# Curvature implying Closedness in N dimensions

1. Jul 31, 2015

### jfizzix

A two-dimensional surface with everywhere positive curvature is a closed surface with no boundary (isomprphic to a sphere).

Is this true for higher dimensional surfaces as well?
Would a three-dimensional surface, with everywhere positive curvature be a closed hypersurface isomorphic to a hypersphere?

2. Jul 31, 2015

### micromass

Unless you are speaking very loosely, I don't think this is true. What do you mean with "isomorphic" in this case anyway?

3. Jul 31, 2015

### micromass

If you are refering to the Stoker-Hadamard theorem, it states that any closed surface with positive Gaussian curvature is positive everywhere is either
1) Diffeomorphic to the sphere if it is compact
2) Diffeomorphic to the graph on an open, convex subset of the plane.
So the conclusion is not that the surface is closed in $\mathbb{R}^3$, rather it is one of the hypotheses.

4. Aug 11, 2015

### HallsofIvy

I think you are misquoting that theorem. A surface with positive constant curvature is homeomorphic (not "isomorphic") to a sphere, but you can have a non-close surface, such as a paraboloid, that has positive curvature everywhere, going to 0 as the distance from a fixed point goes to infinity.

5. Aug 11, 2015

### lavinia

This is not true for surfaces. One can give the projective plane a metric of constant positive Gauss curvature. There are many surfaces with boundary that have positive Gauss curvature.

A closed orientable smooth surface without boundary of positive Gauss curvature must be a sphere. This is because the integral of the Gauss curvature is 2π time the Euler characteristic and all surfaces other than the sphere have non-positive Euler characteristic.

Last edited: Aug 29, 2015