1.5.6 The Matrix-Matrix Exponential and Smoothing the Tangent Space

In Sec­tion 1.5.1, we ap­pre­ci­at­ed that the sym­met­ric conic group was path-con­nect­ed by way of the mul­ti­pli­ca­tion op­er­a­tion be­tween el­e­ments of the Lie al­ge­bra $\in$ $\mathfrak{s}_{3\mathbb{C}}^{\Delta}$. As the sym­met­ric conic group en­com­pass­es mul­ti­ple types of geome­tries, the effect of the mul­ti­pli­ca­tion op­er­a­tion was to move around the sphere of tan­gent spaces, en­abling the abil­i­ty to trans­form from one geo­met­ric space to an­oth­er once the Lie al­ge­bra is ex­po­nen­tial­ly mapped. The prod­uct of Lie al­ge­bra el­e­ments is quite dif­fer­ent than the prod­uct of group-like el­e­ments, but by uti­liz­ing matrix-matrix ex­po­nen­ti­a­tion, we can de­scribe a math­e­mat­i­cal rep­re­sen­ta­tion of path-con­nect­ed­ness in the tan­gent space for all the group gen­er­a­tors $\in S_{3\mathbb{C}}^{\Delta}$. This path-con­nect­ed­ness has its basis in moving in the tan­gent space by chang­ing angles $\in \mathbb{C}^{3}$, which rep­re­sents a de­par­ture from move­ment in a geo­met­ric space by ro­ta­tions and boosts, and this move­ment in the angle space can be de­fined as fol­lows:

De­f­i­n­i­tion (1.5.78) – The Matrix Group-Tan­gent Space Ex­po­nen­tial Map­ping – For normal and non-sin­gu­lar $n \times n$matrix tan­gents spaces, $\mathfrak{T}(X),\mathfrak{T}(Y)$, be­long­ing to the Lie al­ge­bra of a group $\in \mathbb{C}^{n}$, con­nect­ed in tan­gent space, and for an angle ar­gu­ment, $t\in \mathbb{R}$:

As matrix-matrix prod­ucts can be non-com­mu­ta­tive, we see that there is a dis­tinc­tion be­tween the prod­uct order of the ex­po­nen­tial, which also ap­plies for the matrix group-tan­gent space ex­po­nen­tial. To remain con­sis­tent, we will tend to uti­lize the right-sided matrix group-tan­gent space ex­po­nen­tial $\left( e^{\ t\mathfrak{T}(X)} \right)^{\mathfrak{T}(Y)}$. Start­ing with the simple ex­am­ple of the $2 \times 2$ hy­per­bol­ic sub­group, with a tan­gent space Lie al­ge­bra el­e­ment

for the angle ar­gu­ment$\ t$, the matrix ex­po­nen­tial yields the group gen­er­a­tor form

This group is path-con­nect­ed in tan­gent space with the com­plex unit circle group, and we can move in the angle space by the matrix group-tan­gent space ex­po­nen­tial

for the tan­gent space $\mathfrak{T}(Y)$ de­fined, as

and

where we arrive at the com­plex unit circle group, ac­com­plished by moving in the tan­gent space. It may not be read­i­ly ob­vi­ous yet, but the result above comes from simply moving in the space of angles, syn­ony­mous with the phase space. If we assign a vari­able $e^{i\theta}t$, for $\theta,t\in \mathbb{R}$, the same trans­for­ma­tion is ac­com­plished by set­ting $\theta = \frac{\pi}{2}$ , re­sult­ing in $e^{i\theta}t = it$, with which we can ex­press

Mind­ful of the trigono­met­ric re­la­tions, $\cosh(it) = \cos(t),\sinh(it) = isin(t)$, we arrive at the same result

To con­sid­er the above ex­am­ples in the con­text of tan­gent spaces and path-con­nect­ed­ness, the Lie al­ge­bra of the sym­met­ric group must be de­fined with a min­i­mal el­e­ment con­sis­tent with the choice of $\theta$, for $e^{i\theta}t.$So, for the ex­am­ple above to be con­sis­tent with tan­gent spaces and Lie theory, the min­i­mal el­e­ment must be $\frac{\pi}{2}$ ra­di­ans or small­er, and the sym­met­ric form of this group would have 32 el­e­ments com­pris­ing its Lie al­ge­bra $\mathfrak{g}$. This cal­cu­la­tion is based upon two forms of el­e­ments $\mathcal{Y},\mathcal{X} \in\mathfrak{g}$

for the en­tries $a,b,c,d$, taking any value from the set $\left\{ 1,i, - 1, - i \right\},$mean­ing there are four pos­si­ble choic­es for any entry, $\mathcal{Y},\mathcal{X} \in\mathfrak{g}$, and there­fore, the el­e­ment total is $2^{2} \times 2^{2} \times 2 = 32.$

The­o­rem (1.5.87) – For finite groups and their square $n \times n$ matrix el­e­ments, with one entry for each row/column (finite), and $n$-number of matrix en­tries for each el­e­ment, the number of total el­e­ments in the sym­met­ric form of the group is

For $p$, the number of dis­tinct el­e­ment types in the finite group, $\mathrm{K}_{\mathcal{me}}$, the number of unique values in the min­i­mal el­e­ment basis set on the com­plex unit circle with a wind­ing number con­sid­ered, and $\begin{pmatrix} \mathrm{K}_{\mathcal{me}} \\ 1 \end{pmatrix}$, a bi­no­mi­al co­ef­fi­cient.

Proof – Finite group el­e­ments have only one entry per row/column com­bi­na­tion, there­fore, for a square $n \times n$ matrix el­e­ment there are $n$ en­tries. Each $n$ can take on any value that is a mul­ti­ple of the min­i­mal el­e­ment up to $2\pi$, for the min­i­mal el­e­ment de­fined as the clos­est el­e­ment coun­ter­clock­wise to the real-axis on the com­plex unit circle. As such, the number of el­e­ments in $\mathrm{K}_{\mathcal{me}}$ is the cir­cum­fer­ence of the unit circle di­vid­ed by the min­i­mal el­e­ment $\frac{2\pi}{\mathcal{me}} = \mathrm{K}_{\mathcal{me}}.$ As there are $n$ en­tries in each matrix el­e­ment, there are $\begin{pmatrix} \mathrm{K}_{\mathcal{me}} \\ 1 \end{pmatrix}^{n}$ dis­tinct matrix el­e­ments for each el­e­ment type, where two of the same valued-en­tries in an el­e­ment are per­mis­si­ble, and mul­ti­ply­ing this by the number of el­e­ment types $p$, gives the total number of el­e­ments in the sym­met­ric form of the finite group con­sid­ered.

$\square$

To clar­i­fy how the wind­ing number re­lates to the min­i­mal el­e­ment, we con­sid­er how many trips around the unit circle are re­quired to return to the start­ing point, en­sur­ing that we land on the iden­ti­ty el­e­ment $\epsilon$, on the real-axis. So, this is some­what dif­fer­ent from wind­ing num­bers of con­tin­u­ous func­tions on a closed curve, since the points can skip dis­tances on the unit circle for each it­er­a­tion. The wind­ing number $k\in \mathbb{Z}$, then de­fines the number of ro­ta­tions around the unit circle re­quired by way of the it­er­a­tive process of suc­ces­sive powers re­quired for the so­lu­tion $\left( e^{i\alpha} \right)^{\mathrm{K}_{\mathcal{me}}} = e^{i\alpha}$.

For the ex­am­ple, $e^{\left( \frac{5i\pi}{7} \right)}$, mul­ti­ple ro­ta­tions around the unit circle are re­quired before re­turn­ing to the start­ing point. The penul­ti­mate it­er­a­tion equals the iden­ti­ty el­e­ment, oc­cur­ring one it­er­a­tion before its it­er­a­tive power number, $\mathrm{K}_{\mathcal{me}} - 1$. This is be­cause the iden­ti­ty el­e­ment occurs at $e^{2i\pi k}$, for $k\in \mathbb{Z}$, equal­ing $\epsilon$ for those $k$ values. For the ex­am­ple, $e^{\left( \frac{5i\pi}{7} \right)} \times \epsilon = e^{\left( \frac{5i\pi}{7} \right)}$, which holds for any choice, $e^{2i\pi\left( \frac{\mathcal{n}}{\mathcal{d}} \right)}$, for $\frac{\mathcal{n}}{\mathcal{d}}$ $\in \mathbb{R},and\ \mathcal{n},\mathcal{d}\in \mathbb{Z}$. We con­sid­er the wind­ing number for $e^{\left( \frac{5i\pi}{7} \right)} = e^{2i\pi\left( \frac{5}{14} \right)}$, as

for $\mathrm{K}_{\mathcal{me}}$, rep­re­sent­ing the ex­po­nen­tial term that gives the first in­te­ger so­lu­tion for $\frac{2\pi k\mathcal{d}}{2\pi\mathcal{n}} = \epsilon$, which is also the value in the bi­no­mi­al co­ef­fi­cient $\begin{pmatrix} \mathrm{K}_{\mathcal{me}} \\ 1 \end{pmatrix}$. This can be de­rived by start­ing with

and by taking the log­a­rithm of both sides. We find for the left-hand side

So, for $e^{2i\pi\left( \frac{5}{14} \right)},$we have $\mathcal{d} = 14$, $\mathcal{n} = 5$, where the above equa­tion has the $\mathrm{K}_{\mathcal{me}}$ equiv­a­lence

and since this is a re­duced ra­tio­nal, we find by set­ting $k = 5$, the first in­te­ger so­lu­tion for $\mathrm{K}_{\mathcal{me}}$, is

So, the wind­ing number is $k = 5$, rep­re­sent­ing the first in­te­ger so­lu­tion, oc­cur­ring for the it­er­a­tive power $\mathrm{K}_{\mathcal{me}} = 14$. This it­er­a­tive process also rep­re­sents the number of unique points on the com­plex unit circle that $\left( e^{2i\pi\left( \frac{\mathcal{n}}{\mathcal{d}} \right)} \right)^{\mathrm{K}_{\mathcal{me}}}$ gen­er­ates. This is stated, as

We can apply the number of el­e­ments theory to the finite sym­me­try groups by using the case for the $D_{3}/S_{3}$ group. The group has the rep­re­sen­ta­tive min­i­mal el­e­ment $2\pi$, which is easily solved to find a wind­ing number of $k = 1$, in­di­cat­ing the group gen­er­ates only one unique point on the unit circle, given $\mathrm{K}_{\mathcal{me}} = 1$

Since there are six dis­tinct el­e­ments $\in D_{3}/S_{3}$, formed by $p = 6$ dis­tinct ma­tri­ces, given by $(n^{2} - n)$, for $n = 3$, we find a total of six el­e­ments are gen­er­at­ed on the unit circle by $D_{3}/S_{3}$,

Since $\mathcal{d} =\frac{{\mathcal{n}\mathrm{K}}_{\mathcal{me}}}{k}$, we can sub­sti­tute this into our ex­po­nen­tial

to con­sid­er min­i­mal el­e­ments ap­proach­ing zero in size, by having $\mathrm{K}_{\mathcal{me}}$ ap­proach in­fin­i­ty, or geo­met­ri­cal­ly speak­ing, by having the min­i­mal el­e­ment ap­proach the pos­i­tive real-axis. This can be de­fined by the limit of the above ex­po­nen­tial re­la­tion as $\mathrm{K}_{\mathcal{me}} \rightarrow \infty$

which for in­fin­i­tes­i­mal min­i­mal el­e­ments, shows that any wind­ing number $k$ gen­er­ates the iden­ti­ty el­e­ment at $e^{2i\pi k}.$As ex­pect­ed, for the wind­ing number $k \in \mathbb{R}^{\pm \ }$, as the min­i­mal el­e­ment ap­proach­es the real-axis, the value ap­proach­es $\pm \infty$ for $\mathrm{K}_{\mathcal{me}}$

and the sign of $\infty$ de­pends on the sign of $\mathcal{n}$. By the group de­f­i­n­i­tion, its el­e­ments are closed under its de­fined op­er­a­tion(s), and for the Lie al­ge­bra $\mathfrak{s}_{3\mathbb{C}}^{\Delta}$, the op­er­a­tion of mul­ti­pli­ca­tion of its Lie al­ge­bra el­e­ments re­sults in an­oth­er Lie al­ge­bra el­e­ment $\in \mathfrak{s}_{3\mathbb{C}}^{\Delta}.$Mind­ful of this, we can state the fol­low­ing the­o­rem as it ap­plies to the matrix group-tan­gent space ex­po­nen­tial:

The­o­rem (1.5.99) – For a finite sym­me­try group $G$, with a basis of a min­i­mal el­e­ments re­lat­ing to the number of matrix el­e­ments $\in \mathfrak{g}$, and for el­e­ments of the sym­me­try group $\mathfrak{T}(X),\mathfrak{T}(Y)$, which are normal and non-sin­gu­lar tan­gent spaces of $G$, for $e^{\mathfrak{T}(X)}$, the matrix ex­po­nen­tial of $\mathfrak{T}(X)$. The matrix group-tan­gent space ex­po­nen­tial re­sults in a matrix ex­po­nen­tial $\in G$, of a tan­gent space $\mathfrak{T}\left( Z_{1,2} \right)\in \mathfrak{g}$ , for both as fol­lows

Proof – As we are work­ing in a formal sym­me­try group space, we are guar­an­teed path-con­nect­ed­ness for the given prod­ucts of any two tan­gent spaces $\mathfrak{T}(X),\mathfrak{T}(Y)$ $\in \mathfrak{g}$, which yield an­oth­er tan­gent space of the group $\in \mathfrak{g}.$ As the sym­me­try group fits the cri­te­ria for a group, all el­e­ments are bi­jec­tive and will be closed under the com­po­si­tions $\mathfrak{T}(X)\circ \mathfrak{T}(Y)$ and $\mathfrak{T}(Y)\circ \mathfrak{T}(X)$ $\forall$ el­e­ments $\in \mathfrak{g}$. As all tan­gent spaces of $G$ are in the neigh­bor­hood of $\epsilon$, the fol­low­ing hold

As the com­po­si­tions $\mathfrak{T}(X)\circ \mathfrak{T}(Y),\mathfrak{T}(Y)\circ \mathfrak{T}(X)$ are com­prised of el­e­ments $\in \mathfrak{g}$ and bi­jec­tive, this guar­an­tees that both com­po­si­tions yield tan­gent space el­e­ments $\in \mathfrak{g}$. Since, $\mathfrak{T}\left( Z_{1} \right)$ and $\mathfrak{T}\left( Z_{2} \right)$ are el­e­ments $\in \mathfrak{g}$, we know that $e^{\mathfrak{T}\left( Z_{1} \right)},e^{\mathfrak{T}\left( Z_{2} \right)}$ will be $\in G$, given they rep­re­sent matrix ex­po­nen­tials formed by el­e­ments of the Lie al­ge­bra $\mathfrak{g}$.

$\square$

The­o­rem (1.5.100) – For the finite sym­me­try group $S_{3\mathbb{C}}^{\Delta}$ with the min­i­mal el­e­ment basis $e^{\left( \frac{i\pi}{4} \right)}$, any two el­e­ments $\mathfrak{T}(X),\mathfrak{T}(Y)$ $\in \mathfrak{s}_{3\mathbb{C}}^{conic}$are normal and non-sin­gu­lar tan­gent spaces of $S_{3\mathbb{C}}^{\Delta}$, for $e^{t\mathfrak{T}(X)}$, the matrix ex­po­nen­tial of the tan­gent space t$\mathfrak{T}(X)$, and for $t$, an angle ar­gu­ment $\in \mathbb{R}.$ The matrix group-tan­gent space ex­po­nen­tial yields a matrix ex­po­nen­tial $\in S_{3\mathbb{C}}^{\Delta}$, for both

where the fol­low­ing hold

Proof – The group, $S_{3\mathbb{C}}^{\Delta},$ is a sym­me­try group with all matrix el­e­ments de­fined as normal and non-sin­gu­lar. There­fore, the left and right matrix group-tan­gent space ex­po­nen­tials hold true by the pre­vi­ous proof. For el­e­ments of the $\mathcal{X}$ type $\in \mathfrak{s}_{3\mathbb{C}}^{\Delta}$, we use the ex­am­ple $t\mathcal{X}_{1}$, with the matrix en­tries $a,b,c$ each taking any value from the set $\left\{ \pm 1, \pm i, \pm ( - 1)^{\left( \frac{1}{4} \right)}, \pm ( - 1)^{\left( \frac{3}{4} \right)} \right\}$ to form its min­i­mal el­e­ment basis. We define $t\mathcal{X}_{1}$,

whose matrix ex­po­nen­tial $e^{t\mathcal{X}_{1}}$ is straight­for­ward

We cal­cu­late $\left. \ e^{t\mathcal{X}_{1}} \right|_{t = 0}$

and for $\frac{d}{dt}\left( e^{t\mathcal{X}_{1}} \right)$, we find the matrix result

To cal­cu­late the tan­gent space in the neigh­bor­hood of the iden­ti­ty, let $t = 0$, and the above be­comes

which is equiv­a­lent to our orig­i­nal matrix el­e­ment $\mathcal{X}_{1}$. As the $\mathcal{X}_{2},\mathcal{X}_{3}$ matrix el­e­ments are anal­o­gous to the $\mathcal{X}_{1}$ matrix el­e­ment, the same holds true for them. Every matrix el­e­ment $\in \mathfrak{s}_{3\mathbb{C}}^{\Delta}$is a tan­gent space of the $S_{3\mathbb{C}}^{\Delta}$ group. There­fore, the fol­low­ing is true for the $\mathcal{X}$-type el­e­ments

We know that all the matrix el­e­ments $\in \mathfrak{s}_{3\mathbb{C}}^{\Delta}$ are normal, and we define the matrix el­e­ment $t\mathcal{Y}_{2}$, like ${t\mathcal{X}}_{1}$, as

We can use the spec­tral the­o­rem to take ad­van­tage of the di­ag­o­nal­iza­tion process to find the eigen­val­ue $(D)$ and eigen­vec­tor $(V)$ma­tri­ces of ${t\mathcal{Y}}_{2}$, as

noting for $V^{- 1}$, that

Taking the matrix ex­po­nen­tial of the di­ag­o­nal­ized $D$ matrix is straight­for­ward, where $e^{D}$ equals

We com­pute the grou­p­like form $e^{t\mathcal{Y}_{2}}$

and arrange the re­sults by the column vec­tors $e^{t\mathcal{Y}_{2}}(j,1,2,3)$:

For the co­ef­fi­cients of the ex­po­nen­tials in the $e^{t\mathcal{Y}_{2}}$ matrix above, we see a sim­i­lar pat­tern for the non-di­ag­o­nal el­e­ments, for $k_{j}$, the co­ef­fi­cient out­side of the paren­the­sis

and it is read­i­ly seen that for $t = 0$, $e^{t{(0)\mathcal{Y}}_{2}}$ will have values of zero for all non-di­ag­o­nal en­tries, and have di­ag­o­nal en­tries equal to

which are equal to $1$, and there­fore,

Since$\ \mathcal{Y}_{3}$ is simply the trans­pose of $\mathcal{Y}_{2}$, we would find sim­i­lar re­sults. The matrix el­e­ment ${t\mathcal{Y}}_{1}$ is simply a di­ag­o­nal matrix that will have the iden­ti­ty as its eigen­vec­tor matrix $I$, and eigen­val­ues $ta,tb,tc$ on the main di­ag­o­nal for $D$. When ex­po­nen­ti­at­ed, $e^{t\mathcal{Y}_{1}} = Ve^{D}V^{- 1}$ is simply the ex­po­nen­ti­at­ed eigen­val­ue matrix

and it is easily ver­i­fied that $\left. \ e^{t\mathcal{Y}_{1}} \right|_{t = 0} = \epsilon$. As each matrix el­e­ment $\in \mathfrak{s}_{3\mathbb{C}}^{\Delta}$ is a tan­gent space of the group $S_{3\mathbb{C}}^{\Delta}$. For $\mathcal{Y}$-type el­e­ments the fol­low­ing is also true as it was for the $\mathcal{X}$ el­e­ments

Below we cal­cu­late $\frac{d}{dt}{(e}^{t\mathcal{Y}})$, ar­rang­ing them by the column vec­tors ${\frac{d}{dt}(e}^{t\mathcal{Y}}(j,1,2,3))$:

For $t = 0,$ all the co­ef­fi­cients which are not in $\mathcal{Y}_{2}$-en­tries cancel and as a result, $\frac{d}{dt}\left( \left. \ e^{t\mathcal{Y}_{2}} \right|_{t = 0} \right) = \mathcal{Y}_{2}$

As $\mathcal{Y}_{3}$ is simply the trans­pose of $\mathcal{Y}_{2}$, we would find a sim­i­lar result, and once again, since ${t\mathcal{Y}}_{1}$ is simply a di­ag­o­nal matrix, it will have the iden­ti­ty as its eigen­vec­tor matrix $I$, eigen­val­ues of $ta,tb,tc$ on the main di­ag­o­nal for D, and for $\frac{d}{dt}\left( \left. \ e^{t\mathcal{Y}_{2}} \right|_{t = 0} \right)$, it is straight­for­ward that

As each matrix el­e­ment $\in \mathfrak{s}_{3\mathbb{C}}^{\Delta}$ is a tan­gent space of the group $S_{3\mathbb{C}}^{\Delta},$ for the $\mathcal{Y}$-type el­e­ments, the fol­low­ing is true just as it was for the $\mathcal{X}$ el­e­ments

There­fore, we can glob­al­ly state that for all tan­gent space el­e­ments of the Lie al­ge­bra $\mathfrak{s}_{3\mathbb{C}}^{\Delta}$, where we sub­sti­tute $Z$ with the entire Lie al­ge­bra $\mathfrak{s}_{3\mathbb{C}}^{\Delta}$ to em­pha­size this for all tan­gent spaces, that the fol­low­ing are true

The matrix group-tan­gent space ex­po­nen­tial gives us a way to math­e­mat­i­cal­ly use the prod­uct com­po­si­tion of a group gen­er­a­tor with a group tan­gent space, to arrive at an­oth­er group gen­er­a­tor $\in S_{3\mathbb{C}}^{\Delta}$. In this sec­tion we learned that ap­ply­ing the ex­po­nen­tial map­ping to el­e­ments in the tan­gent space is con­sis­tent with chang­ing the lo­ca­tion of a matrix group's func­tions on the unit circle, ac­com­plished by chang­ing each angle to an­oth­er in the min­i­mal el­e­ment basis. For the $S_{3\mathbb{C}}^{\Delta}$ group, this min­i­mal el­e­ment basis is any angle co­ef­fi­cient taking the form $t \times \left\{ \pm 1, \pm i, \pm ( - 1)^{\left( \frac{1}{4} \right)}, \pm ( - 1)^{\left( \frac{3}{4} \right)} \right\},$for $t\in \mathbb{R}$, the angle vari­able, and is taken as a prod­uct with a member of the min­i­mal el­e­ment basis set. This should be read­i­ly ap­par­ent to the reader for the matrix ex­po­nen­tial of the $X_{1}$-type tan­gent space, for the map­ping $e^{t\mathfrak{T}\left( X_{1} \right)}$

Above, we ef­fec­tive­ly have a matrix ex­po­nen­tial tem­plate that will hold for all angle co­ef­fi­cients from the min­i­mal el­e­ment basis. If we were to choose angle co­ef­fi­cients $a = i,b = ( - 1)^{\left( \frac{1}{4} \right)},c = ( - 1)^{\left( \frac{3}{4} \right)}$, where­by plug­ging and re­ar­rang­ing these into the matrix tem­plate, we find for $e^{t\mathfrak{T}\left( X_{1} \right)}$

Since $\sqrt{( - 1)^{\left( \frac{1}{4} \right)} \times ( - 1)^{\left( \frac{3}{4} \right)}} = \sqrt{- 1} = i$, $e^{t\mathfrak{T}\left( X_{1} \right)}$ be­comes

and mind­ful that $\cosh(it) = \cos(t),\sinh(it) = isin(t)$, the group above sim­pli­fies to

Taking the matrix log­a­rithm of $e^{t\mathfrak{T}\left( X_{1} \right)}$, ver­i­fies that we have the ex­pect­ed tan­gent space with the same angle co­ef­fi­cients that were ini­tial­ly se­lect­ed

The case for choos­ing angle co­ef­fi­cients that are equal for $b,c$, $\in e^{t\mathfrak{T}\left( X_{1} \right)}$, de­not­ed by $bc = d^{2}:d \in \mathcal{me}$, the process is much sim­pler and in­volves just chang­ing the angle co­ef­fi­cient for each trigono­met­ric func­tion

which means that we can simply change the angle co­ef­fi­cient di­rect­ly in the group and arrive at a dif­fer­ent matrix group in the tan­gent space of func­tions!