Let $(U, (\theta, \varphi))$ be the spherical coordinate chart on the sphere $\mathbb{S}^2$, and consider the vector field on $U$ defined by $\dfrac{\partial}{\partial \varphi}$. Then in this chart, the metric tensor $g_{ij}$ can be expressed as follows:
$$g_{ij} = d\varphi^2 + \sin^2\varphi \, d\theta^2. $$
This means that the vector $\dfrac{\partial}{\partial \varphi} \Bigr|_{p} \in T_p\mathbb{S}^2$ has norm $1$ for any point $p \in U$. But this seems to contradict the Hairy Ball theorem, which says that every smooth vector field on $\mathbb{S}^2$ has to vanish somewhere. What's going on? What is wrong with this logic?
I think the problem is that the spherical coordinate chart on $\mathbb{S}^2$ is not a global chart. So perhaps this vector field defined on $U$ cannot be extended to a smooth vector field on the whole manifold. My question is: is there an intuitive geometric reason why we would expect that this vector field cannot be smoothly extended? Something involving pictures of vector fields would be helpful. Thanks.
In the hope a picture is worth a thousand words: