All critical PSD equilibria synchronize on positive regular expanders

The norm inequality and $d>0$ force $\alpha \ge 0$, and imply the same inequality with $a_0=0.0816$ in place of $\alpha$. The graph is connected: otherwise subtracting its mean from the indicator of a nontrivial component gives a nonzero vector $v$ with $Jv=0$ and $Av=dv$, whence $\mathrm{\parallel }(A-(d\mathrm{/}n)J)v\mathrm{\parallel }=d\mathrm{\parallel }v\mathrm{\parallel }>a_{0}d\mathrm{\parallel }v\mathrm{\parallel }$, contrary to the norm bound. This component calculation uses regularity, including the loop convention.

Let $\theta$ be any critical PSD state. Rotate its phase average to be real and nonnegative and choose principal representatives. The regular PSD mass lemma gives a strictly positive first order parameter, so this normalization is legitimate. The certified PSD amplification lemma gives $C_{\pi \mathrm{/}2}=\mathrm{\varnothing }$. Every absolute phase is therefore below $\pi \mathrm{/}2$, and the PSD half-circle lemma on the connected graph gives all normalized phases equal zero. Undoing the common rotation leaves every pair of phases equal modulo $2\pi$, which is full synchronization as defined in [F2]. No local-minimum conclusion has been used.