A positive transverse Hessian gives an open nonsynchronous basin

Choose real representatives of $q$ in an angular chart and put $n=\mathrm{\mid }V\mathrm{\mid }$ and $E={\mathbf{1}}^{\perp }$. Differentiating $U(\theta +c\mathbf{1})=U(\theta )$ in $c$ and in the coordinates of $\theta$, using [F6], gives ${\mathbf{1}}^T\mathrm{\nabla }U(\theta )=0$ and $\mathrm{\nabla }U(\theta +c\mathbf{1})=\mathrm{\nabla }U(\theta )$. Differentiating the latter identity in $c$ gives $H\mathbf{1}=0$. By [F4], $H$ is symmetric, so ${\mathbf{1}}^{T}Hx=(H\mathbf{1}{)}^{T}x=0$ for $x\in E$: its restriction to $E$ is a self-adjoint endomorphism. The nonsynchrony hypothesis excludes $n=1$, so $E$ has positive dimension. By [F5], choose an orthonormal eigenbasis of this restriction. Every eigenvalue is positive by the assumed positive quadratic form; let $m>0$ be their minimum. Expansion in this basis gives $x^{T}Hx\ge m\mathrm{\parallel }x{\mathrm{\parallel }}^2$ for $x\in E$.

Continuity of the Hessian gives $r>0$ such that for $\mathrm{\parallel }z\mathrm{\parallel }\le r$, all entries of $A(z)={\mathrm{\nabla }}^{2}U(q+z)-H$ have absolute value less than $m\mathrm{/}(2n)$ and these points lie in the chosen chart. Indeed, $\mathrm{\mid }{x}^{T}A(z)x\mathrm{\mid }\le {\max }_{i,j}\mathrm{\mid }A(z{)}_{ij}\mathrm{\mid }({\sum }_i\mathrm{\mid }{x}_i\mathrm{\mid }{)}^2\le n{\max }_{i,j}\mathrm{\mid }A(z{)}_{ij}\mathrm{\mid }\mathrm{\parallel }x{\mathrm{\parallel }}^2$; the last inequality follows by expanding ${\sum }_{i<j}(\mathrm{\mid }{x}_i\mathrm{\mid }-\mathrm{\mid }{x}_j\mathrm{\mid }{)}^2\ge 0$. Thus $x^T{\mathrm{\nabla }}^{2}U(q+z)x\ge (m\mathrm{/}2)\mathrm{\parallel }x{\mathrm{\parallel }}^2$ for $x\in E$. For $y\in E$ with $\mathrm{\parallel }y\mathrm{\parallel }<r$, the scalar function $s\mapsto y^T\mathrm{\nabla }U(q+sy)$ is continuously differentiable on an open interval containing $[0,1]$, by smoothness and [F6]. Its derivative is $y^T{\mathrm{\nabla }}^{2}U(q+sy)y$, and its value at zero is zero because $q$ is an equilibrium. Applying both parts of [F7] yields $y^T\mathrm{\nabla }U(q+y)={\int }_0^1{y}^T{\mathrm{\nabla }}^{2}U(q+sy)yds\ge (m\mathrm{/}2)\mathrm{\parallel }y{\mathrm{\parallel }}^2$.

Shrink $r$ if necessary and choose $\delta >0$ so that the closed cylinder $q+c\mathbf{1}+y$, $\mathrm{\mid }c\mathrm{\mid }\le \delta$, $y\in E$, $\mathrm{\parallel }y\mathrm{\parallel }\le r$, lies inside the angular chart. A solution initially $q+c\mathbf{1}+y_0$ with $\mathrm{\mid }c\mathrm{\mid }<\delta$ and $\mathrm{\parallel }{y}_0\mathrm{\parallel }<r$ has constant mean coordinate $c$ and satisfies $\dot{y}=-\mathrm{\nabla }U(q+y)$ by the rotation identities in step 1.1. As long as it stays in the ball, step 2.1 and [F6] give $d\mathrm{\parallel }y{\mathrm{\parallel }}^2\mathrm{/}dt\le -m\mathrm{\parallel }y{\mathrm{\parallel }}^2$. By [F6] and [F8], $g(t)=e^{mt}\mathrm{\parallel }y(t){\mathrm{\parallel }}^2$ has $g^{\mathrm{\prime }}(t)\le 0$. On any such interval $[0,t]$, [F7] gives $g(t)-g(0)={\int }_0^t{g}^{\mathrm{\prime }}(s)ds\le 0$; hence $\mathrm{\parallel }y(t){\mathrm{\parallel }}^2\le e^{-mt}\mathrm{\parallel }{y}_0{\mathrm{\parallel }}^2$. Since $e^{mt}$ has positive derivative and $e^0=1$ (the reciprocal identity at zero and positivity in [F8]), [F7] also gives $e^{mt}\ge 1$ for $t\ge 0$. Thus the estimate prevents a first exit from the ball by continuity. The closed cylinder stays inside the chart, and the complete flow in [F1] continues this estimate for all positive times. Finally [F8] gives $y(t)\to 0$, so the trajectory converges to $R_{c}q$.

Allowing $c$ to range over $(-\delta ,\delta )$ and $y_0$ over a smaller open ball in $E$ gives a nonempty open set in the angular chart: $\mathbb{R}\mathbf{1}\oplus E={\mathbb{R}}^V$, with continuous inverse coordinates $c=n^{-1}{\sum }_v({\theta }_v-q_v)$ and $y=\theta -q-c\mathbf{1}$. Step 3.1 makes each such solution converge to a rotation of $q$. No such limit is synchronous by hypothesis. The synchronous set is closed, being the zero fibre of the continuous relative-phase map in [F2]; an open metric ball about each nonsynchronous limit is therefore disjoint from that set. Convergence to this limit prevents the solution from approaching the synchronous set. By [F3] the open initial set has positive phase volume and cannot be contained in a phase-null exception set. This rules out almost-everywhere synchronous convergence.