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Let $\Sigma$ be a $C^{\infty}$ compact surface in $R^3$.

(1)If the tangent space of every point lies the same side of $\Sigma$, we call $\Sigma$ convex surface.

(2)If the Guass Curvature $K>0$, we call $\Sigma$ ovaloid.

(3)If $\Sigma$ is homeomorphic to $S^2$, and the union of interior of $\Sigma$ and $\Sigma$ is convex, we call it a (3)-surface.

What is the relationship between convex surface and ovaloid? How is the situation in higher dimension?

In $R^n$, if $S$ is a compact convex set, then the boundary of $S$ is homeomorphic to a sphere. So is the definition (3) equal to the definition (1)?

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gaoxinge
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    A capsule is a convex surface that is not an ovaloid, since the Gaussian curvature is equal to zero along the cylindrical surface. Note that a capsule is not a $C^\infty$ surface (it is only $C^1$), but it can be modified slightly to make a $C^\infty$ surface by "smoothing" the transition from cylinder to hemisphere. – Jim Belk Jan 01 '14 at 04:11
  • Thank you. A convex surface is not necessarily an ovaloid. Is an ovaloid is a convex surface? – gaoxinge Jan 01 '14 at 04:44
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    I suspect that the answer is yes. I also suspect that (1) is equivalent to (3) as long as the surface is homeomorphic to $S^2$. – Jim Belk Jan 01 '14 at 05:25
  • I think that an ovaloid is a strict convex surface i.e it does not contain line segments. – Semsem Jan 01 '14 at 21:58
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    For example the cylinder is a convex surface and is not an ovaloid.I think that every ovaloid is a convex surface. – Semsem Jan 01 '14 at 22:01
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    Finally, a convex set is not a convex surface and vice versa. The convex surface is a certain subset of the boundary of a convex set with non-empty interior. A convex set means every pair of points are joined by a line segment in the set. – Semsem Jan 01 '14 at 22:06
  • I find a paper "On the Total Curvature of Immersed Manifold" by S.S Chern. If $\Sigma$ is a compact surface in $R^3$ with nonnegative Guass Curvature, then it is a convex surface. However, it cannot be generalized into higher dimension. If Guass Curvature is positive, we can make it in any dimension. And this is called Hadamard Theorem. – gaoxinge Jan 02 '14 at 01:17
  • I have some more specific questions. Is a Guass Curvature of convex surface nonnegitive?@Jim Belk @Sane – gaoxinge Jan 02 '14 at 01:25
  • Can you take an example about your statement: Finally, a convex set is not a convex surface and vice versa.@Sane – gaoxinge Jan 02 '14 at 01:28
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    The sphere $S^{2}$(the boundary of a closed ball in $\mathbb{R}^{3}$) is a convex surface but as a set it is not a convex set since the line segment joining any pair of points of $S^{2}$ does lie in $S^{2}$ while the closed ball is a convex set. – Semsem Jan 02 '14 at 06:10
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    This closed ball is a good example of a compact convex set and hence its boundary is homeomorphic to a sphere. – Semsem Jan 02 '14 at 06:14
  • So is (3) equal to (1)? @Sane – gaoxinge Jan 02 '14 at 08:02
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    for compact surface, I think so. – Semsem Jan 02 '14 at 08:40

1 Answers1

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(1) and (3) are equivalent (this also holds in higher dimensions).

Suppose (1) holds. Each tangent plane divides the space into two half-spaces, one of which contains $\Sigma$. Take the intersection of all such halfspaces: it is a convex set containing $\Sigma$ and its interior. Show that it is equal to the union of $\Sigma$ and its interior. To prove that $\Sigma $ is homeomorphic to $S^2$, use radial projection from an interior point.

Suppose (3) holds. Convexity implies that for each $p$ there is a supporting plane: a plane that passes through $p$ and has the set on one side. Show that this plane is the tangent plane at $p$.

(2) implies the other properties (in 3 dimensions), but this is not trivial. The key words are Hadamard's ovaloid theorem

(1)-(3) do not imply (2); counterexamples are found in comments.

(1)-(3) imply $K\ge 0$. Indeed, suppose $K<0$ at some point $p$. Choose a system of coordinates so that $p$ is the origin and the tangent plane at $p$ is the $xy$-plane. Then the surface $\Sigma$ is locally the graph of a function $z=f(x,y)$ with a saddle point, because the Hessian determinant of $f$ is negative (see here). This contradicts (1).

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