1.5.2 The Symmetric Cone Group Lie Algebra

The $S_{3\mathbb{C}}^{\Delta}$ group's tan­gent space, its Lie brack­et op­er­a­tions, and its vector space struc­ture is the Lie al­ge­bra of the group de­fined as $\mathfrak{s}_{3\mathbb{C}}^{\Delta}$. The Lie al­ge­bra $\mathfrak{s}_{3\mathbb{C}}^{\Delta}$ is a matrix Lie al­ge­bra by way of Ado's the­o­rem¹, which states that every finite-di­men­sion­al Lie al­ge­bra can be en­com­passed as a Lie al­ge­bra of square ma­tri­ces under the com­mu­ta­tor brack­et (Lie brack­et). A matrix Lie al­ge­bra is a vector space of ma­tri­ces that is closed under a Lie brack­et $\lbrack X,Y\rbrack = XY - YX.$² For an el­e­ment $X$ in the Lie al­ge­bra $\mathfrak{s}_{3\mathbb{C}}^{\Delta}$, the Lie brack­et with itself equals zero,

We can rep­re­sent our matrix el­e­ments $\mathcal{X}_{i},\mathcal{Y}_{i}$ $\in \mathfrak{s}_{3\mathbb{C}}^{\Delta}$, as fol­lows

which for $\mathcal{X}_{j}$, the $a,b,c$ en­tries remain row con­sis­tent, as do the en­tries $d,e,f$ of $\mathcal{Y}_{j}$. The axes of ro­ta­tion for the $\mathcal{X}_{j}$ el­e­ments are de­fined as $\mathcal{X}_{1} = a,\mathcal{X}_{2} = b$, and $\mathcal{X}_{3} = c$. The reader should recall that the set of en­tries, $\{ a,b,c,d,e,f\}$, may take any value from the set $\left\{ \pm 1, \pm i, \pm ( - 1)^{\left( \frac{1}{4} \right)}, \pm ( - 1)^{\left( \frac{3}{4} \right)} \right\}$. Lie brack­ets of the form $\left\lbrack \mathcal{X}_{j},\mathcal{X}_{k} \right\rbrack$ yield re­sults $\in \mathcal{Y}_{l}$. Start­ing with $\left\lbrack \mathcal{X}_{1},\mathcal{X}_{2} \right\rbrack$

we find the Lie brack­et result takes the form $\mathcal{Y}_{3} - \mathcal{Y}_{2}$, and switch­ing the brack­et order, $\left\lbrack \mathcal{X}_{2},\mathcal{X}_{1} \right\rbrack$, re­sults in the neg­a­tive of $\left\lbrack \mathcal{X}_{1},\mathcal{X}_{2} \right\rbrack$

This neg­a­tive re­la­tion­ship holds for any Lie brack­et of el­e­ments $\in \mathcal{X}_{i}$, $\left\lbrack \mathcal{X}_{i},\mathcal{X}_{j} \right\rbrack = - \left\lbrack \mathcal{X}_{j},\mathcal{X}_{i} \right\rbrack.$For $\left\lbrack \mathcal{X}_{2},\mathcal{X}_{3} \right\rbrack$ and $\left\lbrack \mathcal{X}_{3},\mathcal{X}_{1} \right\rbrack$, the Lie brack­et re­sults are

for which a pat­tern emerges, where the axes of ro­ta­tion for the re­spec­tive el­e­ments $\mathcal{X}_{j}$ appear as squared values in the column rep­re­sent­ing the $j^{th}$ column number not present in the brack­et op­er­a­tion. The axis in the left-po­si­tion of the Lie brack­et op­er­a­tion is pos­i­tive, while the axis in the right-po­si­tion of the Lie brack­et is neg­a­tive. To fur­ther clar­i­fy, we take the $\left\lbrack \mathcal{X}_{2},\mathcal{X}_{3} \right\rbrack$ brack­et op­er­a­tion, where $\mathcal{X}_{1}$ is not present in the op­er­a­tion's cal­cu­la­tion, having $\mathcal{X}_{2}$ on the left in the Lie brack­et and $\mathcal{X}_{3}$ on the right. We then expect to see the squared en­tries of the $\mathcal{X}_{2},\mathcal{X}_{3}$-axes $\left( b^{2}, - c^{2} \right)$ in the 1^st axis column of the matrix result from the re­spec­tive $X_{2},X_{3}$, rows. In bold font the squared axis num­bers are shown as they appear in the $X_{1}$ column, from the Lie brack­et op­er­a­tion $\left\lbrack \mathcal{X}_{2},\mathcal{X}_{3} \right\rbrack$

The el­e­ments $\in \mathcal{Y}_{2},\mathcal{Y}_{3}$ are trace­less, since no en­tries are present on the main di­ag­o­nal where they would typ­i­cal­ly appear as the axes of ro­ta­tion for the $\mathcal{X}_{j}$ el­e­ments. To for­mal­ize this fur­ther, we can say that for the left-most el­e­ment $\in \mathcal{X}_{j}$, its axis of ro­ta­tion forms a prod­uct with the right-most el­e­ment $\in \mathcal{X}_{k}$, for its co­in­cid­ing $\mathcal{X}_{j}$ row-com­po­nent of the $\mathcal{X}_{k}$ el­e­ment. This prod­uct will appear in either of the re­sult­ing $\mathcal{Y}_{2},\mathcal{Y}_{3}$, el­e­ments in the cor­re­spond­ing $\mathcal{X}_{j}$ row-com­po­nent of the column $\notin \left\{ \mathcal{X}_{j},\mathcal{X}_{k} \right\}$. Using the prod­ucts ex­am­ples $\mathcal{X}_{1} \circ \mathcal{X}_{2}$, $\mathcal{X}_{1} \circ \mathcal{X}_{3}$,

we see the $a^{2}$ en­tries in the first row ($\mathcal{X}_{1}$ com­po­nent row) in both re­sults are found in the column row of the el­e­ment $\notin \left\{ \mathcal{X}_{j},\mathcal{X}_{k} \right\}$. For el­e­ment prod­ucts $\in \mathcal{X}_{j}$, or­dered $\mathcal{X}_{j} \circ \mathcal{X}_{j + 1}$, for $j$mod 3, the result is an el­e­ment $\in \mathcal{Y}_{3}$, and the result for el­e­ment prod­ucts in the re­versed order $\mathcal{X}_{j} \circ \mathcal{X}_{j - 1}$, yields an el­e­ment $\in \mathcal{Y}_{2}$. For any Lie brack­et with el­e­ments $\in Y_{j}$, the result will be an­oth­er el­e­ment $\in Y_{j}$, $\left\lbrack \mathcal{Y}_{j},\mathcal{Y}_{k} \right\rbrack = \mathcal{Y}_{l}$, seen for the ex­am­ple $\left\lbrack \mathcal{Y}_{2},\mathcal{Y}_{3} \right\rbrack$,

The $\left\lbrack \mathcal{Y}_{2},\mathcal{Y}_{3} \right\rbrack$ result takes the form $\mathcal{Y}_{1} - \mathcal{Y}_{1}^{'},$where the su­per­script in $\mathcal{Y}_{1}^{'}$, de­notes the same $Y_{1}$ form, having po­ten­tial­ly dif­fer­ent entry values. For the Cartan sub­al­ge­bra cases for $d = e = f$, the Lie brack­et result would be abelian and equal to zero. For the Lie brack­et $\left\lbrack \mathcal{Y}_{3},\mathcal{Y}_{2} \right\rbrack$, we also find the neg­a­tive of the for­ward-ori­ent­ed Lie brack­et coun­ter­part $\left\lbrack \mathcal{Y}_{3},\mathcal{Y}_{2} \right\rbrack = - \left\lbrack \mathcal{Y}_{2},\mathcal{Y}_{3} \right\rbrack$, which is also abelian, im­ply­ing $\left\lbrack \mathcal{Y}_{3},\mathcal{Y}_{2} \right\rbrack + \left\lbrack \mathcal{Y}_{2},\mathcal{Y}_{3} \right\rbrack = 0.$Con­cern­ing the Lie brack­ets of el­e­ments with the same finite matrix entry lo­ca­tions $\left\lbrack \mathcal{Y}_{j},\mathcal{Y}_{j}^{'} \right\rbrack$, for $\mathcal{Y}_{j} = \mathcal{Y}_{2}$, and ${d^{'},e}^{'},f' \in \mathcal{Y}_{j}^{'}$, the result would equal

For $d = d^{'},e = e^{'},f = f'$, the result above would com­mute as it does for the Cartan sub­al­ge­bra. Fi­nal­ly, for Lie brack­ets of the form $\left\lbrack \mathcal{X}_{j},\mathcal{Y}_{k} \right\rbrack$, we find re­sults $\in \mathcal{X}$, as $\left\lbrack \mathcal{X}_{j},\mathcal{Y}_{k} \right\rbrack = - \left\lbrack \mathcal{Y}_{k},\mathcal{X}_{j} \right\rbrack$, seen by the case ex­am­ple $\mathcal{X}_{j} = \mathcal{X}_{1}$, $\mathcal{Y}_{k} = \mathcal{Y}_{2}$

The result $\left\lbrack \mathcal{X}_{1},\mathcal{Y}_{2} \right\rbrack = \mathcal{X}_{3} - \mathcal{X}_{2}$, makes it ap­par­ent that the com­po­si­tion $\mathcal{X}_{1} \circ \mathcal{Y}_{2} = \mathcal{X}_{3}$, acts to right­ward shift the orig­i­nal el­e­ments $\in \mathcal{X}_{1}$, and the com­po­si­tion $\mathcal{Y}_{2} \circ \mathcal{X}_{1} = \mathcal{X}_{2}$ acts to upward shift the orig­i­nal el­e­ments $\in \mathcal{X}_{1}$.

1. Ado, Igor D. – "Note on the rep­re­sen­ta­tion of finite con­tin­u­ous groups by means of linear sub­sti­tu­tions", Izv. Fiz.-Mat. Obsch. (Kazan'), 7: 1–43, 1935.
2. *Still­well, John – "Naive Lie Theory", Springer 2008.*