TFPT scientific source excerpt Original repository path: documents/archive--tfpt-45--source_extracts--03_em_flavor_source/sections/005-source.tex Original file SHA-256: 786d2619e775775d70adf177794138485b261968042a880b2f0f617bb0111814 Inclusive original lines: 1-100 Line numbers below refer to the original source. This bounded excerpt is source evidence; it is not a new proof/checker run. Conversation metadata and personal execution details are omitted where indicated. 000001 | 000002 | Because $U_6$ has eigenphases $e^{in\pi/3}$, the regularized transport spectrum is 000003 | \begin{equation} 000004 | \label{eq:transport-spectrum} 000005 | \lambda_{y,n}^{-1} 000006 | = 000007 | y^2+\delta_{\mathrm{ph}}^2-2y\delta_{\mathrm{ph}}\cos\!\left(\frac{n\pi}{3}\right)+\varepsilon^2, 000008 | \qquad 000009 | n=0,\dots,5, 000010 | \end{equation} 000011 | for the eigenvalues $\lambda_{y,n}$ of $\mathcal Y_y^{(\varepsilon)}$. In particular, 000012 | \begin{lemma}[Finite admissible hexagon gap] 000013 | \label{lem:finite-admissible-hexagon-gap} 000014 | On the admissible cusp set 000015 | \[ 000016 | y\in\left\{1,\frac13,\frac23\right\}, 000017 | \] 000018 | the smallest singular value of the unregulated hexagon denominator is attained at $y=\frac23$ and $n=0$. Equivalently, 000019 | \[ 000020 | \epsilon_f 000021 | := 000022 | \min_{\substack{y\in\{1,\frac13,\frac23\}\\ n=0,\dots,5}} 000023 | \left|y-\delta_{\mathrm{ph}}e^{in\pi/3}\right| 000024 | = 000025 | \frac23-\delta_{\mathrm{ph}}. 000026 | \] 000027 | Hence the finite admissible hexagon has a strictly positive spectral gap $\epsilon_f$. 000028 | \end{lemma} 000029 | 000030 | \begin{proof} 000031 | For $\varepsilon=0$, the eigenvalues in \Cref{eq:transport-spectrum} reduce to 000032 | $|y-\delta_{\mathrm{ph}}e^{in\pi/3}|^{-2}$. By \Cref{thm:transport-pole-algebraic}, 000033 | $\delta_{\mathrm{ph}}$ is the unique admissible root in $(1/3,2/3)$. Hence the positive-real cusp 000034 | $y=2/3$ with $n=0$ is the nearest admissible point, while every other cusp or nonzero phase has 000035 | strictly larger distance. Therefore 000036 | \[ 000037 | \epsilon_f=\frac23-\delta_{\mathrm{ph}}, 000038 | \] 000039 | which is positive because $\delta_{\mathrm{ph}}<2/3$. 000040 | \end{proof} 000041 | 000042 | \begin{definition}[Gap-stable admissibility] 000043 | \label{def:gap-stable-admissibility} 000044 | Let 000045 | \[ 000046 | D_0:=D_{2/3}. 000047 | \] 000048 | We say that the admissible transport sector is gap-stable if the physical transport operator has the form 000049 | \[ 000050 | D_f=D_0+V_{\mathrm{adm}}, 000051 | \qquad 000052 | [P_{\mathrm{str}},V_{\mathrm{adm}}]=0, 000053 | \qquad 000054 | \|V_{\mathrm{adm}}\|<\frac{\epsilon_f}{2}. 000055 | \] 000056 | \end{definition} 000057 | 000058 | \begin{proposition}[Structural gap lift] 000059 | \label{prop:gap-lift-full-operator} 000060 | If admissibility is gap-stable in the sense of \Cref{def:gap-stable-admissibility}, then for every 000061 | \[ 000062 | \psi\in\mathrm{Ran}(P_{\mathrm{str}}) 000063 | \] 000064 | one has 000065 | \[ 000066 | \|D_f\psi\| 000067 | \ge 000068 | \frac{\epsilon_f}{2}\|\psi\|. 000069 | \] 000070 | Consequently 000071 | \begin{equation} 000072 | \label{eq:gap-lift-full-operator} 000073 | P_{\mathrm{str}} D_f^\dagger D_f P_{\mathrm{str}} 000074 | \ge 000075 | \frac{\epsilon_f^2}{4}\,P_{\mathrm{str}} 000076 | > 000077 | 0. 000078 | \end{equation} 000079 | In particular, the final admissible restriction to $\mathrm{Ran}(\Padm)$ is strictly positive. 000080 | \end{proposition} 000081 | 000082 | \begin{proof} 000083 | Since $\psi\in\mathrm{Ran}(P_{\mathrm{str}})$ and $[P_{\mathrm{str}},V_{\mathrm{adm}}]=0$, 000084 | \[ 000085 | \|D_f\psi\| 000086 | \ge 000087 | \|D_0\psi\|-\|V_{\mathrm{adm}}\psi\| 000088 | \ge 000089 | \bigl(\epsilon_f-\|V_{\mathrm{adm}}\|\bigr)\|\psi\| 000090 | \ge 000091 | \frac{\epsilon_f}{2}\|\psi\|. 000092 | \] 000093 | Squaring gives \Cref{eq:gap-lift-full-operator}. 000094 | \end{proof} 000095 | 000096 | \begin{equation} 000097 | \label{eq:mobius-quotient} 000098 | \frac{\lambda_{y,0}}{\lambda_{y,3}} 000099 | = 000100 | \frac{(y+\delta_{\mathrm{ph}})^2+\varepsilon^2}{(y-\delta_{\mathrm{ph}})^2+\varepsilon^2}.