Discrepancy between curl/grad in cartesian and cylindrical coordinates I have a scalar function $V$ and a 1-form $A$ defined as
$$
V = q \left(\frac{1}{r_+} - \frac{1}{r_-}\right)\\
A = q  \left(\frac{z-a}{r_+}-\frac{a+z}{r_-}\right)d\phi
$$
where
$$
r_+ = \sqrt{\rho^2 + (z-a)^2}\\
r_- = \sqrt{\rho^2 + (z+a)^2}\\
$$
And have to verify the equation
$$
\vec \nabla \cdot V = \vec \nabla \times \vec A
$$
As you can see $A$ suddenly becomes a vector. This is not a mistake on my part but is exactly what is done in the paper I am following. I believe that $\vec A$ is just the vector containing the components of the 1-form $A$ since in the paper the metric is actually $5$ dimensional and only the components of $A$ in a flat $\mathbb{R}^3$ sub manifold are used. However, I don't think this is important for my question (but I just added it to make sure that my error is not in my misinterpretation of this part).
Where my problem lies is when I calculate the gradient of $V$ together with the curl of $\vec A$. I use mathematica to calculate this using the commands
curlA = Curl[A, {rho, phi, z}, "Cylindrical"]
gradV = Grad[V, {rho, phi, z}, "Cylindrical"]

And find that these two are not the same.
The strange part comes when I transform $V$ and $\vec A$ to Cartesian coordinates using the transformation
$$
\rho \to \sqrt{x^2+y^2}\\
\phi \to \arctan{\frac{x}{y}}\\
d\rho \to \frac{x}{\sqrt{x^2+y^2}}dx + \frac{y}{\sqrt{x^2+y^2}}  dy\\
d\phi \to -\frac{y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy
$$
If I now calculate the gradient and curl using
curlA = Curl[A, {x, y, z}, "Cartesian"]
gradV = Grad[V, {x y, z}, "Cartesian"]

I do get that they are equal. Can anyone help me as to why I don't find this in cylindrical coordinates.
PS: I did also verify that their is no strange behaviour in the Curl and Grad function of Mathematica but this is not the case since it get the same result if I manually implement them.
 A: Converting between forms and vectors is not trivial in case of curvilinear coordinates.
It helps to remember that musical isomorphism from $\mathbf{d}\phi$ to vectors will give you $\mathbf{e}_\phi=\frac{d}{d\phi}$ i.e. a vector tangent to curve that is corresponds to fixed $\rho,z$ and changing $\phi$. As far as I remember, this is true by definition of $\mathbf{d}\phi$, i.e. $\mathbf{d}\phi$ is a form such that:
$$
\begin{align}
\mathbf{d}\phi\left(\frac{d}{d\rho}\right)&=0\\
\mathbf{d}\phi\left(\frac{d}{d\phi}\right)&=1\\
\mathbf{d}\phi\left(\frac{d}{dz}\right)&=0
\end{align}
$$
You can then use:
$$
\mathbf{e}_\phi=\frac{d}{d\phi}=\frac{\partial x}{\partial \phi}\frac{d}{dx}+\frac{\partial y}{\partial \phi}\frac{d}{dy}+\frac{\partial z}{\partial \phi}\frac{d}{dz}=\frac{\partial x}{\partial \phi}\mathbf{\hat{x}}+\frac{\partial y}{\partial \phi}\mathbf{\hat{y}}+\frac{\partial z}{\partial \phi}\mathbf{\hat{z}}=-\rho\sin\phi\mathbf{\hat{x}}+\rho\cos\phi\mathbf{\hat{y}}
$$
It follows that:
$$
\mathbf{e}_\phi.\mathbf{e}_\phi=\rho^2
$$
Thus $\mathbf{e}_\phi=\rho\boldsymbol{\hat{\phi}}$, where $\boldsymbol{\hat{\phi}}$ is the normalized basis-vector expected by Mathematica
The vector equivalent of the form $\mathbf{A}=A_\phi \mathbf{d}\phi$ is $\mathbf{A}=A_\phi\mathbf{e}_\phi$, so:
$$
\mathbf{A}=A_\phi\mathbf{e}_\phi=A_\phi \rho \:\boldsymbol{\hat{\phi}}
$$
And the vector you should give to Mathematica is:
$$
\left(\boldsymbol{\hat{\rho}}.\mathbf{A},\,\boldsymbol{\hat{\phi}}.\mathbf{A},\,\mathbf{\hat{z}}.\mathbf{A}\right)=\left(0,\,A_\phi\,\rho,\,0\right)
$$
Does this fix the issue?
