TITLE: Existence of point with zero mean curvature
QUESTION [5 upvotes]: I'm a physicist studying differential geometry for my GR research, and I come up with the following claim (not sure if it's true or not):
For any compact surface $S$ that's not homeomorphic to a sphere, $S$ must have a point with vanishing mean curvature, i.e $H=0$
Is this claim true or not? I know that by applying Gauss-Bonnet theorem, we can prove an analogous claim for the Gaussian curvature $K$

REPLY [7 votes]: A more pedestrian example: let $(u,v)\in [0,2\pi)^2$ and parametrize the torus $\mathbb{T}_{a,r} = \{ ( (a + r \cos u) \cos v, (a+r\cos u) \sin v, r \sin u) \}$.
The mean curvature is given by
$$ H = \frac1r + \frac{\cos u}{a + r \cos u} $$
When $r \ll a$ (this is something that looks like a bike tire: large overall radius and small cross section) then $\frac1r > \frac{1}{a - r} \geq |\frac{\cos u}{a + r \cos u}| $ and hence the mean curvature is never zero.