A generalization of the "Four vertex theorem"

Question

Assume that $S$ is a compact surface in $\mathbb{R}^3$. Its Gaussian curvature is denoted by $\kappa$.

Is it true to say that $\kappa$ has more than two critical points?

Answer 1

You can prescribe Gaussian curvature function on the smooth sphere as long as it is positive (Kazdan and Warner, 1980s) and then embed the surface isometrically as a convex surface in $E^3$: Alexandrov, Pogorelov, Nierenberg. The history of the embedding result is a bit convoluted: It was first proven by Alexandrov in 1940s without any smoothness properties; smoothness (as well as uniqueness of convex embedding) was established shortly thereafter by Pogorelov and few years later by Nierenberg. Nierenberg's proof is probably the most readable. I will chase the references when I have time.

Thus, the answer to your question is negative, you can have exactly two critical points. It would be interesting though to find out if your question about critical points has a positive answer for the mean curvature function.

Answer 2

Here's a concrete example of a 2-vertex surface.

TL;DR: glue together two half-ellipsoids.

Consider the surface of revolution parameterized by $$ S(u, v) = (\varphi(v) \cos u, \varphi(v) \sin u, \psi(v)). $$ The "profile curve" $(\varphi, \psi)$ completely determines the curvature. Computing the second fundamental form and writing things out yields \begin{align} K &= \frac{N}{D}\\ N &= -(\psi'^2)\varphi'' + \varphi' \psi' \psi'' \\ D &= \varphi (\varphi'^2 + \psi'^2)^2 \end{align} Let's look at one quarter of an ellipse, $E_{a,r}$ parameterized via \begin{align} \varphi(t) &= ar \sin t \\ \psi(t) &= r \cos t, \end{align} for $t \in [0, \pi/2]$. Direct computation shows that \begin{align} N &= ar^3 \sin^2 t \sin t + ar^3 \cos^2 t \sin t \\ &= ar^3 \sin t ~~(\sin^2 t + \cos^2 t) \\ &= ar^3 \sin t \\ D &= ar \sin t ~~(a^2 r^2 \cos^2 t + r^2 \sin'^2 t)^2 \\ &= ar^5 \sin t ~~(a^2 \cos^2 t + \sin'^2 t)^2 \end{align} so that the Gaussian curvature of the associated surface of revolution is \begin{align} K&= \frac{N}{D} \\ &= \frac{1}{r^2 \left(a^2 \cos^2 t + \sin^2 t\right)^2}. \end{align}

Letting \begin{align} u &= a^2 \cos^2 t + \sin^2 t. \end{align} we see (via some calculus that I'm not going to write out here, because I know you can do it) that for $a > 1$, $u$ is a decreasing function on $[0, \pi/2]$, and for $0 < a < 1$, $u$ is increasing on the same interval. This means that the Gaussian curvature, as a function of $v$, is either increasing ($a < 1)$ or decreasing ($a > 1$).

Furthermore, at $t = \pi/2$, the point on our ellipse has horizontal tangent, is at location $$ P = (\psi(\pi/2)), \varphi(\pi/2)) = (0, ar) $$ and the surface there has curvature $$ K = \frac{1}{r^2}, $$ while at $t = 0$, the point is at $$ Q = (r, 0) $$ with a vertical tangent, and the surface there has curvature $$ K = \frac{1}{r^2 a^4}. $$

So if we choose $a = \frac{1}{2}$ and $r = 2$, we get an half-egg-like surface whose curvature goes from $4$ (at $t = 0$) to $1$ (at $t = \pi/2$).

On the other hand, if we choose $a = 2$ and $r = \frac{1}{2}$, the curvature goes from $\frac{1}{4}$ (at $t = 0$) to $1$ (at $t = \pi/2$).

If we flip this second curve across the $y$-axis, or simply parametrize it by $$ \pi /2 \le t \le \pi, $$ and join it to the first, we end up with a surface of revolution whose curvature increases steadily along its axis, from $1/4$ to $4$.

It seems only fair to include a picture of this surface, badly created in Matlab:

and, for the record, the matlab code that created it:

function ctest()
clear;

n = 100;
k = 50;

t = linspace(0, pi/2, n);
a = 0.5; 
r = 2;
psi1 = r * cos(t); 
phi1 = a * r * sin(t); 

a = 2; 
r = 0.5;
psi2 = -r * cos(t(end:-1:1)); 
phi2 = a * r * sin(t(end:-1:1)); 

z = [psi1, psi2];
r = [phi1, phi2];

plot(z, r); figure(gcf); 
figure(2);
u = linspace(0, 2*pi, 50); 

[U,V] = ndgrid(u, 1:(2*n));
X = sin(U) .* r(V);
Y = cos(U) .* r(V); 
Z = z(V); 

colormap summer
surf(X, Y, -Z, 'LineStyle', 'none', 'FaceLighting', 'gouraud', 'BackFaceLighting', 'lit'); figure(gcf);
set(gca, 'DataAspectRatio', [1 1 1]);
camlight
view -56 26