commit algebra

In $\C\otimes\Oct$ fix an imaginary unit $e_p$ and put $D=-iL_{e_p}$. For each of the six imaginary units $e_a$ orthogonal to $e_p$, $L_{e_a}^2=-I$ and $L_{e_a}$ anticommutes with $L_{e_p}$, hence with $D$; the product of the six is $\pm L_{e_p}$. These statements do not depend on the signs of the units.

So along a word of moves of the tetrahedron $T_0$, each paired with the left multiplication of a unit orthogonal to $e_p$ and acting on $\C^2\otimes(\C\otimes\Oct)$, the first factor keeps the class of the spinor system $V$ while $D$ changes sign at each move: the product of the class and $D$ is reversed by every move, and after $n$ moves the preserved operator is $(-1)^nD$.

Fix a point $p$ of the Fano plane, with unit $e_p$, and the table $A$. For the three lines $\{p,q,r\}$ through $p$, ordered so that $e_qe_r=e_p$, put $a_j=(L_{e_q}+iL_{e_r})/2$ and $N=\sum_ja_j^\dagger a_j$, and on a second factor $\C^2$ write $X$ and $Z$ for the Pauli matrices. The commit algebra at $e_p$ is

and $D_{\mathrm{age}}=D\otimes Z$.

For each of the seven points: (1) the six $L_{e_a}$, $a\neq p$, generate $M_8(\C)$, with commutant $\C$; (2) the seven generators of $\mathcal A_p$ anticommute in pairs, six square to $-I$ and one to $+I$, and their product is $\pm L_{e_p}\otimes Z$, so $\mathcal A_p$ is the complex Clifford algebra $\mathrm{Cl}_7$, of dimension 128; (3) the commutant of $\mathcal A_p$ is its centre, $\mathrm{span}\{I,D_{\mathrm{age}}\}$, and $\mathcal A_p\cong M_8(\C)\oplus M_8(\C)$, with central projections $P_\pm=\tfrac12(I\pm D_{\mathrm{age}})$ of rank 8; (4) $(-1)^N=\pm D$, so $D_{\mathrm{age}}=\pm(-1)^N\otimes Z$; (5) complex conjugation of $\C\otimes\Oct$ fixes every generator and exchanges $P_+$ with $P_-$.

(1) Each $L_{e_a}$, $a\neq p$, changes $N$ by $\pm1$. A function of $N$, tensored with $I$ or with $Z$, commutes with $\mathcal A_p$ only if it lies in $\mathrm{span}\{I,(-1)^N\otimes Z\}$, and the products $L_{e_a}L_{e_b}$ generate $M_4(\C)$ on each eigenspace of $D$. (2) With $u=e_0+\dots+e_6$, $L_{e_a}L_uL_{e_a}^{-1}=2L_{e_a}-L_u$ for every imaginary unit, so no $L_{e_a}$ sends $J_0=L_u/\sqrt7$ to $\pm J_0$; $J_0\otimes I$ does not commute with $D_{\mathrm{age}}$, so it is not in $\mathcal A_p$. (3) $L_uL_{e_p}+L_{e_p}L_u=-2I$ for every point, so the $+i$-eigenspaces $W$ of $J_0$ and $Q$ of $L_{e_p}$ are isoclinic: $P_WP_QP_W=\tfrac12\bigl(1\pm\tfrac1{\sqrt7}\bigr)P_W$, the sign depending on conventions.

With a further factor $\C^2$ for the spinor, each of the two sectors has dimension 16. In the gauge in which $(-1)^N=-D$, the sector $D_{\mathrm{age}}=-1$ holds $Z=+1$, $D=-1$, $N\in\{0,2\}$, which is $\mathbf1\oplus\mathbf3$ for $\mathrm{SU}(3)_{e_p}$, together with $Z=-1$, $D=+1$, $N\in\{1,3\}$, which is $\bar{\mathbf3}\oplus\mathbf1$; the sector $D_{\mathrm{age}}=+1$ holds the other two combinations. Neither $D$ nor $N$ is constant on a sector.

At $e_p$, with the quartet $\mathbf4$ the $-1$-eigenspace of $D$, put $B=\tfrac13(P_{N=2}-P_{N=1})$ and $L=P_{N=0}-P_{N=3}$. (1) The fifteen products $L_{e_a}L_{e_b}$, $a,b\neq p$, span $\mathfrak{su}(4)$ and commute with $D$; those annihilating 1 span $\mathfrak{su}(3)_{e_p}$. (2) On $\mathbf4\otimes\bar{\mathbf4}$ the $\mathfrak{su}(3)_{e_p}$-invariants have dimension 2 and the $\mathfrak{su}(4)$-invariants dimension 1, spanned by $\sum_ie_i\otimes\bar e_i$. (3) Every $\mathfrak{su}(3)_{e_p}$-invariant vector of $\mathbf4\otimes\bar{\mathbf4}$ is annihilated by $B\otimes I+I\otimes B$, $L\otimes I+I\otimes L$ and $D\otimes I+I\otimes D$.

So an element of $(V\otimes V)\otimes(\mathbf4\otimes\bar{\mathbf4})$ invariant under $\SL(2,\Z[\omega])$ and $\mathfrak{su}(4)$ is unique up to scale, $\varepsilon\otimes\sum_ie_i\otimes\bar e_i$.

A bilinear pairing realized by operators as $[\psi(f),\psi(g)]_\pm=\langle f,g\rangle$ is symmetric for the anticommutator and antisymmetric for the commutator, so a symmetric pairing admits only the canonical anticommutation relations. On the stabilizer of $I$ in $\SL(2,\Z[\omega])$, the dicyclic group of order twelve, $V$ is irreducible with Frobenius–Schur indicator $-1$, so its invariant bilinear forms are the multiples of the antisymmetric $\varepsilon$: the pairing of the spinor alone selects no statistics.

Identify $(\C\otimes\Oct)^{\otimes2}$ with the complex $8\times8$ matrices and put $D_p=-iL_{e_p}$. A tensor $\tau$ with $(D_p\otimes I+I\otimes D_p)\tau=0$ for all seven $p$ is a multiple of $\tau_0=1\otimes1+\sum_ke_k\otimes e_k$, the tensor of the octonion norm, which is symmetric; at one point the $\mathfrak{su}(4)$-invariant ones are spanned by $\tau_0$ and the antisymmetric tensor $\tau_D$ of $D_p$, and $\sum_ie_i\otimes\bar e_i=\tfrac12(\tau_0-\tau_D)$.

Let $\tau$ have symmetry $s_\tau=\pm1$ under the swap of its factors and put $\kappa=\varepsilon\otimes\tau$, a tensor on the sixteen modes of $V\otimes(\C\otimes\Oct)$. In the Fock space of these modes with exchange sign $\eta$, $-1$ for fermions and $+1$ for bosons, the pair $\sum_{A,B}\kappa_{AB}\,c_A^\dagger c_B^\dagger\,\Omega$ is nonzero if and only if $\eta=-s_\tau$. As a two-particle tensor $\kappa$ has eigenvalue $-s_\tau$ under the swap and $-1$ under a rotation by $2\pi$ of one particle, and the two agree exactly when $\tau$ is symmetric. So the invariant pair $\varepsilon\otimes\sum_ie_i\otimes\bar e_i$ has a half that makes pairs only for fermions and a half only for bosons, and a pair neutral for all seven points is $\varepsilon\otimes\tau_0$ up to scale and makes pairs only for fermions.

At $e_p$, let $P_\ell$ and $P_q$ be the projections onto the plane $\langle1,e_p\rangle$ and onto the six units orthogonal to it. The symmetric pair tensors that are $\mathfrak{su}(3)_{e_p}$-invariant and annihilated by $D_p\otimes I+I\otimes D_p$ are spanned by $\tau_0$ and $\tau_0\circ(P_\ell-\tfrac13P_q)$, and $P_\ell-\tfrac13P_q=-\sigma\,(B-L)D_p$, where $D_p=\sigma(-1)^N$. For each of the fifteen generators $X$ of $\mathfrak{su}(4)$, and for $X=B-L$, the tensor $\tau_0\circ X$ is antisymmetric, and $(B-L)D_p$ commutes with no $L_{e_q}$, $q\neq p$.

If each generation’s lepton and quark pair coefficients are equal, and the lepton and quark terms are dressed by factors that do not depend on the generation, then $(m_s/m_b)/(m_\mu/m_\tau)=1$. If instead the second generation carries only the second structure and the third only the first, the ratio of ratios is $1/3$ in absolute value.