I read somewhere people write gradient in covariant form because of their proposes. I think gradient expanded in covariant basis $i$, $j$, $k$, so by invariance nature of vectors, component of gradient must be in contravariant form. However we know by transformation properties and chain rule we find it is a covariant vector. What is wrong with my reasoning?
My second question is: if gradient has been written in covariant form, what is the contravariant form of gradient?
Most of the answers posted here are incorrect. The Wikipedia page for the gradient says
The gradient of $f$ is defined as the unique vector field whose dot product with any vector $v$ at each point $x$ is the directional derivative of $f$ along $v$.
A look at Theodore Frankel's The Geometry of Physics confirms this. Other posters have said that the components of the gradient of $f$ are given by $\partial_i f$; these are in fact the components of the differential of $f$, which is a covector. The gradient is this with the index raised.
Let's now calculate both sides of the expression from Wikipedia. The inner product of $\mathrm{grad}(f)$ with a vector $v$ is $$ \mathrm{grad}(f)^{i} g_{i j} v^j =\mathrm{grad}(f)^i v_i. $$ The directional derivative of $f$ along $v$ is $$ D_v f = v^i \partial_i f = g^{ij} \partial_i f ~v_j. $$ We can clearly identify $$ \mathrm{grad}(f)^{i} = g^{ij} \partial_j f $$ or $$ \mathrm{grad}f = g^{-1} \mathrm{d} f. $$
Gradient is covariant! Why?
The components of a vector contravariant because they transform in the inverse (i.e. contra) way of the vector basis. It is customary to denote these components with an upper index. So, if your coordinates are called $q$'s, they are denoted $q^i$.
Therefore, the gradient (or a derivative if you prefer) is $$\partial_i = \frac{\partial}{\partial q^i},$$ which transform as the inverse of the component transformation ( 1 / contra-variant = co-variant ).
If you're still not convinced... try this!
Cheers! ;-)
Recall the integral definition of the gradient:
$$\nabla \varphi = \lim_{V \to 0} \frac{1}{V} \oint_{\partial V} \varphi \hat n \, dS$$
This should tell you the gradient's components transform the same way as those of the normal vector $\hat n$, which is known to have covariant components.
You can verify that the normal vector has covariant components by recalling that the normal can be defined through a cross product of tangent vectors (which have contravariant components; the cross product of true vectors is a pseudovector, which has covariant components), for instance.
A covariant vector is commonly a vector whose components are written with ``downstairs" index, like $x_{\mu}$. Now, the gradient is defined as $\partial_\mu := \dfrac{\partial}{\partial x^\mu}$. As you can see the covariant vector $\partial_\mu$ is the derivative with respect to the contravariant vector $x^\mu$. the contravariant form of $\partial_\mu$ is $\partial^\mu := g^{\mu\nu}\partial_\nu$ - and in case the metric is constant $\partial^\mu = \frac{\partial}{\partial x_\mu}$.
Sometimes the vector $\partial_\mu$ is used to indicate a coordinate basis of the tangent vector space in some point of a manifold. In that case the index $\mu$ is not the index of the component (that for a vector should be upstairs), but it indicates the $\mu$th vector of the basis.
Gradient is covariant. Let's consider gradient of a scalar function. The reason is that such a gradient is the difference of the function per unit distance in the direction of the basis vector.
We often treat gradient as usual vector because we often transform from one orthonormal basis into another orthonormal basis. And in this case matrix transpose and inverse are the same.
Let $E, E'$ be matrices of basis vectors and $A$ be the transformation matrix between them.
$$ (E')^T = \begin{bmatrix} \hat{e}'_1 & \hat{e}'_2 & \vdots & \hat{e}'_n \end{bmatrix} = A \begin{bmatrix} \hat{e}_1 \\ \hat{e}_2 \\ \dots \\ \hat{e}_n \end{bmatrix} = A\ E $$
Let $\hat{v}, \hat{v}'$ be vectors in respective bases. Then $$\label{1} \tag{1} \hat{v} = v^i\hat{e}_i = E^T\begin{bmatrix}v_1\\v_2\\ \vdots \\v_n\end{bmatrix}$$ $$\label{2} \tag{2} \hat{v}' = v^i\hat{e}'_i = (E')^T\begin{bmatrix}v'_1\\v'_2\\ \vdots \\v'_n\end{bmatrix} = (A\ E)^T\begin{bmatrix}v'_1\\v'_2\\ \vdots \\v'_n\end{bmatrix} = E^TA^T\begin{bmatrix}v'_1\\v'_2\\ \vdots \\v'_n\end{bmatrix} $$ From $\ref{1}$ and $\ref{2}$ we have: $$A^T\hat{v}' = \hat{v}$$ and finally $$\boxed{\hat{v}' = (A^T)^{-1}\hat{v}}$$
I'll offer a simple explanation that relies only on the slope.
Suppose we have a line with slope 5, so for every 5 units up there corresponds one unit to the right. Now let us dilate the y axis by 3, that is, the spaces between each increment of 1 is now 3 times larger. In order for our slope to be invariant (equal 5), the slope's y component must also dilate by a factor of 3, that is, co-vary with the y axis.
Allow me to try to provide the simplest explanation of why the gradient is a covariant vector.
By definitions, the components of a covariant vector transform obey the law : $$ \overline A_i = \sum_{j=1}^n \frac {\partial x^j} {\partial \overline x^i} A_j \qquad \qquad (1) $$
and the the components of a contravariant vector transform obey the law : $$ \overline A^i = \sum_{j=1}^n \frac {\partial \overline x^j} {\partial x^i} A^j \qquad \qquad (2)$$
If the components of gradient of a scalar field in coordinate system $ \Bbb {\mathit {x_j}} $ , namely $ \frac {∂f} {∂x_j} $ , are known, then we can find the components of the gradient in coordinate system $ \Bbb { \overline { \mathit {x_i}}}$, namely $ \frac {∂f} {∂ \overline x_j} $, by the chain rule:
$$ \frac {\partial f} {\partial \overline x_i} = \frac {\partial f} {\partial x_1} \frac {\partial x_1} {\partial \overline x_i} + \frac {\partial f} {\partial x_2} \frac {\partial x_2} {\partial \overline x_i} + \cdot\cdot\cdot+ \frac {\partial f} {\partial x_n} \frac {\partial x_n} {\partial \overline x_i} = \sum_{j=1}^n \frac {\partial x_j} {\partial \overline x_i}\frac {\partial f} {\partial x_j} $$
Obviously $ \quad \frac {\partial f} {\partial \overline x_i}=\overline A_i \quad $ , and $ \quad \frac {\partial f} {\partial x_j} = A_j \quad $, then we get the equation same as $ (1) \quad $
Therefore, the gradient is a covariant vector.
