On a topological sphere, the minimal set of locations of null tangent vectors?

The Hairy Ball Theorem tells us that it is impossible to have a continuous tangent vector field on the sphere which has no zeros.

It is easy to find a continuous tangent vector field with two zeros - consider the vector field that, at each point, points East with magnitude $\cos\lambda$ , where $\lambda$ is the latitude. Then this continuous vector field only vanishes at the North and South poles. Thus it is possible to find a tangent vector field with two zeros. For the purposes of this question, this is enough, since the only acceptable option is for there to be just $\boxed{1}$ zero. However, we need to do more to show that it is possible to find a continuous tangent vector field with exactly one zero.

Stereographic projection (from the North pole) gives a parametrisation of the unit sphere, minus the north pole $(0,0,1)$ , as $\mathbf{r}(u,v) \; = \; \left(\dfrac{2u}{u^2 + v^2 + 1},\dfrac{2v}{u^2 + v^2 + 1}, \dfrac{u^2 + v ^2 - 1}{u^2 + v^2 + 1}\right) \hspace{2cm} u,v \in \mathbb{R}$ and if we define the vector field $\mathbf{F}$ on the sphere by setting $\mathbf{F}(\mathbf{r}) \; = \; \left\{ \begin{array}{lll} \left(\dfrac{1 + v^2 - u^2}{(u^2 + v^2 + 1)^2}, - \dfrac{2uv}{(u^2 + v^2 + 1)^2},\dfrac{2u}{(u^2 + v^2 + 1)^2}\right) & \hspace{2cm} & \mathbf{r} = \mathbf{r}(u,v) \\ (0,0,0) & & \mathbf{r} = (0,0,1) \end{array} \right.$ then $\mathbf{r} \cdot \mathbf{F} = 0$ everywhere, and $\mathbf{F}$ is a continuous tangent vector field which is continuous on the whole sphere, but only vanishes at $(0,0,1)$ (parametrically, this involves seeing what happens as $u,v \to \infty$ ). Thus a single zero is indeed possible.

See the hairy ball theorem .