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Let A be a closed (compact no boundary), embedded (no self intersections), smooth surface of R^3. We say that the interior of A is star shaped if there exists a point p in A, such that for any point q in A, the line segment joining p and q lies entirely within A. My questions is this:

Let (S^2,g) be the sphere with a given smooth Riemannian metric (may not be the constant curvature one), and assume there exists a smooth isometric embedding f: S^2 -> R^3. Does there exist a condition on the Gaussian curvature of g that ensures the interior of f(S^2) (the image of S^2 under f) is star shaped?

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Positive Gauss curvature certainly suffices, by the theorem of Cohn-Vossen, which implies that the embedded surface is convex. Beyond that, I'm not aware of any other intrinsic characterization of the boundary of a star-shaped domain. It sounds quite difficult to me, since neither existence nor uniqueness of the isometric embedding is known, if the curvature is not everywhere non-negative (positive?).

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  • $\begingroup$ I was thinking maybe a sufficient condition is that the region of positive curvature ( > 0 ) is connected, and if K = 0 then $|\nabla K| \neq 0$ (i.e. curvature changes sign cleanly) $\endgroup$
    – Thomas
    Commented Oct 27, 2009 at 14:08
  • $\begingroup$ Do you have an example of what you have in mind? My limited geometric intuition can't come up with a closed surface where the region of positive curvature is connected but there are no open regions of zero curvature. $\endgroup$
    – Deane Yang
    Commented Oct 27, 2009 at 19:02
  • $\begingroup$ Take a surface of negative curvature in R^3, and intersect it with the upper hemisphere of a sphere. Throw away all of the negatively curved surface apart from the region cut out by the sphere (i.e. when you press out a round cookie from flat pastry, you through away the remaining pastry). Also get rid of the region of the sphere lying below the negatively curved surface. So you are left with an upper hemisphere with a negatively curved surface joined to its bottom. Now smoothly round out the edges connecting the surface to the sphere (does this introduce curvature of both + and - ?) $\endgroup$
    – Thomas
    Commented Oct 27, 2009 at 19:54
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Here is an example of a smooth sphere in R^3 that is not star shaped, and such that both positive and negative cuvature parts are connected. Take in R^3 a thin rotation-invariant torus. Its positive and negative curvature parts are long thin annuli. Now cut a half of it by a plane going through the axe of rotation. Take one of the halthes (a long, thin, curved tube) and cup its two little holes with two little half-spheres -- so you get a sort of banana. This can be done in a smooth way so that positive curvature part is connected.

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This seems very related to the Minkowski problem and its converse which you can read about in this nice survey by Herman Gluck.

I'll see if I can find more recent results and post them here.

I haven't read this book by Alexandrov, but it ought to be informative as well: Intrinsic Geometry of Convex Surfaces

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