A standard embedded torus has generic height directions with four Morse critical points, but symmetry directions create degenerate or nongeneric behavior

Write a unit direction as $\mathbf{q}=(A\cos \varphi ,A\sin \varphi ,c)$ with $A\ge 0$ and $A^2+c^2=1$. Its height on the torus is $h_{\mathbf{q}}(u,v)=A(R+r\cos u)\cos (v-\varphi )+cr\sin u\mathrm{.}$ If $A>0$, then $R+r\cos u>0$ makes the critical-point equations equivalent to $\sin (v-\varphi )=0,-A\sin u\cos (v-\varphi )+c\cos u=0.$ For each of the two signs $s=\cos (v-\varphi )\in \{1,-1\}$, the second equation has exactly two solutions modulo $2\pi$. Hence every nonvertical direction has exactly four critical points.

At a critical point from step 1.1 the mixed second derivative vanishes, while the two diagonal Hessian entries are $-r(As\cos u+c\sin u),-As(R+r\cos u)\mathrm{.}$ The first has absolute value $r\sqrt{A^2+c^2}=r$, and the second is nonzero because $A>0$ and $R+r\cos u>0$. Thus all four critical points are nondegenerate, so every nonvertical height is Morse by [F1]. In particular the $x$-height is the case $(A,c,\varphi )=(1,0,0)$.

If $A=0$, then $c=\pm 1$ and the height is $k(u,v)=\pm r\sin u$. Its critical set is the union of the two circles $u=\pi \mathrm{/}2$ and $u=3\pi \mathrm{/}2$, so both vertical directions are degenerate and not Morse.

The two vertical poles form a null subset of the direction sphere, while every direction in their complement has the four Morse critical points from steps 1.1 and 2.1. This proves directly that generic directions have four critical points and identifies the exceptional symmetry directions, consistently with [L1].