Convergence of the universal sum-of-squares hierarchy
Prove that the universal sum-of-squares hierarchy for generalised incompatibility robustness satisfies \lim_{t\to\infty}\chi^g_{k,t}(n)=\lambda_{k,n}/k for every k,n\geq2, and consequently establish \chi^g_3(n)=\lambda_{3,n}/3 whenever a 3-fold unbiased measurement with n outcomes exists, including maximal incompatibility of the Hadamard--Clifford triples.
References
The broader pattern motivates the following conjecture. For every $k,n\geq2$, \begin{equation} \lim_{t\to\infty}\chi_{k,t}{g}(n)=\frac{\lambda_{k,n}}{k}, \qquad\text{so that}\qquad \chi_3{g}(n)=\frac{\lambda_{3,n}}{3} \end{equation} whenever a 3-UM with $n$ outcomes exists, and every 3-UM with those parameters is maximally incompatible.
— $k$-fold unbiased measurements and maximal incompatibility
(2609.20728 - Designolle et al., 17 Sep 2026) in Conjecture 1, Section 6.2, 'Universal sum-of-squares hierarchy'