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?

  • 3
    $\begingroup$ What properties do you want for the generalized curl Do you want it to be a vector field? What should it do for you? $\endgroup$ – Deane Yang Jan 22 '11 at 15:14
  • 6
    $\begingroup$ The curl in $\mathbb{R}^3$ is the exterior derivative of a $1$-form under the identification: vector fields = $1$-forms=$*$2-forms given by the Euclidean metric. So in this sense, it surely does generalize. $\endgroup$ – Donu Arapura Jan 22 '11 at 16:39
  • $\begingroup$ Answers to this question might be relevant: mathoverflow.net/questions/10574 $\endgroup$ – José Figueroa-O'Farrill Jan 22 '11 at 18:11

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

  • $\begingroup$ What does bullet dot mean? any article or book about it? which field of math focus on this topic? many thanks. $\endgroup$ – ar2015 Jan 15 '18 at 0:01
  • 2
    $\begingroup$ @ar2015: It just means scalar product. I should have used $\cdot$ instead of the bullet. $\endgroup$ – Christian Blatter Jan 15 '18 at 10:21

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.

  • $\begingroup$ your second component for $\alpha$ is wrong, should be $R_x - P_z$ (orientation matters) $\endgroup$ – bthmas Jan 16 at 17:02
  • $\begingroup$ (cont.) $\star dy = dz \wedge dx$ but you reversed the orientation $\endgroup$ – bthmas Jan 16 at 17:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.