Broken trajectories in an index-two torus moduli space

Choose a smooth positive periodic function $\mu (\theta )$ equal to $4\mathrm{/}(1+\cos \theta )$ near $0$ and $4\mathrm{/}(1-\cos \theta )$ near $\pi$. It exists by [F1]: use disjoint cutoff neighbourhoods of the poles and take the convex combination of these positive local functions with the constant $2$ outside them. Put $Y(\theta )=\mu (\theta )\sin \theta$ and $X=(Y(u),Y(v))$. Then $df(X)=-\mu (u){\sin }^{2}u-2\mu (v){\sin }^{2}v<0$ off the four critical points. Near $0$, the signed coordinate $w=\mathrm{sgn}(\theta )\sqrt{1-\cos \theta }$ is smooth and satisfies $\dot{w}=2w$; near $\pi$, $w=\mathrm{sgn}(\theta -\pi )\sqrt{1+\cos \theta }$ satisfies $\dot{w}=-2w$. Multiplying the second coordinate by $\sqrt{2}$ gives exactly the Morse coordinates for the weighted second cosine. Thus $X$ satisfies [F2]. The Hessian $\mathrm{diag}(-\cos u,-2\cos v)$ has indices $2,1,1,0$ at $a,b,c,d$.

In each coordinate, stable and unstable sets are a pole or the circle with the other pole removed. Their products in $M$ are transverse: every nonempty coincidence has, in each coordinate, at least one full tangent direction; the only potential point/point intersection for distinct equilibria is empty. The smooth field is complete on compact $M$ by [F2], so the pair is Morse--Smale. For each adjacent-index pair, one coordinate is constant and the other follows either of the two complementary arcs, giving exactly two orbit classes. These are all adjacent-index pairs.

On each of the four open rectangles between the coordinate separatrices, both coordinates run from $0$ to $\pi$. On either chosen arc the time coordinate $\tau (\theta )={\int }_{{\theta }_{\ast }}^{\theta }d\eta \mathrm{/}Y(\eta )$ is a diffeomorphism onto $\mathbb{R}$: its derivative has the sign of $Y$, and the simple zeros of $Y$ make the endpoint times infinite. Every trajectory is therefore $u(t)={\tau }_u^{-1}(t-A)$, $v(t)={\tau }_v^{-1}(t-B)$. Common time translation changes $A$ and $B$ equally, leaving the relative delay $A-B\in \mathbb{R}$ as the unique parameter. Its two infinite ends break through $b$ and $c$, respectively, with the arc choices fixed. Hence there are four open intervals, each with its two distinct broken endpoints.

By [F3] these are exactly the compactifying endpoints, with one collar branch each; the fixed arc choices identify each broken pair with the appropriate rectangle, so no endpoints are identified across intervals. There are $2\cdot 2+2\cdot 2=8$ pairs in $(\mathcal{M}(a,b)\times \mathcal{M}(b,d))\bigsqcup (\mathcal{M}(a,c)\times \mathcal{M}(c,d))$, two per closed interval. By [F4] the differential is $\mathrm{\partial }a=2b+2c=0$, $\mathrm{\partial }b=\mathrm{\partial }c=2d=0$ and $\mathrm{\partial }d=0$.