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When I read the Four Vertex Theorem, I think that there may be analogue for smooth Riemannian 2-sphere $(S^2,g)$. Namely, assume the Gauss curvature of $(S^2,g)$ is $K$, Then, K has at least six extreme points on $(S^2,g)$, is it ?

Enhao Lan
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    H.Gluck, "The Generalized Minkowski Problem in Differential Geometry in the Large", Annals of Mathematics, 1972, Vol. 96, No. 2, pp. 245-276 – Moishe Kohan Dec 03 '23 at 10:11
  • @MoisheKohan So, if the Gauss curvature change sign, it is open ? – Enhao Lan Dec 03 '23 at 10:35
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    Chang, Sy.A., Yang, P.C. Prescribing Gaussian curvature on $S^2$ . Acta Math. 159, 215–259 (1987) might be of interest – Didier Dec 03 '23 at 11:13
  • @Didier Thanks, prescring curvature problem seemly be popular in the past. – Enhao Lan Dec 03 '23 at 12:06
  • @EnhaoLan It still is a very popular topic (see this, or this, or this...) – Didier Dec 03 '23 at 12:10
  • Yes, the existence of a metric on $S^2$ with prescribed Gaussian curvature is called "Nierenberg's Problem" and it is open, even though much is known. A discrete version of this problem (spherical metrics with conical singularities) was recently solved, but the answer is complicated as there are obstructions beyond the Gauss-Bonnet. – Moishe Kohan Dec 03 '23 at 13:25

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One way of showing that the Gaussian curvature function on the 2-sphere may have as few as $2$ critical points is to use Gluck's result from

Gluck, Herman. The generalized Minkowski problem in differential geometry in the large. Ann. of Math. (2) 96 (1972), 245–276.

Gluck showed that any positive function on the sphere can be realized as the Gaussian curvature of an embedding in 3-space (in fact his result is more general). Thus, if one chooses a rotationally-symmetric function which is $1$ at the south pole and from there increases to the value $2$ at the north pole (in such a way that the gradient is nonzero everywhere except the two poles), one will obtain a metric with only two critical points (a maximum and a minimum).

Mikhail Katz
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  • Thank you, I was just finishing Gluck's result when I saw your answer. If the Gauss curvature change sign, there must be at least four vertex ? (in terms of geometric intuition) – Enhao Lan Dec 03 '23 at 10:38
  • That's a good question. Further results on realisation of curvature functions are in Kazdan-Warner (Curvature Functions for Compact 2-Manifolds, from '74). I suggest you check there. – Mikhail Katz Dec 03 '23 at 10:44
  • Great paper, I intuitively feel why using PDE for geometry. Thanks again. – Enhao Lan Dec 03 '23 at 11:00
  • As far as your question is concerned, one can use rotationally symmetric metrics in which case the PDE problem reduces to an ODE, so the corresponding answers probably don't need the full strength of the results by Gluck and others. @EnhaoLan] – Mikhail Katz Dec 03 '23 at 11:02
  • Yes, But I've rarely seen geometric problems directly translated into PDE problems before. The paper of '74 also state that for $S^2$, the sufficient condition of existence of metric for perscribed Gauss curvature is unknow. Chow (The Ricci flow on the 2-sphere) has a paper state that the Gauss curvature will become positive under normalized Ricci flow. It might work. (I just thought of it. ) – Enhao Lan Dec 03 '23 at 11:22
  • Yes but the Ricci flow will certainly not preserve the critical points of $K$. So I don't know what the answer would be if one requires that $K$ is negative somewhere. @EnhaoLan – Mikhail Katz Dec 03 '23 at 11:28
  • Yes. I know so little. These questions are not suitable for me now, I have to accumulate enough first. Thanks. – Enhao Lan Dec 11 '23 at 06:55