1.4.4 Null Curves and the Cone Group

As pre­vi­ous­ly dis­cussed, in tra­di­tion­al conic sec­tions, the null curve rep­re­sents lines des­ig­nat­ing the bound­ary of the cone. In the com­plex plane, these lines have slopes taking the form of the pos­i­tive and neg­a­tive square roots of $i$, which can be rep­re­sent­ed by the ex­po­nen­tials $e^{\left( \frac{\pm i\pi}{4} \right)},e^{\left( \frac{\pm 3i\pi}{4} \right)}$, and their in­ter­sec­tions with the unit circle. Nu­mer­i­cal­ly, these ex­po­nen­tials rep­re­sent the com­plex num­bers,

with $2 \times 2$ matrix rep­re­sen­ta­tions

each of which, squares to

There exists a more fun­da­men­tal rep­re­sen­ta­tion of these num­bers, where we argue that these are fun­da­men­tal and anal­o­gous to the imag­i­nary and split com­plex num­bers $\in \mathbb{C}^{2},\mathbb{R}^{2}$

These num­bers take the forms

where $\pm i, \pm 1$, are arranged on the anti-di­ag­o­nal, having more geo­met­ric sig­nif­i­cance, given that the matrix en­tries al­ter­nate be­tween real and imag­i­nary values. The matrix rep­re­sen­ta­tion for $( - 1)^{\left( \frac{1}{4} \right)}$, will square to $i$, and when raised to se­quen­tial powers, it will cycle around the unit circle points, $( - 1)^{\left( \frac{3}{4} \right)},{- ( - 1)}^{\left( \frac{1}{4} \right)},{- ( - 1)}^{\left( \frac{3}{4} \right)}$, and the axes of the circle

In Figure 1.4.10, the geom­e­try of the cone num­bers is shown at their in­ter­sec­tion points with the com­plex unit circle.

When com­par­ing the matrix ex­po­nen­tials for the two rep­re­sen­ta­tions of el­e­ments, they differ in the $\sinh$ trigono­met­ric en­tries

Both rep­re­sen­ta­tions are es­sen­tial­ly hy­brids of the trigono­met­ric func­tions of the circle and the hy­per­bo­la, and these can just as easily be rep­re­sent­ed in the forms

For matrix de­riv­a­tives, we find­­­­­

re­veal­ing that they will differ by the en­tries on their anti-di­ag­o­nals. When re­lat­ing di­he­dral forms $\in \mathbb{C}^{3}$ to the cone num­bers, being con­sid­er­ing, we define the cone number el­e­ments $\in \mathbb{C}^{3}$, by the fol­low­ing,

where each el­e­ment is a cone number with re­spec­tive ro­ta­tion­al axes from the set $( - 1)^{\frac{1}{4}},( - 1)^{\frac{3}{4}},{- ( - 1)}^{\frac{1}{4}},{- ( - 1)}^{\frac{3}{4}}$. The prod­ucts of these cone num­bers form the el­e­ments of a Cartan al­ge­bra $Y$ of the cone group, a few of which we cal­cu­late prod­ucts with, for $X_{j},X_{k}$, in­di­cat­ing the set of prod­ucts:

The $X_{j},X_{k}$ prod­uct re­sults are sim­i­lar in form to the $Y_{2},Y_{3}$ Cartan sub­al­ge­bras from the other di­he­dral groups

and for the $Y_{1}$ el­e­ments

In the case of a $Y_{1}$ per­mu­ta­tion matrix, for which, $a,b,c = 1$, we have the mul­ti­plica­tive iden­ti­ty $\epsilon$ of the group. The sim­plest means for a cone number el­e­ment to trans­form to the iden­ti­ty $\epsilon$, is by suc­ces­sive self-prod­ucts

For entry values $\{ a,b,c\}$, sim­i­lar rep­re­sen­ta­tions in­clu­sive of all $X_{j}$-type per­mu­ta­tions, take the gen­er­al form

The group being con­sid­er­ing is com­prised of a much larger set of per­mu­ta­tion el­e­ments in com­par­i­son to the other di­he­dral groups cov­ered up to this point. The values for each entry $a,b,c$ can be any com­plex number from the set of eight el­e­ments $\left\{ 1,( - 1)^{\left( \frac{1}{4} \right)},i,( - 1)^{\left( \frac{3}{4} \right)}, - 1,{- ( - 1)}^{\left( \frac{1}{4} \right)}, - i,{- ( - 1)}^{\left( \frac{3}{4} \right)} \right\}.$The $X_{j}$ el­e­ments are a sub­group of the larger $D_{3\mathbb{c}}$ group, where for in­stance, we can derive $X_{1,1} \in D_{3\mathbb{c}}$, by way of the fol­low­ing com­bi­na­tion of prod­ucts $\left( X_{3,1}X_{2,1}X_{3,1}X_{1,1}^{6} \right)$

In like fash­ion, the re­main­ing two el­e­ments $X_{2,1},X_{3,1} \in D_{3\mathbb{c}}$, can be gen­er­at­ed by the fol­low­ing prod­ucts

The fa­mil­iar Cartan sub­al­ge­bra forms $Y_{j},$can be gen­er­at­ed by the prod­ucts ${X_{2,1}X_{3,1}^{n},X_{1,1}X}_{2,1}X_{2,1}X_{1,1}$, for $n = 1,3,5,7$,

Con­cern­ing cone number per­mu­ta­tion ma­tri­ces of the $Y_{j}$-type*,* they are not nec­es­sar­i­ly com­mu­ta­tive as they were for the Cartan sub­al­ge­bras from pre­vi­ous stud­ies. This is be­cause the per­mu­ta­tion ma­tri­ces $\in Y_{j}$ can now vary in their in­di­vid­ual entry values, which are drawn from the set $\left\{ 1,( - 1)^{\left( \frac{1}{4} \right)},i,( - 1)^{\left( \frac{3}{4} \right)}, - 1,{- ( - 1)}^{\left( \frac{1}{4} \right)}, - i,{- ( - 1)}^{\left( \frac{3}{4} \right)} \right\}.$For the $Y_{2},Y_{3}$-type ma­tri­ces, a cyclic shift in the entry values occurs for the Lie brack­et op­er­a­tion

More suc­cinct­ly, we can de­scribe the com­mu­ta­tor re­la­tion­ships, as

which im­plies that for the $Y_{2},Y_{3}$ ma­tri­ces to com­mute, we must have that $ae = cd = bf$.

By these con­sid­er­a­tions, we can now define the cone group $\in \mathbb{C}^{3}$, as fol­lows:

De­f­i­n­i­tion (1.4.84) – The di­he­dral cone group $\mathbf{D}_{\mathbf{3}\mathbb{C}}^{\boldsymbol{\Delta}}$, is de­fined as a group under com­po­si­tion that sits in the com­plex geo­met­ric space $\mathbb{C}^{3 \times 3}$, with a set $A^{\Delta}$ of per­mu­ta­tion el­e­ments, with the su­per­script $\Delta,$ rep­re­sent­ing the cone.

The­o­rem (1.4.85) – The set of per­mu­ta­tions $A^{\Delta}$ forms a group under com­po­si­tion

Proof – Ver­i­fi­ca­tion of group axioms

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

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

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

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

$\square$

The­o­rem (1.4.86) – The di­he­dral cone group $D_{3\mathbb{C}}^{\Delta}$ con­tains a total of 1,536 per­mu­ta­tion el­e­ments $\in \mathbb{C}^{3}$, where el­e­ments of $X_{j}\in \mathcal{X}$ are each com­prised of 256 per­mu­ta­tions with cone num­bers for their de­ter­mi­nants, and the el­e­ments $Y_{j}\in \mathcal{Y}$ are also each com­prised of 256 el­e­ments with axis num­bers for their de­ter­mi­nants.

$\mathbf{Proof}\mathbf{-}$ $\forall\ X_{j,k}$ el­e­ments, de­ter­mi­nants are com­plex num­bers on the unit circle, lo­cat­ed on cone points from the set $\left\{ 1,( - 1)^{\left( \frac{1}{4} \right)},i,( - 1)^{\left( \frac{3}{4} \right)}, - 1,{- ( - 1)}^{\left( \frac{1}{4} \right)}, - i,{- ( - 1)}^{\left( \frac{3}{4} \right)} \right\}$, and specif­i­cal­ly, the el­e­ments $\in X_{j,1}$ have de­ter­mi­nants equal to $- ( - 1)^{\left( \frac{3}{4} \right)}$. For el­e­ments $\in Y_{j,k}$, they also have com­plex number de­ter­mi­nants on the unit circle, lo­cat­ed at axis points from the set $\left\{ 1, - i, - 1,i \right\}$, and specif­i­cal­ly, the el­e­ments $\in Y_{j,1}$ have de­ter­mi­nants equal to $- i$ . From linear al­ge­bra, the prod­uct of de­ter­mi­nants is equal to de­ter­mi­nant of the matrix-matrix prod­uct

and the fol­low­ing prod­uct re­la­tions for the $D_{3}$ di­he­dral groups, hold con­gru­ent modulo 3,

From these prod­uct re­la­tion­ships of group el­e­ments and de­ter­mi­nants, the $\mathcal{Y}$-el­e­ment de­ter­mi­nants will always remain on the axes of the unit circle, and the $\mathcal{X}$-el­e­ment de­ter­mi­nants will always be cone number points on unit circle. Fur­ther, the prod­uct re­sults of any even powers in­volv­ing $\mathcal{X}$, will always be on the axes and $\in$ $\mathcal{Y}$, for ex­am­ple

The sum of two even ex­po­nents yield an even ex­po­nent result and based on the prod­uct re­la­tion­ship $\mathcal{X}_{j}\mathcal{X}_{k}\in \mathcal{Y}$, any prod­uct com­bi­na­tion of el­e­ments $\in \mathcal{X}$ that re­sults in an even ex­po­nent is always $\in \mathcal{Y}$. All odd power ex­po­nent sums of el­e­ment prod­uct of com­bi­na­tions $\in \mathcal{X}$, result in a cone number de­ter­mi­nant matrix, based on the unit circle's sym­me­try and the el­e­ment prod­uct re­la­tion­ships, which are always $\in \mathcal{X}$. For ex­am­ple,

has an odd num­bered sum for the prod­uct ex­po­nents $3 + 6 = 9$, and always holds, since any $even + odd = odd$. This is the case for the re­sults of prod­ucts $\in \mathcal{X}$, and not prod­ucts $\in \mathcal{Y}$, and/or the com­bi­na­tion of prod­ucts $\in \mathcal{X},\mathcal{Y}.$ The reason is the fol­low­ing de­ter­mi­nant holds, $\det\left( X_{j}^{m}X_{k}^{n} \right) = \det\left( X_{j}^{m} \right) \times \det\left( X_{k}^{n} \right)$. This im­plies that for $m,n = odd,even$ and/or even/odd, the result is a matrix de­ter­mi­nant that is a cone number and there­fore $\in \mathcal{X}$. While any com­bi­na­tions, m$,n = odd,odd$ and/or even/even result in an axis num­bered matrix de­ter­mi­nant, and is there­fore, $\in \mathcal{Y}$. The de­ter­mi­nants are cal­cu­lat­ed by the prod­ucts of the three en­tries in each el­e­ment $\in \mathcal{X},\mathcal{Y}$. For cone num­bered de­ter­mi­nants, these en­tries must have values that are an arrange­ment of $ACC$ or $CAA$, where $C$, de­fines a cone number, and $A$, an axis number. For ex­am­ple, the $X_{1,1}$matrix has $CAA$ for its en­tries, and its de­ter­mi­nant is

where the prod­uct $CAA$, gives a cone number result. When taking the de­ter­mi­nant prod­uct of ma­tri­ces $\in D_{3\mathbb{C}}^{\Delta}$, this re­la­tion­ship will always hold and de­ter­mines if the result is $\in \mathcal{X}$, for cone num­bers, and $\in \mathcal{Y}$, for axis num­bers. For clar­i­fi­ca­tion of these num­bers, we see in Figure 1.4.12 the arrange­ment of cone num­bers and axis num­bers on the unit circle and their sym­me­try.

For all ma­tri­ces $X_{j,k}\in \mathcal{X}$, the arrange­ment of cone num­bers and axis num­bers can be one of four pos­si­ble forms for each $X_{j,k}$:

And for the ma­tri­ces with el­e­ments $\in \mathcal{Y}$, each $Y_{j,k}$ form will also have four pos­si­ble forms:

As there are four dif­fer­ent num­bers in each group $A\left\{ 1,i, - 1, - i \right\},\ \ C\left\{ ( - 1)^{\frac{1}{4}},( - 1)^{\frac{3}{4}}, - ( - 1)^{\frac{1}{4}}, - ( - 1)^{\frac{3}{4}} \right\}$, there are $4^{3} = 64$ pos­si­ble per­mu­ta­tions for each matrix above, and given there are a total of 24 matrix forms above, there are $64 \times 24 = 1536$ el­e­ments total in the group.

The brute force por­tion of this proof now fol­lows. Since we must show that all these per­mu­ta­tion pos­si­bil­i­ties are indeed achiev­able by mul­ti­plica­tive clo­sure of the el­e­ments $X_{1,1},X_{2,1},X_{3,1}$. We start by defin­ing per­mu­ta­tion el­e­ments $\left( M_{j = 1,2,3}^{n} \right)$ in the form of $\ Y_{1,k}$ ma­tri­ces with main di­ag­o­nals en­tries

and fur­ther define the matrix el­e­ment $Y_{2} = M_{4}^{n}$, as

We start by work­ing on the $X_{1,k}$-per­mu­ta­tions for the number arrange­ments

while taking $X_{1,1}$, as

Since all $M_{j}^{n}$ have el­e­ment $a_{1,1} = 1$, all prod­ucts $M_{j}^{n}X_{1,1}$ will yield the same value $( - 1)^{\frac{1}{4}}\ ,$ for $a_{1,1}$ in the $X_{1,k}$-matrix result. To save space, we dis­play the re­sults of these prod­uct trans­for­ma­tions in a $3 \times 1$ vector no­ta­tion,

For the forms, $\begin{pmatrix} C & 0 & 0 \\ 0 & 0 & C \\ 0 & C & 0 \end{pmatrix}$, we pro­ceed to find the six­teen com­bi­na­tions for our fixed cone number, C, at $a_{1,1}$, using the fol­low­ing prod­ucts

For the forms, $\begin{pmatrix} C & 0 & 0 \\ 0 & 0 & A \\ 0 & A & 0 \end{pmatrix}$, the re­main­ing six­teen per­mu­ta­tion com­bi­na­tions for $C$ at $a_{1,1}$, we use the fol­low­ing prod­ucts

In total there are fif­teen dis­tinct per­mu­ta­tions above, where $X_{1,1}$ has the same form, $\begin{pmatrix} C & 0 & 0 \\ 0 & 0 & A \\ 0 & A & 0 \end{pmatrix}$, and the six­teenth group per­mu­ta­tion is simply our orig­i­nal $X_{1,1}$ el­e­ment

To find the re­main­ing $X_{1,k}$ per­mu­ta­tions, re­quires find­ing all forms of the 32 ma­tri­ces above, having a cone number for its $a_{1,1}$-el­e­ment. To do so, we need to iden­ti­fy the fol­low­ing per­mu­ta­tions ma­tri­ces,

having the same $2 \times 2$ bottom-right sub­ma­trix as the $X_{1,1}$-per­mu­ta­tions, and then simply apply all the same trans­for­ma­tions pre­vi­ous­ly done for the $X_{1,1}$ matrix. Find­ing per­mu­ta­tion prod­ucts to yield these ma­tri­ces re­quires a map­ping $X_{1,(2,3,4)} \rightarrow X_{1,k}$, de­not­ed in the bold font in the fol­low­ing ma­tri­ces,

By using the prod­uct com­bi­na­tions of our pre­vi­ous ma­tri­ces ${M_{1}^{m}M}_{2}^{n}$, we find so­lu­tions for these map­ping ma­tri­ces

and after ap­ply­ing the same map­pings for the first 32 per­mu­ta­tions, this brings us to a total of 128 per­mu­ta­tions for the $X_{1,k}$ el­e­ments.

The other $X_{1,k}$-el­e­ment per­mu­ta­tions, come by con­sid­er­ing the second half of the matrix number forms

for $a_{1,1}$, an axis number, and the two en­tries in the $2 \times 2$ bottom-right sub­ma­trix will al­ter­nate be­tween axis and cone num­bers. To arrive at the form above, the en­tries of the left matrix must be mapped by the row/column trans­for­ma­tions: $a_{3,2} \rightarrow a_{1,1},\ \ a_{1,1} \rightarrow a_{2,3},\ \ a_{2,3} \rightarrow a_{3,2}$, and onto the right matrix. For the matrix $X_{1,1}$, this map­ping rep­re­sents

To ac­com­plish this, we use the fol­low­ing matrix prod­ucts

To de­ter­mine the $X_{1,1}^{'}$ trans­for­ma­tions, we pro­ceed by ap­ply­ing the same $X_{1,1}$ trans­for­ma­tions to find six­teen el­e­ments of the form

by way of the fol­low­ing prod­ucts:

The re­main­ing six­teen el­e­ments of the form

are then found by the closed mul­ti­pli­ca­tion trans­for­ma­tions:

which com­prise fif­teen dif­fer­ent per­mu­ta­tions, and the six­teenth is simply our $X_{1,1}^{'}$ matrix

To arrive at the re­main­ing per­mu­ta­tions of ${X'}_{1}$, it re­quires iden­ti­fy­ing the fol­low­ing per­mu­ta­tions

having a $2 \times 2$ bottom-right sub­ma­trix equal to that found in the ${X'}_{1,1}$per­mu­ta­tion. We then we apply the same ${X'}_{1,1}$ trans­for­ma­tions, giving us the fol­low­ing ma­tri­ces de­not­ed in bold below, which map ${X'}_{1,1} \rightarrow X_{1,k}^{'}$:

By using the prod­uct com­bi­na­tions $M_{2}^{n}$, $X_{j}^{n}$, so­lu­tions are found by eval­u­at­ing the prod­ucts

With com­plete so­lu­tions for all cone group per­mu­ta­tions $X_{1,k},{X'}_{1,j} \in \mathcal{X}_{1}$, we find the values for all re­main­ing $X_{2},{X'}_{2,}X_{3},X_{3}^{'}$, per­mu­ta­tions by using the Cartan sub­al­ge­bra el­e­ment $Y_{3}$, found by using the prod­uct trans­for­ma­tion

So, the fol­low­ing trans­for­ma­tions yield $\mathcal{X}_{2},\mathcal{X}_{3}$, as fol­lows

All per­mu­ta­tions in $\mathcal{X}$ have now been found, to­tal­ing $64 \times 4 \times 3 = 768$ per­mu­ta­tions. Fi­nal­ly, to prove that we have all the per­mu­ta­tion el­e­ments $\in \mathcal{Y}_{1},\mathcal{Y}_{2},\mathcal{Y}_{3}$, we recall that

which when acting on an­oth­er per­mu­ta­tion $\in \mathcal{X}$ , will result in a per­mu­ta­tion matrix $\in \mathcal{Y}$, having uni­form­ly ro­tat­ed el­e­ments by $( - 1)^{\left( \frac{1}{4} \right)}$ . Ap­ply­ing this linear trans­for­ma­tion to the $\mathcal{X}$ per­mu­ta­tion el­e­ments, yields the $\mathcal{Y}_{1},\mathcal{Y}_{2},\mathcal{Y}_{3}$, el­e­ments

We now have all per­mu­ta­tion el­e­ments $\in \mathcal{Y}$, $4^{3} \times 4 \times 3 = 768$, making the stated the­o­rem's total of $2 \times 768 = 1536 \in \left\{ \mathcal{X},\mathcal{Y} \right\}$

$\square$

It is also im­por­tant to point out that there are many more pos­si­ble com­bi­na­tions when con­sid­er­ing matrix groups that are a com­po­si­tion of el­e­ments. An ex­am­ple would be all the pos­si­ble com­bi­na­tions of the $X_{1,k} + X_{2,k} + X_{3,k}$ el­e­ments in a Lie al­ge­bra group of the form

which can be cal­cu­lat­ed by find­ing prod­uct of all pos­si­ble per­mu­ta­tions for each $X_{1,k},X_{2,k},X_{3,k}$. The result is would then be equal to $256^{3} = 16,777,216$ dif­fer­ent pos­si­bil­i­ties! After con­sid­er­ing the di­he­dral cone group $D_{3\mathbb{C}}^{\Delta},$we have a much better idea of how many per­mu­ta­tions are pos­si­ble in com­plex spaces and, more im­por­tant­ly, we have also con­sid­ered where de­ter­mi­nants sit in com­plex space. At this junc­tion, it is im­por­tant to point out that dis­tance func­tions will tend to change as we move from one sub­group of the cone group to an­oth­er. Our next goals are to con­sid­er the sym­me­try group of this di­he­dral space, which is called the sym­met­ric/di­he­dral conic group $S_{3\mathbb{C}}^{\Delta}/D_{3\mathbb{C}}^{\Delta}$, and to expand on the other group el­e­ments more thor­ough­ly. Before defin­ing this sym­me­try group, we first con­sid­er its el­e­ments in more detail by min­i­mal slic­ing.