Morse Functions Critical Values and Genericity — Examples

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].

Parametrize the circle by $x(\theta )=(\cos \theta ,\sin \theta )$. Then $d_p(\theta )=\mathrm{\parallel }x(\theta )-p{\mathrm{\parallel }}^2=1+a^2+b^2-2(a\cos \theta +b\sin \theta )\mathrm{.}$ Therefore $d_p^{\mathrm{\prime }}(\theta )=2(a\sin \theta -b\cos \theta )\text{and}{d}_p^{\mathrm{\prime }\mathrm{\prime }}(\theta )=2(a\cos \theta +b\sin \theta )\mathrm{.}$

If $p\mathrm{\ne }0$, the equation $a\sin \theta =b\cos \theta$ has exactly two solutions modulo $2\pi$, corresponding to the two points where the radius through $p$ meets the circle. At those two points one has $d_p^{\mathrm{\prime }\mathrm{\prime }}(\theta )=\pm 2\sqrt{a^2+b^2}\mathrm{\ne }0$, so both critical points are nondegenerate and $d_p$ is Morse. This matches the generic-center theorem [L1].

If $p=0$, then $d_0(\theta )\equiv 1$ is constant. Every point of the circle is critical, so the function is degenerate and not Morse.

Thus centers off the medial axis yield Morse squared-distance functions, while the center itself is the exceptional degenerate case.

The partial derivatives are $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }x=-2\pi \sin (2\pi x)$ and $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }y=-2\pi \sin (2\pi y)$, so the critical points are exactly the four points with $x,y\in \{0,\frac{1}{2}\}$ modulo $\mathbb{Z}$. At the two saddles $(0,\frac{1}{2})$ and $(\frac{1}{2},0)$ the critical value is $0$.

The Hessian is diagonal with entries $-4{\pi }^2\cos (2\pi x)$ and $-4{\pi }^2\cos (2\pi y)$, so all four critical points are nondegenerate. By [F1], $f$ is Morse. Choose pairwise disjoint neighbourhoods of the two saddles and bump functions ${\rho }_1,{\rho }_2$ that are identically $1$ near the corresponding saddle and supported away from the other one.

For sufficiently small $\epsilon >0$, define $g:=f+\epsilon {\rho }_1-\epsilon {\rho }_2\mathrm{.}$ The compact perturbation argument from [L1] applies to these two fixed bumps: for small enough $\epsilon$, the function $g$ has the same critical points as $f$ and the same Hessians at those critical points. Near the first saddle one has $g=f+\epsilon$, and near the second one has $g=f-\epsilon$, so the two saddle critical values become $\epsilon$ and $-\epsilon$.

Hence two equal critical levels can be separated by disjoint bump perturbations without changing the local Hessians.

The partial derivatives are $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }x=-2\pi \sin (2\pi x)$ and $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }y=-2\pi \sin (2\pi y)$, so the critical points are exactly the four points with $x,y\in \{0,\frac{1}{2}\}$ modulo $\mathbb{Z}$.

The Hessian is diagonal with entries $-4{\pi }^2\cos (2\pi x)$ and $-4{\pi }^2\cos (2\pi y)$. At each of the four critical points these entries are nonzero, so every critical point is nondegenerate. By [F1], the function $f$ is Morse.

The two saddle points $(0,\frac{1}{2})$ and $(\frac{1}{2},0)$ both have critical value $0$, so distinct critical points can share one critical level. This is exactly the boundary described in [L1], and by [F1] it shows that $f$ is not excellent. Therefore the displayed universal claim is false.

One has $\mathrm{\parallel }{g}_n-f{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1\mathrm{/}n$, so the perturbations are uniformly tiny in value. Their supports lie in $[n-1\mathrm{/}{n}^2,n+1\mathrm{/}{n}^2]$, hence for every fixed compact set $K\subseteq \mathbb{R}$ one has $g_n=f$ on $K$ once $n$ is large enough.

Differentiating gives $g_n^{\mathrm{\prime }}(x)=1+n{\beta }^{\mathrm{\prime }}(n^2(x-n))$. In particular $g_n^{\mathrm{\prime }}(n)=1+n{\beta }^{\mathrm{\prime }}(0)=1-n\le 0$, with equality only when $n=1$, while outside the support interval one has $g_n^{\mathrm{\prime }}(x)=1$. Thus $g_1^{\mathrm{\prime }}(1)=0$, and for every $n\ge 2$ continuity gives some $x_n\in [n-1\mathrm{/}{n}^2,n+1\mathrm{/}{n}^2]$ with $g_n^{\mathrm{\prime }}(x_n)=0$. Thus each $g_n$ has a new critical point near $x=n$.

Step 1.1 shows that the perturbations are tiny on every fixed compact set, while step 2.1 shows that they still create new far-out critical points. This is exactly the noncompact failure mode recorded in [L1]. Therefore the displayed claim is false.