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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

UPADTE 1 : it may be that I do confusions between covariant/contravariant coordinates of a vector and curvilinear coordinates (curviliear coordinates don't have covariant/contravariants properties)

UPDATE 2 : I am stucked on the following expression :

$$\vec{e_{i}}=g_{ij}\vec{e^{j}} = \partial_{i}=g_{ij} \partial^{j} = = g_{ij}dx^{j} = \dfrac{\partial}{\partial x^{i}}\dfrac{\partial}{\partial x^{j}} dx^{j} \neq \dfrac{\partial}{\partial x^{i}}\quad(1)$$

In the same time, I can write :

$$\vec{e_{i}}=g_{ij}\vec{e^{j}}=\dfrac{\partial}{\partial x^{i}}\dfrac{\partial}{\partial x^{j}} dx^{j}=\delta_{ij} \dfrac{\partial}{\partial x^{j}} = \dfrac{\partial}{\partial x^{i}}\quad(2)$$

What's one of these two above expressions, (1) and (2), is correct ?

The main issue is that I don't know if I can express the basis vector $\vec{e_{i}}$ as a linear combination of 1-forms $\vec{e^{i}}=\partial/\partial x^{i}$ (dual basis vectors) : at first sight, it doesn't make any sense. Anyone could help me with this contradiction ?

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    $\begingroup$ I have no idea what you mean by "covariant coordinates." $\endgroup$ – Ted Shifrin Jun 17 '18 at 21:04
  • $\begingroup$ -@TedShifrin indeed, this is the confusion I do, one says "contravariant/covariant" coordinates for a vector but for curvilinear coordinates, can we distinguish $\partial^{i}$ and $\partial_{i}$ by considering $\partial_{i} = (g^{ij})^{-1}\partial^{j}$ and introducing a pseudo "covariant curvilinear coordinates" $x_{i}$ ?, i.e can one find a way to check or define a coordinate "$x_{i}$" that could allow to write : $\partial^{i} = \dfrac{\partial}{\partial x_{i}}$, I hope I am clear. $\endgroup$ – youpilat13 Jun 17 '18 at 21:30
  • $\begingroup$ Your conventions are quite confusing. Usually, when one distinguishes between covariant and contravariant components, writes indices for vector components up ($v^i $) and indices for vectors themselves down ($\partial_i $). Of course you can change the criterion, but be consecuent. If you write $v_i$ for vector components then $\partial/\partial x^i $ must be denote as $\partial^i$. $\endgroup$ – Dog_69 Jun 17 '18 at 21:36
  • $\begingroup$ -@Dog_69 . So, can I have a relation between basis and dual basis vectors, i.e a type of relation like : $\vec{e_{i}}=g_{ij}\vec{e^{j}} = \partial_{i}=g_{ij} \partial^{j} = = g_{ij}dx^{j} = \dfrac{\partial}{\partial x^{i}}\dfrac{\partial}{\partial x^{j}} dx^{j} \neq \dfrac{\partial}{\partial x^{i}}$ since I have with Einstein summing convention : $\vec{e_{i}}=\sum_{j=1}^{n}\dfrac{\partial}{\partial x^{i}}\dfrac{\partial}{\partial x^{j}} dx^{j} \neq \dfrac{\partial}{\partial x^{i}}$ ? any clarification is welcome $\endgroup$ – youpilat13 Jun 17 '18 at 21:53
  • $\begingroup$ Well, the problem I see here is that $g_{ij}dx^j$ is a linear combination of 1-forms and hence just a 1-form. I mean, you can rise and down indices, but you have to be cautious. It is true the metric induces a bundle isomorphism $\phi_g:TM\rightarrow T^M$. This map takes a vector $X$ and gives you a 1-form $\alpha=\Phi_g(X)$ which in components can be express as $\alpha_i=g_{ij}X^j$. This is the operation of downing indices. With the inverse map $\phi_g^{-1}$ you can rise the indices of a 1-form $\alpha$: $X=\phi_g^{-1}(\alpha)$; $X^i = g^{ij}\alpha_j$. But this is different than $g_{ij}dx^j$. $\endgroup$ – Dog_69 Jun 18 '18 at 10:32

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