I was puzzeling with the pseudosphere , a surface that except for a cusp has a constant negative Gaussian curvature, but has not everywhere the same mean curvature. (https://en.wikipedia.org/wiki/Pseudosphere )

This made me wonder are there also surfaces that have (for a largish area) a constant positive Gaussian curvature but not a constant mean curvature?

Something like a surface that has principle curvatures $K_1=1 ,K_2=1$ for a single point or limited set of curves while for the other points in this neightboorhood the Gaussian curvature $K_1K_2 =1$ but $ K_1 \not = K_2$ and this not just at a small part of the surface but at a largish area, like an area bounded by a cusp or boundary.

The surface does not need to be closed.

  • $\begingroup$ What do you mean by "large area"? Do you mean that the surface may have constant positive GC at some "area", and different GC elsewhere? Because wherever there is constant positive curvature, there is something that looks like a sphere, which has constant mean curvature. $\endgroup$ – uniquesolution Nov 16 '15 at 9:08
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    $\begingroup$ My answer to this question might be helpful. $\endgroup$ – Jack Lee Nov 16 '15 at 16:12
  • $\begingroup$ @uniquesolution I am looking for a surface that has principle curvatures $K_1 , K_2 = 1 $ while at other points $K_1 K_2 =1 $ and $K_1 \not = K_2 $ and that not just at a small part of the surface but at a reasonable area, like an area bounded by a cusp, the surface does not need to be closed. $\endgroup$ – Willemien Nov 16 '15 at 22:24

Theorem: If $M$ is a $C^{2}$ surface in $E^{3}$ having constant positive Gaussian curvature, and if $M$ is not part of a sphere, then the mean curvature of $M$ is non-constant.

Proof: Assume, contrapositively, that $M$ has constant Gaussian and mean curvatures. If $k_{1}$ and $k_{2}$ denote the principal curvature functions, then the sum and product $k_{1} + k_{2}$, $k_{1}k_{2}$ are both constant, so $$ (k_{1} - k_{2})^{2} = (k_{1} + k_{2})^{2} - 4k_{1}k_{2} $$ is constant, and therefore $k_{1} - k_{2}$ is constant. Since $k_{1} \pm k_{2}$ are constant, the principal curvatures themselves are constant.

Since $k_{1}k_{2} > 0$ by hypothesis, it follows that $k_{1} \equiv k_{2}$, i.e., $M$ is totally umbilic, and hence part of a sphere.

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  • $\begingroup$ I don't understand your answer, I think you grasp my question. My question is more: Is there a surface that has constant positive Gaussian curvature and which mean curvature is non-constant. you seem to proof that such a surface does not exist but i cannot follow the proof. $\endgroup$ – Willemien Nov 20 '15 at 21:41
  • $\begingroup$ Actually, the claim is the opposite: Every non-spherical surface of constant positive Gaussian curvature in $E^{3}$ automatically has non-constant mean curvature. :) (There are many such surfaces, including the surfaces of rotation Jack Lee mentioned, and surfaces having helical symmetry.) $\endgroup$ – Andrew D. Hwang Nov 21 '15 at 0:48
  • $\begingroup$ I thing you mean the two surfaces depicted in math.stackexchange.com/a/1521686/88985 , but then I think the surfaces have a different but constant principal curvatures, and by this they do have a constant mean curvature (except the inner surface at its extremities) so that would not qualify as a surface as I intended. $\endgroup$ – Willemien Nov 21 '15 at 20:05
  • $\begingroup$ Those are indeed the surfaces I meant. :) The principal directions for those surfaces are tangents to the meridians and parallels; the meridians do not have constant radii of curvature (i.e., the meridians are not arcs of circles), so the principal curvatures are not constant. Since those surfaces have constant Gaussian curvature, the argument in my post implies the mean curvatures are non-constant. $\endgroup$ – Andrew D. Hwang Nov 21 '15 at 20:53

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