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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 $K$

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    $\begingroup$ I assume that compact surface is interpreted to be embedded in $\mathbb{R}^3$? $\endgroup$ Mar 16 '21 at 19:52
  • $\begingroup$ umm...probably yes? I guess just the usual definition of compactness (closed and bounded by Hein Borel) should suffice here $\endgroup$ Mar 16 '21 at 19:56
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    $\begingroup$ mean curvature is only defined relative to an embedding/immersion. An abstract RIemannian manifold in and of itself does not carry a notion of mean curvature. $\endgroup$ Mar 16 '21 at 19:56
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    $\begingroup$ Wouldn't a thin enough torus around an embedded loop be mean convex, that is have positive mean curvature? The large principal curvature coming from the short 'meridians' would compensate for any negative contributions from the loop, heuristically speaking. $\endgroup$
    – Leo Moos
    Mar 16 '21 at 20:03
  • $\begingroup$ @LeoMoos: yeah, I just finished typing that up below. $\endgroup$ Mar 16 '21 at 20:07
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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.

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  • $\begingroup$ Cool! Thank you so much Willie! I remember that such a "thin torus" model actually has many applications in physics. $\endgroup$ Mar 16 '21 at 20:10
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No, it isn’t true. For example, there are tori with constant mean curvature (Wente tori). This also holds for higher-genus surfaces.

Those are immersed surfaces, but not embedded; they have self-intersections. I’m not sure off the top of my head if it is true for embedded surfaces.

Here is a nice animation of the simplest Wente torus: https://m.youtube.com/watch?v=FjFKLPYAi_U&feature=youtu.be

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