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  The Physical Interpretation for Curl 2 form of a vector field

+ 1 like - 0 dislike

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.


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

As a related post see the  following MO post:


Thank  you.

asked Jan 28, 2018 in Mathematics by AliTaghaviMath (145 points) [ no revision ]
recategorized Jan 28, 2018 by Dilaton

1 Answer

+ 2 like - 0 dislike

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.

answered Jan 28, 2018 by Ryan Thorngren (1,925 points) [ revision history ]
edited Jan 29, 2018 by Ryan Thorngren

@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.

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 + ...

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