Twisted cycle equilibria and transverse positivity

Set $\delta =2\pi \mathrm{/}n$ and $v(t)=1-\cos t$. By [F2], $v$ is even and periodic, with $v^{\mathrm{\prime }}(t)=\sin t$ and $v^{\mathrm{\prime }\mathrm{\prime }}(t)=\cos t$; it is $C^2$ by those derivative formulas. Use weight one on each cycle edge and zero otherwise. Since $n\ge 5$, the two neighbors of each vertex are distinct and the pair-potential expression in [F3] counts each edge once. At ${\theta }^{\ast }$ the two incident differences at vertex $j$ are congruent to $\delta$ and $-\delta$ modulo $2\pi$, including across the edge from $n-1$ to $0$. Thus [F2] and [F3] give $(\mathrm{\nabla }U({\theta }^{\ast }){)}_j=\sin \delta +\sin (-\delta )=0$. Therefore the negative-gradient vector field vanishes.

Use the relative-coordinate section with vertex $0$ fixed. For a sufficiently small relative-coordinate displacement $h=(h_1,\dots ,h_{n-1})$, put $h_0=0$ and represent the nearby class by ${\theta }^{\ast }+h$. The descended potential in this chart is $C^2$ by [F1] and [F2]. For each $j$, telescoping along the path from $0$ to $j$ and [F6] give $h_j^2\le n{\sum }_{k=0}^{n-1}(h_{k+1}-h_k{)}^2$, with indices modulo $n$. Summing over $j$ yields $\mathrm{\parallel }h{\mathrm{\parallel }}^2\le n^2{\sum }_k(h_{k+1}-h_k{)}^2$.

At every edge, the cosine of the phase difference is $c=\cos \delta$. For every real $f$, [F3] gives $f^T{\mathrm{\nabla }}^{2}U({\theta }^{\ast })f=c{\sum }_j(f_{j+1}-f_j{)}^2$. The cycle is connected because consecutive edges give a path between any two vertices; [F4] identifies the sum as $f^{T}Lf$. Both matrices are symmetric, so equality of their quadratic forms for every $f$ implies equality of matrices by evaluating on coordinate vectors and their pairwise sums. Hence ${\mathrm{\nabla }}^{2}U({\theta }^{\ast })=cL$.

Since $n\ge 5$, $0<\delta <\pi \mathrm{/}2$; [F5] therefore gives $c>0$. By [F4], the connected cycle Laplacian has nonnegative quadratic form and kernel consisting of constants. Consequently $f^T{\mathrm{\nabla }}^{2}U({\theta }^{\ast })f>0$ for each nonzero $f\in {\mathbf{1}}^{\perp }$, since a constant vector with zero coordinate sum is zero.

The first-order Taylor term vanishes by step 1.1. Combining [F6], steps 2.1 and 1.2 gives $U({\theta }^{\ast }+h)-U({\theta }^{\ast })\ge c\mathrm{\parallel }h{\mathrm{\parallel }}^2\mathrm{/}(2n^2)+o(\mathrm{\parallel }h{\mathrm{\parallel }}^2)$. Because $c>0$, the right side is positive for all sufficiently small nonzero $h$. The relative-coordinate homeomorphism in [F1] identifies these displacements with nearby distinct rotation classes, proving strict local minimality. The argument establishes no basin or measure assertion.