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

migrated from physics.stackexchange.comJun 17 '18 at 21:03

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• Would Mathematics be a better home for this question? – Qmechanic Jun 17 '18 at 20:06
• -@Qmechanic you are right, I am going to transfert it on Math exchange. – youpilat13 Jun 17 '18 at 21:00
• Coordinates are neither covariant nor contravariant. A tangent vector has contravariant components, and a covector has covariant components. If you are using upper indices to label your coordinates, this leads to labeling the components of your vectors with upper indices and the components of your covectors with lower indices. You can then make the definition $\partial^i := g^{ij}\frac{\partial}{\partial x^j}$, but the $x$ in the derivative operator labels coordinates, not the components of a vector; which means you cannot write $\frac{\partial}{\partial x_i}$ – Jackozee Hakkiuz Jun 18 '18 at 21:15
• -@JackozeeHakkiuz that's what I suspected, the confusion came from the fact that I can't express $\partial^{i}$ by considering a new "pseudo" curvilinear coordinate, the only thing I can write is : $\partial^{i}=g^{ik}\partial_{k} = g^{ik}\partial/\partial x^{k}$, can't I ? – youpilat13 Jun 18 '18 at 21:29
• That's right. You can use $g^{ij}$ and $g_{ij}$ to raise and lower indices that label the components of vectors, covectors and other tensors. Not the indices of the vectors themselves. – Jackozee Hakkiuz Jun 19 '18 at 3:32

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 $T_p^{*}M$.

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

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 $$\dfrac{\partial}{\partial x^{i}}\dfrac{\partial}{\partial x^{j}} dx^{j} = \dfrac{\partial}{\partial x^{i}}\otimes\dfrac{\partial}{\partial x^{j}}\otimes dx^{j} \tag{4.3}$$

• I want to emphasize that this is a really good answer, a perfect example of what answers on this website are expected to be. – Yuri Vyatkin Jul 14 '18 at 7:41
• @YuriVyatkin Hey, that is very nice of you. Thanks :) ! – Jackozee Hakkiuz Jul 18 '18 at 6:53

$$du=\vec{\textrm{grad}}u\cdot d\vec{r}$$ 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): $$g^i_k=\left(\vec{e}^l\frac{\partial \xi^i}{\partial x^l}\right)\cdot \left(\frac{\partial x^m}{\partial\xi^k}\vec{e}_m\right)$$
• -@HBR 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 – youpilat13 Jun 17 '18 at 22:41
• -@HBR 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, regards – youpilat13 Jun 17 '18 at 22:44
• -@HBR how to qualify this pseudo "covariant coordinate" $x_{i}$ in the expression $\partial^{i} = \dfrac{\partial}{\partial x_{i}}$ ? the classical expression is $\partial_{i} = \dfrac{\partial}{\partial x^{i}}$ and $\partial_{i}$ is considered as a classical basis vector $\vec{e_{i}}$. I don't know how to distinguish the basis vectors assimilated to $\vec{e_{i}}=\dfrac{\partial}{\partial x^{i}}$ and $\vec{e^{i}}=\text{d}x^{i}$ from the definition of gradient which implies term $\dfrac{\partial}{\partial x^{i}}$ too . Any help is welcome – youpilat13 Jun 18 '18 at 8:41
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 $\partial^i=\partial_i$.