Generalization of Curl to higher dimensions In terms of vector field analogies to closed and exact differential forms, conservative and incompressible vector fields (gradient and divergence) generalize to higher dimensions, but curl and irrotational fields do not. Why? 
Cross product doesn't generalize either but one can use exterior products and hodge duals to fullfill the need. In differential geometry, there is a duality between the boundary operator on chains and the exterior derivative as expressed by the general Stokes' theorem. 
By a theorem of De Rham, the exterior derivative is the dual of the boundary map on singular simplices. From this perspective, there should be a generalized curl. Perhaps, there is an explanation in terms of the Poincare Lemma where for n>3 (but perhaps not 7), curl fails for higher dimensions? 
 A: 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.
A: Let ${\bf K}$ be a vector field in the neighbourhood of ${\bf p}\in{\mathbb R}^n$, and let ${\bf X}$ and ${\bf Y}$ be two tangent vectors at ${\bf p}$. These two vectors span a parallelogram $P$ with one vertex at ${\bf p}$. The "circulation" of ${\bf K}$ around $P$ computes to
$$
\int_{\partial P}{\bf K}\cdot \mathrm{d}{\bf x}= (L\,{\bf X})\cdot{\bf Y}- (L\,{\bf Y})\cdot{\bf X} + o(|P|^2)
$$
with $L:=\mathrm{d}{\bf K}({\bf p})$ and $|P|:= \mathrm{diam}(P)$. It follows that there is a certain skew bilinear function ${\rm Rot}{\bf K}({\bf p}):T_{\bf p}\times T_{\bf p}\to{\mathbb R}$ with
$$
\int_{\partial P}{\bf K}\cdot \mathrm{d}{\bf x}={\rm Rot}{\bf K}({\bf p})({\bf X},{\bf Y})+ o(|P|^2) \quad (|P|\to 0).
$$
In the case $n=3$ the bilinear form ${\rm Rot}$ can be represented by the vector ${\rm curl}{\bf K}$ in the form
$$ 
{\rm Rot}{\bf K}({\bf p})({\bf X},{\bf Y}) =  {\rm curl}{\bf K}({\bf p})\cdot({\bf X}\times{\bf Y}).
$$
