Let $X$ be a compact smooth 2-dimensional Riemannian manifold with boundary. Assume that the Gauss curvature of $X$ is at least $\kappa$, the diameter is at most $D$, and the second fundamental form of the boundary is at least $\lambda$.

Does there exist an upper bound on the area of $X$ in terms of $\kappa, D,\lambda$ only?

Remarks. (1) In the case of the empty boundary the answer is known to be positive (e.g. it follows from the Bishop inequality).

(2) If the boundary is non-empty and $\lambda\geq 0$, then the answer is also positive. In this case $X$ is locally geodesically convex and hence it is an Alexandrov space of curvature at most $\kappa$ in the sense of Alexandrov, and for such spaces there is a generalization of the Bishop inequality due to Burago-Gromov-Perelman. (Although there is a little bit more direct but longer explanation.)


Yes, there is a bound; your problem can be reduced to the case of convex boundary using the following trick.

Without loss of generality, we can assume that $\lambda=-\tfrac1{10}$ and $\kappa=-1$.

In this case you can attach a collar to your surface locally isometric to the tubular neighborhood in of line in the Lobachevsky plane which has curvature of boundary $\tfrac1{10}$. In the obtained surface boundary is convex and it has curvature $\ge -1$ in the sense of Alexandrov (by Alexandrov's gluing theorem). The diameter, area and perimeter of the obtained surface can be bounded in terms of the diameter area and perimeter of the original surface and the other way around.

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  • $\begingroup$ What version of the Alexandrov gluing theorem you are using? Should it apply to locally geodesically convex surfaces? $\endgroup$ – makt Aug 27 '17 at 13:59
  • $\begingroup$ One more question on your argument. Why the curvature of the boundary of the tubular neighborhood of a line in Lobachevsky plane is positive? That should imply that this tubular neighborhood is locally geodesically convex. Is it indeed the case? (By comparison, on the 2-sphere the analogous statement is not true.) Or just a sign is missing in your answer? $\endgroup$ – makt Aug 27 '17 at 14:13
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    $\begingroup$ The gluing theorem states that if one cuts a hole in a surface and instead a patch with the same geodesic curvature or larger then as the result we get a surface with the same lower curvaure bound as the original surface and the patch. $\endgroup$ – Anton Petrunin Aug 27 '17 at 15:22
  • $\begingroup$ Sorry, I still do not understand the gluing theorem. Let us take a 2-sphere and cut out of it a very small cap. The boundary has a negative constant geodesic curvature. Let us take doubling of this surface. If I understand correctly, the new surface has unbounded below curvature. $\endgroup$ – makt Aug 27 '17 at 18:23
  • $\begingroup$ @smd right. We need to glue in something very convex, but it is OK to glue something with negative (but bounded below) curvature. If you want to apply it fro sphere without small cap you need to glue a collar with very negative constant curvature. $\endgroup$ – Anton Petrunin Aug 28 '17 at 1:15

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