Coordinate Functions and Coordinate Bases

Question

Let $M$ be a smooth manifold of n dimension. And let $(U,\varphi)$ and $(V,\psi)$ be two smooth charts on $M$.

Define the coordinate functions of $\varphi$ and $\psi$ as $(x^i)$ and $(\tilde{x}^i)$ respectively, i.e. $\forall p \in U\cap V$, $\varphi(p)=(x^1(p), ..., x^n(p)\,)$ and $\psi(p)=(\tilde{x}^1(p), ..., \tilde{x}^n(p)\,)$.

And then the author says that any tangent vector at $p \in U\cap V$ can be represented with respect to either basis $(\partial\,/ \, \partial x^i \vert _p)$ or $(\partial\,/ \, \partial x^i \vert _p)$. Later he also uses notation like $\frac{\partial}{\partial \widetilde{x}^j}$.

So I guess the author here identifies the coordinate bases with the coordinate functions. But how is such identification valid? I.e. Is there a one-to-one correspondence between the set of coordniate bases and the set of coordinate functions?

I'm really confused. Thanks for help.

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Edit: What I really wanted to ask was that, if $f\in C^\infty[\varphi(U \cap V)]$, what does $\partial _{x^i} \vert _{\varphi(p)}f$ mean?

Answer 1

The set of tangent vector at $p$ noted $T_pM$ is a vector space of dimension $n$. You can have a lot of basis, and what is said is simply that you can use both basis to decompose any tangent vectors. Remember that $\partial_{x_i}$ is a notation: for $f$ smooth, we have $$\partial_{x_i}f=\partial_i(f\circ x^{-1})(x(p))$$ where $x=(x_i)$ is the diffeomorphism that have $x_i$ as coordinate functions.

Answer 2

This answer is for my own future reference as well as someone who gets stuck here too.

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What I really wanted to ask was that, if $f\in C^\infty[\varphi(U \cap V)]$, what does $\partial _{x^i} \vert _{\varphi(p)}f$ mean?

It turns out that we can regard $\partial _{x^i} \vert _{\varphi(p)} f$ as a partial derivative w.r.t the coordinate axis determined by $x^i $(recall that $x^i$ is a function but not an element of a basis for each $i$) at $\varphi(p)$.

Lemma: Suppose we have two bases $\{a_1,a_2, ..., a_n\}$ and $\{b_1,b_2, ..., b_n\}$ for $\mathbb R^n$ and for each $\alpha$ in some small open ball centered at the origin, the coordinates of $\alpha$ w.r.t these two bases are identical. Then these two bases are the same.

Proof of lemma: Let $i\in \{ 1,2, ..., n\}$. For each $a_i$, there exists some $k_i >0$ s.t. $k_i a_i$ lies in the open ball. Its coordinate w.r.t the basis $\{a_1,a_2, ..., a_n\}$ is clearly $(0,...,0,k_i,0,...,0)$, where $k_i$ is at the $i$-th slot. By our assumption, its coordinate w.r.t the basis $\{b_1,b_2, ..., b_n\}$ should also be $(0,...,0,k_i,0,...,0)$, where $k_i$ is at the $i$-th slot. Hence, $k_i a_i = k_i b_i$, which implies $a_i=b_i$. Since $i$ is arbitrary, we are done.

By subtracting the constant vector $\varphi (p)$ if necessarily, we may assume that the coordinate chart $(U,\varphi)$ is centered at $p$, which means $\varphi(p)=0$. A chart actually gives a way to assign coordinates to each point in a small ball centered at $\varphi(p)$, which is the origin of $\mathbb{R}^n$. Since we can write $\varphi (p)$ and every point in some small open neighborhood of $\varphi (p)$ as n-tuples, it implicitly implies that there exists at least one local basis. By the lemma, we see a chart does produce a unique (local) basis.

An n-tuple has $n$ slots, each of which corresponds to an axis. Recall that we define $x^i$ by $\varphi(p)=(x^1(p), ..., x^n(p)\,)$.

So for each $i$, $x^i$ can represents an axis. And so we can regard $\partial _{x^i} \vert _{\varphi(p)} f$ as a partial derivative w.r.t the coordinate axis determined by $x^i$ at $\varphi(p)$.