# Definition of the curl in general coordinates

I have been unable to find the definition of the curl in general coordinates. Most sources that I have checked, including the math SE, only provide the expression for the curl in orthogonal coordinates. I suspect that the definition, written in the notation used in the post

Trouble with the definition of the cross product

ought to be

$$\nabla \times F := \frac{1}{\sqrt{\det[g_x]_{\gamma_x}}}\epsilon_{ijk}\frac{\partial\Big(\sqrt{\det[g_x]_{\gamma_x}}\big([F \circ \phi^{-1}]_{\gamma_x}\big)_j\Big)}{\partial{q_i}}b_l$$

where $$\epsilon_{ijk}$$ is the Levi-Civita symbol (not to be confused with the Levi-Civita tensor) and summation is implied over the repeated indices. My reason for suspecting this definition follows from examining the Voss-Weyl formula for the divergence. That is,

$$\nabla \cdot F := \frac{1}{\sqrt{\det[g_x]_{\gamma_x}}}\frac{\partial\Big(\sqrt{\det[g_x]_{\gamma_x}}\big([F \circ \phi^{-1}]_{\gamma_x}\big)_i\Big)}{\partial{q_i}}$$

Seeing that the Voss-Weyl formula retains the form of the usual "dot product", I figured that the expression for the curl ought to retain the form of the usual "cross product". Are my suspicions correct? If not, can someone please provide the proper expression? I apologize for the rather clunky notation.

Conceptual (but perhaps not practical) answer:

Are you familiar with the language of differential forms? They are really the right way to understand the curl. By "right" I mean that this understanding works in any coordinate system. If you can read John Lee's Smooth Manifolds then you can find a more complete version of this answer there.

One way to say what the curl is would be "exterior derivative + a three-dimensional coincidence." I'll sketch what I mean, but it might only make sense if you've seen these concepts before.

Step 1 is to take your vector field and replace it with a 1-form (so-called "flat" operation). If you've never done this before it should be thought of as taking a column vector and replacing it with a row vector. Doing so requires taking into account the inner product, so this will introduce the metric into the expression.

Step 2: take the exterior derivative of that 1-form. This produces a 2-form, because the exterior derivative of a $$k$$-form is a $$(k+1)$$-form. If exterior derivative is unfamiliar then I'm sure you could find some good posts on this site about it. This step will introduce partial derivatives in the coordinate directions.

Step 3: the 3-dimensional coincidence is that 2-forms can be identified with 1-forms. This step will introduce the $$\varepsilon_{ijk}$$ symbol and will again involve the metric. The more general phenomenon that's going on here is that $$k$$-forms on an $$n$$-manifold (equipped with a metric) can be identified with $$(n-k)$$-forms. Look up Hodge duality.

Step 4: Turn that 1-form back into a vector field. This is the curl of the vector field you started with.

Now all of those steps are easiest if you are on $$\mathbb R^3$$ and are using Cartesian coordinates, but they can be carried out in any coordinate system.

Like I said, maybe not the immediately useful, practical answer you were looking for. Turning it into a formula will require you to understand how to do each of those steps in your general coordinate system. Nonetheless, this is the way to define curl in a general coordinate system (and in fact on a general Riemannian 3-manifold).