Acording to Fortney's A Visual Introduction to Differential Forms and Calculus on Manifolds, the differential operator curl can be rewritten as the composite function $\sharp \circ \star \circ d \circ \flat$, where $\sharp$ and $\flat$ are the musical isomorphisms, $\star$ is the Hodge star operator and $d$ is the exterior derivative operator. I conjecture that this equality can be extended to higher dimensions and so I use it to derive a formula (in Cartesian coordinates) for a $4$-dimensional curl. But before we can do this, we have to extend the scope of the musical isomorphisms from vector fields and $1$-forms to $k$-vector fields and $k$-forms by defining
$(v_1 \wedge \ldots \wedge v_k)^\flat = v_1{}^\flat \wedge \ldots \wedge v_k{}^\flat, \,\,\,\,\, (\alpha_1 \wedge \ldots \wedge \alpha_k)^\sharp = \alpha_1{}^\sharp \wedge \ldots \wedge \alpha_k{}^\sharp$
where $v_i$ are vector fields and $\alpha_i$ are $1$-forms.
Given a vector field in a $4$-dimensional space $v = v_1\frac{\partial}{\partial x_1} + v_2\frac{\partial}{\partial x_2} + v_3\frac{\partial}{\partial x_3} + v_4\frac{\partial}{\partial x_4}$, we first compute $v^\flat = v_1dx_1 + v_2dx_2 + v_3dx_3 + v_4dx_4$. Then compute
\begin{eqnarray*} (d \circ \flat)v & = & \left(\frac{\partial v_2}{\partial x_1} - \frac{\partial v_1}{\partial x_2}\right)dx_1 \wedge dx_2 + \left(\frac{\partial v_3}{\partial x_1} - \frac{\partial v_1}{\partial x_3}\right)dx_1 \wedge dx_3 \\ & & + \left(\frac{\partial v_4}{\partial x_1} - \frac{\partial v_1}{\partial x_4}\right)dx_1 \wedge dx_4 + \left(\frac{\partial v_3}{\partial x_2} - \frac{\partial v_2}{\partial x_3}\right)dx_2 \wedge dx_3 \\ & & + \left(\frac{\partial v_4}{\partial x_2} - \frac{\partial v_2}{\partial x_4}\right)dx_2 \wedge dx_4 + \left(\frac{\partial v_4}{\partial x_3} - \frac{\partial v_3}{\partial x_4}\right)dx_3 \wedge dx_4 \end{eqnarray*}
Then apply $\star$ to the above to obtain
\begin{eqnarray*} (\star \circ d \circ \flat)v & = & \left(\frac{\partial v_2}{\partial x_1} - \frac{\partial v_1}{\partial x_2}\right)dx_3 \wedge dx_4 + \left(\frac{\partial v_3}{\partial x_1} - \frac{\partial v_1}{\partial x_3}\right)(-dx_2 \wedge dx_4) \\ & & + \left(\frac{\partial v_4}{\partial x_1} - \frac{\partial v_1}{\partial x_4}\right)dx_2 \wedge dx_3 + \left(\frac{\partial v_3}{\partial x_2} - \frac{\partial v_2}{\partial x_3}\right)dx_1 \wedge dx_4 \\ & & + \left(\frac{\partial v_4}{\partial x_2} - \frac{\partial v_2}{\partial x_4}\right)(-dx_1 \wedge dx_3) + \left(\frac{\partial v_4}{\partial x_3} - \frac{\partial v_3}{\partial x_4}\right)dx_1 \wedge dx_2 \end{eqnarray*}
Finally, apply $\sharp$ (the extended version) to the above and obtain
\begin{eqnarray*} \mbox{curl }v & = & (\sharp \circ \star \circ d \circ \flat)v \\ & = & \left(\frac{\partial v_2}{\partial x_1} - \frac{\partial v_1}{\partial x_2}\right)\frac{\partial}{\partial x_3} \wedge \frac{\partial}{\partial x_4} + \left(\frac{\partial v_1}{\partial x_3} - \frac{\partial v_3}{\partial x_1}\right)\frac{\partial}{\partial x_2} \wedge \frac{\partial}{\partial x_4} \\ & & + \left(\frac{\partial v_4}{\partial x_1} - \frac{\partial v_1}{\partial x_4}\right)\frac{\partial}{\partial x_2} \wedge \frac{\partial}{\partial x_3} + \left(\frac{\partial v_3}{\partial x_2} - \frac{\partial v_2}{\partial x_3}\right)\frac{\partial}{\partial x_1} \wedge \frac{\partial}{\partial x_4} \\ & & + \left(\frac{\partial v_2}{\partial x_4} - \frac{\partial v_4}{\partial x_2}\right)\frac{\partial}{\partial x_1} \wedge \frac{\partial}{\partial x_3} + \left(\frac{\partial v_4}{\partial x_3} - \frac{\partial v_3}{\partial x_4}\right)\frac{\partial}{\partial x_1} \wedge \frac{\partial}{\partial x_2} \end{eqnarray*}
Thus, a $4$-dimensional curl is an operator that acts on a vector field and returns a $2$-vector field with $6$ components. I would like to know if the above computations are reasonable and if the result is consistent with any established results for $4$-dimensional curls.