An explicit eight-phase weighted stability certificate

Every vertex has one incident edge of each of types $a,c,d$ and two of type $b$. At an even vertex the sine sum is $a\sin \alpha -c\cos \alpha -d\sin \alpha$; at an odd vertex it is its negative. The two $b$ edges have differences $\pi \mathrm{/}2$ and $-\pi \mathrm{/}2$ and cancel. Since $(a-d)\sin \alpha =c\cos \alpha$, these sums vanish. Vertices $0$ and $2$ differ by $\pi \mathrm{/}2$, proving nonsynchrony.

Put $A=a\cos \alpha$, $C=c\sin \alpha$, $D=d\cos \alpha$ and $h=A+C-D$. The Hessian edge coefficients are $A,C,-D,0$ on the four types $a,c,d,b$. Pair vertices as $(2r,2r+1)$ with $r\in {\mathbb{Z}}_4$. For each $z\in \{1,-1,i,-i\}$ the two-dimensional complex subspace with even coordinates $u{z}^r$ and odd coordinates $v{z}^r$ is invariant. Substitution into the Hessian gives the Hermitian block with diagonal entries $h$ and off-diagonal entries $-(A+C{z}^{-1}-D{z}^2)$ and its conjugate. These four subspaces span the whole complex vertex space: the four sequences $z^r$ are pairwise orthogonal since each nontrivial fourth root has geometric sum zero. Each block therefore has eigenvalues $h\pm \mathrm{\mid }A+C{z}^{-1}-D{z}^2\mathrm{\mid }$.

For $z=1$, the eigenvalues are $0,2h$; for $z=-1$, they are $2C,2(A-D)$; for $z=\pm i$, they are $h\pm \sqrt{(A+D{)}^2+C^2}$. Here $A-D=(a-d)\cos \alpha >0$, $C=c^2\cos \alpha \mathrm{/}(a-d)>0$ and $h>0$. Moreover $h^2-((A+D{)}^2+C^2)=2C(A-D)-4AD=2{\cos }^2\alpha (c^2-2ad)>0$. Thus all eigenvalues except the single zero are strictly positive. The zero vector mode has equal coordinates in both components and corresponds to constants; [F2] now gives the claimed positivity on the real subspace ${\mathbf{1}}^{\perp }$.