The symmetric torus height flow is not Morse--Smale

Put $A=R+r\cos v>0$. Differentiating the embedding gives $g=A^{2}d{u}^2+r^{2}d{v}^2$, and hence $\dot{u}=-\cos u\mathrm{/}A$, $\dot{v}=\sin v\sin u\mathrm{/}r$. This smooth field is complete because the torus is compact.

The critical equations are $\cos u=0$ and $\sin v=0$, giving exactly four points. At each, the Hessian in $(u,v)$ is diagonal with entries $-A\sin u$ and $-r\cos v\sin u$. Both entries are nonzero. Thus $(\pi \mathrm{/}2,0)$ is a maximum, $(3\pi \mathrm{/}2,0)$ a minimum, and $a=(\pi \mathrm{/}2,\pi )$ and $b=(3\pi \mathrm{/}2,\pi )$ are saddles; in particular $f$ is Morse everywhere.

The circle $v=\pi$ is invariant. On its interval $\pi \mathrm{/}2<u<3\pi \mathrm{/}2$, $\dot{u}=-\cos u\mathrm{/}(R-r)>0$, so every point has backward limit $a$ and forward limit $b$. Locally at $a$, writing $w=v-\pi$ gives $\dot{w}=-\sin w\sin u\mathrm{/}r$; for $w\mathrm{\ne }0$ small this forces $\mathrm{\mid }w\mathrm{\mid }$ to increase backwards while $u$ stays near $\pi \mathrm{/}2$. Thus a backward-converging trajectory must have $w=0$. At $b$ the same equation forces nonzero $\mathrm{\mid }w\mathrm{\mid }$ to increase forwards, so a forward-converging trajectory must also have $w=0$. Uniqueness and flow transport show that the indicated unstable and stable branches are precisely arcs of this circle.

Consequently, at every point of this open interval, $T{W}^u(a)=T{W}^s(b)=\mathbb{R}{\mathrm{\partial }}_u$. Their tangent sum has dimension one, whereas the torus has dimension two. The metric pair $(f,g)$ therefore fails the Morse--Smale condition.