\subsection{Exact optimal readout on the oscillator slice} \label{app:oscillatoroptimal} The oscillator endpoint admits a stronger result than a gap estimate. For \(u=0\), we determine the exact minimum risk and an attaining eight-effect POVM for every \(1\le k\le2\) and \(0<\eta\le1/6\). The lower bound covers all measurements; only the attaining construction uses four equally spaced phases at each of two radii. Retain the moments in Eq.~\eqref{eq:familymoments}, now with \(\ell_1=1/\sqrt3\), \(\ell_2=\sqrt{2/3}\), and put \[ w_0=p/\sqrt3,\qquad m_0=p\sqrt{2/3},\qquad n_0=q+t. \] For real \(\xi,\zeta\), define \begin{align*} d&=p^2\zeta(n_0+\zeta)-qt\xi(w_0+\xi),\\ A&=\frac{\xi m_0pt(n_0+\zeta)}{d},\qquad D=\frac{\xi m_0qt(w_0+\xi)}{d},\\ R_k(\xi,\zeta)&=\begin{pmatrix} 0&\xi&0&A\\ -w_0-\xi&0&D&0\\ 0&-m_0-D&0&\zeta\\ -A&0&-n_0-\zeta&0 \end{pmatrix},\qquad c_i=(R_kS^{-1}R_k^T)_{ii}. \end{align*} These are the weighted-Gram formulas of Appendix~\ref{app:nonscalaroptimal} with different, explicitly stated ladder entries. Their algebra does not require that appendix's fixed value of \(k\). Indeed, \(R_k+R_k^T=-(B+B^\dagger)\), and the only possible off-diagonal weighted row products are \[ -\frac{\xi(m_0+D)}p+\frac{A\zeta}t=0,\qquad \frac{A(w_0+\xi)}p-\frac{D(n_0+\zeta)}q=0. \] Consequently, with \(A_0=T_0-\operatorname{diag}(c_i)\), \begin{equation} F(r\omega)-A_0= D(\omega)(rS+R_k)S^{-1}(rS+R_k)^TD(\omega)^\dagger\succeq0 \label{eq:oscillatorglobalfactor} \end{equation} for every \(r\ge0\), \(|\omega|=1\), including the origin. Here \(D(\omega)=\operatorname{diag}(1,\omega,\omega^2,\omega^3)\), distinct from the scalar entry \(D\) in \(R_k\). We specify the exact branch through normalization, not stationarity. Reorder \(R_kS^{-1}\) as \(1,3,2,4\), obtaining off-diagonal blocks \(K,L\), and set \[ \sigma=\Tr(KL),\qquad \kappa=\det(KL),\qquad h(x)=x^2-\sigma x+\kappa. \] Apply the tridiagonal null-vector recurrence in Appendix~\ref{app:nonscalaroptimal} with these \(w_0,m_0,n_0,p,q,t,c_i\). At the roots of \(h\), write the squared components as \(\nu_i(x)^2=E_i+X_i x\), with \(\nu_1=1\). The explicit rational formulas there give \(y=(1-E_2)/X_2\) and the two equations \begin{equation} G_3(k;\xi,\zeta)=E_3+X_3y-1=0,\qquad G_4(k;\xi,\zeta)=E_4+X_4y-1=0. \label{eq:oscillatornormalization} \end{equation} All their denominators are nonzero on the boxes used below. On each box, \[ \sigma>0,\quad\kappa>0,\quad\sigma^2>4\kappa, \quad X_2\ne0,\quad h(y)<0. \] Thus the roots \(x_-