How to obtain real vector from abstract tangent vector in the case of the manifold $\mathbb R^n$

Question

I know that for every $p\in\mathbb{R}^n$ the map \begin{align} \Phi_p\colon\mathbb{R}^n&\to T_p\mathbb{R}^n\\ v&\mapsto D_{v,p} \end{align} is an isomorphism, where \begin{align} D_{v,p}\colon C^{\infty}(\mathbb{R}^n,\mathbb{R})&\to\mathbb{R}\\ f&\mapsto D_{v,p}f \end{align} is the directional derivative.

But is there an explicit formula for ${\Phi_p}^{-1}(v)$ for a given $v\in T_p\mathbb{R}^n$?

Answer 1

Yes, there is: \begin{align} {\Phi_p}^{-1}(v)=e_iv(e^i), \end{align}

where $(e_1,\dots,e_n)$ is the standard basis of $\mathbb{R}^n$ and $(e^1,\dots,e^n)$ is the dual basis.

The proof (which involves Taylor's theorem) is part of the proof of Proposition 3.2 in John Lee's book "Introduction to smooth manifolds".

Answer 2

HINT: Suppose $\Phi_p \in T_p(\Bbb R^n)$(that is, it's a derivation of the algebra $C^{\infty}(\Bbb R^n.\Bbb R)$). Now let $\alpha_i = \Phi_p(x^i)$ where $x^i$ is the coordinate projection. Now show that $\Phi_p = D_{v,p}$ where $v = (\alpha_1,\dots,\alpha_n)$.