Length of tangent vector on sphere

Question

The metric on sphere $\mathbb{S}^2$ with radius $r$ is given by $$ g = r^2 \, d\varphi \otimes d\varphi + r^2 \sin^2 \varphi \, d\theta \otimes d\theta$$ in the spherical coordinates. Let $N$ be the north pole and consider a tangent vector $\dfrac{\partial}{\partial \theta} \in T_N \mathbb{S}^2$. Then its length is \begin{align*} \left \Vert \frac{\partial}{\partial \theta} \right \Vert^2 = \left\langle \frac{\partial}{\partial \theta}, \frac{\partial}{\partial \theta} \right\rangle = r^2 \sin^2 \varphi. \end{align*} Since $\varphi = 0$ on the north pole, I get $\left \Vert \dfrac{\partial}{\partial \theta} \right\Vert = 0$. I think I went wrong somewhere but I couldn't find any mistake. Would you please check if it is right? Thank you.

Answer 1

Your calculation is correct. It shows that the parametrisation $(\sin{\varphi}\cos{\theta},\sin{\varphi}\sin{\theta},\cos{\theta})$ is not a regular parametrisation of the unit sphere, since the tangent vectors are not everywhere linearly independent. (By the way, it is more standard to have $\theta$ as the colatitude and $\varphi$ as the longitudinal angle.) It is a regular parametrisation of the region of the sphere where $0<\theta<\pi$. (You can also see that problems occur with $\partial/\partial\varphi$ at the pole: which direction does it point in? Symmetry implies it can't point in any direction!)

In fact, there is no regular parametrisation of the whole of the sphere: this is a consequence of the hairy ball theorem.