# Given a smooth manifold $M$, how does local coordinates affect the basis of $T_p(M)$?

Let's say I have a smooth manifold $$M$$ of dimension $$n$$ and a smooth chart $$(U, \phi)$$ consisting of open set $$U$$ of $$M$$ along with a homeomorphism $$\phi : U \to \widehat{U} =\phi[U] \subseteq \mathbb{R}^n$$.

Recall that the component functions $$x^i : U \to \mathbb{R}$$ of $$\phi$$ defined by $$\phi(p) = (x^1(p), \dots, x^n(p))$$ are called the local coordinates on $$U$$.

Now the basis for $$T_p(M)$$ is given by $$\left\{\frac{\partial}{\partial x^1}\bigg|_p, \dots, \frac{\partial}{\partial x^n}\bigg|_p\right\}$$ where each $$\frac{\partial}{\partial x^i}\bigg|_p$$ is the derivation defined by $$\frac{\partial}{\partial x^i}\bigg|_p(f) = \frac{\partial}{\partial x^i}\bigg|_{\phi(p)}\left(f \circ \phi^{-1} \right) \ \ \ \ \ (*)$$ for $$f \in C^{\infty}(U)$$.

Now in Introduction to Smooth Manifolds by John Lee, it's said that these tangent vectors $$\frac{\partial}{\partial x^i}\bigg|_p$$ are called the coordinate vectors at $$p$$ assosciated with the given coordinate system.

But I don't exactly see how the basis vectors $$\frac{\partial}{\partial x^i}\bigg|_p$$ depend on a given coordinate system other than simply the homeomorphism $$\phi$$. To me it seems that the right hand side of $$(*)$$, that being $$\frac{\partial}{\partial x^i}\bigg|_{\phi(p)}\left(f \circ \phi^{-1} \right)$$, is just the usual partial derivative operator in $$\mathbb{R}^n$$, so by that I mean that $$\frac{\partial}{\partial x^i}\bigg|_{\phi(p)}\left(f \circ \phi^{-1} \right) = \lim_{t \to 0} \frac{(f \circ \phi^{-1})(\phi(p) + t e_i) - (f\circ \phi^{-1})(\phi(p))}{t}$$

where $$e_i$$ is the element $$(0, \dots, 1, \dots, 0) \in \mathbb{R}^n$$ with a $$1$$ in the $$i$$-th position of the $$n$$-tuple. Am I correct in saying that? I think I am wrong, because what if I have some $$2$$-dimensional manifold $$N$$, and a chart $$(V, \psi)$$ where points of $$\widehat{V}$$ are best expressed in polar coordiantes and I wanted to compute $$\frac{\partial}{\partial x^i}\bigg|_{q}(f)$$ for some $$q \in N$$ and $$f \in C^{\infty}(V)$$. I don't see how my interpretation of $$(*)$$ would make sense then there.

Maybe I'm getting lost in notation but for example I don't understand what the relation between the symbolic $$x^i$$'s in the notation of the component functions $$x^i$$ of $$\phi$$, the basis vectors $$\frac{\partial}{\partial x^i}\bigg|_{p}$$, and $$\frac{\partial}{\partial x^i}\bigg|_{\phi(p)}\left(f \circ \phi^{-1} \right)$$ are other than they are simply a reminder for the latter two that we're taking the $$i$$-th partial derivative in $$\widehat{U}$$. The notation used suggests there must be some stronger relationship between these.

I'm not shure I have fully understood your question, but hopefully what I say can help.

1) When you say it's said that these tangent vectors $$\frac{\partial}{\partial x^i}\bigg|_p$$ are called the coordinate vectors at $$p$$ assosciated with the given coordinate system.

Here by the given coordinate system we mean simply the chart $$\phi$$. If $$\phi=(x^1,\dots,x^n)$$ then $$(x^1,\dots,x^n)$$ are exactly the coordinates we are referring to.

2) As regards the symbolic $$x^i$$'s, they are present in the notation of the basis vectors $$\frac{\partial}{\partial x^i}\bigg|_{p}$$, beacause this definition depends on $$\phi$$ and so on the function $$x^i$$'s, but they should not be in the notation of $$\frac{\partial}{\partial x^i}\bigg|_{\phi(p)}\left(f \circ \phi^{-1} \right)$$.

Beacuse, as you sad, $$\frac{\partial}{\partial x^i}\bigg|_{\phi(p)}\left(f \circ \phi^{-1} \right) = \lim_{t \to 0} \frac{(f \circ \phi^{-1})(\phi(p) + t e_i) - (f\circ \phi^{-1})(\phi(p))}{t}$$ and in the right hand side we have no link to the coordinate functions of $$\phi$$, i.e. to te $$x^i$$'s.

Maybe it could be better to write $$\partial_i\bigg|_{\phi(p)}\left(f \circ \phi^{-1} \right) = \lim_{t \to 0} \frac{(f \circ \phi^{-1})(\phi(p) + t e_i) - (f\circ \phi^{-1})(\phi(p))}{t}$$

Definitely, we should define $$\frac{\partial}{\partial x^i}\bigg|_{p}f=\partial_i\bigg|_{\phi(p)}\left(f \circ \phi^{-1} \right)$$