I made confusions with the definition of function differential and notations with basis vectors $\{e_{i}=\dfrac{\partial}{\partial x^{i}}\}$ and dual basis vectors $\{e^{i}=\text{d}x^{i}\}$.
Let's take the differential of a function $f(x,y,z)$ :
$$df=\dfrac{\partial f}{\partial x}\text{d}x+\dfrac{\partial f}{\partial y}\text{d}y+\dfrac{\partial f}{\partial z}\text{d}z$$
$df$ Differential of the function $f$ is so an application from $\mathbb{R}^3$ to $\mathbb{R}$.
Now I would like to do the link between the expression of $df$ and its expression with basis vectors and dual basis vectors.
How can I express $df$ with dual basis ? this may be :
$$df=a_1\vec{e^1}+a_2\vec{e^2}+a_3\vec{e^3}$$
with $a_i=\dfrac{\partial f}{\partial x^{i}}$ and $\vec{e^{i}}=\text{d}x^{i}$
??
How to do the link between $df$ and its expression in starting basis vectors, i.e $\{e_{i}=\dfrac{\partial}{\partial x^{i}}\}$ ?
Surely, I made confusion supposing $df$ is a linear form and so not a classic vector.
Thanks for your help
What you want to do in fact is to use the Riesz lemma with the standard dot product for each linear functional $df(p)$, obtaining then a vector ${\rm grad}\,f(p)$ satisfying $$df(p)(v) = \langle {\rm grad}\,f(p),v\rangle,$$for all $v$. We have $$\begin{align} df &= \frac{\partial f}{\partial x}\,dx + \frac{\partial f}{\partial y}\,dy + \frac{\partial f}{\partial z}\, dz \\ {\rm grad}\,f &= \frac{\partial f}{\partial x}\,\frac{\partial }{\partial x} + \frac{\partial f}{\partial y}\,\frac{\partial }{\partial y} + \frac{\partial f}{\partial z}\, \frac{\partial }{\partial z}. \end{align}$$Bear in mind this duality, because $df$ is not a vector per se (you can call it a covector, if you want).