a question about differential geometry(Gauss-bonnet theorem and isolated singular point in the surface)

Question

Let C be a regular closed simple curve on a sphere $S^2$. Let v be a differentiable vector field on $S^2$ such that the trajectories of v are never tangent to C. prove that each of the two regions determined by C contains at least one singular point of v.

My thoughts:Based on poincare's theorem,we know that $$\sum I_{i}={1\over 2\pi}\iint_{s}Kds$$,where $I_{i}$ is the index of v at the isolated singular point.Then,since the Euler-poincare characteristic of a sphere is 2,so we have $\sum I_{i}=2$. Then. I don't know how to contine to analyze the question,since I think $I_{i}$ is not necessary to be 1. So any help?

Answer 1

You should just think about a simple closed curve $C$ in the plane, bounding a region $\Omega$. I don't know what your background is, but you should be able to show that if $\mathbf v$ has no zero inside the curve, then it must be somewhere tangent to the curve. In particular, if $\mathbf v$ is nowhere tangent, it always points out or it always points in, and so you can show that the map $\mathbf v/\|\mathbf v\|\colon C\to S^1$ has degree $1$. On the other hand, if this map extends to $\Omega$, it must have degree $0$.