1.4.7 The Axis Group

Having de­ter­mined the sym­met­ric conic group and its cone sub­group, it would be easy to think of the el­e­ments that are not in the cone sub­group, as a set dif­fer­ence be­tween the conic group and the cone group

but it turns out that the union of the axis group and the cone group is not nec­es­sar­i­ly the sym­met­ric conic group

We recall that the $\mathcal{X},\mathcal{Y}$, el­e­ments in the cone group have de­ter­mi­nants bound to cone num­bers for $\mathcal{X}$, and axis number de­ter­mi­nants for $\mathcal{Y}$

While the el­e­ments of the sym­met­ric conic group have cone num­bers for de­ter­mi­nants and de­ter­mi­nants of their prod­ucts in the cone and axis number sets for $\mathcal{X},\mathcal{Y}$,

Stat­ing these re­la­tion­ships by cat­e­go­riz­ing odd and even de­ter­mi­nants, dis­cussed in Sec­tion 1.4.4, the cone group had odd el­e­ments for $\mathcal{X}$, and even el­e­ments for $\mathcal{Y}$. The sym­met­ric conic group had odd and even el­e­ments for $\mathcal{X},\mathcal{Y}$, and the axis group $D_{3\mathbb{C}}^{\boxplus}$ had el­e­ments with even de­ter­mi­nants for $\mathcal{X},\mathcal{Y}$. We gen­er­al­ize the el­e­ment com­bi­na­tions for cone num­bers and axis num­bers using the no­ta­tion $ACC,AAA$ for $C$, a cone number, and $A$, an axis number, with which we can ex­press $\mathcal{X}$ in the rep­re­sen­ta­tive forms

All matrix prod­ucts for el­e­ments $\in \mathcal{X}$, result in even-valued de­ter­mi­nants $\in \mathcal{Y}$, which when gen­er­al­ized to the com­bi­na­tions of $ACC,AAA$ reveal the same el­e­ments that we found for the cone group! This is be­cause the prod­ucts of $\mathcal{X}$ that are even, yield $\mathcal{Y}$, which are also even, and of the rep­re­sen­ta­tive $Y_{j,k}$ forms:

Pri­ma­ry gen­er­a­tors in the axis group can be rep­re­sent­ed, as

where the $X_{1,1},X_{3,1}$ gen­er­a­tors are equal to the iden­ti­ty for their 4^th powers, $X_{1,1}^{4} = \epsilon,X_{3,1}^{4} = \epsilon$, but the $X_{2,1}$ gen­er­a­tor equals the iden­ti­ty for its 8^th power. The de­ter­mi­nants differ as well, where $X_{1,1},X_{3,1}$ have de­ter­mi­nants equal to $i,$and $X_{2,1}$, a de­ter­mi­nant equal to $1.$The asym­me­try of the $X_{2,1}$ gen­er­a­tor is the key to find­ing the axis group, given that we can find de­ter­mi­nants $\in \mathcal{X}$ on the real-axis, and de­ter­mi­nants $\in \mathcal{Y}$ on the imag­i­nary-axis. To prove this, the fol­low­ing is true con­cern­ing de­ter­mi­nants:

Where an even power of an el­e­ment $\in \mathcal{X}$ with itself, or an­oth­er el­e­ment $\in \mathcal{X}$, re­sults in an el­e­ment $\in \mathcal{Y}$. If the three gen­er­a­tors had imag­i­nary-valued de­ter­mi­nants, then for odd powers, all el­e­ments in $\in \mathcal{Y}$ have real-valued de­ter­mi­nants. This would have been the case if we set $X_{2,1}$, like the $X_{1,1},X_{3,1}$ gen­er­a­tors, to be

This is akin to the cone group el­e­ments $\in \mathcal{X}$, having com­plex num­bered de­ter­mi­nants as cone num­bers, and el­e­ments $\in \mathcal{Y}$, having imag­i­nary and real de­ter­mi­nants as axis num­bers. The conic group had de­ter­mi­nant values for $\mathcal{X},\mathcal{Y}$, that took on both cone and axis num­bers for values. To fur­ther em­pha­size this, powers of $X_{1,1}$ have the fol­low­ing axis num­bers for their de­ter­mi­nants

for $X_{1,1}^{2},X_{1,1}^{4}\in \mathcal{Y}$, and all $X_{2,1}$ de­ter­mi­nants equal one, given the fol­low­ing holds for de­ter­mi­nants of matrix prod­ucts

If we now take the de­ter­mi­nant of the prod­ucts ${X_{1,1} \times X}_{2,1}$, ${X_{2,1} \times X}_{1,2}$, we know that this will have an imag­i­nary valued de­ter­mi­nant, which in essence un­locks the axis group to find all its el­e­ments.

De­f­i­n­i­tion (1.4.122) – We define the Di­he­dral Cone Group $\mathbf{D}_{\mathbf{3}\mathbb{C}}^{\mathbf{\boxplus}}$ as a group under com­po­si­tion, which sits in the com­plex geo­met­ric space $\mathbb{C}^{3 \times 3}$, with a set of per­mu­ta­tion el­e­ments $A^{\boxplus}$, where the symbol $\boxplus$ geo­met­ri­cal­ly rep­re­sents axes.

The­o­rem (1.4.123) – For $A^{\boxplus}$, with a set of per­mu­ta­tions $\in D_{3\mathbb{C}}^{\boxplus}$, forms a group under com­po­si­tion.

Proof – Ver­i­fi­ca­tion of group axioms

• Clo­sure: All el­e­ments $\in A^{\boxplus}$are bi­jec­tive, and the com­po­si­tion of any el­e­ments $\phi \circ \psi\ \in A^{\boxplus}$ is also bi­jec­tive.

• As­so­cia­tiv­i­ty: Per­mu­ta­tions are func­tions, and their com­po­si­tions are as­so­cia­tive.

• Iden­ti­ty: The iden­ti­ty func­tion $\epsilon_{A^{\boxplus}}$ bi­jec­tive­ly maps all el­e­ments of the set $A^{\boxplus}$to them­selves $\left( \epsilon_{A^{\boxplus}}:A^{\boxplus} \rightarrow A^{\boxplus} \right)$.

• In­verse: $\phi$ is a per­mu­ta­tion and is thus bi­jec­tive, and there exists a func­tion $\phi^{- 1}$ that is also bi­jec­tive.

$\square$

The­o­rem (1.4.124) – The Di­he­dral Axis Group $\mathbf{D}_{\mathbf{3}\mathbb{C}}^{\mathbf{\boxplus}}$ con­tains a total of 1,536 per­mu­ta­tion el­e­ments $\in \mathbb{C}^{3}$, where el­e­ments of $\mathcal{X}_{j}\in \mathcal{X}$ have 256 per­mu­ta­tions, el­e­ments of $\mathcal{Y}_{j}\in \mathcal{Y}$ have 256 per­mu­ta­tions, and axis num­bers for de­ter­mi­nant values.

Proof - Like the cone group, we will define the ma­tri­ces $M_{j}$, cal­cu­late their de­ter­mi­nants, and these will act on the $X_{1,1}$ el­e­ment. The first goal is to find all el­e­ments in the forms

where the bottom-right $2 \times 2$ sub­ma­trix will have either two axis num­bers, or two cone num­bers for its matrix en­tries. Since the ini­tial $M_{j}$ ma­tri­ces will be acting on $X_{1,1}$ , the axis number in the $a_{1,1}$ po­si­tion re­mains un­changed, we look to find $32$ el­e­ments in this sub­group $\subset$ $\mathcal{X}.$ The fol­low­ing $M_{j}$ ma­tri­ces are de­fined by the fol­low­ing prod­ucts:

Using the prod­ucts of the $M_{j}$ ma­tri­ces with the $X_{1,1},X_{1,imag}$ el­e­ments, we find the fol­low­ing group of el­e­ments, rep­re­sent­ing 32 of the 128 el­e­ments $\in X_{1,k}^{'}$. For $A$, in the po­si­tion $a_{1,1}$, fixed at $a_{1,1} = i$, we want to solve for the fol­low­ing matrix forms:

The subset $A_{1,k} \subset X_{1,k}^{'}$ is cal­cu­lat­ed as fol­lows, for the sub­script value $n$

find­ing, as ex­pect­ed, 32 dif­fer­ent matrix values in the subset. To find the full sub­group, we use the $M_{8}^{n}$ matrix raised to powers of $n$, by taking the sum of the fol­low­ing prod­ucts of the subset $A_{1,k}$

with which, the fol­low­ing set of prod­ucts gives us the $X_{1,k}^{'}$ sub­group

To find the other 128 el­e­ments $\in X_{1,k}^{'}$, we solve for all el­e­ments with the fol­low­ing matrix forms:

in which, a cone number rep­re­sents the $a_{1,1}$ entry. We will want to take ad­van­tage of the real-valued el­e­ments $Y_{2},Y_{3}$, first seen in the $S_{3}/D_{3}$ group, where the $Y_{2}$ el­e­ment can be cal­cu­lat­ed by the fol­low­ing prod­uct com­bi­na­tion

and the $Y_{3}$ el­e­ment is simply­­

It is im­por­tant to con­sid­er what each of the $Y_{2},Y_{3}$, ma­tri­ces do when they act by mul­ti­pli­ca­tion on an el­e­ment in $\in \mathcal{X}_{1}$. For the $Y_{2}$ el­e­ment, mul­ti­pli­ca­tion on the left-side with an el­e­ment $\in \mathcal{X}_{1}$, com­posed of one axis number and two cone num­bers, will shift the matrix en­tries in $X_{1,k}^{'}$ up by one po­si­tion, re­sult­ing in an el­e­ment $\in \mathcal{X}_{2}$

For mul­ti­pli­ca­tion by the el­e­ment $Y_{2}$ on the right-side, we find a po­si­tion shift of the matrix en­tries to the right, re­sult­ing in the trans­for­ma­tion of $\mathcal{X}_{1}$ onto an el­e­ment $\in \mathcal{X}_{3}$

Com­bin­ing both the up­shift and the right-shift re­sults in an el­e­ment $\in \mathcal{X}_{1}$, with its orig­i­nal en­tries re­arranged, while re­main­ing $\in \mathcal{X}_{1}$

There is a sim­i­lar re­la­tion­ship con­cern­ing $Y_{3}$, where in­stead, there is a down­shift and a left-shift of the en­tries $\in \mathcal{X}_{1}$, trans­form­ing it to el­e­ments $\in \mathcal{X}_{3},\mathcal{X}_{2}$, re­spec­tive­ly

The com­bi­na­tion of trans­for­ma­tions yields the second arrange­ment of the el­e­ment $\mathcal{X}_{1}$, having the $C_{2}$ cone number in the matrix po­si­tion $a_{1,1}$

With this in mind, we can use either of the $Y_{2},Y_{3}$ el­e­ments to find the other 128 el­e­ments $\in \mathcal{X}_{1}$

and the com­bi­na­tion of the two sub­sets yields the 256 el­e­ments $\in \mathcal{X}_{1}$

The re­main­ing sub­sets $\mathcal{X}_{2},\mathcal{X}_{3}$ are simply found by $Y_{2}\mathcal{X}_{1} = \mathcal{X}_{2}$, $Y_{3}\mathcal{X}_{1} = \mathcal{X}_{3}$, im­ply­ing the sub­group $\mathcal{X}$ is the set

To find the $\mathcal{Y}$-sub­group, we first find the hy­per­bol­ic pre­sen­ta­tion of the $X_{1}$ el­e­ment, by the prod­uct,

then de­ter­mine $\mathcal{Y}_{j}$, by the prod­ucts

lead­ing to $\mathcal{Y}$, as the set

Since there are 256 el­e­ments in the six el­e­ment types $\mathcal{X}_{j},\mathcal{Y}_{j}$, this gives us a total of 1,536 el­e­ments in the axis group, which is the same number we found in the cone group. There­fore, the axis group can be de­fined as the union of the el­e­ments $\mathcal{X},\mathcal{Y}$,

$\square$