The Physical Interpretation for Curl 2 form of a vector field

Question

Let  M  be  a  Riemannian  manifold. Then TM has a natural  structure of  a  symplectic manifold. Let w be the  symplectic  2-form imposed on TM. Assume that X: M  ---->TM  is  a  vector  field  on M.  The  Curl 2-form of X is  defined  as  X*(w).

In 3  dimensional Euclidean space this definition  is  closely related to the  standard Curl vector field. So the  above  definition enable us to  consider a concept of  "Curl" in arbitrary higher dimension.

Question:  

What is the  physical interpretations  for  Curl in dimension 4,5,...etc?

As a related post see the  following MO post:

https://mathoverflow.net/questions/291099/a-generalization-of-gradient-vector-fields-and-curl-of-vector-fields

Thank  you.

Answer 1

It's better to work in 1-forms to see what's going on. We can think of a one-form $\lambda$ as a section $s:X \to T^*X$ by pulling back the Liouville form $s^*\theta = \lambda$. The symplectic form is $d\theta$ so the "curl" from this roundabout construction is $s^*d\theta = ds^*\theta = d\lambda$, so curl is just dual (in the sense that 1-forms correspond to vector fields by line integrals) to the exterior derivative, which is ubiquitous in geometry and physics.

Here is a geometric interpretation of the curl: it's a 2-form, so it takes a value on each tangent plane. If one projects the vector field onto this plane and computes the usual 2d curl it will be the value of the 2-form curl on this tangent plane.

Comments

@RyanThorngren  Thank you  and (+1)  for  your  answer. Very interesting point. But just  some  questions:  By  projection of the  vector field on the  2-plane do you mean we use the map EXP locally?If not what do you mean by projection?  Moreover what is  the  2  dim. curl? The  curl is  a  vector  field  not a  scalar field. I guess that you mean Q_x  -P_y  where the  2 dim vector  field is  (P,Q).  Yes? Did I understand well? May you more  explain your answer?

Thank you.

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No need to use the exponential map. The vector field is already valued in the tangent plane at that point (everything is already linearized). Yes Qx - Py is the formula for the 2d curl. Think about the formula for the 3d curl in this way: (Qx - Py)dxdy + ...