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I came across this old thread here:

"... Actually, there is a generalization of curl to any dimension. If you have a vector field, you can take its dual. So, the dual of $4 i + 2 j + 3 k$ would just be $4 dx + 2 dy + 3 dz$. So, that operation allows us to turn vector fields into 1-forms. The curl of the vector field corresponds to the exterior derivative. You take the dual, then exterior derivative, then the dual of that. That gives you curl. This process works on a vector field in any dimension. All you need is a smooth manifold...."

So I wanted to try this and I tried to calculate the following example:

$F = x^2 y \cdot i + xyz \cdot j - x^2 y^2 \cdot k$.

I found the dual of $F$: $F = x^2 y dx+ xyz dy - x^2 y^2 dz$ and I calculated the exterior derivative:

$$ dF = {\partial \over \partial y}(x^2 y)dy\wedge dx + {\partial \over \partial z} (xyz) dz \wedge dy + {\partial \over \partial x}(xyz) dx \wedge - {\partial \over \partial x}(x^2 y^2) dx \wedge dz - {\partial \over \partial y} (x^2 y^2) dy \wedge dz$$

where I used that $dx \wedge dx = 0$, etc.

Calculating the derivatives in this expression I got:

$$ dF = (x^2 - yz) dy \wedge dx + (xy + 2yx^2) dz \wedge dy - 2xy^2 dx \wedge dz$$

Now there is only one step left towards the solution: taking the dual of $dF$.

But what is the dual of a $2$-form like $dx \wedge dy$?

self-learner
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  • There is a trick for defining exterior derivative over tenzor product and multiple other algebra components. I think that may help. If no answer until later i will try answering :) – daniels_pa Apr 12 '16 at 06:13
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    In three dimensions, $dx^i\land dx^j$ is the dual of $x^i\times x^j$. – amd Apr 12 '16 at 06:33
  • I’d guess that you end up with an antisymmetic rank-2 tensor. In three dimensions, this can be identified with the usual definition of curl via cross product, as per my previous comment. – amd Apr 12 '16 at 07:38
  • @amd Great! Thank you for your comment! This is so helpful to me! – self-learner Apr 13 '16 at 03:25
  • You might also want to take a look at the discussion in this related question. – amd Apr 13 '16 at 10:50

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