Masteries · Jason GlowneyArithmoí Book 1

Masteries

1.5.8 The Infinite Symmetry Group and the Subtangent Space

In the last sec­tion, we saw that by taking the limit of the min­i­mal el­e­ment in­fin­i­tes­i­mal­ly close to the iden­ti­ty el­e­ment, we were able to smooth the min­i­mal el­e­ment yield­ing a con­tin­u­ous func­tion space that was in­fi­nite­ly dif­fer­en­tiable. All the di­he­dral groups that we have de­scribed in this book have gen­er­al­ized per­mu­ta­tions in six forms, which we arrange as the fol­low­ing two sub­sets 𝒳k\mathcal{X}_{k}, 𝒴k\mathcal{Y}_{k}:

𝒳k={(a0000b0c0),   (00a0b0c00),   (0a0b0000c)},\mathcal{X}_{k} = \left\{ \begin{pmatrix} a & 0 & 0 \\ 0 & 0 & b \\ 0 & c & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & a \\ 0 & b & 0 \\ c & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & a & 0 \\ b & 0 & 0 \\ 0 & 0 & c \end{pmatrix} \right\},
𝒴k={(a000b000c),   (0a000bc00),   (00ab000c0)}.\begin{array}{r} \mathcal{Y}_{k} = \left\{ \begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix},\ \ \ \begin{pmatrix} 0 & a & 0 \\ 0 & 0 & b \\ c & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & a \\ b & 0 & 0 \\ 0 & c & 0 \end{pmatrix} \right\}. \end{array}
(1.5.125)

Up until this point, the matrix en­tries a,b,ca,b,c were from the fol­low­ing number set {±1,±i,±(−1)(14),±(−1)(34)}\left\{ \pm 1, \pm i, \pm ( - 1)^{\left( \frac{1}{4} \right)}, \pm ( - 1)^{\left( \frac{3}{4} \right)} \right\}, having its basis de­ter­mined by the min­i­mal el­e­ment choice (−1)(14)( - 1)^{\left( \frac{1}{4} \right)}. For a min­i­mal el­e­ment taken as the in­fin­i­tes­i­mal of the C∞C^{\infty} func­tion, de­scribed in the last sec­tion, the matrix en­tries a,b,ca,b,c can locate on any point on the unit circle for spe­cif­ic values in the in­fi­nite set U(1)U(1). We can recast the el­e­ment sub­sets with the U(1):C∞U(1):C^{\infty} func­tion in mind, where the set A∞={𝒳k∞,𝒴k∞}A^{\infty} = \left\{ \mathcal{X}_{k}^{\infty},\mathcal{Y}_{k}^{\infty} \right\} is com­posed of the fol­low­ing sub­sets

𝒳k∞={(eia0000eib0eic0),(00eia0eib0eic00),(0eia0eib0000eic)},\mathcal{X}_{k}^{\infty} = \left\{ \begin{pmatrix} e^{ia} & 0 & 0 \\ 0 & 0 & e^{ib} \\ 0 & e^{ic} & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 & e^{ia} \\ 0 & e^{ib} & 0 \\ e^{ic} & 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & e^{ia} & 0 \\ e^{ib} & 0 & 0 \\ 0 & 0 & e^{ic} \end{pmatrix} \right\},
𝒴k∞={(eia000eib000eic),(0eia000eibeic00),(00eiaeib000eic0)}.\begin{array}{r} \mathcal{Y}_{k}^{\infty} = \left\{ \begin{pmatrix} e^{ia} & 0 & 0 \\ 0 & e^{ib} & 0 \\ 0 & 0 & e^{ic} \end{pmatrix},\begin{pmatrix} 0 & e^{ia} & 0 \\ 0 & 0 & e^{ib} \\ e^{ic} & 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 & e^{ia} \\ e^{ib} & 0 & 0 \\ 0 & e^{ic} & 0 \end{pmatrix} \right\}. \end{array}
(1.5.126)

We can see from the above that the prod­ucts of any el­e­ments from these sub­sets would yield an­oth­er el­e­ment with all en­tries ∈U(1)\in U(1), for a,b,ca,b,c ∈ℝ\in \mathbb{R}. All these el­e­ments can be fur­ther con­sid­ered as the in­fi­nite tan­gent space of the S3ℂS_{3\mathbb{C}} group ∈ℂ3\in \mathbb{C}^{3}, de­not­ed as

𝔗(S3ℂ∞),\mathfrak{T}\left( S_{3\mathbb{C}}^{\infty} \right),

and this in­fi­nite tan­gent space is an in­fi­nite Lie al­ge­bra as well. Con­cern­ing this in­fi­nite tan­gent space, we can now define the in­fi­nite sym­me­try group S3ℂ∞S_{3\mathbb{C}}^{\infty}:

De­f­i­n­i­tion (1.5.127) – The uni­tary In­fi­nite Sym­me­try Group 𝐒𝟑ℂ∞\mathbf{S}_{\mathbf{3}\mathbb{C}}^{\mathbf{\infty}} is a group under com­po­si­tion sit­ting in the com­plex geo­met­ric space ℂ3×3\mathbb{C}^{3 \times 3}, which is com­prised of C∞C^{\infty} func­tions for its in­fin­i­tes­i­mal min­i­mal el­e­ment basis ∈ℂ3,\in \mathbb{C}^{3}, yield­ing the in­fi­nite set A∞A^{\infty} of uni­tary el­e­ments in tan­gent space.

S3ℂ∞ ={A∞∈GL(3,ℂ)|A∞∙A∞†=I }.\begin{array}{r} S_{3\mathbb{C}}^{\infty}\ = \left\{ A^{\infty} \in GL\left( 3,\mathbb{C} \right)|A^{\infty} \bullet {A^{\infty}}^{\dagger} = I\ \right\}. \end{array}

The­o­rem (1.5.128) – For A∞A^{\infty}as a set, the set of per­mu­ta­tions ∈S3ℂ∞\in S_{3\mathbb{C}}^{\infty} forms a group under com­po­si­tion

Proof – Ver­i­fi­ca­tion of group axioms

  • Clo­sure: All el­e­ments ∈A∞\in A^{\infty}are bi­jec­tive, and the com­po­si­tion of any el­e­ments ϕ∘ψ ∈A∞\phi \circ \psi\ \in A^{\infty} is also bi­jec­tive.
  • As­so­cia­tiv­i­ty: Per­mu­ta­tions are func­tions, and their com­po­si­tions are as­so­cia­tive.
  • Iden­ti­ty: The iden­ti­ty func­tion ϵA∞\epsilon_{A^{\infty}} bi­jec­tive­ly maps all el­e­ments of the set A∞A^{\infty}to them­selves (ϵA∞:A∞→A∞)\left( \epsilon_{A^{\infty}}:A^{\infty} \rightarrow A^{\infty} \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

When con­sid­er­ing the in­fi­nite tan­gent space of el­e­ments and how the el­e­ments are com­posed of C∞C^{\infty}, we can math­e­mat­i­cal­ly con­sid­er the ex­is­tence of a sub­tan­gent space to the tan­gent space, de­fined as:

De­f­i­n­i­tion (1.5.129) – The Sub­tan­gent Space 𝒯(𝔰𝟑ℂ∞)\mathcal{T}\left( \mathfrak{s}_{\mathbf{3}\mathbb{C}}^{\mathbf{\infty}} \right) is the matrix log­a­rithm of the in­fi­nite tan­gent space, 𝔗(S3ℂ∞)\mathfrak{T}\left( S_{3\mathbb{C}}^{\infty} \right),

𝒯(𝔗(S3ℂ∞))=ln⁡(𝔗(S3ℂ∞))=𝒯(𝔰3ℂ∞).\mathcal{T}\left( \mathfrak{T}\left( S_{3\mathbb{C}}^{\infty} \right) \right) = \ln{\left( \mathfrak{T}\left( S_{3\mathbb{C}}^{\infty} \right) \right)= \mathcal{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right)}.

In the de­f­i­n­i­tion above, the sub­tan­gent space, when paired down, op­er­ates as the matrix log­a­rithm of the Lie al­ge­bra el­e­ments ∈𝔰3ℂ∞\in \mathfrak{s}_{3\mathbb{C}}^{\infty}, where­as the tan­gent space op­er­ates as the matrix log­a­rithm of the Lie group S3ℂ∞.S_{3\mathbb{C}}^{\infty}.It turns out that the sub­tan­gent space is very con­sis­tent with what we al­ready know about Lie al­ge­bras in gen­er­al, where many of the same prop­er­ties hold in the sub­tan­gent space, and this opens up many pos­si­bil­i­ties about how we may go about con­sid­er­ing group in­ter­re­la­tion­ships. We start by first defin­ing the iden­ti­ty group as the S3ϵS_{3}^{\epsilon} sym­me­try group:

De­f­i­n­i­tion (1.5.130) – The Iden­ti­ty Group 𝐒𝟑𝛜\mathbf{S}_{\mathbf{3}}^{\mathbf{\epsilon}}, of the in­fin­i­ty sym­me­try group S3ℂ∞S_{3\mathbb{C}}^{\infty} is the set of el­e­ments Aϵ,A^{\epsilon}, com­pris­ing the S3ϵS_{3}^{\epsilon} sym­me­try group

Aϵ={(100001010),   (001010100),   (010100001),(100010001),   (010001100),   (001100010)}.\begin{array}{r} A^{\epsilon} = \left\{ \begin{array}{r} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \\ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \end{array} \right\}. \end{array}

We take the tan­gent space 𝔗(u×𝔰3ℂ∞)\mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right) to be the fol­low­ing set of in­fin­i­ty el­e­ments

𝔗(u×𝔰3ℂ∞)={(eu(ia)0000eu(ib)0eu(ic)0),   (00eu(ia)0eu(ib)0eu(ic)00),   (0eu(ia)0eu(ib)0000eu(ic)),(eu(ia)000eu(ib)000eu(ic)),   (0eu(ia)000eu(ib)eu(ic)00),   (00eu(ia)eu(ib)000eu(ic)0)},\begin{array}{r} \mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right) = \left\{ \begin{array}{r} \begin{pmatrix} e^{u(ia)} & 0 & 0 \\ 0 & 0 & e^{u(ib)} \\ 0 & e^{u(ic)} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & e^{u(ia)} \\ 0 & e^{u(ib)} & 0 \\ e^{u(ic)} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & e^{u(ia)} & 0 \\ e^{u(ib)} & 0 & 0 \\ 0 & 0 & e^{u(ic)} \end{pmatrix}, \\ \begin{pmatrix} e^{u(ia)} & 0 & 0 \\ 0 & e^{u(ib)} & 0 \\ 0 & 0 & e^{u(ic)} \end{pmatrix},\ \ \ \begin{pmatrix} 0 & e^{u(ia)} & 0 \\ 0 & 0 & e^{u(ib)} \\ e^{u(ic)} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & e^{u(ia)} \\ e^{u(ib)} & 0 & 0 \\ 0 & e^{u(ic)} & 0 \end{pmatrix} \end{array} \right\}, \end{array}
(1.5.131)

to define the matrix log­a­rithm ln⁡(𝔗(u×𝔰3ℂ∞))\ln\left( \mathfrak{T}\left( {u \times \mathfrak{s}}_{3\mathbb{C}}^{\infty} \right) \right), as 𝒯(𝔰3ℂ∞)\mathcal{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right)

𝒯(𝔰3ℂ∞)={(ia0000ib0ic0),   (00ia0ib0ic00),   (0ia0ib0000ic),(ia000ib000ic),   (0ia000ibic00),   (00iaib000ic0)},\begin{array}{r} \mathcal{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right) = \left\{ \begin{array}{r} \begin{pmatrix} ia & 0 & 0 \\ 0 & 0 & ib \\ 0 & ic & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & ia \\ 0 & ib & 0 \\ ic & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & ia & 0 \\ ib & 0 & 0 \\ 0 & 0 & ic \end{pmatrix}, \\ \begin{pmatrix} ia & 0 & 0 \\ 0 & ib & 0 \\ 0 & 0 & ic \end{pmatrix},\ \ \ \begin{pmatrix} 0 & ia & 0 \\ 0 & 0 & ib \\ ic & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & ia \\ ib & 0 & 0 \\ 0 & ic & 0 \end{pmatrix} \end{array} \right\}, \end{array}
(1.5.132)

then state the fol­low­ing the­o­rem:

The­o­rem (1.5.133) – The dif­fer­en­tial sub­tan­gent space of S3ℂ∞S_{3\mathbb{C}}^{\infty}, de­not­ed by 𝒯(𝔰3ℂ∞)\mathcal{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right), con­sists of the com­plex 3×33 \times 3 ma­tri­ces which rep­re­sent the dif­fer­en­tial angle space of the tan­gent space, for all 𝔗(u×𝔰3ℂ∞)\mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right), as equal to an el­e­ment of the iden­ti­ty group S3ϵS_{3}^{\epsilon}, for u=0u = 0, and the de­riv­a­tive of the tan­gent space u𝔗(𝔰3ℂ∞)u\mathfrak{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right), equals the sub­tan­gent space 𝒯(𝔰3ℂ∞)\ \mathcal{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right), in the neigh­bor­hood of the iden­ti­ty group and its el­e­ments S3ϵS_{3}^{\epsilon} , for u=0u = 0.

 𝔗(u×𝔰3ℂ∞)|u=0=S3ϵ,\left. \ \mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right) \right|_{u = 0} = S_{3}^{\epsilon},
𝒯(𝔗(S3ℂ∞))=ln⁡(ddu( e𝔗(u×𝔰3ℂ∞)|u=0))=ddu( 𝔗(u×𝔰3ℂ∞)|u=0)=𝒯(𝔰3ℂ∞).\mathcal{T}\left( \mathfrak{T}\left( S_{3\mathbb{C}}^{\infty} \right) \right) = \ln\left( \frac{d}{du}\left( \left. \ e^{\mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right)} \right|_{u = 0} \right) \right) = \frac{d}{du}\left( \left. \ \mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right) \right|_{u = 0} \right) = \mathcal{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right).

Proof – The el­e­ments of the tan­gent space 𝔗(u×𝔰3ℂ∞)\mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right) have three matrix en­tries in the form eu(iθk)e^{u\left( i\theta_{k} \right)}, re­sult­ing from the ex­po­nen­ti­a­tion of the in­di­vid­ual matrix en­tries and not the matrix ex­po­nen­tial. As such, the matrix dif­fer­en­tial will take the form com­prised of el­e­ments iθki\theta_{k}, for u=0,u = 0, rep­re­sent­ing the angle ar­gu­ments for each ex­po­nen­tial entry ∈𝔗(u×𝔰3ℂ∞)\in \mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right). These angle ar­gu­ments, arranged in their re­spec­tive ma­tri­ces, have been de­fined as the sub­tan­gent space 𝒯(𝔰3ℂ∞)\mathcal{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right), and there­fore, dd𝓊( 𝔗(u×𝔰3ℂ∞)|u=0)=𝒯(𝔰3ℂ∞).\frac{d}{d\mathcal{u}}\left( \left. \ \mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right) \right|_{u = 0} \right) = \mathcal{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right). For any ex­po­nen­tial entry in a tan­gent space matrix of the form eu(iθk)e^{u\left( i\theta_{k} \right)}, for u=0u = 0, we find that eu(iθk)=1e^{u\left( i\theta_{k} \right)} = 1, and all ma­tri­ces take the form of an el­e­ment in the iden­ti­ty group S3ϵS_{3}^{\epsilon}.

□\square

Mind­ful as we con­tin­ue, stated in the the­o­rem is that the sub­tan­gent space basis is made up of the space of angles from the group's tan­gent space. Ex­pand­ing the de­f­i­n­i­tion of how the iden­ti­ty is a pro­found con­cept, help­ing to link pre­vi­ous­ly dis­parate groups to­geth­er. Pre­vi­ous­ly we con­sid­ered the iden­ti­ty el­e­ment of our di­he­dral groups, as

ϵ=(100010001),\begin{array}{r} \epsilon = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \end{array}
(1.5.134)

and in the con­text of the sub­tan­gent space, we can see that the iden­ti­ty el­e­ment ϵ\epsilon is just an el­e­ment in the S3ϵS_{3}^{\epsilon} iden­ti­ty group. We may con­sid­er our orig­i­nal iden­ti­ty ϵ\epsilon, as the iden­ti­ty of an iden­ti­ty group, where the con­cept of nest­ing iden­ti­ties nec­es­sar­i­ly arises, which leads to the fol­low­ing de­f­i­n­i­tion:

De­f­i­n­i­tion (1.5.135) – The Iden­ti­ty Group 𝐒𝟑𝛜\mathbf{S}_{\mathbf{3}}^{\mathbf{\epsilon}} of the sub­tan­gent space 𝒯(𝔰3ℂ∞)\mathcal{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right) is the group of el­e­ments from the de­fined tan­gent space 𝔗(u×𝔰3ℂ∞)\mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right)

𝔗(u×𝔰3ℂ∞)={(eu(ia)0000eu(ib)0eu(ic)0),   (00eu(ia)0eu(ib)0eu(ic)00),   (0eu(ia)0eu(ib)0000eu(ic)),(eu(ia)000eu(ib)000eu(ic)),   (0eu(ia)000eu(ib)eu(ic)00),   (00eu(ia)eu(ib)000eu(ic)0)},\mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right) = \left\{ \begin{array}{r} \begin{pmatrix} e^{u(ia)} & 0 & 0 \\ 0 & 0 & e^{u(ib)} \\ 0 & e^{u(ic)} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & e^{u(ia)} \\ 0 & e^{u(ib)} & 0 \\ e^{u(ic)} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & e^{u(ia)} & 0 \\ e^{u(ib)} & 0 & 0 \\ 0 & 0 & e^{u(ic)} \end{pmatrix}, \\ \begin{pmatrix} e^{u(ia)} & 0 & 0 \\ 0 & e^{u(ib)} & 0 \\ 0 & 0 & e^{u(ic)} \end{pmatrix},\ \ \ \begin{pmatrix} 0 & e^{u(ia)} & 0 \\ 0 & 0 & e^{u(ib)} \\ e^{u(ic)} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & e^{u(ia)} \\ e^{u(ib)} & 0 & 0 \\ 0 & e^{u(ic)} & 0 \end{pmatrix} \end{array} \right\},

re­al­ized for u=0u = 0

S3ϵ= 𝔗(u×𝔰3ℂ∞)|u=0={(100001010),   (001010100),   (010100001),(100010001),   (010001100),   (001100010)},S_{3}^{\epsilon} = \left. \ \mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right) \right|_{u = 0} = \left\{ \begin{array}{r} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \\ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \end{array} \right\},

for S3ϵS_{3}^{\epsilon}, equal to D3/S3D_{3}/S_{3}, and is there­fore a group under com­po­si­tion as proven pre­vi­ous­ly.

As de­fined, the dif­fer­en­tial sub­tan­gent space is the ar­gu­ment com­po­si­tion of the angle space. From the log­a­rith­mic stand­point, these angle space ar­gu­ments would rep­re­sent the el­e­ment-by-el­e­ment prin­ci­pal log­a­rithms of the tan­gent space for a,b,c∈ℝ+a,b,c \in \mathbb{R}^{+}, and not the matrix log­a­rithm func­tion. From the per­spec­tive of the taking the log­a­rithm of in­di­vid­ual en­tries in the matrix el­e­ments ∈𝔗(u×𝔰3ℂ∞)\in \mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right), neg­a­tive in­fini­ties are the result for the zero-valued en­tries. An ex­am­ple of­fered is the el­e­ment-by-el­e­ment prin­ci­pal log­a­rithm of 𝒳1∞∈𝔗(u×𝔰3ℂ∞)\mathcal{X}_{1}^{\infty}\in \mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right), which we define using the no­ta­tion lnelement⁡(𝒳1∞){\ln_{element}(\mathcal{X}}_{1}^{\infty})

Figure
Figure 1.5.1 – The Ex­po­nen­tial Func­tion: Zero and In­fin­i­ty
lnelement⁡(𝒳1∞)=lnelement⁡(e(ia)0000e(ib)0e(ic)0)=(ia−∞−∞−∞−∞ib−∞ic−∞),   (a,b,c∈ℝ+).\begin{array}{r} {\ln_{element}(\mathcal{X}}_{1}^{\infty}) = \ln_{element}{\begin{pmatrix} e^{(ia)} & 0 & 0 \\ 0 & 0 & e^{(ib)} \\ 0 & e^{(ic)} & 0 \end{pmatrix} =}\begin{pmatrix} ia & - \infty & - \infty \\ - \infty & - \infty & ib \\ - \infty & ic & - \infty \end{pmatrix},\ \ \ \left( a,b,c \in \mathbb{R}^{+} \right). \end{array}
(1.5.136)

We know that these neg­a­tive in­fini­ties arise as so­lu­tions for xx in the equa­tion ex=0e^{x} = 0. In Figure 1.5.1, we plot the func­tion, exe^{x} for x∈ℝx\in \mathbb{R}, where we see the iden­ti­ty is the point where the func­tion cross­es the ver­ti­cal axis, and the func­tion goes to +∞+ \infty, as x→∞x \rightarrow \infty, while for x→−∞x \rightarrow - \infty, the exe^{x} func­tion goes to zero e−∞=0.e^{- \infty} = 0.

Like the prin­ci­pal el­e­ment-by-el­e­ment log­a­rithm, the dif­fer­en­tial sub­tan­gent al­ge­bra is trans­formed to the al­ge­bras of 𝔗(𝔰3ℂ∞)\mathfrak{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right) tan­gent space by the ex­po­nen­tial func­tion, and not the matrix ex­po­nen­tial. Since the dif­fer­en­tial sub­tan­gent space is de­fined as a de­riv­a­tive with re­spect to uu for dd𝓊( 𝔗(u×𝔰3ℂ∞)|u=0),\frac{d}{d\mathcal{u}}\left( \left. \ \mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right) \right|_{u = 0} \right),it is con­sis­tent with the de­f­i­n­i­tion for the tan­gent space in Lie theory, dif­fer­ing from the matrix log­a­rithm op­er­a­tion, which de­fines the tan­gent space 𝔗(𝔰3ℂ∞)\mathfrak{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right). We demon­strate this with an ex­am­ple using the el­e­ment 𝒳1∞\mathcal{X}_{1}^{\infty} ∈𝔗(𝔰3ℂ∞)\in \mathfrak{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right)

𝒳1∞=(e(ia)0000e(ib)0e(ic)0),\begin{array}{r} \mathcal{X}_{1}^{\infty} = \begin{pmatrix} e^{(ia)} & 0 & 0 \\ 0 & 0 & e^{(ib)} \\ 0 & e^{(ic)} & 0 \end{pmatrix}, \end{array}
(1.5.137)

and cal­cu­late its matrix log­a­rithm ln⁡(𝒳1∞)\ln\left( \mathcal{X}_{1}^{\infty} \right)

ln⁡(𝒳1∞)=(ln⁡(eia)00012(ln⁡(e(i(b+c)2))+ln⁡(−e(i(b+c)2)))e( i(b−c)2)2(ln⁡(e(i(b+c)2))−ln⁡(−e( i(b+c)2)))0e(− i(b−c)2)2(ln⁡(e(i(b+c)2))−ln⁡(−e( i(b+c)2)))12(ln⁡(e(i(b+c)2))+ln⁡(−e(i(b+c)2)))),\begin{array}{r} \ln\left( \mathcal{X}_{1}^{\infty} \right) = \begin{pmatrix} \ln\left( e^{ia} \right) & 0 & 0 \\ 0 & \frac{1}{2}\left( \ln\left( e^{\left( \frac{i(b + c)}{2} \right)} \right) + \ln\left( - e^{\left( \frac{i(b + c)}{2} \right)} \right) \right) & \frac{e^{\left( \ \frac{i(b - c)}{2} \right)}}{2}\left( \ln\left( e^{\left( \frac{i(b + c)}{2} \right)} \right) - \ln\left( - e^{\left( \ \frac{i(b + c)}{2} \right)} \right) \right) \\ 0 & \frac{e^{\left( - \ \frac{i(b - c)}{2} \right)}}{2}\left( \ln\left( e^{\left( \frac{i(b + c)}{2} \right)} \right) - \ln\left( - e^{\left( \ \frac{i(b + c)}{2} \right)} \right) \right) & \frac{1}{2}\left( \ln\left( e^{\left( \frac{i(b + c)}{2} \right)} \right) + \ln\left( - e^{\left( \frac{i(b + c)}{2} \right)} \right) \right) \end{pmatrix}, \end{array}
(1.5.138)

and by as­sum­ing a prin­ci­pal valued matrix log­a­rithm for a,b,c∈ℝ+a,b,c \in \mathbb{R}^{+}, this sim­pli­fies to

ln⁡(𝒳1u×∞)=(ia000i(b+c−π)2iπ×e(i(b−c)2)20 iπ×e(i(c−b)2)2i(b+c−π)2),   (a,b,c∈ℝ+).\begin{array}{r} \ln\left( \mathcal{X}_{1}^{u \times \infty} \right) = \begin{pmatrix} ia & 0 & 0 \\ 0 & \frac{i(b + c - \pi)}{2} & \frac{{i\pi \times e}^{\left( \frac{i(b - c)}{2} \right)}}{2} \\ 0 & \frac{\ i\pi \times e^{\left( \frac{i(c - b)}{2} \right)}}{2} & \frac{i(b + c - \pi)}{2} \end{pmatrix},\ \ \ \left( a,b,c \in \mathbb{R}^{+} \right). \end{array}
(1.5.139)

A sim­i­lar ap­proach ap­plied for the other el­e­ment 𝒳1∞\mathcal{X}_{1}^{\infty} ∈𝔗(𝔰3ℂ∞)\in \mathfrak{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right), re­sults in the 𝒳k∞\mathcal{X}_{k}^{\infty} subset of el­e­ments based upon the prin­ci­pal matrix log­a­rithms for a,b,c∈ℝ+a,b,c \in \mathbb{R}^{+},

ln⁡(𝒳k∞)={(ia000i(b+c−π)2iπ×e(i(b−c)2)20iπ×e(i(c−b)2)2i(b+c−π)2),(i(a+c−π)20iπ×e(i(a−c)2)20ib0iπ×e(i(c−a)2)20i(a+c−π)2),(i(a+b−π)2iπ×e(i(a−b)2)20iπ×e(i(b−a)2)2i(a+b−π)2000ic)}.\begin{array}{r} {\ln(\mathcal{X}}_{k}^{\infty}) = \left\{ \begin{pmatrix} ia & 0 & 0 \\ 0 & \frac{i(b + c - \pi)}{2} & \frac{{i\pi \times e}^{\left( \frac{i(b - c)}{2} \right)}}{2} \\ 0 & \frac{i\pi \times e^{\left( \frac{i(c - b)}{2} \right)}}{2} & \frac{i(b + c - \pi)}{2} \end{pmatrix},\begin{pmatrix} \frac{i(a + c - \pi)}{2} & 0 & \frac{{i\pi \times e}^{\left( \frac{i(a - c)}{2} \right)}}{2} \\ 0 & ib & 0 \\ \frac{i\pi \times e^{\left( \frac{i(c - a)}{2} \right)}}{2} & 0 & \frac{i(a + c - \pi)}{2} \end{pmatrix},\begin{pmatrix} \frac{i(a + b - \pi)}{2} & \frac{{i\pi \times e}^{\left( \frac{i(a - b)}{2} \right)}}{2} & 0 \\ \frac{{i\pi \times e}^{\left( \frac{i(b - a)}{2} \right)}}{2} & \frac{i(a + b - \pi)}{2} & 0 \\ 0 & 0 & ic \end{pmatrix} \right\}. \end{array}
(1.5.140)

For the 𝒴k∞\mathcal{Y}_{k}^{\infty} in­fin­i­ty el­e­ments the prin­ci­pal matrix log­a­rithm of 𝒴k∞\mathcal{Y}_{k}^{\infty} is straight­for­ward

ln⁡(𝒴1∞)=ln⁡(eia000eib000eic)=(ia000ib000ic),   (a,b,c∈ℝ+).\begin{array}{r} \ln{\left( \mathcal{Y}_{1}^{\infty} \right) =}\ln\begin{pmatrix} e^{ia} & 0 & 0 \\ 0 & e^{ib} & 0 \\ 0 & 0 & e^{ic} \end{pmatrix} = \begin{pmatrix} ia & 0 & 0 \\ 0 & ib & 0 \\ 0 & 0 & ic \end{pmatrix},\ \ \ \left( a,b,c \in \mathbb{R}^{+} \right). \end{array}
(1.5.141)

For the eval­u­a­tion of the in­fin­i­ty el­e­ments 𝒴2∞,𝒴3∞\mathcal{Y}_{2}^{\infty},\mathcal{Y}_{3}^{\infty}, we apply the eigen­de­com­po­si­tion for 𝒴2∞\mathcal{Y}_{2}^{\infty}

ln⁡(𝒴2,3∞)=VDV−1,\begin{array}{r} \ln{\left( \mathcal{Y}_{2,3}^{\infty} \right) = VDV^{- 1},} \end{array}
(1.5.142)

find­ing the eigen­val­ue, eigen­vec­tor ma­tri­ces, and in­verse eigen­vec­tor ma­tri­ces D,V,V−1\ D,V,V^{- 1}

D=((ei(a+b+c))13000(−1+i3)2(ei(a+b+c))13000(−1−i3)2(ei(a+b+c))13),D = \begin{pmatrix} \left( e^{i(a + b + c)} \right)^{\frac{1}{3}} & 0 & 0 \\ 0 & \frac{\left( - 1 + i\sqrt{3} \right)}{2}\left( e^{i(a + b + c)} \right)^{\frac{1}{3}} & 0 \\ 0 & 0 & \frac{\left( - 1 - i\sqrt{3} \right)}{2}\left( e^{i(a + b + c)} \right)^{\frac{1}{3}} \end{pmatrix},
V=(e−ic(ei(a+b+c))13−(1+i3)2e−ic(ei(a+b+c))13(−1+i3)2e−ic(ei(a+b+c))13e−i(a+c)(ei(a+b+c))23(−1+i3)2e−i(a+c)(ei(a+b+c))23−(1+i3)2e−i(a+c)(ei(a+b+c))23111),V = \begin{pmatrix} e^{- ic}\left( e^{i(a + b + c)} \right)^{\frac{1}{3}} & - \frac{\left( 1 + i\sqrt{3} \right)}{2}e^{- ic}\left( e^{i(a + b + c)} \right)^{\frac{1}{3}} & \frac{\left( - 1 + i\sqrt{3} \right)}{2}e^{- ic}\left( e^{i(a + b + c)} \right)^{\frac{1}{3}} \\ e^{- i(a + c)}\left( e^{i(a + b + c)} \right)^{\frac{2}{3}} & \frac{\left( - 1 + i\sqrt{3} \right)}{2}e^{- i(a + c)}\left( e^{i(a + b + c)} \right)^{\frac{2}{3}} & - \frac{\left( 1 + i\sqrt{3} \right)}{2}e^{- i(a + c)}\left( e^{i(a + b + c)} \right)^{\frac{2}{3}} \\ 1 & 1 & 1 \end{pmatrix},
V−1=(e−i(a+b)(ei(a+b+c))233e−ib(ei(a+b+c))13313(−1+i3)e−i(a+b)(ei(a+b+c))236−(1+i3)e−ib(ei(a+b+c))13613−(1+i3)e−i(a+b)(ei(a+b+c))236(−1+i3)e−ib(ei(a+b+c))13613).\begin{array}{r} V^{- 1} = \begin{pmatrix} \frac{e^{- i(a + b)}\left( e^{i(a + b + c)} \right)^{\frac{2}{3}}}{3} & \frac{e^{- ib}\left( e^{i(a + b + c)} \right)^{\frac{1}{3}}}{3} & \frac{1}{3} \\ \frac{{\left( - 1 + i\sqrt{3} \right)e}^{- i(a + b)}\left( e^{i(a + b + c)} \right)^{\frac{2}{3}}}{6} & - \frac{\left( 1 + i\sqrt{3} \right)e^{- ib}\left( e^{i(a + b + c)} \right)^{\frac{1}{3}}}{6} & \frac{1}{3} \\ - \frac{{\left( 1 + i\sqrt{3} \right)e}^{- i(a + b)}\left( e^{i(a + b + c)} \right)^{\frac{2}{3}}}{6} & \frac{\left( - 1 + i\sqrt{3} \right)e^{- ib}\left( e^{i(a + b + c)} \right)^{\frac{1}{3}}}{6} & \frac{1}{3} \end{pmatrix}. \end{array}
(1.5.143)

The prin­ci­pal matrix log­a­rithm ln⁡(D)\ln(D) is then easily cal­cu­lat­ed

D=(i(a+b+c)3000i(a+b+c+2π)3000i(a+b+c−2π)3),   (a,b,c∈ℝ+),\begin{array}{r} D = \begin{pmatrix} \frac{i(a + b + c)}{3} & 0 & 0 \\ 0 & \frac{i(a + b + c + 2\pi)}{3} & 0 \\ 0 & 0 & \frac{i(a + b + c - 2\pi)}{3} \end{pmatrix},\ \ \ \left( a,b,c \in \mathbb{R}^{+} \right), \end{array}
(1.5.144)

with which we de­ter­mine for a,b,c∈ℝ+\ a,b,c \in \mathbb{R}^{+}, the prod­uct 𝒴2∞=VDV−1\mathcal{Y}_{2}^{\infty} = VDV^{- 1}

ln⁡(𝒴2∞)=(i(a+b+c)313(23π3)e−i(b+c)(ei(a+b+c))23−13(23π3)e−i(c)(ei(a+b+c))13−13(23π3)e−ia(ei(a+b+c))13i(a+b+c)313(23π3)e−i(a+c)(ei(a+b+c))2313(23π3)e−i(a+b)(ei(a+b+c))23−13(23π3)e−i(b)(ei(a+b+c))13i(a+b+c)3).\begin{array}{r} \ln{\left( \mathcal{Y}_{2}^{\infty} \right) =}\begin{pmatrix} \frac{i(a + b + c)}{3} & \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- i(b + c)}\left( e^{i(a + b + c)} \right)^{\frac{2}{3}} & - \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- i(c)}\left( e^{i(a + b + c)} \right)^{\frac{1}{3}} \\ - \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- ia}\left( e^{i(a + b + c)} \right)^{\frac{1}{3}} & \frac{i(a + b + c)}{3} & \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- i(a + c)}\left( e^{i(a + b + c)} \right)^{\frac{2}{3}} \\ \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- i(a + b)}\left( e^{i(a + b + c)} \right)^{\frac{2}{3}} & - \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- i(b)}\left( e^{i(a + b + c)} \right)^{\frac{1}{3}} & \frac{i(a + b + c)}{3} \end{pmatrix}. \end{array}
(1.5.145)

The sim­i­lar prin­ci­ple matrix log­a­rithm ln⁡(𝒴3∞)\ln\left( \mathcal{Y}_{3}^{\infty} \right), using the eigen­de­com­po­si­tion, would then round out the 𝒴k∞\mathcal{Y}_{k}^{\infty} in­fin­i­ty matrix log­a­rithm subset, hold­ing for a,b,c∈ℝ+a,b,c \in \mathbb{R}^{+}

ln⁡(𝒴k∞)={(ia000ib000ic),(i(a+b+c)313(23π3)e−i(b+c)(ei(a+b+c))23−13(23π3)e−i(c)(ei(a+b+c))13−13(23π3)e−ia(ei(a+b+c))13i(a+b+c)313(23π3)e−i(a+c)(ei(a+b+c))2313(23π3)e−i(a+b)(ei(a+b+c))23−13(23π3)e−i(b)(ei(a+b+c))13i(a+b+c)3),(i(a+b+c)3−13(23π3)e−i(b)(ei(a+b+c))1313(23π3)e−i(b+c)(ei(a+b+c))2313(23π3)e−i(a+c)(ei(a+b+c))23i(a+b+c)3−13(23π3)e−i(c)(ei(a+b+c))13−13(23π3)e−ia(ei(a+b+c))1313(23π3)e−i(a+b)(ei(a+b+c))23i(a+b+c)3)}.\begin{array}{r} \ln{\left( \mathcal{Y}_{k}^{\infty} \right) =}\left\{ \begin{array}{r} \begin{pmatrix} ia & 0 & 0 \\ 0 & ib & 0 \\ 0 & 0 & ic \end{pmatrix}, \\ \begin{pmatrix} \frac{i(a + b + c)}{3} & \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- i(b + c)}\left( e^{i(a + b + c)} \right)^{\frac{2}{3}} & - \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- i(c)}\left( e^{i(a + b + c)} \right)^{\frac{1}{3}} \\ - \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- ia}\left( e^{i(a + b + c)} \right)^{\frac{1}{3}} & \frac{i(a + b + c)}{3} & \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- i(a + c)}\left( e^{i(a + b + c)} \right)^{\frac{2}{3}} \\ \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- i(a + b)}\left( e^{i(a + b + c)} \right)^{\frac{2}{3}} & - \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- i(b)}\left( e^{i(a + b + c)} \right)^{\frac{1}{3}} & \frac{i(a + b + c)}{3} \end{pmatrix}, \\ \begin{pmatrix} \frac{i(a + b + c)}{3} & - \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- i(b)}\left( e^{i(a + b + c)} \right)^{\frac{1}{3}} & \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- i(b + c)}\left( e^{i(a + b + c)} \right)^{\frac{2}{3}} \\ \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- i(a + c)}\left( e^{i(a + b + c)} \right)^{\frac{2}{3}} & \frac{i(a + b + c)}{3} & - \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- i(c)}\left( e^{i(a + b + c)} \right)^{\frac{1}{3}} \\ - \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- ia}\left( e^{i(a + b + c)} \right)^{\frac{1}{3}} & \frac{1}{3}\left( \frac{2\sqrt{3}\pi}{3} \right)e^{- i(a + b)}\left( e^{i(a + b + c)} \right)^{\frac{2}{3}} & \frac{i(a + b + c)}{3} \end{pmatrix} \end{array} \right\}. \end{array}
(1.5.146)

With the matrix log­a­rithm sub­tan­gent space forms de­ter­mined, we state the fol­low­ing the­o­rem:

The­o­rem (1.5.147) – The log­a­rith­mic sub­tan­gent space of S3ℂ∞S_{3\mathbb{C}}^{\infty}, de­not­ed 𝒯ln⁡ (𝔰3ℂ∞)\mathcal{T}_{\ln}\ \left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right), is com­prised of com­plex 3×33 \times 3 ma­tri­ces which rep­re­sent the angle func­tion space of the tan­gent space for all 𝔗(u×𝔰3ℂ∞)\mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right), equal to an el­e­ment of the iden­ti­ty group S3ϵS_{3}^{\epsilon} for u=0u = 0. The prin­ci­pal log­a­rithm of the tan­gent space 𝔗(𝔰3ℂ∞)\mathfrak{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right), for angle ar­gu­ments ∈ℝ\in \mathbb{R}, equals the log­a­rith­mic sub­tan­gent space 𝒯ln⁡(𝔰3ℂ∞)\ \mathcal{T}_{\ln}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right), in the neigh­bor­hood of the iden­ti­ty group and its el­e­ments S3ϵS_{3}^{\epsilon}, for u=0u = 0.

 𝔗(u×𝔰3ℂ∞)|u=0=S3ϵ,\left. \ \mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right) \right|_{u = 0} = S_{3}^{\epsilon},
𝒯ln⁡(𝔗(S3ℂ∞))=ln⁡(ln⁡(e𝔗(𝔰3ℂ∞)))=ln⁡(𝔗(𝔰3ℂ∞))=𝒯ln⁡(𝔰3ℂ∞).\mathcal{T}_{\ln}\left( \mathfrak{T}\left( S_{3\mathbb{C}}^{\infty} \right) \right) = \ln\left( \ln\left( e^{\mathfrak{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right)} \right) \right) = \ln\left( \mathfrak{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right) \right) = \mathcal{T}_{\ln}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right).

Proof – The el­e­ments of the tan­gent space 𝔗(𝔰3ℂ∞)\mathfrak{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right) have three matrix en­tries e(iθk)e^{\left( i\theta_{k} \right)}, re­sult­ing from the matrix ex­po­nen­tial. As such, the matrix log­a­rithm will take the form of al­ge­bra­ic func­tion groups,, which rep­re­sent the matrix log­a­rithm of the angle ar­gu­ments of the tan­gent space ma­tri­ces 𝔗(u×𝔰3ℂ∞)\mathfrak{T}\left( u \times \mathfrak{s}_{3\mathbb{C}}^{\infty} \right). These angle func­tions arranged in their re­spec­tive ma­tri­ces are de­fined as the log­a­rith­mic sub­tan­gent space 𝒯ln⁡(𝔰3ℂ∞)\mathcal{T}_{\ln}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right) , and there­fore, ln⁡(𝔗(𝔰3ℂ∞))=𝒯ln⁡(𝔰3ℂ∞).\ln\left( \mathfrak{T}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right) \right) = \mathcal{T}_{\ln}\left( \mathfrak{s}_{3\mathbb{C}}^{\infty} \right). Taking any matrix group 𝒳k∞,𝒴k∞{\mathcal{X}_{k}^{\infty},\mathcal{Y}}_{k}^{\infty}, for all angle vari­ables θk=0\theta_{k} = 0, it holds ∀ eln⁡(𝒳k∞,𝒴k∞)\forall\ e^{\ln\left( {\mathcal{X}_{k}^{\infty},\mathcal{Y}}_{k}^{\infty} \right)} that they take the form of a el­e­ment in the iden­ti­ty group S3ϵS_{3}^{\epsilon}.

□\square

Math­e­mat­i­cal­ly, we have seen that a con­sis­tent sub­tan­gent space exists and is log­a­rith­mi­cal­ly re­lat­ed to the tan­gent space lo­cat­ed on the sphere of tan­gent spaces. The de­fined dif­fer­en­tial sub­tan­gent space, presents an arrange­ment com­prised of finite angle el­e­ments, taking form as a result of the zero-point e−∞e^{- \infty}. Where­as the de­fined log­a­rith­mic sub­tan­gent space is a well-de­fined angle space func­tion, ori­ent­ed at a math­e­mat­i­cal cross­roads where in­fini­ties take finite forms.