Masteries · Jason GlowneyArithmoí Book 1

Masteries

1.1.26 The Symmetric/Dihedral Cone Group

De­f­i­n­i­tion (1.1.198) – We now define the Sym­met­ric/Di­he­dral group 𝐃𝟑𝕔\mathbf{D}_{\mathbf{3}\mathbb{c}}, as a group under com­po­si­tion that sits in the com­plex geo­met­ric space ℂ3×3\mathbb{C}^{3 \times 3}, having a set of per­mu­ta­tion el­e­ments A𝕔A^{\mathbb{c}}

D3𝕔 ={A𝕔∈U(3,ℂ)|A𝕔∙(A𝕔)∗=I }.\begin{array}{r} D_{3\mathbb{c}}\ = \left\{ A^{\mathbb{c}} \in U\left( 3,\mathbb{C} \right)|A^{\mathbb{c}} \bullet \left( A^{\mathbb{c}} \right)^{*} = I\ \right\}. \end{array}

The set of per­mu­ta­tion el­e­ments X∈A𝕔X \in A^{\mathbb{c}} of D3𝕔D_{3\mathbb{c}} have multi-valued en­tries ∈ℂ3×3\in \mathbb{C}^{3 \times 3} for the Xj,kX_{j,k} per­mu­ta­tions, and values ∈ℂ3×3\in \mathbb{C}^{3 \times 3} for the cyclic sub­group Yj,kY_{j,k}, seen below.

Y1,1=(100010001),   Y1,2=(i000i000i),   Y1,3=(−1000−1000−1),   Y1,4=(−i000−i000−i).\begin{array}{r} Y_{1,1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\ \ \ Y_{1,2} = \begin{pmatrix} i & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & i \end{pmatrix},\ \ \ Y_{1,3} = \begin{pmatrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \end{pmatrix},\ \ \ Y_{1,4} = \begin{pmatrix} - i & 0 & 0 \\ 0 & - i & 0 \\ 0 & 0 & - i \end{pmatrix}. \end{array}
(1.1.199)
Y2,1=(010001100),   Y2,2=(0i000ii00),   Y2,3=(0−1000−1−100),   Y2,4=(0−i000−i−i00).\begin{array}{r} Y_{2,1} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix},\ \ \ Y_{2,2} = \begin{pmatrix} 0 & i & 0 \\ 0 & 0 & i \\ i & 0 & 0 \end{pmatrix},\ \ \ Y_{2,3} = \begin{pmatrix} 0 & - 1 & 0 \\ 0 & 0 & - 1 \\ - 1 & 0 & 0 \end{pmatrix},\ \ \ Y_{2,4} = \begin{pmatrix} 0 & - i & 0 \\ 0 & 0 & - i \\ - i & 0 & 0 \end{pmatrix}. \end{array}
(1.1.200)
Y3,1=(001100010), Y3,2=(00ii000i0),Y3,3=(00−1−1000−10),Y3,4=(00−i−i000−i0).\begin{array}{r} Y_{3,1} = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix},\ Y_{3,2} = \begin{pmatrix} 0 & 0 & i \\ i & 0 & 0 \\ 0 & i & 0 \end{pmatrix},Y_{3,3} = \begin{pmatrix} 0 & 0 & - 1 \\ - 1 & 0 & 0 \\ 0 & - 1 & 0 \end{pmatrix},Y_{3,4} = \begin{pmatrix} 0 & 0 & - i \\ - i & 0 & 0 \\ 0 & - i & 0 \end{pmatrix}. \end{array}
(1.1.201)
{X1,1,X1,2,X1,3,X1,4}={(1+i200001+i201+i20),   (i−120000i−120i−120), (−(1+i)20000−(1+i)20−(1+i)20),   (−i+120000−i+120−i+120)}.\begin{array}{r} \left\{ X_{1,1},X_{1,2},X_{1,3},X_{1,4} \right\} = \left\{ \begin{array}{r} \begin{pmatrix} \frac{1 + i}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & \frac{1 + i}{\sqrt{2}} \\ 0 & \frac{1 + i}{\sqrt{2}} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} \frac{i - 1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & \frac{i - 1}{\sqrt{2}} \\ 0 & \frac{i - 1}{\sqrt{2}} & 0 \end{pmatrix}, \\ \ \begin{pmatrix} \frac{- (1 + i)}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & \frac{- (1 + i)}{\sqrt{2}} \\ 0 & \frac{- (1 + i)}{\sqrt{2}} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} \frac{- i + 1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & \frac{- i + 1}{\sqrt{2}} \\ 0 & \frac{- i + 1}{\sqrt{2}} & 0 \end{pmatrix} \end{array} \right\}. \end{array}
(1.1.202)
{X2,1,X2,2,X2,3,X2,4}={(001+i201+i201+i200),   (00i−120i−120i−1200), (00−(1+i)20−(1+i)20−(1+i)200),   (00−i+120−i+120−i+1200)}\begin{array}{r} \left\{ X_{2,1},X_{2,2},X_{2,3},X_{2,4} \right\} = \left\{ \begin{array}{r} \begin{pmatrix} 0 & 0 & \frac{1 + i}{\sqrt{2}} \\ 0 & \frac{1 + i}{\sqrt{2}} & 0 \\ \frac{1 + i}{\sqrt{2}} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & \frac{i - 1}{\sqrt{2}} \\ 0 & \frac{i - 1}{\sqrt{2}} & 0 \\ \frac{i - 1}{\sqrt{2}} & 0 & 0 \end{pmatrix}, \\ \ \begin{pmatrix} 0 & 0 & \frac{- (1 + i)}{\sqrt{2}} \\ 0 & \frac{- (1 + i)}{\sqrt{2}} & 0 \\ \frac{- (1 + i)}{\sqrt{2}} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & \frac{- i + 1}{\sqrt{2}} \\ 0 & \frac{- i + 1}{\sqrt{2}} & 0 \\ \frac{- i + 1}{\sqrt{2}} & 0 & 0 \end{pmatrix} \end{array} \right\} \end{array}
(1.1.203)
{X3,1,X3,2,X3,3,X3,4}={(01+i201+i200001+i2),   (0i−120i−120000i−12), (0−(1+i)20−(1+i)20000−(1+i)2),   (0−i+120−i+120000−i+12)}\begin{array}{r} \left\{ X_{3,1},X_{3,2},X_{3,3},X_{3,4} \right\} = \left\{ \begin{array}{r} \begin{pmatrix} 0 & \frac{1 + i}{\sqrt{2}} & 0 \\ \frac{1 + i}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & \frac{1 + i}{\sqrt{2}} \end{pmatrix},\ \ \ \begin{pmatrix} 0 & \frac{i - 1}{\sqrt{2}} & 0 \\ \frac{i - 1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & \frac{i - 1}{\sqrt{2}} \end{pmatrix}, \\ \ \begin{pmatrix} 0 & \frac{- (1 + i)}{\sqrt{2}} & 0 \\ \frac{- (1 + i)}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & \frac{- (1 + i)}{\sqrt{2}} \end{pmatrix},\ \ \ \begin{pmatrix} 0 & \frac{- i + 1}{\sqrt{2}} & 0 \\ \frac{- i + 1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & \frac{- i + 1}{\sqrt{2}} \end{pmatrix} \end{array} \right\} \end{array}
(1.1.204)

Our de­fined group S3𝕔/D3𝕔S_{3\mathbb{c}}/D_{3\mathbb{c}} has 2424 el­e­ments, is non-abelian, and any per­mu­ta­tion ϕ\phi is both an in­jec­tive and sur­jec­tive (bi­jec­tive) func­tion from the set of three vari­ables, rep­re­sent­ing the equi­lat­er­al tri­an­gle sym­me­tries {A,B,C}∈ℂ3\{ A,B,C\} \in \mathbb{C}^{3} onto itself. For our abelian cyclic group, we have el­e­ments that rotate to values ± i,±1\pm \ i, \pm 1 on the axes of the com­plex unit circle.

The­o­rem (1.1.205) – The set of per­mu­ta­tions A𝕔∈D3𝕔A^{\mathbb{c}} \in D_{3\mathbb{c}} forms a group under com­po­si­tion

Proof – Ver­i­fi­ca­tion of group axioms

  • Clo­sure: All el­e­ments ∈A𝕔\in A^{\mathbb{c}} are bi­jec­tive, and the com­po­si­tion of any el­e­ments ϕ∘ψ ∈A𝕔\phi \circ \psi\ \in A^{\mathbb{c}}is also bi­jec­tive.
  • As­so­cia­tiv­i­ty: Per­mu­ta­tions are func­tions, and their com­po­si­tions are as­so­cia­tive.
  • Iden­ti­ty: The iden­ti­ty func­tion ϵA𝕔\epsilon_{A^{\mathbb{c}}} bi­jec­tive­ly maps all el­e­ments of the set A𝕔A^{\mathbb{c}} to them­selves (ϵA𝕔:A𝕔→A𝕔).\left( \epsilon_{A^{\mathbb{c}}}:A^{\mathbb{c}} \rightarrow A^{\mathbb{c}} \right).
  • In­verse: ϕ\phi is a per­mu­ta­tion and thus bi­jec­tive, and a func­tion exists ϕ−1\ \phi^{- 1} that is also bi­jec­tive.
□\square

Much like the Figure 1.1.5 plot of the S3S_{3} iden­ti­ty in hy­per­bol­ic space, the iden­ti­ty el­e­ment ϵ(ABC)\epsilon(ABC) has its ver­tices aligned along the real axes, shown in Figure 1.1.17.

Figure
Figure 1.1.17 –The iden­ti­ty el­e­ment ϵ (ABC)\epsilon\ (ABC).
Figure
Figure 1.1.18 – The ϕ(X1,1):ϵ\phi\left( X_{1,1} \right):\epsilon map­ping.

Taking the prod­uct X1,1X_{1,1} with the iden­ti­ty el­e­ment ϵ=Y1,1\epsilon = Y_{1,1}, yields the com­pos­ite func­tion ϕ(X1,1):ϵ\phi\left( X_{1,1} \right):\epsilon, which re­flects the ver­tices B,CB,C along the axis AA, and ro­tates the equi­lat­er­al tri­an­gle in all three of the com­plex XjX_{j}-planes to the co­or­di­nates ACB=(1+i)2(1,1,1)ACB = \frac{(1 + i)}{\sqrt{2}}(1,1,1) (Figure 1.1.18). We can con­tin­ue by taking the com­pos­ite func­tion from above and the Y2,2Y_{2,2} per­mu­ta­tion, to see a cyclic order shift ACB→BACACB \rightarrow BAC occurs for the tri­an­gle, having ro­tat­ed to the ad­ja­cent quad­rant values BAC=(i−1)2(1,1,1)BAC = \frac{(i - 1)}{\sqrt{2}}(1,1,1), seen in Figure 1.1.19 below.

Figure
Figure 1.1.19 –The ϕ(X1,2Y2,2):ϵ\phi\left( X_{1,2}Y_{2,2} \right):\epsilon map­ping.

In gen­er­al, the com­po­si­tion of func­tions iso­lat­ed to the Yj,kY_{j,k} sub­group will result in ro­ta­tions and re­flec­tions of the equi­lat­er­al tri­an­gle that lay along the imag­i­nary and real axes of the ℂ3×3\mathbb{C}^{3 \times 3} space. Com­pare this to the com­po­si­tion of per­mu­ta­tion func­tions from the Xj,kX_{j,k} sub­group, which will rotate and re­flect the ver­tices of the equi­lat­er­al tri­an­gle to values which are n={0,±1,±2,±3,…},n = \left\{ 0, \pm 1, \pm 2, \pm 3,\ldots \right\}, (π+2πn)4\frac{(\pi + 2\pi n)}{4} ra­di­ans away from any real and imag­i­nary-axis in the geo­met­ric space.

This sec­tion has ex­pand­ed our di­he­dral group's de­f­i­n­i­tion to in­clude com­plex-valued per­mu­ta­tions with el­e­ments that per­fect­ly bal­ance real and imag­i­nary values. The D3𝕔D_{3\mathbb{c}} group has Lie al­ge­bra el­e­ments com­posed of an even bal­ance of real and imag­i­nary num­bers, which will be cen­tral to the study of light in the up­com­ing physics volume.