Suppose a 1-form $A$ of $\mathbb{R}^3$ is represented as $A= A_r (r,\theta,z)dr + A_\theta (r,\theta,z)d\theta + A_z(r,\theta,z)dz$ using cylindrical coordinate system $(r,\theta, z)$.
The external derivative of $A$ is
$dA = \Bigl( \dfrac{\partial A_z}{\partial \theta} - \dfrac{\partial A_\theta}{\partial z} \Bigr) d\theta\wedge dz
+\Big( \dfrac{\partial A_r}{\partial z} - \dfrac{\partial A_z}{\partial r} \Big) dz\wedge dr
+ \Big( \dfrac{\partial A_\theta}{\partial r} - \dfrac{\partial A_r}{\partial \theta} \Big) dr\wedge d\theta
$.
$*dA=\dfrac{1}{r} \Bigl( \dfrac{\partial A_z}{\partial \theta} - \dfrac{\partial A_\theta}{\partial z} \Bigr)dr + r \Big( \dfrac{\partial A_r}{\partial z} - \dfrac{\partial A_z}{\partial r} \Big) d\theta +\dfrac{1}{r} \Big( \dfrac{\partial A_\theta}{\partial r} - \dfrac{\partial A_r}{\partial \theta} \Big) dz $.
$(*dA)^\sharp = \dfrac{1}{r} \Bigl( \dfrac{\partial A_z}{\partial \theta} - \dfrac{\partial A_\theta}{\partial z} \Bigr) \dfrac{\partial}{\partial r} + \dfrac{1}{r} \Big( \dfrac{\partial A_r}{\partial z} - \dfrac{\partial A_z}{\partial r} \Big) \dfrac{\partial}{\partial \theta} +\dfrac{1}{r} \Big( \dfrac{\partial A_\theta}{\partial r} - \dfrac{\partial A_r}{\partial \theta} \Big) \dfrac{\partial}{\partial z} $.
But, for $\textbf{A}:= A_r (r,\theta,z)\dfrac{\partial}{\partial r} + A_\theta (r,\theta,z)\dfrac{\partial}{\partial \theta} + A_z(r,\theta,z)\dfrac{\partial}{\partial z}$, according to wikipedia, $\nabla \times \textbf{A} \neq (*dA)^\sharp$. I would like to know why.
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I should have written 1-form $A$ as $A=A_rdr + A_\theta rd\theta + A_zdz$.
The formulas in Wikipedia assume an orthonormal basis $$ \{\partial_r,\textstyle\frac{1}{r}\partial_\theta,\partial_z\} $$ (in their notation $\{\hat{\boldsymbol{\rho}},\hat{\boldsymbol{\varphi}},\hat{\boldsymbol{z}}\}$) w.r.t. the well-known cylindrical metric $$ g=\begin{pmatrix}1&0&0\\0&r^2&0\\0&0&1 \end{pmatrix}\,. $$ The dual basis is obviously $$ \{dr,r\,d\theta,dz\}\,. $$ OP's ansatz $$ A=A^r\,dr+A^\theta\,d\theta+A^z\,dz $$ tells us that the components of $A$ are in the coordinate basis $\{\partial_r,\partial_\theta,\partial_z\}\,.$ To get this into the the orthonormal basis we modify $A^\theta\,,$ that is, we make the better ansatz \begin{align} A&=A^r\,dr+A^\theta\,\color{red}{r}\,d\theta+A^z\,dz\,. \end{align} Then, \begin{align} dA&=(\partial_\theta A^z-r\,\partial_zA^\theta)\,d\theta\wedge dz+ (\partial_zA^r-\partial_rA^z)\,dz\wedge dr+(A^\theta+r\,\partial_rA^\theta-\partial_\theta A^r)\,dr\wedge d\theta\,. \end{align} Since the Hodge dual requires the orthonormal basis lets write this as \begin{align} dA&=\Big(\frac{\partial_\theta A^z}{r}-\partial_zA^\theta\Big)\,r\,d\theta\wedge dz+ (\partial_zA^r-\partial_rA^z)\,dz\wedge dr\\&\quad+\Big(\frac{A^\theta}{r}+\partial_rA^\theta-\frac{\partial_\theta A^r}{r}\Big)\,dr\wedge (r\,d\theta)\,. \end{align} Then, \begin{align} *dA&=\Big(\frac{\partial_\theta A^z}{r}-\partial_zA^\theta\Big)\,dr+ (\partial_zA^r-\partial_rA^z)\,r\,d\theta+\Big(\frac{A^\theta}{r}+\partial_rA^\theta-\frac{\partial_\theta A^r}{r}\Big)\,dz\,, \end{align} and therefore \begin{align} (*dA)^\sharp&=\Big(\frac{\partial_\theta A^z}{r}-\partial_zA^\theta\Big)\,\partial_r+ (\partial_zA^r-\partial_rA^z)\,\frac{\partial_\theta}{r}+\Big(\frac{A^\theta}{r}+\partial_rA^\theta-\frac{\partial_\theta A^r}{r}\Big)\,\partial_z\,. \end{align} The $\partial_z$-component can obviously be written as $$ \frac{A^\theta+r\,\partial_rA^\theta-\partial_\theta A^r}{r}\,\partial_z = \frac{\partial_r(rA^\theta)-\partial_\theta A^r}{r}\,\partial_z\,. $$ This shows that $(*dA)^\sharp$ equals the formula for $\nabla\times A$ from Wikipedia.