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Is there a solution or progress of the following problem (maybe old conjecture): Is An immersed surface with constant mean curvature and with a circle as a boundary part of a sphere??. If we replace "immersed" by "embedded" I think the problem was solved by Alexandroff kind of long time ago. Is someone could enlighten about what exactly Alexandroff solves and what is to do, it would help me a lot.

Thanks Mario

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There are lots of immersed surfaces with boundary the circle. i think you are asking about immersed surfaces with constant mean curvature. As far as I know this is still an open question. – Rbega Jan 9 '13 at 0:13
Yes, you're right. I didn't ask the question correctly but it fixes now. Do you know some references or progress about this question?. – Mario Jan 9 '13 at 0:19
By the way thanks Rbega. – Mario Jan 9 '13 at 0:20
up vote 3 down vote accepted

Check out this paper of Rafael Lopez and references therein.

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Right, R. Lopez has several papers with progress towards the conjectures. See theorem 9.2 in the survey I linked to for a comprehensive list of all known conditions which guarantee an immersed surface with circular boundary to be a spherical cap. – Gjergji Zaimi Jan 9 '13 at 1:05

Assuming your interest is in constant mean curvature surfaces with circular boundary, I found the survey, "Surfaces with constant mean curvature in Euclidean space" by R. Lopez to be a great introduction, and it contains the state of the art, and several references.

Hopf proved that the only constant mean curvature closed surfaces of genus 0 are spheres. Alexandrov's theorem says that constant mean curvature closed surfaces that are embedded are spheres. Unfortunately when we allow boundaries both analogs are conjectural:

Conjecture 1: The only constant mean curvature compact surfaces with circular boundary that are topological disks are spherical caps.

Conjecture 2: The only constant mean curvature compact surfaces with circular boundary that are embedded are spherical caps.

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Thanks a lot. I will read the survey. – Mario Jan 9 '13 at 8:38

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