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In the classical Euclidean setting $\mathbb{R}^n$ with cartesian coordinates $x_i$ one can identify a vector field with a map $\mathbb{R^n}\to\mathbb{R}^n$, because every tangent space at any point can be identified with $\mathbb{R}^n$ without ambiguity.

Suppose now that one has on $\mathbb{R}^n$ (or on an open subset of it, a ball for instance) a metric different from the classical one. My question is:

up to renormalise the basis $\frac{\partial}{\partial x_i}$ with respect to the new metric, can one identify $\frac{\partial}{\partial x_i}$ with $x_i$ so that a vector field can be identified once again with a map $\mathbb{R}^n\to\mathbb{R}^n$?

The identification will be something like $b_i\frac{\partial}{\partial x^i}\mapsto b_ix^i$.

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    $\begingroup$ What does this have to do with a metric? This just depends on the structure of ${\mathbb R}^n$ as a differentiable manifold. As long as this structure is standard, the identification makes perfect sense. What is relevant here is that the tangent bundle is trivial. $\endgroup$ Nov 10, 2019 at 13:01

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Suppose that $M$ is a differentiable $n$-manifold with trivial tangent bundle, $TM\cong M\times R^n$. For instance, $M$ can be any open subset of $R^n$.

A vector field on $M$ is just a smooth section of $TM$; in view of triviality of $TM$, it is just a smooth map $M\to TM, x\mapsto (x, v(x)), x\in M, v(x)\in R^n$. But it is the same thing as to say that you have a smooth map $x\mapsto v(x)$, i.e. a smooth map $M\to R^n$.

You do not need a metric for this, you also do not need explicit coordinates for this.

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