The Curl of a vector field $X=P\partial_x+Q\partial_y+R\partial_z$ is equal to $$Curl X= (R_y-Q_z)\partial_x +(P_z-R_x)\partial_y+ (Q_x-P_y)\partial_z$$

For the moment we replace $\partial_x,\partial_y,\partial_z$ with $dy\wedge dz, dx\wedge dz, dx\wedge dy$ respectively.(In fact we apply the Hodge star operator to the dual of the base $\partial_x,\partial_y,\partial_z$.

So actually the component of $Curl X$ is identical to the compenents of the $2$ form $$\alpha=(R_y-Q_z)dy\wedge dz +(P_z-R_x)dx\wedge dz +(Q_x-P_y)dx\wedge dy$$

On the other hand the vector field $X$, being a section of the tangent bundle $T\mathbb{R}^3$, can be considered as a map $$X:\mathbb{R}^3 \to \mathbb{R}^3\times \mathbb{R}^3\\X(x,y,z)=(x,y,z, P(x,y,z),Q(x,y,z),R(x,y,z))$$. Note that the following equality hold:

$$\alpha =X^* \omega$$ where $\omega$ is the natural symplectic structure of $\mathbb{R}^3 \times \mathbb{R}^3$ with $\omega= dx\wedge dp +dy\wedge dq+dz \wedge dr$.

The situation described above is a motivation to consider the following generalization of the concept of the Curl of a vector field on an arbitrary Riemannian manifold.

A generalized Curl: Let $(M,g)$ be a Riemannian manifold. The metric $g$ gives an isomorphism(hence diffeomorphism) between $TM$ and $T^* M$. So the standard intrinsic symplectic structure of the cotangant bundle is carried to a symplectic structure $\omega$ on $TM$. Now assume that $X:M \to TM$ is a vector field. We define the Curle of $X$ as a $2$-form with the following formula $$Curl X=X^* \omega$$

This is already mentioned at this MO question.

A generalization of Gradient vector fields and Curl of vector fields

The curl of a vector field $X=P\partial_x+Q\partial_y+R\partial_z$ is equal to
$$ \mathrm{Curl}(X)= (R_y-Q_z)\,\partial_x +(P_z-R_x)\,\partial_y+ (Q_x-P_y)\,\partial_z $$

For the moment we replace $\partial_x,\partial_y,\partial_z$ with $dy\wedge dz,\ dx\wedge dz$ and $dx\wedge dy$ respectively  (In fact we apply the Hodge star operator to the dual of the basis $\partial_x,\partial_y,\partial_z$).

So actually the component of $\mathrm{Curl}(X)$ is identical to the components of the $2$ form $$ \alpha=(R_y-Q_z)\,dy\wedge dz +(P_z-R_x)\,dx\wedge dz +(Q_x-P_y)\,dx\wedge dy $$

On the other hand the vector field $X$, being a section of the tangent bundle $T\mathbb{R}^3$, can be considered as a map 
\begin{align} X\colon \mathbb{R}^3& \to \mathbb{R}^3\times \mathbb{R}^3\\ (x,y,z)&\mapsto (x,y,z, P(x,y,z),Q(x,y,z),R(x,y,z)). \end{align} Note that the following equality holds: $$ \alpha =X^* \omega, $$
where $\omega$ is the natural symplectic structure of $\mathbb{R}^3 \times \mathbb{R}^3$ with $\omega= dx\wedge dp +dy\wedge dq+dz \wedge dr$.

The situation described above is a motivation to consider the following generalization of the concept of the curl of a vector field on an arbitrary Riemannian manifold.

A generalized curl: Let $(M,g)$ be a Riemannian manifold. The metric $g$ gives an isomorphism  (hence diffeomorphism) between $TM$ and $T^* M$. So the standard intrinsic symplectic structure of the cotangent bundle is carried to a symplectic structure $\omega$ on $TM$. Now assume that $X:M \to TM$ is a vector field. We define the curl of $X$ as a $2$-form with the following formula: $$ \mathrm{Curl}(X):=X^* \omega. $$

This was already mentioned at the MO question  A generalization of Gradient vector fields and Curl of vector fields.