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Consider the following non-convex function $$E(\theta):=-\sum_{i,j}A_{ij}\cos(\theta_i-\theta_j)$$ where $A$ is a symmetric, diagonal-free matrix whose non-diagonal element are $\pm 1$. In other words, $A_{ii}=0$ and $A_{ij}= \pm 1$ for $i \neq j$ .

The second-order critical points of this function satisfy:

(1) $\sum_{j}A_{ij}\sin(\theta_i-\theta_j)=0$

(2) The Hessian matrix $H$, whose $H_{ii}=\sum_jA_{ij}\cos(\theta_i-\theta_j)$ and $H_{i\neq j}=-A_{ij}\cos(\theta_i-\theta_j)$ is PSD.

My question: Under what conditions on $A$ will it be the case that all second-order critical points satisfy the inequality

$$\sum_j\cos(\theta_i-\theta_j)>0,\forall i\in[n].$$

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    $\begingroup$ Added the statistical physics tag because looks like a spin glass version of the XY or rotator model. $\endgroup$
    – Abdelmalek Abdesselam
    Commented Feb 28 at 22:05

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