Masteries · Jason GlowneyArithmoí Book 1

Masteries

1.4.6 The Conic Group

At this point, we de­scribe the sym­me­try group rep­re­sen­ta­tive of all pos­si­ble ways to per­mute the equi­lat­er­al tri­an­gle in the ℂ3\mathbb{C}^{3} space, in­clud­ing the eight points on the unit circle point set {1,(−1)(14),i,(−1)(34),−1,−(−1)(14),−i,−(−1)(34)}\left\{ 1,( - 1)^{\left( \frac{1}{4} \right)},i,( - 1)^{\left( \frac{3}{4} \right)}, - 1, - ( - 1)^{\left( \frac{1}{4} \right)}, - i, - ( - 1)^{\left( \frac{3}{4} \right)} \right\}, for three or­thog­o­nal­ly arranged unit cir­cles. To do this, we use gen­er­a­tors with cone number de­ter­mi­nants. To start, we choose two el­e­ments, the first ∈𝒳\in \mathcal{X}, and the second ∈𝒴\in \mathcal{Y}

X2,1=(00i0(−1)140100),   Y2,1=(0i0001(−1)1400),\begin{array}{r} X_{2,1} = \begin{pmatrix} 0 & 0 & i \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \\ 1 & 0 & 0 \end{pmatrix},\ \ \ Y_{2,1} = \begin{pmatrix} 0 & i & 0 \\ 0 & 0 & 1 \\ ( - 1)^{\frac{1}{4}} & 0 & 0 \end{pmatrix}, \end{array}
(1.4.103)

for which, det⁡(X2,1)=−(−1)(34),\det\left( X_{2,1} \right) = - ( - 1)^{\left( \frac{3}{4} \right)}, det⁡(Y2,1)=(−1)(34)\det\left( Y_{2,1} \right) = ( - 1)^{\left( \frac{3}{4} \right)}. The ra­tio­nale for choos­ing el­e­ments having cone number de­ter­mi­nants is that the prod­uct re­la­tion­ships for 𝒴,𝒳\mathcal{Y},\mathcal{X}, result in el­e­ments having de­ter­mi­nants which are either a cone number or an axis number. After con­sid­er­ing how we went about find­ing all el­e­ment per­mu­ta­tions in the cone group, it might seem like an even more daunt­ing task to do so for the sym­me­try group the cone group be­longs to. How­ev­er, it turns out that it can be much more straight­for­ward, given that el­e­ments ∈𝒴\in \mathcal{Y} can be cone num­bers. The el­e­ments that we will be study­ing are in the Y1,kY_{1,k} group. Defin­ing a set 𝒞\mathcal{C} with el­e­ments having the fol­low­ing arrange­ment of num­bers:

𝒞={((−1)1400010001),   (1000(−1)140001),   (10001000(−1)14)}.\begin{array}{r} \mathcal{C} =\left\{ \begin{pmatrix} ( - 1)^{\frac{1}{4}} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & 1 \end{pmatrix},\ \ \ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \end{pmatrix} \right\}. \end{array}
(1.4.104)

Since we are deal­ing with finite el­e­ments, using this set amounts to turn the dial on a com­bi­na­tion lock to arrive at any pos­si­ble per­mu­ta­tion in the sym­me­try group. Now, the task is to show how to do just this.

De­f­i­n­i­tion (1.4.105) – We define the Sym­met­ric Conic Group 𝐒𝟑ℂ𝚫\mathbf{S}_{\mathbf{3}\mathbb{C}}^{\boldsymbol{\Delta}}, as group under com­po­si­tion that sits in the com­plex geo­met­ric space ℂ3×3\mathbb{C}^{3 \times 3}, with a set of per­mu­ta­tion el­e­ments AΔA^{\Delta} with su­per­script no­ta­tion Δ\Delta, rep­re­sent­ing the cone.

S3ℂΔ ={AΔ∈U(3,ℂ)|AΔ∙AΔ†=I }.\begin{array}{r} S_{3\mathbb{C}}^{\Delta}\ = \left\{ A^{\Delta} \in U\left( 3,\mathbb{C} \right)|A^{\Delta} \bullet {A^{\Delta}}^{\dagger} = I\ \right\}. \end{array}

The­o­rem (1.4.106) – For AΔA^{\Delta}as a set, the set of per­mu­ta­tions ∈S3ℂΔ{\in S}_{3\mathbb{C}}^{\Delta}, forms a group under com­po­si­tion

Proof – Ver­i­fi­ca­tion of group axioms

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

The­o­rem (1.4.107) – The Di­he­dral Conic Group 𝐒𝟑ℂ𝚫\mathbf{S}_{\mathbf{3}\mathbb{C}}^{\boldsymbol{\Delta}} con­tains 3,072 unique per­mu­ta­tion el­e­ments ∈ℂ3\in \mathbb{C}^{3}, where el­e­ments of Xj∈𝒳X_{j}\in \mathcal{X} have 512 per­mu­ta­tions, el­e­ments of Yj∈𝒴Y_{j}\in \mathcal{Y} have 512 per­mu­ta­tions, and all de­ter­mi­nants are cone number/axis number-valued.

𝐏𝐫𝐨𝐨𝐟−\mathbf{Proof}\mathbf{-} Star­ing with the el­e­ments

X2,1=(00i0(−1)140100), Y2,1=(0i0001(−1)1400),X_{2,1} = \begin{pmatrix} 0 & 0 & i \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \\ 1 & 0 & 0 \end{pmatrix},\ Y_{2,1} = \begin{pmatrix} 0 & i & 0 \\ 0 & 0 & 1 \\ ( - 1)^{\frac{1}{4}} & 0 & 0 \end{pmatrix},

we find all three el­e­ments in the 𝒞\mathcal{C}-subset, where mul­ti­ple com­bi­na­tions of per­mu­ta­tions can be con­sid­ered, of which, we present the fol­low­ing prod­ucts that result in the el­e­ments ∈𝒞\in \mathcal{C}:

C1=(X2,12Y2,13(X2,1Y2,1)2)3=((−1)1400010001),{C_{1} = \left( X_{2,1}^{2}Y_{2,1}^{3}\left( X_{2,1}Y_{2,1} \right)^{2} \right)}^{3} = \begin{pmatrix} ( - 1)^{\frac{1}{4}} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},
C2=C37(C1X2,1Y2,1)6=(1000(−1)140001),C_{2} = C_{3}^{7}\left( C_{1}X_{2,1}Y_{2,1} \right)^{6} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & ( - 1)^{\frac{1}{4}} & 0 \\ 0 & 0 & 1 \end{pmatrix},
C3=X2,16(C17(C1X2,1)2)=(10001000(−1)14).C_{3} = X_{2,1}^{6}\left( C_{1}^{7}\left( C_{1}X_{2,1} \right)^{2} \right) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \end{pmatrix}.

Since each of the three matrix en­tries in the per­mu­ta­tion may take any value in the set

{1,(−1)14,i,(−1)34,−1,−(−1)14,−i,−(−1)34}\left\{ 1,( - 1)^{\frac{1}{4}},i,( - 1)^{\frac{3}{4}}, - 1, - ( - 1)^{\frac{1}{4}}, - i, - ( - 1)^{\frac{3}{4}} \right\},

and given there are six per­mu­ta­tion forms, a simple cal­cu­la­tion shows the total number of per­mu­ta­tions is

83×6=3,072 total permutations.8^{3} \times 6 = 3,072\ total\ permutations.
□\begin{array}{r} \end{array}\square

To show the power of these el­e­ments using the el­e­ments of 𝒞\mathcal{C}, it be­comes pos­si­ble to arrive at all our orig­i­nal el­e­ments from the cone group, by the fol­low­ing prod­ucts

C23C12(X2,1Y2,1)5=((−1)140000i010),\begin{array}{r} C_{2}^{3}C_{1}^{2}\left( X_{2,1}Y_{2,1} \right)^{5} = \begin{pmatrix} ( - 1)^{\frac{1}{4}} & 0 & 0 \\ 0 & 0 & i \\ 0 & 1 & 0 \end{pmatrix}, \end{array}
(1.4.108)
C3C17C35(Y2,1X2,1)=(0i010000(−1)14),\begin{array}{r} {C_{3}C}_{1}^{7}C_{3}^{5}\left( Y_{2,1}X_{2,1} \right) = \begin{pmatrix} 0 & i & 0 \\ 1 & 0 & 0 \\ 0 & 0 & ( - 1)^{\frac{1}{4}} \end{pmatrix}, \end{array}
(1.4.109)

for X2,1X_{2,1}, a prod­uct el­e­ment.