Relations between curvature and topology

[joseph0086]

Hello, all, the most important results that I know in this topic is the Gauss-Bonnet Theorem (and hence the classification of compact orientable surfaces) and also the Poincare-Hopf index theorem.

But there are still some fundamental problems I don't understand.

For example, is the following problem true?
Let C be a simple closed curve on a sphere. Let v be a vector field on it such that v is never tangent to C. Is it true that each of the regions determined by C contains at least one singular point of v.

I feel it is true, but I dont know how to prove it...

Moreover, it is intuitivlely clear that all compact orientable surfaces with negative euler characteristics have some regions which have Gaussian curvatures positive, negative and zero. The fact that it has regions that have -ve curvature is obvious by Guass-Bonnet. But how to show mathematically that it must have some regions that have positive curvature? (In other words, my question is: Show that compact orientable surface has negative curvature at all points.)

I learned from a textbook that for all such surfaces with -ve Euler characteristics, then two geodesics which start from the same point wont meet again. (Easy to show by Guass-Bonnet), I don't know if it is related to my question..


Hope somebody here can tell me his understanding.. Thanks a lot.

[lavinia]

Hello, all, the most important results that I know in this topic is the Gauss-Bonnet Theorem (and hence the classification of compact orientable surfaces) and also the Poincare-Hopf index theorem.

But there are still some fundamental problems I don't understand.

For example, is the following problem true?
Let C be a simple closed curve on a sphere. Let v be a vector field on it such that v is never tangent to C. Is it true that each of the regions determined by C contains at least one singular point of v.

I feel it is true, but I dont know how to prove it...

If the vector field always points into or out of the disc shaped region then it must have a singularity in the interior. I am sorry that I can not quickly give you a picture but here is an attempt at a proof.

Imagine a disk in the plane with a vector field on it that always points outwards along its bounding circle. If the vector field has no singularity then it can be normalized to have length 1 i.e. it can be turned into a map from the disk into the unit circle. This means that the map restricted to the boundary is null homotopic. But if it always points outwards then it can be smoothly deformed into the field that is perpendicular to the boundary and this map is not null homotopic.

It occurs to me that this is really the Brower Fixed Point theorem in disguise but not sure. It would be fun to think about this if you like.

I would love to see a more intuitive proof.



Moreover, it is intuitivlely clear that all compact orientable surfaces with negative euler characteristics have some regions which have Gaussian curvatures positive, negative and zero. The fact that it has regions that have -ve curvature is obvious by Guass-Bonnet. But how to show mathematically that it must have some regions that have positive curvature? (In other words, my question is: Show that compact orientable surface has negative curvature at all points.)

In 3 space every surface must have a region of positive Gauss curvature. This was proved by Hilbert and is easy to see. Try a proof.

However, a surface of negative Euler characteristic can be given a metric of constant negative Gauss curvature. I think the proof is that all such surfaces are covered by the unit disk in the plane under the action of a group of Mobius transformations that preserve the Poincare metrx which has curvature -1.

The torus has Euler characteristic zero so must have equal total positive and negative curvature. The torus can be given a metric of everywhere zero Gauss curvature. this because it is covered by the action of a lattice on the plane. An explicit realization of the flat torus is

(x,y) -> (sinx,cosx,siny,cosy) but this is in four dimensions.