{u}_{j} ma­tri­ces for 𝓊0,𝓊3\mathcal{u}_{0},\mathcal{u}_{3}, by having the bottom-right 2×22 \times 2 sub­ma­tri­ces flip signs for each entry. The el­e­ments 𝓊1,𝓊2,𝓊3\mathcal{u}_{1},\mathcal{u}_{2},\mathcal{u}_{3} all square to -1, in terms of their bottom-right el­e­ments, which is con­sis­tent with being imag­i­nary number forms. The prod­ucts of the ma­tri­ces are re­versed com­pared to the pre­vi­ous, where

𝓊12=𝓊22=𝓊32=−1,   𝓊1𝓊2𝓊3=1\mathcal{u}_{1}^{2} = \mathcal{u}_{2}^{2} = \mathcal{u}_{3}^{2} = - 1,\ \ \ \mathcal{u}_{1}\mathcal{u}_{2}\mathcal{u}_{3} = 1
𝓊1𝓊2=−𝓊3,𝓊2𝓊3=−𝓊1,𝓊3𝓊1=−𝓊2,\mathcal{u}_{1}\mathcal{u}_{2} = - \mathcal{u}_{3},\mathcal{u}_{2}\mathcal{u}_{3} = - \mathcal{u}_{1},\mathcal{u}_{3}\mathcal{u}_{1} = - \mathcal{u}_{2},
𝓊2𝓊1=𝓊3,   𝓊3𝓊2=𝓊1,   𝓊1𝓊3=𝓊2.\begin{array}{r} \mathcal{u}_{2}\mathcal{u}_{1} = \mathcal{u}_{3},\ \ \ \mathcal{u}_{3}\mathcal{u}_{2} = \mathcal{u}_{1},\ \ \ \mathcal{u}_{1}\mathcal{u}_{3} = \mathcal{u}_{2}. \end{array}
(1.4.147)

The re­ver­sal of the chi­ral­i­ty occurs be­cause we have swapped the a2,2,a3,3a_{2,2},a_{3,3} el­e­ments ∈ΔX2,1\in \overline{\Delta X_{2,1}}, from their po­si­tions ∈ΔX2,1\in \Delta X_{2,1}. In essence, this en­tails going from a coun­ter­clock­wise ori­en­ta­tion in the se­quence of el­e­ments to a clock­wise one.

In Figure 1.4.19, a graph­ic de­picts this chi­ral­i­ty change for the SU(2)SU(2) group, where the el­e­ments of the matrix ΔX1,2\Delta X_{1,2} are in the X1X_{1}-plane, for A=(−1)(34),B=(−1)(14)A = ( - 1)^{\left( \frac{3}{4} \right)},B = ( - 1)^{\left( \frac{1}{4} \right)} . The el­e­ment AA is coun­ter­clock­wise from the center point, while the matrix ΔX2,1\overline{\Delta X_{2,1}} for C=−(−1)(34),D=(−1)(14)C = - ( - 1)^{\left( \frac{3}{4} \right)},D = ( - 1)^{\left( \frac{1}{4} \right)}, the el­e­ment DD is coun­ter­clock­wise from the center point in the X2X_{2}-plane. To make this clear­er the ma­tri­ces ΔX2,1,ΔX2,1\Delta X_{2,1},\overline{\Delta X_{2,1}} are pre­sent­ed below with the same values for CC and DD

ΔX1,2=(10000(−1)340(−1)140)=(10000A0B0),   ΔX2,1=(1000−(−1)34000(−1)14)=(1000C000D).\begin{array}{r} \Delta X_{1,2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & ( - 1)^{\frac{3}{4}} \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & A \\ 0 & B & 0 \end{pmatrix},\ \ \ \overline{\Delta X_{2,1}} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & {- ( - 1)}^{\frac{3}{4}} & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & C & 0 \\ 0 & 0 & D \end{pmatrix}. \end{array}
(1.4.148)
Figure
Figure 1.4.19 – Chi­ral­i­ty change by switch­ing el­e­ments C,BC,B, in matrix, ΔX2,1\overline{\Delta X_{2,1}}

A com­par­i­son of the Lie al­ge­bras of the two sub­groups (ΔX1,1,ΔX3,2),(ΔX1,2,ΔX2,1)\left( \Delta X_{1,1},\Delta X_{3,2} \right),\left( \Delta X_{1,2},\overline{\Delta X_{2,1}} \right), shows

ΔX1,1,ΔX3,2=(3k000ia−z0z−ia)={M:a,k∈ℝ, z∈ℂ},\begin{array}{r} \Delta X_{1,1},\Delta X_{3,2} = \begin{pmatrix} 3k & 0 & 0 \\ 0 & ia & - \overline{z} \\ 0 & z & - ia \end{pmatrix} = \left\{ M:a,k\in \mathbb{R},\ z\in \mathbb{C} \right\}, \end{array}
(1.4.149)
ΔX1,2,ΔX2,1=(3k000−ia−z0zia)={M:a,k∈ℝ, z∈ℂ},\begin{array}{r} \Delta X_{1,2},\overline{\Delta X_{2,1}} = \begin{pmatrix} 3k & 0 & 0 \\ 0 & - ia & - \overline{z} \\ 0 & z & ia \end{pmatrix} = \left\{ M:a,k\in \mathbb{R},\ z\in \mathbb{C} \right\}, \end{array}
(1.4.150)

where the change in sign for the aa el­e­ments on the main di­ag­o­nal become ap­par­ent as the dif­fer­ence be­tween the two,

ΔX1,k=(0e−iπ4eiπ40),   (0eiπ4e−iπ40),   (0e−i3π4ei3π40),   (0ei3π4e−i3π40),\Delta X_{1,k} = \begin{pmatrix} 0 & e^{- \frac{i\pi}{4}} \\ e^{\frac{i\pi}{4}} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & e^{\frac{i\pi}{4}} \\ e^{- \frac{i\pi}{4}} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & e^{- \frac{i3\pi}{4}} \\ e^{\frac{i3\pi}{4}} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & e^{\frac{i3\pi}{4}} \\ e^{- \frac{i3\pi}{4}} & 0 \end{pmatrix},
ΔX3,k=(eiπ400ei3π4),   (ei3π400eiπ4),   (e−iπ400e−i3π4),   (e−i3π400e−iπ4).\begin{array}{r} \Delta X_{3,k} = \begin{pmatrix} e^{\frac{i\pi}{4}} & 0 \\ 0 & e^{\frac{i3\pi}{4}} \end{pmatrix},\ \ \ \begin{pmatrix} e^{\frac{i3\pi}{4}} & 0 \\ 0 & e^{\frac{i\pi}{4}} \end{pmatrix},\ \ \ \begin{pmatrix} e^{- \frac{i\pi}{4}} & 0 \\ 0 & e^{- \frac{i3\pi}{4}} \end{pmatrix},\ \ \ \begin{pmatrix} e^{- \frac{i3\pi}{4}} & 0 \\ 0 & e^{- \frac{i\pi}{4}} \end{pmatrix}. \end{array}
(1.4.25)

Gen­er­al­iz­ing the ma­tri­ces that pro­duce the SU(2)SU(2) sub­group rep­re­sen­ta­tions, de­pends on using a common axis of ro­ta­tion for both the main di­ag­o­nal ma­tri­ces and the ΔXj{\Delta X}_{j} forms. These can be rep­re­sent­ed by el­e­ments of the sym­met­ric conic group S3ℂΔS_{3\mathbb{C}}^{\Delta}, as

ImΔX1,2=(i0000(−1)340(−1)140),   ReΔX1X2=(1000(−1)14000−(−1)34),Im\Delta X_{1,2} = \begin{pmatrix} i & 0 & 0 \\ 0 & 0 & ( - 1)^{\frac{3}{4}} \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \end{pmatrix},\ \ \ Re\Delta X_{1}X_{2} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & - ( - 1)^{\frac{3}{4}} \end{pmatrix},
ImΔX2,2=(00(−1)340i0(−1)1400),   ReΔX2X3=((−1)140001000−(−1)34),Im\Delta X_{2,2} = \begin{pmatrix} 0 & 0 & ( - 1)^{\frac{3}{4}} \\ 0 & i & 0 \\ ( - 1)^{\frac{1}{4}} & 0 & 0 \end{pmatrix},\ \ \ Re\Delta X_{2}X_{3} = \begin{pmatrix} ( - 1)^{\frac{1}{4}} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - ( - 1)^{\frac{3}{4}} \end{pmatrix},
ImΔX3,2=(0(−1)340(−1)140000i),   ReΔX3X1=((−1)14000−(−1)340001),\begin{array}{r} Im\Delta X_{3,2} = \begin{pmatrix} 0 & ( - 1)^{\frac{3}{4}} & 0 \\ ( - 1)^{\frac{1}{4}} & 0 & 0 \\ 0 & 0 & i \end{pmatrix},\ \ \ Re\Delta X_{3}X_{1} = \begin{pmatrix} ( - 1)^{\frac{1}{4}} & 0 & 0 \\ 0 & - ( - 1)^{\frac{3}{4}} & 0 \\ 0 & 0 & 1 \end{pmatrix}, \end{array}
(1.4.151)

where the ImΔXj,2Im\Delta X_{j,2} ma­tri­ces on the left rep­re­sent the fa­mil­iar form of the XjX_{j}-type ma­tri­ces, for ii, the center of ro­ta­tion in each. The ma­tri­ces ReΔXjXkRe\Delta X_{j}X_{k} on the right rep­re­sent the re­spec­tive cone num­bers of the or­thog­o­nal plane, with the iden­ti­ty en­tries de­not­ed by the number 11, co­in­cid­ing with the cen­ters of ro­ta­tion for the ImΔXj,2Im\Delta X_{j,2} ma­tri­ces. Taking the ex­am­ples ImΔX3,1,ReΔX3X1Im\Delta X_{3,1},Re\Delta X_{3}X_{1}, they form matrix prod­ucts rep­re­sent­ing an SU(2)SU(2) sub­group with its tan­gent space po­si­tioned at the in­ter­sec­tion of the imag­i­nary X3X_{3}-plane, and the or­thog­o­nal rep­re­sen­ta­tion of the X1X_{1}-plane

(ImΔX3,2)2=(−1000−1000−1)=𝓊0,  ImΔX3,2×ReΔX3X1=(0i0i0000i)=𝓊1,\left( Im\Delta X_{3,2} \right)^{2} = \begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix} = \mathcal{u}_{0},\ \ Im\Delta X_{3,2} \times Re\Delta X_{3}X_{1} = \begin{pmatrix} 0 & i & 0 \\ i & 0 & 0 \\ 0 & 0 & i \end{pmatrix} = \mathcal{u}_{1},
ReΔX3X1×ImΔX3,2=(0−1010000i)=𝓊2,  (ReΔX3X1)2=(i000−i0001)=𝓊3.\begin{array}{r} Re\Delta X_{3}X_{1} \times Im\Delta X_{3,2} = \begin{pmatrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & i \end{pmatrix} = \mathcal{u}_{2},\ \ \left( Re\Delta X_{3}X_{1} \right)^{2} = \begin{pmatrix} i & 0 & 0 \\ 0 & - i & 0 \\ 0 & 0 & 1 \end{pmatrix} = \mathcal{u}_{3}. \end{array}
(1.4.152)

We can see the ImΔX3,1,ReΔX3X1Im\Delta X_{3,1},Re\Delta X_{3}X_{1} prod­ucts have the fa­mil­iar SU(2)SU(2) sub­group form, dif­fer­ing in the po­si­tion for the axis of ro­ta­tion a3,3a_{3,3} for the 𝓊j\mathcal{u}_{j} ma­tri­ces. From the Lie al­ge­bra stand­point, this sub­group can be de­fined as

ImΔX3,2,ReΔX3X1=(ia−z0z−ia0003k)={M:a,k∈ℝ, z∈ℂ}.\begin{array}{r} Im\Delta X_{3,2},Re\Delta X_{3}X_{1} = \begin{pmatrix} ia & - \overline{z} & 0 \\ z & - ia & 0 \\ 0 & 0 & 3k \end{pmatrix} = \left\{ M:a,k\in \mathbb{R},\ z\in \mathbb{C} \right\}. \end{array}
(1.4.153)

These arrange­ments have a nice geo­met­ric in­tu­ition, which un­der­scores that we can rotate the ImΔX3,2,ReΔX3X1Im\Delta X_{3,2},Re\Delta X_{3}X_{1} el­e­ments to arrive at the other chiral forms of SU(2)SU(2), which is what was oc­cur­ring for the ΔX1,2,ΔX2,1\Delta X_{1,2},\overline{\Delta X_{2,1}} ex­am­ple. To demon­strate, using ImΔX3,2Im\Delta X_{3,2}, ReΔX3X1Re\Delta X_{3}X_{1}, we rotate both ma­tri­ces by rais­ing them to the odd power of three

(ImΔX3,2)3=(0(−1)340(−1)140000i)3=(0−(−1)340−(−1)140000−i),\begin{array}{r} \left( Im\Delta X_{3,2} \right)^{3} = \begin{pmatrix} 0 & ( - 1)^{\frac{3}{4}} & 0 \\ ( - 1)^{\frac{1}{4}} & 0 & 0 \\ 0 & 0 & i \end{pmatrix}^{3} = \begin{pmatrix} 0 & - ( - 1)^{\frac{3}{4}} & 0 \\ {- ( - 1)}^{\frac{1}{4}} & 0 & 0 \\ 0 & 0 & - i \end{pmatrix}, \end{array}
(1.4.154)
(ReΔX3X1)3=((−1)14000−(−1)340001)3=((−1)34000−(−1)140001),\begin{array}{r} \left( Re\Delta X_{3}X_{1} \right)^{3} = \begin{pmatrix} ( - 1)^{\frac{1}{4}} & 0 & 0 \\ 0 & - ( - 1)^{\frac{3}{4}} & 0 \\ 0 & 0 & 1 \end{pmatrix}^{3} = \begin{pmatrix} ( - 1)^{\frac{3}{4}} & 0 & 0 \\ 0 & - ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & 1 \end{pmatrix}, \end{array}
(1.4.155)

seeing that ImΔX3,2Im\Delta X_{3,2} has ro­tat­ed to its an­tipo­dal point on the sphere of tan­gent spaces, and ReΔX3X1Re\Delta X_{3}X_{1} has mapped to its neg­a­tive com­plex con­ju­gate, ReΔX3X1→−ReΔX3X1Re\Delta X_{3}X_{1} \rightarrow - \overline{Re\Delta X_{3}X_{1}}. The ma­tri­ces of the re­sult­ing SU(2)SU(2) group become

((ImΔX3,2)3)2=(−1000−1000−1)=𝓊0,\left( \left( Im\Delta X_{3,2} \right)^{3} \right)^{2} = \begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix} = \mathcal{u}_{0},
(ImΔX3,2)3×(ReΔX3X1)3=(0−1010000−i)=𝓊2,\left( Im\Delta X_{3,2} \right)^{3} \times \left( Re\Delta X_{3}X_{1} \right)^{3} = \begin{pmatrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & - i \end{pmatrix} = \mathcal{u}_{2},
(ReΔX3X1)3×(ImΔX3,2)3=(0i0i0000−i)=𝓊1,\left( Re\Delta X_{3}X_{1} \right)^{3} \times \left( Im\Delta X_{3,2} \right)^{3} = \begin{pmatrix} 0 & i & 0 \\ i & 0 & 0 \\ 0 & 0 & - i \end{pmatrix} = \mathcal{u}_{1},
((ReΔX3X1)3)2=(−i000i0001)=𝓊3,\begin{array}{r} \left( \left( Re\Delta X_{3}X_{1} \right)^{3} \right)^{2} = \begin{pmatrix} - i & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & 1 \end{pmatrix} = \mathcal{u}_{3}, \end{array}
(1.4.156)

show­ing that the prod­uct (ImΔX3,2)3×(ReΔX3X1)3\left( Im\Delta X_{3,2} \right)^{3} \times \left( Re\Delta X_{3}X_{1} \right)^{3} now maps to the 𝓊2\mathcal{u}_{2} matrix. Fur­ther, (ReΔX3X1)3×(ImΔX3,2)3\left( Re\Delta X_{3}X_{1} \right)^{3} \times \left( Im\Delta X_{3,2} \right)^{3} now maps to 𝓊1\mathcal{u}_{1}, in­stead of the pre­vi­ous 𝓊2\mathcal{u}_{2}. The over­all effect is to re­verse the sub­group's chi­ral­i­ty, lead­ing to the chiral Lie al­ge­bra

(ImΔX3,2)3,(ReΔX3X1)3=(−ia−z0zia0003k)={M:a,k∈ℝ, z∈ℂ}.\begin{array}{r} \left( Im\Delta X_{3,2} \right)^{3},\left( Re\Delta X_{3}X_{1} \right)^{3} = \begin{pmatrix} - ia & - \overline{z} & 0 \\ z & ia & 0 \\ 0 & 0 & 3k \end{pmatrix} = \left\{ M:a,k\in \mathbb{R},\ z\in \mathbb{C} \right\}. \end{array}
(1.4.157)

The essence of the trans­for­ma­tion maps ImΔX3,2Im\Delta X_{3,2} to its neg­a­tive and an­tipo­dal po­si­tion, where ReΔX3X1Re\Delta X_{3}X_{1} be­comes its neg­a­tive com­plex con­ju­gate, and there­fore, we can rep­re­sent the trans­for­ma­tion as

ImΔX3,2,ReΔX3X1→−ImΔX3,2,− (ReΔX3X1)†.\begin{array}{r} Im\Delta X_{3,2},Re\Delta X_{3}X_{1} \rightarrow - Im\Delta X_{3,2}, - \ \left( Re\Delta X_{3}X_{1} \right)^{\dagger}. \end{array}
(1.4.158)

In Figure 1.4.20, we show a geo­met­ric ren­der­ing of this trans­for­ma­tion, where the ma­tri­ces ImΔX3,2,ReΔX3X1Im\Delta X_{3,2},Re\Delta X_{3}X_{1} trans­form for co­in­cid­ing points A,B,C,D→A′,B′,C′,D′A,B,C,D \rightarrow A^{'},B^{'},C^{'},D', as fol­lows:

ImΔX3,2→−ImΔX3,2=(0A=(−1)340B=(−1)140000i)→(0A′=−(−1)340B′=−(−1)140000−i),\begin{array}{r} Im\Delta X_{3,2} \rightarrow - Im\Delta X_{3,2} = \begin{pmatrix} 0 & A = ( - 1)^{\frac{3}{4}} & 0 \\ {B = ( - 1)}^{\frac{1}{4}} & 0 & 0 \\ 0 & 0 & i \end{pmatrix} \rightarrow \begin{pmatrix} 0 & A^{'} = - ( - 1)^{\frac{3}{4}} & 0 \\ {B^{'} = - ( - 1)}^{\frac{1}{4}} & 0 & 0 \\ 0 & 0 & - i \end{pmatrix}, \end{array}
(1.4.159)
ReΔX3X1→−(ReΔX3X1)∗=(C=(−1)14000D=−(−1)340001)→(C′=(−1)34000D′=−(−1)140001).\begin{array}{r} Re\Delta X_{3}X_{1} \rightarrow - \left( Re\Delta X_{3}X_{1} \right)^{*} = \begin{pmatrix} {C = ( - 1)}^{\frac{1}{4}} & 0 & 0 \\ 0 & D = - ( - 1)^{\frac{3}{4}} & 0 \\ 0 & 0 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} {C^{'} = ( - 1)}^{\frac{3}{4}} & 0 & 0 \\ 0 & D^{'} = - ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & 1 \end{pmatrix}. \end{array}
(1.4.160)

The point to not over­look, is that ImΔX3,2Im\Delta X_{3,2} ro­tates in the X3X_{3}-plane, while ReΔX3X1Re\Delta X_{3}X_{1} trans­forms in the X1X_{1}-plane.

The final thing we con­sid­er, are the ex­po­nen­tial map­pings of the SU(2)SU(2) sub­group al­ge­bras as forms based ∈ℂ3\in \mathbb{C}^{3}. It turns out that since the axis prop­er­ties of the Eu­clid­ean dis­tance metric re­mains intact, the finite group re­la­tion­ship re­mains intact, seen by using de­not­ing the 𝔰𝔲′(2)\mathfrak{su}'(2) Lie al­ge­bra, as

𝔰𝔲′(2)=(x1+x2+x3000ix1−x2+ix30x2+ix3−x1)={M:xi∈ℝ},\begin{array}{r} \mathfrak{s}\mathfrak{u}^{'}(2) = \begin{pmatrix} x_{1} + x_{2} + x_{3} & 0 & 0 \\ 0 & ix_{1} & - x_{2} + ix_{3} \\ 0 & x_{2} + ix_{3} & - x_{1} \end{pmatrix} = \left\{ M:x_{i}\in \mathbb{R} \right\}, \end{array}
(1.4.161)

then pro­ceed­ing to find the matrix ex­po­nen­tial exp⁡(𝔰𝔲′(2))\exp\left( \mathfrak{s}\mathfrak{u}^{'}(2) \right), re­sult­ing in the Lie group rep­re­sen­ta­tion

(e(x1+x2+x3)000cos⁡(x12+x22+x32)+ix1sin⁡(x12+x22+x32)x12+x22+x32(−x2+ix3)sin⁡(x12+x22+x32)x12+x22+x320(x2+ix3)sin⁡(x12+x22+x32)x12+x22+x32cos⁡(x12+x22+x32)ix1sin⁡(x12+x22+x32)x12+x22+x32).\begin{array}{r} \begin{pmatrix} e^{\left( x_{1} + x_{2} + x_{3} \right)} & 0 & 0 \\ 0 & \cos{\left( \sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}} \right) +}\frac{ix_{1}\sin\left( \sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}} \right)}{\sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}}} & \frac{\left( - x_{2} + ix_{3} \right)\sin\left( \sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}} \right)}{\sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}}} \\ 0 & \frac{\left( x_{2} + ix_{3} \right)\sin\left( \sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}} \right)}{\sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}}} & \cos\left( \sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}} \right)\frac{ix_{1}\sin\left( \sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}} \right)}{\sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}}} \end{pmatrix}. \end{array}
(1.4.162)

The group above shares the same result for its bottom-right 2×22 \times 2 sub-matrix, with the matrix ex­po­nen­ti­a­tion of the Lie al­ge­bra 𝔰𝔲(2)\mathfrak{su}(2). In ad­di­tion, the e(x1+x2+x3)e^{\left( x_{1} + x_{2} + x_{3} \right)} func­tion lo­cat­ed in the a1,1a_{1,1} entry acts to scale the axis of ro­ta­tion for the group.

Figure
Figure 1.4.20 – Chi­ral­i­ty change by map­ping: ImΔX3,2,ReΔX3X1→−ImΔX3,2,−(ReΔX3X1)∗Im\Delta X_{3,2},Re\Delta X_{3}X_{1} \rightarrow - Im\Delta X_{3,2}, - \left( Re\Delta X_{3}X_{1} \right)^{*}.