Two cliques joined by one edge synchronize despite poor expansion

Regard $G$ as the union of three complete graphs: the two cliques and the joining edge, the latter being a copy of $K_2$. Their edge sets are disjoint; the joining edge meets each clique at exactly its endpoint. By [F1], the energy of $G$ is the sum of their energies, each depending only on differences of phases within its own vertex set.

Let $\theta$ be a critical point of the energy of $G$ with positive-semidefinite Hessian. Write $V_1,V_2$ for the disjoint clique vertex sets, and $a\in V_1$, $b\in V_2$ for the joining endpoints. Fix either clique $C$, and let $c$ be its joining endpoint. For any real variation $h$ on $C$, set $k_x=h_x-h_c$ on $C$ and $k_x=0$ outside $C$. Along the path $\theta +tk$, every difference in $C$ changes by $t(h_x-h_y)$, while the joining edge and the other clique have constant differences. Thus the whole-graph energy along this path equals the energy of $C$ along its variation $h$, plus a constant. Its first derivative at $t=0$ is zero and its second derivative is nonnegative. Since $h$ was arbitrary, the restriction to each clique is critical with positive-semidefinite Hessian. This also covers a singleton clique, whose energy is constant.

For an arbitrary real variation $h$ on the joining edge $\{a,b\}$, instead set $k_x=h_a$ for every $x\in V_1$ and $k_x=h_b$ for every $x\in V_2$. Along $\theta +tk$, all within-clique differences remain constant, whereas the joining-edge difference changes by $t(h_a-h_b)$. Hence the whole-graph energy along this path equals the joining-edge energy along $h$, plus a constant. Whole-graph criticality and Hessian nonnegativity therefore make the first derivative zero and the second derivative nonnegative for every $h$. The joining-edge restriction is consequently critical with positive-semidefinite Hessian.

By [F2], each clique has a common phase and the endpoints of the joining edge have equal phases. Hence all vertices of $G$ have the same phase. Every equilibrium with Hessian nonnegative on real vectors is therefore fully synchronized. For the real symmetric Hessian $H$, writing $z=u+iv$ with real $u,v$ gives $z^{\ast }Hz=u^{T}Hu+v^{T}Hv$, so nonnegativity on real vectors is equivalent to nonnegativity on complex vectors as in [F3]. Thus [F3] gives global synchrony.

For equal clique sizes $m\ge 2$, let $X$ be one clique. Exactly the joining edge crosses from $X$ to $X^c$, so [F4] gives $e(X,X^c)=1$ and $\mathrm{\mid }X\mathrm{\mid }=m$. Defining edge expansion as the minimum of $e(S,S^c)\mathrm{/}\mathrm{\mid }S\mathrm{\mid }$ over nonempty $S$ with $\mathrm{\mid }S\mathrm{\mid }\le \mathrm{\mid }V\mathrm{\mid }\mathrm{/}2$, the admissible set $X$ gives the upper bound $1\mathrm{/}m$, which tends to zero. This proves the stated contrast between global synchrony and a uniform positive edge-expansion bound.