Masteries · Jason GlowneyArithmoí Book 1

Masteries

1.3.3 Vector Representations in Complex-Angled Space

The polar vi­su­al­iza­tion of the com­plex unit circle nicely en­com­pass­es that a radial vector is a com­po­si­tion of an imag­i­nary-valued sine func­tion and a real-value cosine func­tion, ori­ent­ed re­spec­tive­ly along the ver­ti­cal and hor­i­zon­tal axes. A sim­i­lar geo­met­ric vi­su­al­iza­tion is pos­si­ble for the di­he­dral Lie groups using radial vector values to rep­re­sent a unique Lie group on the bound­ary ∂B\partial \mathrm{B} of the com­plex-angle space. The com­po­nents of this radial vector stem from the imag­i­nary ex­po­nen­tial func­tion's con­tri­bu­tion along the ver­ti­cal axis and the real ex­po­nen­tial func­tion's con­tri­bu­tion along the hor­i­zon­tal axis (Figure 1.3.9).

Figure
Figure 1.3.9 – Lie group vector com­po­nents sit­u­at­ed in the com­plex-angle space.

Above, the vector uu rep­re­sents the real-valued ex­po­nen­tial func­tion e(X2)e^{\left( \frac{X}{\sqrt{2}} \right)}, the vector vv, the imag­i­nary-valued ex­po­nen­tial func­tion e(iX2)e^{\left( \frac{iX}{\sqrt{2}} \right)}, and the prod­uct of the two vector func­tions u∙vu \bullet v, re­sults in the ex­po­nen­tial func­tion ww, that lies on the bound­ary of the disc

u∙v=w→e(X2)×e(iX2)=e(X(1+i)2).\begin{array}{r} u \bullet v = w \rightarrow e^{\left( \frac{X}{\sqrt{2}} \right)} \times e^{\left( \frac{iX}{\sqrt{2}} \right)} = e^{\left( \frac{X(1 + i)}{\sqrt{2}} \right)}. \end{array}
(1.3.17)

This simple case for a two-di­men­sion­al rep­re­sen­ta­tion holds for the di­he­dral Lie group gen­er­a­tors. Taking a sim­i­lar ex­am­ple in ℂ3\mathbb{C}^{3} space, the prod­uct of the ex­po­nen­tial func­tion vec­tors yields

G3ℂ(X12)∙G3ℝ(X12)=G(X1(1+i)2)=G3𝕔(X1),\begin{array}{r} G^{3\mathbb{C}}\left( \frac{X_{1}}{\sqrt{2}} \right) \bullet G^{3\mathbb{R}}\left( \frac{X_{1}}{\sqrt{2}} \right) = G\left( \frac{X_{1}(1 + i)}{\sqrt{2}} \right) = G^{3\mathbb{c}}\left( X_{1} \right), \end{array}
(1.3.18)

which has the matrix rep­re­sen­ta­tion form

(eix12000cos⁡(x12)isin(x12)0isin(x12)cos⁡(x12))×(ex12000cosh⁡(x12)sinh⁡(x12)0sinh⁡(x12)cosh⁡(x12))=(ex1(1+i)2000cosh⁡(x1(1+i)2)sinh⁡(x1(1+i)2)0sinh⁡(x1(1+i)2)cosh⁡(x1(1+i)2)).\begin{array}{r} \begin{pmatrix} e^{\frac{ix_{1}}{\sqrt{2}}} & 0 & 0 \\ 0 & \cos\left( \frac{x_{1}}{\sqrt{2}} \right) & isin\left( \frac{x_{1}}{\sqrt{2}} \right) \\ 0 & isin\left( \frac{x_{1}}{\sqrt{2}} \right) & \cos\left( \frac{x_{1}}{\sqrt{2}} \right) \end{pmatrix} \times \begin{pmatrix} e^{\frac{x_{1}}{\sqrt{2}}} & 0 & 0 \\ 0 & \cosh\left( \frac{x_{1}}{\sqrt{2}} \right) & \sinh\left( \frac{x_{1}}{\sqrt{2}} \right) \\ 0 & \sinh\left( \frac{x_{1}}{\sqrt{2}} \right) & \cosh\left( \frac{x_{1}}{\sqrt{2}} \right) \end{pmatrix} = \begin{pmatrix} e^{\frac{x_{1(1 + i)}}{\sqrt{2}}} & 0 & 0 \\ 0 & \cosh\left( \frac{x_{1}(1 + i)}{\sqrt{2}} \right) & \sinh\left( \frac{x_{1}(1 + i)}{\sqrt{2}} \right) \\ 0 & \sinh\left( \frac{x_{1}(1 + i)}{\sqrt{2}} \right) & \cosh\left( \frac{x_{1}(1 + i)}{\sqrt{2}} \right) \end{pmatrix}. \end{array}
(1.3.19)

Con­sid­er­ing higher-di­men­sion­al tan­gent spaces using vector rep­re­sen­ta­tions proves useful for their geo­met­ric vi­su­al­iza­tion. We take the two gen­er­a­tors on the left-hand side of the equa­tion above, which have tan­gent space rep­re­sen­ta­tions found through the matrix log­a­rithm func­tion

𝔗3ℂ(X12)=ln⁡(G3ℂ(X12))=(ix120000ix120ix120),𝔗3ℝ(X12)=ln⁡(G3ℝ(X12))=(x120000x120x120).\begin{array}{r} \mathfrak{T}^{3\mathbb{C}}\left( \frac{X_{1}}{\sqrt{2}} \right) = \ln{\left( G^{3\mathbb{C}}\left( \frac{X_{1}}{\sqrt{2}} \right) \right) = \begin{pmatrix} \frac{ix_{1}}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & \frac{ix_{1}}{\sqrt{2}} \\ 0 & \frac{ix_{1}}{\sqrt{2}} & 0 \end{pmatrix}}, \\ \mathfrak{T}^{3\mathbb{R}}\left( \frac{X_{1}}{\sqrt{2}} \right) = \ln{\left( G^{3\mathbb{R}}\left( \frac{X_{1}}{\sqrt{2}} \right) \right) = \begin{pmatrix} \frac{x_{1}}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & \frac{x_{1}}{\sqrt{2}} \\ 0 & \frac{x_{1}}{\sqrt{2}} & 0 \end{pmatrix}}. \end{array}
(1.3.20)

We eval­u­at­ed the prod­uct of these func­tion vec­tors for the case with ex­po­nen­tial func­tion gen­er­a­tors, but given that the log­a­rithm func­tion yields the tan­gent vec­tors of the Lie groups, the tan­gent space on the bound­ary of our com­plex-angle spac**e, ∂B3ℂ\partial \mathrm{B}^{3\mathbb{C}}, is the result of vector ad­di­tion

𝔗3ℂ(X12)+𝔗3ℝ(X12)=𝔗3𝕔(X1).\begin{array}{r} \mathfrak{T}^{3\mathbb{C}}\left( \frac{X_{1}}{\sqrt{2}} \right) + \mathfrak{T}^{3\mathbb{R}}\left( \frac{X_{1}}{\sqrt{2}} \right) = \mathfrak{T}^{3\mathbb{c}}\left( X_{1} \right). \end{array}
(1.3.21)

Ex­pand­ing the ad­di­tion of the tan­gent space vec­tors above, we find

(ix120000ix120ix120)+(x120000x120x120)=(x1(1+i)20000x1(1+i)20x1(1+i)20).\begin{array}{r} \begin{pmatrix} \frac{ix_{1}}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & \frac{ix_{1}}{\sqrt{2}} \\ 0 & \frac{ix_{1}}{\sqrt{2}} & 0 \end{pmatrix} + \begin{pmatrix} \frac{x_{1}}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & \frac{x_{1}}{\sqrt{2}} \\ 0 & \frac{x_{1}}{\sqrt{2}} & 0 \end{pmatrix} = \begin{pmatrix} \frac{x_{1}(1 + i)}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & \frac{x_{1}(1 + i)}{\sqrt{2}} \\ 0 & \frac{x_{1}(1 + i)}{\sqrt{2}} & 0 \end{pmatrix}. \end{array}
(1.3.22)

This is shown in Figure 1.3.10 for the two tan­gent spaces 𝔗3ℂ(X12)\mathfrak{T}^{3\mathbb{C}}\left( \frac{X_{1}}{\sqrt{2}} \right), 𝔗3ℝ(X12)\mathfrak{T}^{3\mathbb{R}}\left( \frac{X_{1}}{\sqrt{2}} \right), em­a­nat­ing from points on a sphere of radius r=12r = \frac{1}{\sqrt{2}}. The vector ad­di­tion of these tan­gent spaces re­sults in the tan­gent space 𝔗3𝕔(X1)\mathfrak{T}^{3\mathbb{c}}\left( X_{1} \right), which sits on the bound­ary of the com­plex-angle space ∂B3ℂ\partial \mathrm{B}^{3\mathbb{C}}, a sphere of uni­tary mod­u­lus. La­beled on the 𝔗3𝕔(X1)\mathfrak{T}^{3\mathbb{c}}\left( X_{1} \right) tan­gent space are the com­po­nent vec­tors x1,x2,x3x_{1},x_{2},x_{3}. These com­po­nent vec­tors reveal that x1x_{1} stays in the plane of the ge­o­des­ic circle eixe^{ix} for all tan­gent spaces 𝔗3ℂ(X12),𝔗3ℝ(X12), 𝔗3𝕔(X1)\mathfrak{T}^{3\mathbb{C}}\left( \frac{X_{1}}{\sqrt{2}} \right),\mathfrak{T}^{3\mathbb{R}}\left( \frac{X_{1}}{\sqrt{2}} \right),\ \mathfrak{T}^{3\mathbb{c}}\left( X_{1} \right), while main­tain­ing an or­thog­o­nal re­la­tion­ship to the sphere's bound­ary along its ge­o­des­ic path. This con­trasts with the vec­tors x2,x3x_{2},x_{3}, which orient along the X3X_{3}, X2\ X_{2}-axes re­spec­tive­ly, and always main­tain an or­thog­o­nal ori­en­ta­tion to their proper X2,X3X_{2},X_{3}-axes. These ori­en­ta­tions and re­la­tions should make sense, re­call­ing that the X1X_{1} el­e­ment in the D3D_{3} di­he­dral group re­flects the two points B,C (x2,x3)B,C\ (x_{2},x_{3}), leav­ing the point A(x1)A\left( x_{1} \right) fixed.

Figure
Figure 1.3.10 – Tan­gent Space Vector Ad­di­tion: 𝔗3ℂ(X12)+𝔗3ℝ(X12)=𝔗3𝕔(X1)\mathfrak{T}^{3\mathbb{C}}\left( \frac{X_{1}}{\sqrt{2}} \right) + \mathfrak{T}^{3\mathbb{R}}\left( \frac{X_{1}}{\sqrt{2}} \right) = \mathfrak{T}^{3\mathbb{c}}\left( X_{1} \right)