Curl: invariant under change of basis or not? I wondered how the curl$$\text{rot}\mathbf{F}=\left( \begin{array}{ccc}\partial_y F_3-\partial_z F_2 \\ \partial_z F_1-\partial_x F_3 \\ \partial_x F_2-\partial_y F_1 \end{array} \right)$$of a vector field $\mathbf{F}=(F_1,F_2,F_3)$ changes when the basis $\mathbb{R}^3$ is changed. I would have thought that it is invariant, because of the intuitive and physical interpretation of the curl and because, if I correctly understand other quantities typical of vector fields, like divergence, are (if I am not wrong $\text{div}\mathbf{F}$ is the trace of the Jacobian matrix $J_{\mathbf{F}}$ of $\mathbf{F}$, and the trace of $E J_{\mathbf{F}} E^{\text{T}}$ - see below - is the same of $J_{\mathbf{F}}$), but I have got serious problem to prove it.

Trial:
If I am not wrong, if $E\in\text{O}(3)$ is the basis change matrix, and if we define the function $\mathbf{G}$ as $\mathbf{y}\mapsto E\mathbf{F}(^t E \mathbf{y})$, invariancy is equivalent to $$E\text{rot}\mathbf{F}=\text{rot}\mathbf{G}$$please correct me if I am wrong. The Jacobian matrix $J_{\mathbf{G}}(\mathbf{y})$ of $\mathbf{G}$ in $\mathbf{y}$ should be:$$J_{\mathbf{G}}(\mathbf{y})=E J_{\mathbf{F}}(\mathbf{x}) E^{\text{T}}$$where $E^{\text{T}}$ is the transpose, and inverse, matrix of $E$.
I have used such an identity to find the expression of $\text{rot}\mathbf{G}(\mathbf{y})$ in terms of the components of $J_{\mathbf{F}}(\mathbf{x})$ and $E$, but my calculations do not give me the expected result: for example, in the first component of the expression of $\text{rot}\mathbf{G}$ calculated in such a way , the coefficient of $\partial_x F_2$ is $(e_{21}e_{32}-e_{22}e_{31})$, while, in the first component of $E\text{rot}\mathbf{F}$, the coefficient of $\partial_x F_2$ is $e_{13}$...

Is the curl invariant under a change of orthogonal basis and, if it is, how can it be correctly proved? Thank you so much for any answer!
 A: Update
In the first version of my answer (reprinted below) I tried to give a "geometrical" or "physical" explanation of the curl vector, but in fact did not address your question, which is the following: How do the coordinates of the curl vector ${\bf c}$ transform when we replace the given orthonormal coordinates  $(x_i)$ by an equally oriented orthonormal coordinates $(\bar x_l)\,$?
Denote the transform matrix from the old to the new coordinates by $T=[t_{il}]\in SO(3)$. The columns of $T$ then are the old coordinates of the new basis vectors $\bar{\bf e}_l$, and the rows of $T$ are the new coordinates of the old basis vectors ${\bf e}_i$. In your setup the coordinates of the points ${\bf x}$ and those of the force field ${\bf F}$  transform in the same way according to formulas like
$$x_i=\sum_l t_{il}\bar x_l,\quad \bar F_l=\sum_i t_{il} F_i\ .$$
Now in any orthonormal system the coordinates of the curl vector are given by
$$c_i={\partial F_{i-1}\over\partial x_{i+1}}-{\partial F_{i+1}\over\partial x_{i-1}}\ .$$
In order to determine the new coordinates $\bar c_l$ of ${\bf c}$ we have to compute partial derivatives of the $\bar F_k$ using the chain rule:
$${\partial \bar F_k\over\partial \bar x_l}=\sum_i t_{ik}\left(\sum_j{\partial F_i\over\partial x_j}\>{\partial x_j\over\partial\bar x_l}\right)=\sum_{i, \>j}{\partial F_i\over\partial x_j}\>t_{ik}t_{jl}\ .$$
We therefore obtain
$$\bar c_l={\partial \bar F_{l-1}\over\partial\bar x_{l+1}}-{\partial \bar F_{l+1}\over\partial \bar x_{l-1}}=\sum_{i, \>j}{\partial F_i\over\partial x_j}\>(t_{i,l-1}t_{j,l+1}-t_{i,l+1}t_{j,l-1})\ .\tag{1}$$
Now
$$t_{i,l-1}t_{j,l+1}-t_{i,l+1}t_{j,l-1}=({\bf e}_j\times {\bf e}_i)_l\ ,$$
which is $=0$ when $i=j$. This allows to write $(1)$ in the form
$$\bar c_l=\sum_i {\partial F_{i-1}\over\partial x_{i+1}}({\bf e}_{i+1}\times {\bf e}_{i-1})_l+
\sum_i {\partial F_{i+1}\over\partial x_{i-1}}({\bf e}_{i-1}\times {\bf e}_{i+1})_l\ ,$$
or
$$\bar c_l=\sum_i \left({\partial F_{i-1}\over\partial x_{i+1}}- {\partial F_{i+1}\over\partial x_{i-1}}\right)({\bf e}_i)_l=\sum_i c_i\>t_{il}\ .$$
This shows that  the coordinates of the curl vector ${\bf c}$ transform in the same way as the coordinates of the vectors ${\bf x}$ and ${\bf F}$.
First version:
Given a force field ${\bf F}$ on a domain $\Omega\subset{\mathbb R}^3$ its curl measures the "local nonconservativity" of ${\bf F}$. Consider a point ${\bf p}\in\Omega$ and a small parallelogram $P$ of diameter $|P|$ with center ${\bf p}$, and spanned by two vectors ${\bf X}$ and ${\bf Y}$. Calculating the integral of ${\bf F}$ around $\partial P$ (which would be $=0$ when ${\bf F}$ is conservative) gives the following:
$$\int_{\partial P}{\bf F}\cdot d{\bf x}= d{\bf F}({\bf p}).{\bf X}\cdot {\bf Y}- d{\bf F}({\bf p}).{\bf Y}\cdot {\bf X}+o\bigl(|P|^2\bigr)\ .$$
Here the main part of the right hand side is a skew bilinear function $R$ of the vectors ${\bf X}$ and ${\bf Y}$ spanning $P$:
$$\int_{\partial P}{\bf F}\cdot d{\bf x}=R({\bf X},{\bf Y})+o\bigl(|P|^2\bigr)\ .$$
This setup makes sense in  euclidean spaces of any dimension. Now in dimension $3$, such a skew bilinear function $R$ can be represented by a vector as follows: There is a well defined vector ${\bf c}$, called the curl of ${\bf F}$ at ${\bf p}$, such that
$$\int_{\partial P}{\bf F}\cdot d{\bf x}={\bf c}\cdot({\bf X}\times{\bf Y})+o\bigl(|P|^2\bigr)\ .$$
My answer to your question would  therefore be the following: As long as you stick to (linear) orthonormal coordinates of positive orientation the usual formula for ${\rm curl\,}{\bf F}$ is valid; but as soon as you introduce more general coordinates you would have to explain more carefully what you mean by the curl in the context.
A: My guess was that this might work only for infinitesimal rotations.
And there seems to be a question  (link) and an answer for this kind of problem: (link)
