 Solution

An odd number of inhibitory connections gives $\Gamma <0$. Write $r=|\Gamma {|}^{1/3}>0$. The cube roots of $-r^3$ have arguments $\pi ,\pi /3,-\pi /3$, soThe real mode is always damped. The pair has negative real part precisely when $r\cos (\pi /3)<1$, givingfor strict linear asymptotic stability. For $r>2$ the origin is unstable to an oscillatory mode.

At $r=2=1/\cos (\pi /3)$ the eigenvalues are $-3/\tau$ and $\pm i\sqrt{3}/\tau$. Thus the linear oscillation frequency is $\boxed{{\omega }_H=\sqrt{3}/\tau }$ and its period is $2\pi \tau /\sqrt{3}$. Varying $r$ crosses the imaginary axis transversely, since $d\mathrm{Re}{\lambda }_{+}/dr=1/(2\tau )>0$.

The hyperbolic tangent has a negative cubic saturation and no quadratic term. To check the nonlinear onset, let $v$ be a critical complex eigenvector of $M=\beta W$ and $p$ its normalized adjoint eigenvector, with $p^{\ast }v=1$. For this weighted cycle, ${\overline{p}}_i{v}_i=1/3$. Since $Mv=(1+i\sqrt{3})v$, the cubic derivative evaluated on $(v,v,\overline{v})$ has projectionIts real part is strictly negative. With no quadratic contributions, this gives a negative first Lyapunov coefficient: the generic onset is a supercritical Hopf bifurcation, with a small attracting periodic orbit just beyond the threshold. Exactly at threshold the linear system has undamped oscillations, but nonlinear cubic damping makes sufficiently small oscillations decay; one must not infer a nonzero sustained cycle at the critical parameter solely from the imaginary eigenvalues.