Following the issue on this link, I want to know the nature of a differential function from a tensor calculus point of view.
In one of the answers, it is said that one has 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 $v$
Then, taking the starting basis $\{e_{i}=\dfrac{\partial}{\partial x^{i}}\}$ and its dual basis $\{e^{i}=\text{d}x^{i}\}$, we can write the differential function $df$ and $\vec{grad}\,f$ as :
$$\begin{align} df &= \frac{\partial f}{\partial x}\,dx + \frac{\partial f}{\partial y}\,dy + \frac{\partial f}{\partial z}\, dz \\ {\rm grad}\,f &= \frac{\partial f}{\partial x}\,\frac{\partial }{\partial x} + \frac{\partial f}{\partial y}\,\frac{\partial }{\partial y} + \frac{\partial f}{\partial z}\, \frac{\partial }{\partial z}. \end{align}$$
So with vectors $e_{i}$ and $e^{i}$ :
$$\begin{align} df &= \frac{\partial f}{\partial x}\,\vec{e^1} + \frac{\partial f}{\partial y}\,\vec{e^2} + \frac{\partial f}{\partial z}\, \vec{e^3} \\ {\rm grad}\,f &= \frac{\partial f}{\partial x}\,\vec{e_1} + \frac{\partial f}{\partial y}\,\vec{e_2} + \frac{\partial f}{\partial z}\, \vec{e_3}. \end{align}$$
So, we could say that ${\rm grad}\,f$ is a classic vector (I mean no covector) and $\text{d}f$ is a co-vector (relatively to basis vectors and dual basis vectors).
My issue is about the nature of $\text{d}f$, I don't understand how the dot product between a vector $\vec{v}$ (maybe a covector with the definition of dot prouct) and the classic vector ${\rm grad}\,f$ could give a differential function under the form of a co-vector, i.e the expression :
$$df(p)(v) = \langle {\rm grad}\,f(p),v\rangle,$$
I thought that the result of a dot product between a vector (${\rm grad\,f}$) and a co-vector $\vec{v}$ is a scalar, not a co-vector.
Maybe I do confusions between the value (scalar value ???) expressed as :
$$df(p)(v) = \langle {\rm grad}\,f(p),v\rangle=$$ $$\begin{align} df(p)(v) &= \frac{\partial f}{\partial x}\,v^x + \frac{\partial f}{\partial y}\,v^y + \frac{\partial f}{\partial z}\,v^z\end{align}$$
and the expression of $df$ as a co-vector (vector expressed in dual basis $\{e^{i}=\text{d}x^{i}\}$) :
\begin{align} df &= \frac{\partial f}{\partial x}\,\vec{e^1} + \frac{\partial f}{\partial y}\,\vec{e^2} + \frac{\partial f}{\partial z}\, \vec{e^3} \end{align}
Someone could help me to understand better or give me clarifications ?
UPDATE 1 :
@Uldreth Your answer is very interesting, especially by your formulation :
$$\langle \text{grad}f,v\rangle=(g^{\mu\nu}\partial_\mu f)g_{\nu\kappa}v^\kappa$$
Then, as I said, this expression is equivalent to :
$$df(\vec{v})=\langle {\rm grad}\,f(p),v\rangle =\partial^\mu f v_\mu$$
because we can swap metric tensor $g_{ij}$ between the 2 equations.
I thought usually that gradient vector was considered like a covector because its components are transformed like the basis vector.
But by noting :
$${\rm grad}\,f = \frac{\partial f}{\partial x}\,\frac{\partial }{\partial x} + \frac{\partial f}{\partial y}\,\frac{\partial }{\partial y} + \frac{\partial f}{\partial z}\, \frac{\partial }{\partial z}.$$
$${\rm grad}\,f = \frac{\partial f}{\partial x}\,\vec{e_1} + \frac{\partial f}{\partial y}\,\vec{e_2} + \frac{\partial f}{\partial z}\, \vec{e_3}.$$
We can see that $\vec{e_{i}}$ are starting basis vectors (not dual basis vectors), so $\frac{\partial f}{\partial x_{i}}$ represent the contravariant components of ${\rm grad}\,f$.
But we can also consider that, with the the expression of $df(v)=\partial_\mu fv^\mu=\partial^\mu f v_\mu$, $\partial^\mu$ represents the contravariant components and $v_\mu$ covariant components.
For example, into 2D polar coordinates, for the $\theta$ component of ${\rm grad}\,f$, we have (taking $x_\theta$ like the covariant component and $x^{\theta}=\theta$ like the contravariant component).
UPDATE 2 :
I wonder why one says that "${\rm \overrightarrow{grad}}\,f$" is a covector (because $({\rm \overrightarrow{grad}}\,f)_{i}$ components are covariant components (relatively to $\{\vec{e_i}\}$ basis vectors since $({\rm \overrightarrow{grad}}\,f)_{i}=({\rm \overrightarrow{grad}}\,f)\cdot\vec{e_i}$) and SO contravariant components relatively to $\{e^i\}$ dual basis vectors since $({\rm \overrightarrow{grad}}\,f)=({\rm \overrightarrow{grad}}\,f)_{i}\vec{e^i}$).
Should I rather call it a "classical vector", i.e its covariant components transform in the same way that $\{e_{i}\}$ basis vectors and its contravariant components in the same way that $\{e^{i}\}$ dual basis vectors ?
By "classical vectors", I mean that I can just express this gradient vector under 2 different forms, i.e :
$$({\rm \overrightarrow{grad}}\,f)=({\rm \overrightarrow{grad}}\,f)_{i}\vec{e^i}=\partial_{i}f\vec{e^i}$$
OR
$$({\rm \overrightarrow{grad}}\,f)=({\rm \overrightarrow{grad}}\,f)^{i}\vec{e_i}=\partial^{i}f\vec{e_i}$$
???
Thanks for your help.