Masteries · Jason GlowneyArithmoí Book 1

Masteries

1.6.8 Matrix Determinants and the Infinite Symmetry Group

In the last chap­ter, we de­fined the tan­gent space 𝔗(𝓊×𝔰3ℂ∞)\mathfrak{T}\left( \mathcal{u} \times\mathfrak{s}_{3\mathbb{C}}^{\infty} \right) of the in­fi­nite sym­me­try group as

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}
(1.6.245)

where we see that the el­e­ments of the Lie al­ge­bra could be points any­where on the com­plex unit circle. For the tan­gent space el­e­ments above, the top row is of the 𝒳\mathcal{X} form and the el­e­ments on the bottom row, the 𝒴\mathcal{Y} form. In order to group the Lie al­ge­bra to­geth­er, we change the vari­ables in the 𝒳\mathcal{X}-type ma­tri­ces as fol­lows

𝔗(𝒳)={(eu(ia1)0000eu(ia2)0eu(ia3)0),   (00eu(ib1)0eu(ib2)0eu(ib3)00),   (0eu(ic1)0eu(ic2)0000eu(ic3))},\begin{array}{r} \mathfrak{T}\left( \mathcal{X} \right) = \left\{ \begin{array}{r} \begin{pmatrix} e^{u\left( ia_{1} \right)} & 0 & 0 \\ 0 & 0 & e^{u\left( ia_{2} \right)} \\ 0 & e^{u\left( ia_{3} \right)} & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & 0 & e^{u\left( ib_{1} \right)} \\ 0 & e^{u\left( ib_{2} \right)} & 0 \\ e^{u\left( ib_{3} \right)} & 0 & 0 \end{pmatrix},\ \ \ \begin{pmatrix} 0 & e^{u\left( ic_{1} \right)} & 0 \\ e^{u\left( ic_{2} \right)} & 0 & 0 \\ 0 & 0 & e^{u\left( ic_{3} \right)} \end{pmatrix} \end{array} \right\}, \end{array}
(1.6.246)

where it be­comes ap­par­ent that each matrix entry can be any point on the com­plex unit circle. Adding the Lie al­ge­bra to­geth­er as one matrix, re­sults in a matrix de­not­ed as ℳ\mathcal{M}

ℳ=(eu(ia1)eu(ic1)eu(ib1)eu(ic2)eu(ib2)eu(ia2)eu(ib3)eu(ia3)eu(ic3)),\begin{array}{r} \mathcal{M} =\begin{pmatrix} e^{u\left( ia_{1} \right)} & e^{u\left( ic_{1} \right)} & e^{u\left( ib_{1} \right)} \\ e^{u\left( ic_{2} \right)} & e^{u\left( ib_{2} \right)} & e^{u\left( ia_{2} \right)} \\ e^{u\left( ib_{3} \right)} & e^{u\left( ia_{3} \right)} & e^{u\left( ic_{3} \right)} \end{pmatrix}, \end{array}
(1.6.247)

which rep­re­sents the entire tan­gent space of the 𝒳\mathcal{X}-type Lie al­ge­bra. The 𝒴\mathcal{Y}-type el­e­ments are de­pen­dent on the 𝒳\mathcal{X}-type el­e­ments and are the result of prod­ucts of the 𝔗(𝒳)\mathfrak{T}\left( \mathcal{X} \right) el­e­ments. The de­ter­mi­nant of the matrix ℳ\mathcal{M} nicely demon­strates this fact, as

det⁡(ℳ)={eu(ia1)eu(ib2)eu(ic3)+eu(ic1)eu(ia2)eu(ib3)+eu(ib1)eu(ic2)eu(ia3)−(eu(ia1)eu(ia2)eu(ia3) + eu(ib1)eu(ib2)eu(ib3)+eu(ic1)eu(ic2)eu(ic3))},\begin{array}{r} \det\left( \mathcal{M} \right) = \left\{ \begin{array}{r} e^{u\left( ia_{1} \right)}e^{u\left( ib_{2} \right)}e^{u\left( ic_{3} \right)} + e^{u\left( ic_{1} \right)}e^{u\left( ia_{2} \right)}e^{u\left( ib_{3} \right)} + e^{u\left( ib_{1} \right)}e^{u\left( ic_{2} \right)}e^{u\left( ia_{3} \right)} \\ - \left( e^{u\left( ia_{1} \right)}e^{u\left( ia_{2} \right)}e^{u\left( ia_{3} \right)}\ {+ \ e}^{u\left( ib_{1} \right)}e^{u\left( ib_{2} \right)}e^{u\left( ib_{3} \right)} + e^{u\left( ic_{1} \right)}e^{u\left( ic_{2} \right)}e^{u\left( ic_{3} \right)} \right) \end{array} \right\}, \end{array}
(1.6.248)

which sim­pli­fies to

det⁡(ℳ)={eiu(a1+b2+c3)+eiu(c1+a2+b3)+eiu(b1+c2+a3)−(eiu(a1+a2+a3)+ eiu(b1+b2+b3)+ eiu(c1+c2+c3))},\begin{array}{r} \det\left( \mathcal{M} \right) = \left\{ \begin{array}{r} e^{iu\left( a_{1} + b_{2} + c_{3} \right)} + e^{iu\left( c_{1} + a_{2} + b_{3} \right)} + e^{iu\left( b_{1} + c_{2} + a_{3} \right)} \\ - \left( e^{iu\left( a_{1} + a_{2} + a_{3} \right)}{+ \ e}^{iu\left( b_{1} + b_{2} + b_{3} \right)}{+ \ e}^{iu\left( c_{1} + c_{2} + c_{3} \right)} \right) \end{array} \right\}, \end{array}
(1.6.249)

and the dis­tinc­tion be­tween the 𝒴,𝒳\mathcal{Y},\mathcal{X}-types be­comes ap­par­ent, as the three ex­po­nen­tials on the top cycle though the a,b,ca,b,c el­e­ments in cyclic fash­ion, which is akin to the 𝒴\mathcal{Y}-type el­e­ments of the Z3Z_{3} group, and the el­e­ments on the bottom are the sub-de­ter­mi­nants of the 𝒳\mathcal{X}-type Lie al­ge­bras. To make this clear­er, we denote the de­pen­dent 𝒴j\mathcal{Y}_{j} el­e­ments, as

𝔗(𝒴 )={(eu(ia1)000eu(ib2)000eu(ic3)),(0eu(ic1)000eu(ia2)eu(ib3)00),(00eu(ib1)eu(ic2)000eu(ia3)0)},\begin{array}{r} \mathfrak{T}\left( \mathcal{Y}\ \right) = \left\{ \begin{array}{r} \begin{pmatrix} e^{u\left( ia_{1} \right)} & 0 & 0 \\ 0 & e^{u\left( ib_{2} \right)} & 0 \\ 0 & 0 & e^{u\left( ic_{3} \right)} \end{pmatrix},\begin{pmatrix} 0 & e^{u\left( ic_{1} \right)} & 0 \\ 0 & 0 & e^{u\left( ia_{2} \right)} \\ e^{u\left( ib_{3} \right)} & 0 & 0 \end{pmatrix},\begin{pmatrix} 0 & 0 & e^{u\left( ib_{1} \right)} \\ e^{u\left( ic_{2} \right)} & 0 & 0 \\ 0 & e^{u\left( ia_{3} \right)} & 0 \end{pmatrix} \end{array} \right\}, \end{array}
(1.6.250)

and note how this co­in­cides with the three el­e­ments on the left side we found in det⁡(ℳ)\det\left( \mathcal{M} \right).