TFPT scientific source excerpt
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000172 | Hence from now on
000173 | \[
000174 | E=E_3\oplus E_2,
000175 | \qquad
000176 | X=-2P_3+3P_2,
000177 | \qquad
000178 | P_3=\frac{\mathbf 1-\epscar}{2},
000179 | \qquad
000180 | P_2=\frac{\mathbf 1+\epscar}{2},
000181 | \]
000182 | and the hypercharge operator is the normalized primitive two-point generator
000183 | \begin{equation}
000184 | \label{eq:explicit-hypercharge}
000185 | Y
000186 | =
000187 | \mathrm{diag}\!\left(-\frac13,-\frac13,-\frac13,\frac12,\frac12\right)
000188 | =
000189 | \frac{X}{6}
000190 | =
000191 | -\frac13 P_3+\frac12 P_2
000192 | =
000193 | \frac1{12}\mathbf 1+\frac5{12}\epscar.
000194 | \end{equation}
000195 | 
000196 | \begin{lemma}[Charge-generator / carrier-polarization inversion]
000197 | On the rigid carrier, the carrier polarization and the primitive two-point generator are recovered by
000198 | \begin{align}
000199 | \epscar&=\frac{2X-\mathbf 1}{5}=\frac{12Y-\mathbf 1}{5},
000200 | \label{eq:hypercharge-seam-inversion}\\
000201 | \qquad
000202 | X&=\frac{5\epscar+\mathbf 1}{2},
000203 | \nonumber\\
000204 | (2X-\mathbf 1)^2&=25\,\mathbf 1,
000205 | \nonumber\\
000206 | X^2-X-6\,\mathbf 1&=0,
000207 | \nonumber\\
000208 | 6Y^2-Y-\mathbf 1&=0.
000209 | \label{eq:carrier-polynomial}
000210 | \end{align}
000211 | Consequently the carrier projectors are
000212 | \[
000213 | P_2=\frac{\mathbf 1+\epscar}{2}=\frac{2(3Y+\mathbf 1)}{5},
000214 | \qquad
000215 | P_3=\frac{\mathbf 1-\epscar}{2}=\frac{3(\mathbf 1-2Y)}{5}.
000216 | \]
000217 | \end{lemma}
000218 | 
000219 | \begin{proof}
000220 | The affine relations between $(X,Y)$ and $\epscar$ are inverted algebraically. Squaring the first
000221 | relation and using $\epscar^2=\mathbf 1$ gives the quadratic generator polynomial
000222 | $X^2-X-6\,\mathbf 1=0$, and dividing by $36$ yields the hypercharge polynomial. The projector
000223 | formulas then follow from $P_2=(\mathbf 1+\epscar)/2$ and $P_3=(\mathbf 1-\epscar)/2$.
000224 | \end{proof}
000225 | 
000226 | \begin{lemma}[Primitive trace-balanced carrier]
000227 | \label{lem:primitive-trace-balanced-carrier}
000228 | Let $Y$ be an endomorphism of a finite-dimensional carrier $E$ such that
000229 | \[
000230 | 6Y^2-Y-\mathbf 1=0,
000231 | \qquad
000232 | \tr_EY=0,
000233 | \]
000234 | and assume the realization is primitive in the sense that $E$ is not a nontrivial direct sum of
000235 | identical trace-balanced $Y$-modules. Then
000236 | \[
000237 | \operatorname{Spec}(Y)=\left\{-\frac13,\frac12\right\},
000238 | \qquad
000239 | \dim\ker\!\left(Y+\frac13\mathbf 1\right)=3,
000240 | \qquad
000241 | \dim\ker\!\left(Y-\frac12\mathbf 1\right)=2.
000242 | \]
000243 | Equivalently,
000244 | \[
000245 | \rank E=5,
000246 | \qquad
000247 | E=E_3\oplus E_2,
000248 | \qquad
000249 | Y=-\frac13P_3+\frac12P_2.
000250 | \]
000251 | \end{lemma}
000252 | 
000253 | \begin{proof}
000254 | The quadratic polynomial factors as
000255 | \[
000256 | (2Y-\mathbf 1)(3Y+\mathbf 1)=0,
000257 | \]
000258 | so the only eigenvalues are
000259 | \[
000260 | y_+=\frac12,
000261 | \qquad
000262 | y_-=-\frac13.
000263 | \]
000264 | Write
000265 | \[
000266 | m_+:=\dim\ker(Y-y_+\mathbf 1),
000267 | \qquad
000268 | m_-:=\dim\ker(Y-y_-\mathbf 1).
000269 | \]
000270 | Then
000271 | \[
000272 | 0=\tr_EY=\frac12\,m_+-\frac13\,m_-,
000273 | \]
000274 | so
000275 | \[
000276 | 3m_+=2m_-.
000277 | \]
000278 | Because $\gcd(2,3)=1$, every positive integer solution has the form
000279 | \[
000280 | (m_+,m_-)=(2k,3k).
000281 | \]
000282 | Primitivity forces $k=1$, hence $(m_+,m_-)=(2,3)$. This is exactly the rigid $3+2$ split, with
000283 | the displayed spectral form of $Y$.