Directional derivative, gradient and metric

Question

Considering the general expression of gradient with the directional derivative operator on $f$ function along $\vec{v}$ vector :

$$df(v)=\langle\text{grad}(f),v\rangle = g^{ij}\partial_{i} f v_{j}=\partial_{i} f v^{i}$$

taking $\partial_{i} = \dfrac{\partial}{\partial x^{i}}$ (with $x^{i}$ contravariant coordinates).

Can I write also :

$$df(v)=\langle\text{grad}(f),v\rangle = \partial^{i} f v_{i}$$

with $\partial^{i} = g^{ij} \partial_{j} =\dfrac{\partial}{\partial x_{i}}$ where $x_{i}$ are covariant coordinates ??

i.e, I don't know if I can raise up the index of $\partial_{j}$ multiplying it by $g^{ij}$ while defining $\partial^{i} = \dfrac{\partial}{\partial x_{i}}$ ?

Regards

EDIT 1 : it may be that I do confusions between covariant/contravariant coordinates of a vector and curvilinear coordinates (curviliear coordinates are always contravariant, aren't they ?)

EDIT 2 : question transferred on https://math.stackexchange.com/questions/2823029/directional-derivative-gradient-and-metric

Answer 1

Suppose you have a $d$-dimensional smooth manifold $M$. A chart is a tuple $(U,x)$ where $U$ is an open set in $M$, and $x:U\to D$ is a homeomorphism to some $D$ in $\mathbb{R}^{n}$.
Since $\mathbb{R}^{n}$ is a cartesian product of $n$ pieces, you can cannonically split $x$ in $n$ maps $x^{i}:U\to\mathbb{R}$. These are the so called coordinate functions. They are neither covariant nor contravariant. The upper index is just a matter of convention.

Now, the tangent space $T_pM$ at a point $p\in M$ is the set of all tangent vectors at $p$. When we use a chart, the coordinate functions induce a basis por $T_pM$. They are denoted by $$\left(\frac{\partial}{\partial x^{i}}\right)_p$$

Since this is a basis, any vector $X\in T_pM$ is be expressible in terms of the $\left(\frac{\partial}{\partial x^{i}}\right)_p$, using certain coefficients: $$X=X^{i}\left(\frac{\partial}{\partial x^{i}}\right)_p$$ when we make a change of coordinates, we make the choice to call the basis vectors covariant. Since the vectors $X\in T_pM$ are invariant, we deduce that the components $X^{i}$ must transform in the opposite fashion as the basis vectors $\left(\frac{\partial}{\partial x^{i}}\right)_p$. That is why say that a vector has covariant components.

By a similar argument of invariance under change of coordinates, we arrive to the conclusion that the basis covectors (the basis of the cotangent space $T^{*}_{p}M$) must transform contravariantly, and hence the components of the covectors must transform covariantly. But remember, all this covariant-contravariant nomenclature is because we choose to put the name covariant to the way the basis of $T_pM$ transforms.

Now, if you are using upper indices to label your coordinates, this leads to labeling the basis of $T_pM$ with lower indices, and the components of your vectors with upper indices. (All because of the notation $\left(\frac{\partial}{\partial x^{i}}\right)_p$, and the desire to use Einstein's summation convention).

You can then make the definition $$∂_i:=\frac{∂}{∂ x^i}$$ but the $x^{i}$ in the derivative operator are the labels of the coordinates, not the components of a vector; which means you cannot write $$\frac{∂}{∂ x_i}\tag{nonsense 1}$$ because there is not an object $x_i$

The lowering-and-raising indices business is done only on the components of vectors. Not on the vectors themselves. This is why the expression $$\vec{e^{i}} = g^{ij}\vec{e_j} \tag{nonsense 2}$$ does not make any sense, since on the left hand side you have an element of $T_pM$ and on the right hand side you have an element of $T_p^{*}M$ (remember $g_{ij}$ are just scalars).

You can still make the definition $$\partial^{i} := g^{ij}\left(\frac{\partial}{\partial x^{i}}\right)_p$$ But since $g_{ij}$ are scalars, both sides of the equation are elements of $T_pM$. I.e., vectors. Hence, $$\partial^{i} = dx^{i}\tag{nonsense 3}$$ makes no sense either: again, you would be equating an element of $T_pM$ with an element of $T_p^{*}M$.

Finally, regarding your expressions

$$\dfrac{\partial}{\partial x^{i}}\dfrac{\partial}{\partial x^{j}} dx^{j} =\delta^{j}_{i} \dfrac{\partial}{\partial x^{j}} = \dfrac{\partial}{\partial x^{i}} \tag{4}$$

they don't make a lot of sense either. If you are performing contraction of a covector with a vector, you need to let your notation express the fact that covector is eating the vector you want it to eat: Remember $$dx^{i}\left(\frac{\partial}{\partial x^{j}}\right) = \delta^{i}_{j}$$ So, coming back to your expression $(4)$, from the right hand side, I deduce you wanted to denote that "$dx^{j}$ is eating $\left(\frac{\partial}{\partial x^{i}}\right)$", in that case then you should write $$dx^{j}\left(\dfrac{\partial}{\partial x^{i}}\right)\dfrac{\partial}{\partial x^{j}} =\delta^{j}_{i} \dfrac{\partial}{\partial x^{j}} = \dfrac{\partial}{\partial x^{i}} \tag{4.1}$$ This is important, because the left hand side of $(4)$ could be also interpreted as "$dx^{j}$ is eating $\left(\frac{\partial}{\partial x^{j}}\right)$", and in that case you'd get $$\dfrac{\partial}{\partial x^{i}}dx^{j}\left(\dfrac{\partial}{\partial x^{j}}\right) = \dfrac{\partial}{\partial x^{i}} \delta^{j}_{j} = n\,\dfrac{\partial}{\partial x^{i}} \tag{4.2}$$ (where $n$ is the dimension of the space).

Just to emphasize once again the importance of the notation, let me say that were it not for the right hand side of $(4)$, I'd have interpreted the left hand side as a tensor product $$\dfrac{\partial}{\partial x^{i}}\dfrac{\partial}{\partial x^{j}} dx^{j} = \dfrac{\partial}{\partial x^{i}}\otimes\dfrac{\partial}{\partial x^{j}}\otimes dx^{j} \tag{4.3}$$

Answer 2

The gradient is defined:

$$du=\vec{\textrm{grad}}u\cdot d\vec{r}$$ with $u=u(\vec{r})$ and $\vec{r}=x^i\vec{e_i}$ since the coordinates $x^i$ transform contravariantly. Thus $d\vec{r}=dx^i \vec{e}_i$ and $$du=\frac{\partial u}{\partial x^i}dx^i=\left(\vec{e}^i\frac{\partial}{\partial x^i}\right)u\cdot\left(dx^k\vec{e}_k\right)=g^i_k\frac{\partial u}{\partial x^i}dx^k$$ since for a cartesian coordinate system $g^i_k=\vec{e}^i\cdot\vec{e}_k=\delta^i_k$ we have $$\vec{\textrm{grad}}= \vec{e}^i\frac{\partial}{\partial x^i}$$

Therefore the coordinates given by $ \frac{\partial}{\partial x^i}$ are always covariant.

You can generalise the result for any system of coordinates given by the transformation $x^i=x^i(\xi^k)$ and the following metric tensor (deduced from the invariance of $du$ w.r.t. the employed coordinate system): $$g^i_k=\left(\vec{e}^l\frac{\partial \xi^i}{\partial x^l}\right)\cdot \left(\frac{\partial x^m}{\partial\xi^k}\vec{e}_m\right) $$

Answer 3

If you are in a flat Euclidean space $\Bbb R^n$ and you are using the euclidean metric then the metric tensor is $g_{ij}=\delta_{ij}$ as well $g^{ij}=\delta^{ij}$. So the law of raising indexes gives that $\partial^i=\partial_i$.