Problem with understanding how abstract tangent space basis vectors are conneted to e.g. polar coordinates

Question

How does the abstract basis of the tangent space $\frac{\partial}{\partial x^i} f := \partial_i (f \circ x^{-1} ) (x(p))$ at a Point $p \in M$ where a function $f$ is empolyed are related to the well-known basis vectors e.g. of the polar coordinates $e_{r} = (cos(\phi), sin(\phi))^T$? What role does $f$ play? Are there any specific asumptions made about the manifold $M$ in this case?

Answer 1

Given a manifold M and local coordinate $\Psi: U\subseteq M \to V\subseteq R^n, \Psi = (x^i)$, the symbol $\frac{\partial}{\partial x^i}$ stands for the vector $d\Psi^{-1}(e_i)$.

In the case of polar coordinates, the manifold is $R^2 \setminus \{0\}$ and the inverse of coordinate map is $\Psi^{-1}:(0,\infty)\times(0,2\pi)=V \to M\setminus \{(t,0)\}=U, (r,\phi) \mapsto (rcos\phi,rsin\phi)$. Hence the differential $d \Psi^{-1} =$ $\begin{bmatrix} cos\phi &-rsin\phi \\sin\phi &rcos\phi\end{bmatrix}$and $e_r=d\Psi^{-1}(1,0)=(cos\phi,sin\phi)$.

There is another formulation of tangent vectors in terms of linear derivations. You can read about them here. {https://en.wikipedia.org/wiki/Tangent_space#Definition_via_derivations}. Also an equivalence of all these definitions is also followed.

Nothing is assumed about the manifold other than it is $C^{\infty}$.