TFPT scientific source excerpt
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000114 | \begin{proposition}[Finite positive transport algebra]
000115 | \label{prop:finite-positive-transport-algebra}
000116 | Let
000117 | \[
000118 | C:=\frac12\bigl(U_6+U_6^\dagger\bigr),
000119 | \qquad
000120 | a:=y^2+\delta_{\mathrm{ph}}^2+\varepsilon^2,
000121 | \qquad
000122 | b:=y\delta_{\mathrm{ph}}.
000123 | \]
000124 | Then
000125 | \[
000126 | D_y^\dagger D_y+\varepsilon^2=a\,\mathbf 1-2b\,C.
000127 | \]
000128 | The cosine operator $C$ has spectrum
000129 | \[
000130 | \mathrm{Spec}(C)=\left\{1,\frac12,-\frac12,-1\right\}
000131 | \]
000132 | and satisfies
000133 | \begin{equation}
000134 | \label{eq:c6-cosine-polynomial}
000135 | 4C^4-5C^2+\mathbf 1=0.
000136 | \end{equation}
000137 | Consequently the regularized positive transport kernel reduces to the cubic polynomial
000138 | \begin{equation}
000139 | \label{eq:finite-positive-transport-kernel}
000140 | \mathcal Y_y^{(\varepsilon)}
000141 | =
000142 | \frac{
000143 | a^3-5ab^2
000144 | +2b(a^2-5b^2)C
000145 | +4ab^2C^2
000146 | +8b^3C^3
000147 | }{
000148 | (a^2-b^2)(a^2-4b^2)
000149 | }.
000150 | \end{equation}
000151 | \end{proposition}
000152 | 
000153 | \begin{proof}
000154 | Because $U_6$ is unitary,
000155 | \[
000156 | D_y^\dagger D_y
000157 | =
000158 | \bigl(y\,\mathbf 1-\delta_{\mathrm{ph}}U_6^\dagger\bigr)
000159 | \bigl(y\,\mathbf 1-\delta_{\mathrm{ph}}U_6\bigr)
000160 | =
000161 | \bigl(y^2+\delta_{\mathrm{ph}}^2\bigr)\mathbf 1
000162 | -y\delta_{\mathrm{ph}}(U_6+U_6^\dagger),
000163 | \]
000164 | which gives the first identity after adding $\varepsilon^2$. The eigenvalues of $C$ are $\cos(n\pi/3)$ for $n=0,\dots,5$, namely $1,\frac12,-\frac12,-1$. Hence $C$ obeys the minimal polynomial \Cref{eq:c6-cosine-polynomial}. The inverse of $a-2bc$ on this four-point spectrum is therefore represented by a unique cubic interpolation polynomial in $c$, and evaluating that interpolation yields \Cref{eq:finite-positive-transport-kernel}.
000165 | \end{proof}
000166 | 
000167 | \begin{proposition}[Strict positivity of the finite admissible $C_6$ Green kernel]
000168 | \label{prop:strict-positive-c6-green}
000169 | For each admissible cusp
000170 | \[
000171 | y\in\left\{1,\frac13,\frac23\right\}
000172 | \]
000173 | and each $\varepsilon\ge 0$, set
000174 | \[
000175 | a_y:=y^2+\delta_{\mathrm{ph}}^2+\varepsilon^2,
000176 | \qquad
000177 | b_y:=y\,\delta_{\mathrm{ph}},
000178 | \qquad
000179 | r_y:=\frac{a_y-\sqrt{a_y^2-4b_y^2}}{2b_y}.
000180 | \]
000181 | Then
000182 | \[
000183 | a_y-2b_y=(y-\delta_{\mathrm{ph}})^2+\varepsilon^2>0,
000184 | \qquad
000185 | 0<r_y<1.
000186 | \]
000187 | If $d=d_{C_6}(i,j)$ denotes the cyclic graph distance on the hexagon, the regularized Green
000188 | kernel satisfies
000189 | \[
000190 | (\mathcal Y_y^{(\varepsilon)})_{ij}
000191 | =
000192 | \frac{r_y^{d}+r_y^{6-d}}
000193 | {\sqrt{a_y^2-4b_y^2}\,(1-r_y^6)}.
000194 | \]
000195 | Hence every matrix entry of $\mathcal Y_y^{(\varepsilon)}$ is strictly positive.
000196 | \end{proposition}
000197 | 
000198 | \begin{proof}[Proof sketch]