Generalization of Curl to higher dimensions

Question

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?

Answer 1

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 $$ {\rm Rot}{\bf K}({\bf p})({\bf X},{\bf Y}) = {\rm curl}{\bf K}({\bf p})\cdot({\bf X}\times{\bf Y}). $$

Answer 2

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.