Vector Field Degrees

Question

Let $F$ be a tangent vector field on the sphere $S^2$, such that it vanishes at a pair of points.

Answer 1

I would agree that under your assumption $F(x)=F(-x)$, we have $$\operatorname{Ind}_{x_0} F = -\operatorname{Ind}_{-x_0} F.$$

The index of the vector field $F$ at an isolated zero $x_0$ is computed as the degree of the map from a nearby circle to the unit circle, defined by normalizing your nearby nonzero vector fields to unit length. This degree is orientation-sensitive, as all degrees are.

Now at $x_0$, say we have such a map $\tilde F: C\to S^1$, where $C$ is a small circle around $x_0$.

At $-x_0$, we similarly have $\tilde F_1: C_1\to S^1$. Here $C_1$ is a small circle around $-x_0$, and we can even identify $$\iota: C\to C_1;\ x\to -x.$$ Then by your assumption $F(x)=F(-x)$, we see that $$\tilde F_1 = \tilde F\circ \iota^{-1}: C_1\to C\to S^1.$$

But if $C$ is a positively oriented around $x_0$ (in the sense if you walk along it, $x_0$ is to your left), the corresponding $C_1=\iota(C)$ is a negatively oriented circle around $-x_0$. By the above, the $\tilde F$ and $\tilde F_1$ are the same set-theoretically, under this identification of $\iota$. We see that they have opposite degrees, hence the conclusion for the indices.

Therefore on $S^2$, vector fields such that $F(x)=F(-x)$ don't exist, since the sum of indices would be 0, but $S^2$ has Euler characteristic 2.

My usual example of vector fields on $S^2$ with index 2 is the vector filed for the rotation around the $z-$axis (or $\frac{\partial}{\partial \theta}$ in spherical coordinates). This vector field satisfies $F(x)=-F(-x)$. It does have indices both 1 at the north and south poles.