TFPT scientific source excerpt
Original repository path: documents/archive--paper-latex--tfpt-42/sections/006-subsection-character-trace-identities-and-the-abelian-in.tex
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000131 | moments of $X$.
000132 | \end{proof}
000133 | 
000134 | \begin{remark}[Table as readout, character as proof device]
000135 | The one-family table remains useful as a reader-facing decoding sheet, but the whole six-sector
000136 | arithmetic is already compressed into the single multiplicative character
000137 | \Cref{eq:multiplicative-family-character}.
000138 | \end{remark}
000139 | 
000140 | \begin{remark}[Six admissible occupancy sectors]
000141 | The six monomials in \Cref{eq:split-weight-enumerator} are exactly the six allowed occupancy pairs $(n_3,n_2)$ of the family packet. This is the shortest combinatorial route from the split carrier to the six Standard Model multiplet sectors listed in \Cref{tab:one-family}.
000142 | \end{remark}
000143 | 
000144 | The carrier packet therefore reproduces the standard trace cancellations
000145 | \[
000146 | \tr_{S^+}Y=0,
000147 | \qquad
000148 | \tr_{S^+}Y^3=0,
000149 | \]
000150 | and packages them together with the quadratic trace in one formula. Since $\tr_{S^+}Y^2=4\gamma=10/3$ and $\tr_\Phi Y^2=1/2$ for one weak doublet with hypercharge $1/2$, the hypercharge trace coefficient organizes as
000151 | \begin{equation}
000152 | \label{eq:abelian-index}
000153 | \bone(N_{\mathrm{fam}},N_\Phi)
000154 | =
000155 | \frac25N_{\mathrm{fam}}\tr_{S^+}Y^2+\frac15N_\Phi\tr_\Phi Y^2
000156 | =
000157 | \frac85N_{\mathrm{fam}}\gamma+\frac1{10}N_\Phi
000158 | =
000159 | \frac43N_{\mathrm{fam}}+\frac1{10}N_\Phi.
000160 | \end{equation}
000161 | On the canonical branch $(N_{\mathrm{fam}},N_\Phi)=(3,1)$ this gives
000162 | \[
000163 | \bone=\frac{41}{10}.
000164 | \]
000165 | The abelian index is therefore not inserted as an isolated Standard Model datum; it is a carrier-level output once the family and Higgs counts are declared.
000166 | 
000167 | \begin{remark}[Canonical-branch Diophantine lock]
000168 | If one imposes $\bone=41/10$ on nonnegative integer data, then
000169 | \[
000170 | 40N_{\mathrm{fam}}+3N_\Phi=123.