# Identification tangent space with manifold

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 $$b_i\frac{\partial}{\partial x^i}\mapsto b_ix^i$$.

• 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. Nov 10, 2019 at 13:01

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 $$M\to R^n$$.