Solenoidal and conservative fields

[TrickyDicky]

I had doubts whether to post this here or in in the physics section but I did here because I'm more interested in a purely mathematical view in this case.
I understand a solenoidal vector field implies the existence of another vector field, of which it is the curl: $$S=\nabla X A$$ because the divergence of the curl of any vector field is zero.
But what if the vector field is conservative instead? I guess in this case it is not necessarly implied the existence of a vector potential.
How about in the case of a laplacian vector field, that is both conservative and solenoidal, does it imply the existence of a vector potential?

 

[TrickyDicky]

no takers?
Is this question more appropriate maybe for the physics section?

 

[homeomorphic]

What do you mean by solenoidal?

Cartan's exterior calculus and De Rham cohomology sheds a lot of light on these things. Locally, every closed form is exact. Closed means its exterior derivative is zero. Exact means it is the exterior derivative of something. Taking the gradient, curl, divergence can be done by taking some sort of exterior derivative. In Euclidean space, the exterior derivative of a function is the gradient, the exterior derivative of the dual of a vector field the hodge dual of the curl, and the exterior derivative of the Hodge dual of the dual to the vector field is divergence. If you believe that this is giving you some sort of cohomology, if you know what that is, it's obvious that, locally, a potential should exist if your curl is zero, and a vector potential should exist if your divergence is zero because cohomology is a topological invariant and a manifold has trivial local topology (this fact is essential what is known as the Poincare Lemma).

Globally, these things actually depend on the topology of the space you are living in.

A conservative vector field is one that is curl-free. That doesn't tell you anything about a vector potential, just a potential. You can easily have a vector field that is curl-free, but has some divergence. That's basically what happens in electostatics all the time. If there are no time-varying magnetic fields, Maxwell's equations tell you that the electric field is curl-free. But it can easily have divergence. Just put down some charges wherever you like.

However, if a vector field has a vector potential, then it must be divergence free because it is the curl of something.

So, vanishing divergence is a requirement to have a vector potential. Luckily, one of Maxwell's equations, again, tells you that that's always true of magnetic fields.

 

[Matterwave]

In not so complicated words, working in Euclidean space, a curl free vector field is the gradient of a scalar field and a divergence free vector field is the curl of a vector field.

 

[TrickyDicky]

Thanks both for your answers.

 

[TrickyDicky]

What do you mean by solenoidal?

Just divergence-free

So, vanishing divergence is a requirement to have a vector potential.

Yes, but is it a sufficient condition too?
Also my question referred to vector fields like the magnetic field that seem to be both divergence free and curl-free, that seems to require for that field to have both a vector potential and a scalar potential, like magnetic fields indeed have. That makes the magnetic field a Laplacian field(meaning the gradient of a Laplace equation solution), right?

 

[TrickyDicky]

working in Euclidean space, a curl free vector field is the gradient of a scalar field

Good that you use the qualifier "in Euclidean space"( I have a question about this in another subforum that got no answers). The "gradient of a scalar field has vanishing curl" rule is not necessarily the case if the space is not flat, right?

 

[Matterwave]

If you have a general manifold on which to work, you have to deal with a somewhat nontrivial extension of what you mean by "curl". I don't want to get too technical, since I'm bound to mess up.

The general result is that the exterior derivative (off of which curl, gradient and divergence are based) satisfies the property that dd=0 (e.g. curl of gradient=0 or divergence of curl =0). The converse, that given df=0 implies that f=dg locally, arises from Poincare's lemma, and that is true in general. If you want to remove the requirement of "locally" and have it hold globally, then your manifold must be contactable (i.e. smoothly deformable to a point - it can't have any holes or w/e).

Flatness is not a requirement. In fact, an exterior algebra can be defined on a manifold without defining a connection (and therefore any sense of curvature).

 

[tiny-tim]

hi homeomorphic!

i assume you don't have the ∂ or Λ symbols?

let's translate …

Cartan's exterior calculus and De Rham cohomology sheds a lot of light on these things. Locally, every closed form is exact. Closed means its exterior derivative is zero. Exact means it is the exterior derivative of something. Taking the gradient, curl, divergence can be done by taking some sort of exterior derivative. In Euclidean space, the exterior derivative of a function is the gradient, the exterior derivative of the dual of a vector field the hodge dual of the curl, and the exterior derivative of the Hodge dual of the dual to the vector field is divergence.

Λ is the wedge product (or exterior product)

∂ Λ is the exterior derivative (also called the boundary, for an intuitively obvious reason which i can't remember ): it converts a scalar to a 1-form (a vector), a 1-form to a 2-form, and so on

∂ Λ ∂ Λ anything is zero ("the boundary of a boundary is 0")

in n dimensions, a k-form has ⁿC_k components (so an n-form is effectively a scalar, and higher forms do not exist)

* is the hodge dual, converting k-forms to (n-k)-forms

** is the identity

F is closed if ∂ Λ F = 0. F is exact if F = ∂ Λ G (and since ∂ Λ ∂ Λ G has to be 0, that means that any exact form is obviously closed). Any closed form is exact.

in ℝ³, a 1-form is a vector, and the dual of a 2-form is a 1-form (in the dual space), and so is a pesudovector or cross-product-vector (a vector in the dual space), and …

∂ Λ f = ∇f = grad(f)
*∂ Λ A = ∇ x A = curl(A)
*(A Λ B) = A x B
*(∂ Λ *A) = ∇.A = div(A)​

in space-time, a 1-form is a 4-vector, the dual of a 3-form is a (pseudo?)4-vector, and a 2-form is a new sort of thing with 6 components, such as the electromagnetic field (technically, the faraday 2-form, whose dual is the maxwell 2-form), and …

an electromagnetic field is ∂ Λ (the electromagnetic potential 4-vector) …

∂ Λ (φ,A) = (E;B) = (-∇φ - ∂A/∂t;∇xA)​

so ∂ Λ (electromagnetic field) = 0 (this is 2 of maxwell's equations) …

∂ Λ (E;B) = *(∇.B, ∂B/∂t + ∇xE) = 0​

and *∂ Λ *(electromagnetic field) = a 1-form, the 4-vector (ρ,J) (this is the other 2) …

∂ Λ *(E;B) = *(∇.E, -∂E/∂t + ∇xB) = *(ρ,J)​

Good that you use the qualifier "in Euclidean space"( I have a question about this in another subforum that got no answers). The "gradient of a scalar field has vanishing curl" rule is not necessarily the case if the space is not flat, right?

the ∂ Λ stuff works in any space, but it doesn't necessarily translate nicely into ∇ stuff …

If you have a general manifold on which to work, you have to deal with a somewhat nontrivial extension of what you mean by "curl".

 

[TrickyDicky]

If you have a general manifold on which to work, you have to deal with a somewhat nontrivial extension of what you mean by "curl". I don't want to get too technical, since I'm bound to mess up.

The general result is that the exterior derivative (off of which curl, gradient and divergence are based) satisfies the property that dd=0 (e.g. curl of gradient=0 or divergence of curl =0). The converse, that given df=0 implies that f=dg locally, arises from Poincare's lemma, and that is true in general. If you want to remove the requirement of "locally" and have it hold globally, then your manifold must be contactable (i.e. smoothly deformable to a point - it can't have any holes or w/e).

Flatness is not a requirement. In fact, an exterior algebra can be defined on a manifold without defining a connection (and therefore any sense of curvature).

Yes, I'm aware of the topological requirements about the space being contractible and simply-connected. And also that in the exterior algebra-differential forms context flatness would not be a requirement. But as tiny tim pointed out the covariant derivative doesn't translate well into the exterior derivative and in the case a curved connection is needed (in the presence of curvature of the space) the "exterior covariant derivative" allows D^2 different than zero.
What I'm saying is that the proof of the vanishing curl of the gradient of a scalar field rests on the fact that the order that the partials are taken doesn't matter, so that every component of the resulting vector should be zero. But I'd say in the case the space has curvature the order of the partials does matter. One way gives the vector and the other the one-form related by the metric, that in this case would not be trivially (1,1,1) as in the Euclidean case.

 

[Matterwave]

How are you generalizing curl to a curved manifold? And are you only considering manifolds of dimension 3 with a connection? I don't know of any "standard" generalizations.

 

[TrickyDicky]

How are you generalizing curl to a curved manifold?

Well, that is precisely what I'm asking help here for.
I can't find any online or in my books, R^3 is always assumed.

And are you only considering manifolds of dimension 3 with a connection? I don't know of any "standard" generalizations.

Unlike divergence, curl doesn't generalize well to dimension other than three,(maybe that is why neither you nor me know any "standard" generalization) so I'm only considering those for the moment.

 

[Matterwave]

Well, one way would be to just define curl as the same thing as it was in the Euclidean case, since the exterior derivative is still defined in a general manifold. Only in the 3-D case though can you identify a 2-form with a 1-form (and vice-versa) through the Hodge duality operator (and if you have a metric, a 1-form can be identified with a vector).

So define $$\nabla \times \mathbf{A}\equiv [*(d\mathbf{A}^\flat)]^\sharp$$

The gradient is then simply: $$\nabla f\equiv (df)^\sharp$$

In this case then, we are asking is the following equality true:
$$0=\nabla \times (\nabla f)=[*(d((df^\sharp)^\flat))]^\sharp=[*(ddf)]^\sharp$$

This is true in general by the definition of the exterior derivative (and the fact that the musical isomorphisms are indeed isomorphisms).

 

[quasar987]

With a riemannian metric g on a 3-manifold M, all kinds of natural isomorphisms pop up. For one thing, the tangent and cotangent bundles TM and T*M get identified via the musical isomorphisms. Also, the Hodge-star * identifies 2-forms to 1-forms and 3-forms to smooth functions. Equivalently, this last identification can also be expressed by saying that every 3-form can be written as fv_g, where v_g is the riemannian volume form. The identification is then f <--> fv_g. To sum up, we have natural identifications

$C^{\infty}(M)$ = $\Omega^0(M)$
$\Gamma (TM)$ <--> $\Omega^1(M)$
$\Gamma (TM)$ <--> $\Omega^2(M)$
$C^{\infty}(M)$ <--> $\Omega^3(M)$

And there is an obvious way to go "differentially" from $\Omega^i(M)$ to $\Omega^{i+1}(M)$, namely the exterior derivative d. So by using the above identifications, you can define grad, curv, and div as the map between functions and vector fields corrersponding to d at the level of differential forms.

In particular, when M=R³ and g = δ_ijdxⁱdx^j is the standard metric, this construction gives the usual grad, curv, div.

Edit: I am saying the same thing as Matterwave but without the nice explicit formulae.

 

[TrickyDicky]

Well, one way would be to just define curl as the same thing as it was in the Euclidean case, since the exterior derivative is still defined in a general manifold. Only in the 3-D case though can you identify a 2-form with a 1-form (and vice-versa) through the Hodge duality operator (and if you have a metric, a 1-form can be identified with a vector).

So define $$\nabla \times \mathbf{A}\equiv [*(d\mathbf{A}^\flat)]^\sharp$$

The gradient is then simply: $$\nabla f\equiv (df)^\sharp$$

In this case then, we are asking is the following equality true:
$$0=\nabla \times (\nabla f)=[*(d((df^\sharp)^\flat))]^\sharp=[*(ddf)]^\sharp$$

This is true in general by the definition of the exterior derivative (and the fact that the musical isomorphisms are indeed isomorphisms).

Yes, I have seen the curl generalized as the hodge dual of the exterior derivative before, however, in a curved manifold the vector field and the one-form are different wrt their covariant derivative unlike the case in R^3 (where the covariant derivative is trivially just the partial derivative without the need for additional connection terms).

 

[TrickyDicky]

I mean that the curl involves the covariant derivative and therefore takes into account not only the covariant one-form but the contravariant vector.
The exterior derivative of the gradient one-form in this case is already zero due to the d^2=0 rule so taking the hodge dual of it here doesn't help us solve the curl in the curved case.

 

[TrickyDicky]

∂ Λ f = ∇f = grad(f)
*∂ Λ *A = ∇ x A = curl(A)
*(A Λ B) = A x B
*(∂ Λ *A) = ∇.A = div(A)​

A minor correction, in terms of dif. forms the divergence is "*d*", and the curl is "*d". With "d" being the exterior derivative.

 

[tiny-tim]

A minor correction, in terms of dif. forms the divergence is "*d*", and the curl is "*d". With "d" being the exterior derivative.

Thanks

The curl equation should be *∂ Λ A = ∇ x A = curl(A) ?

(I don't know how that extra * got in there )

But my div equation is ok, isn't it?

 

[TrickyDicky]

Thanks

The curl equation should be *∂ Λ A = ∇ x A = curl(A) ?

(I don't know how that extra * got in there )

But my div equation is ok, isn't it?

Yes about the curl, and about the div I have always seen it without the parenthesis but I don't think the way is set changes the result at all.

 

[Matterwave]

Yes, I have seen the curl generalized as the hodge dual of the exterior derivative before, however, in a curved manifold the vector field and the one-form are different wrt their covariant derivative unlike the case in R^3 (where the covariant derivative is trivially just the partial derivative without the need for additional connection terms).

This is why I asked you how you were defining curl. The way that I defined it previously is perfectly valid even for a curved 3-manifold since nothing in the definitions depended on curvature. If you want to change the derivatives to co-variant derivatives rather than exterior derivatives, then you have to tell me your definition.

If you want to change it to a covariant exterior derivative, I won't be able to help you much since I myself am not very familiar, at this point, with that operation.

 

[tiny-tim]

Yes about the curl, and about the div I have always seen it without the parenthesis but I don't think the way is set changes the result at all.

Thanks.

I've asked one of the mentors to change the curl, to correct it.

 

[TrickyDicky]

This is why I asked you how you were defining curl. The way that I defined it previously is perfectly valid even for a curved 3-manifold since nothing in the definitions depended on curvature. If you want to change the derivatives to co-variant derivatives rather than exterior derivatives, then you have to tell me your definition.

After a little research online it turns out the reason the curl of the grad is not zero in curved spaces lies at the heart of Riemannian geometry and is a way to get to the Riemann curvature tensor. It goes like thist: the proof that curl (grad f)=0 in R^3 goes like this:
$\nabla$ X$(\nabla f)=0$
Since: $\nabla f=(\frac{∂f}{∂x},\frac{∂f}{∂y},\frac{∂f}{∂z})=F$

$\nabla$X $F=(\frac{∂F_3}{∂y}-\frac{∂F_2}{∂z},\frac{∂F_1}{∂z}-\frac{∂F_3}{∂x},\frac{∂F_2}{∂x}-\frac{∂F_1}{∂y})$
and therefore curl (grad f)=:
$(\frac{∂}{∂y}\frac{∂f}{∂z}-\frac{∂}{∂z}\frac{∂f}{∂y},\frac{∂}{∂z}\frac{∂f}{∂x}-\frac{∂}{∂x}\frac{∂f}{∂z},\frac{∂}{∂x}\frac{∂f}{∂y}-\frac{∂}{∂y}\frac{∂f}{∂x})$
And by the equality of mixed partials everyone knows about from multivariate calculus the result is (0,0,0).
Well according to Riemann's generalization of calculus to manifolds, space is flat if and only if mixed partials are equal, like above. That is the equality of mixed partials implies the Riemann curvature tensor must vanish. And conversely if the Riemann curvature tensor doesn't vanish the mixed partials are not equal and therefore:curl (grad f)≠0
Christoffel formalized this result (due to early death of Riemann) in the form of the well known Christoffel symbols of second class around 1868.

 

[Matterwave]

The Riemann tensor is usually given as the values of the curvature operator:

$$R(X,Y)=\nabla_X\nabla_Y-\nabla_Y\nabla_X-\nabla_{[X,Y]}$$

If you want to call that a curl of a gradient, that's fine with me. I still don't see you giving a definition of a curl in a curved 3-space.

 

[TrickyDicky]

The Riemann tensor is usually given as the values of the curvature operator:

$$R(X,Y)=\nabla_X\nabla_Y-\nabla_Y\nabla_X-\nabla_{[X,Y]}$$

If you want to call that a curl of a gradient, that's fine with me. I still don't see you giving a definition of a curl in a curved 3-space.

No, I woudn't call that curl of a gradient. I call it a Riemann tensor math definition.
I said a result of Riemannian geometry is that when the Riemann tensor vanishes in a space (meaning no curvature or flatness), it implies the equality of mixed partials holds, and that is what makes the curl of the grad be zero in R^3.

 

[TrickyDicky]

Precisely the curvature opearator measures the failure of commutativity of mixed partials.

The definition of curl in curved 3-space just would need to take into account all this thru the corresponding christoffel symbols from the specific metric of the specific curved space that were the case when applying the nabla operator to the specific (grad f), I think.

 

[Matterwave]

The equality of mixed partials always holds (as long as there are no holes in the manifold at least).

I think what you mean is the equality of mixed covariant derivatives. But if you want to formulate a curl using covariant derivatives, then you have to tell me how because I see no obvious way to do it.

 

[homeomorphic]

Yes, but is it a sufficient condition too?

Yes, that's what I was saying about the Poincare Lemma. In Euclidean space, closed forms are always exact, except 0-forms (which is the same thing as saying the De Rham cohomology vanishes, except in dimension 0 it's ℝ). That translates (in ℝ^3) to the fact that if you are divergence free, then you are the curl of something.

Also my question referred to vector fields like the magnetic field that seem to be both divergence free and curl-free, that seems to require for that field to have both a vector potential and a scalar potential, like magnetic fields indeed have. That makes the magnetic field a Laplacian field(meaning the gradient of a Laplace equation solution), right?

The magnetic field is always divergence free (no magnetic monopoles), but it is only curl-free if there is no current or displacement current. It's only locally in the absence of current that the magnetic field has a scalar potential, but it always has a vector potential. Then, it's Laplacian, as you say.

 

[homeomorphic]

What I'm saying is that the proof of the vanishing curl of the gradient of a scalar field rests on the fact that the order that the partials are taken doesn't matter, so that every component of the resulting vector should be zero. But I'd say in the case the space has curvature the order of the partials does matter. One way gives the vector and the other the one-form related by the metric, that in this case would not be trivially (1,1,1) as in the Euclidean case.

Your reasoning is flawed in two ways. The first way is that you insist on not giving a definition of what you mean by curl. The second flaw is that you seem to imply that just because you USED a certain fact in a proof, it is actually necessary to the proof. That itself would require proof and is not true a priori.

If by curl, you mean the ordinary exterior derivative, then yes, the curl of a gradient is zero. It's a matter of definition.

By the way, this definition DOES involve the metric, if you carry it out. You have a vector field. To take the exterior derivative of it, what do you do? You take the dual with respect to the metric, aka, you raise indices, if you like. So, we used the metric. That means we are taking into account the geometry of the manifold. Then, you take the exterior derivative. Then you take the Hodge dual, which also involves the metric because the Hodge dual is the guy that wedges with stuff to give you the volume form that comes from the metric. Then, you take the dual again with respect to the metric.

Until you can give a coherent definition of what you mean by curl, I'm afraid we aren't going to get anywhere.

Edit: Also, we are not talking about the covariant exterior derivative here. On a Riemannian manifold, you still have the ordinary exterior derivative, so you may as well use it.

 

[TrickyDicky]

Your reasoning is flawed in two ways. The first way is that you insist on not giving a definition of what you mean by curl. The second flaw is that you seem to imply that just because you USED a certain fact in a proof, it is actually necessary to the proof. That itself would require proof and is not true a priori.

If by curl, you mean the ordinary exterior derivative, then yes, the curl of a gradient is zero. It's a matter of definition.


Until you can give a coherent definition of what you mean by curl, I'm afraid we aren't going to get anywhere.

Well I tried to make my reasoning clearer in #22, #24 and #25.
The second flaw you mention I don't think that interprets correctly what I was trying to say. But I don't think it deserves further explaining here if you follow what I wrote in #22, #24 and #25.
About the curl definition, no , I don't mean "just" the exterior derivative, I mean the cross product of the $\nabla$ operator (I think that's what a curl is for everybody), in this case acting on a gradient field in a curved 3-space, say a hypersphere for instance.
Is that not possible to do? do you mean the curl is only defined in Euclidean spaces?

 

[TrickyDicky]

The equality of mixed partials always holds (as long as there are no holes in the manifold at least).

Yes in Euclidean spaces, see below

I think what you mean is the equality of mixed covariant derivatives. But if you want to formulate a curl using covariant derivatives, then you have to tell me how because I see no obvious way to do it.

Yes, the covariant derivative is the appropriate generalized notion for the partial derivatives in a curved space. See: http://www.math.ucsd.edu/~ctiee/math20e-w06/grad_n_curl.pdf
in the first page in the note at the bottom.
About using the curl in general manifolds I ask you the same thing I did to homeomorphic, is it an operation not allowed for the curl? Can not the curl be generalized to general 3-manifolds then?

 

[TrickyDicky]

I think it can be generalized like this (using the semicolon notation):
curl of covector A=
$$A_{i;j}-A_{j;i}$$
wich only commutes in the R^3 case and for a gradient field F:

$$F_{i;j}=\partial_jF_i-\Gamma^k_{ij}F_k$$
Please correct if wrong.

 

[TrickyDicky]

From wikipedia page on Vector calculus:
"More generally, vector calculus can be defined on any 3-dimensional oriented Riemannian manifold, or more generally pseudo-Riemannian manifold. This structure simply means that the tangent space at each point has an inner product (more generally, a symmetric nondegenerate form) and an orientation, or more globally that there is a symmetric nondegenerate metric tensor and an orientation, and works because vector calculus is defined in terms of tangent vectors at each point."

So to answer my previous questions the curl is perfectly generalizable to curved spaces.

Also to avoid misunderstandings in what I wrote above, when I referred to mixed partials equality (or lack of) in the context of curved spaces I always meant its generalization for general manifolds (covariant derivative).

 

[quasar987]

From what I gather, you are unsatisfied with the generalization of grad, curl, div provided by Matterwave in post 13 to general riemannian oriented 3-manifolds. I don't understand what is unsatisfactory to you in this, but it seems to have to do with lack of covariant derivatives. So perhaps it would ease your mind to know that the exterior derivative and the covariant derivative are closely related by the formula

$$d\omega(X_0,\ldots,X_p)=\sum_{i=0}^p(-1)^i(\nabla_{X_i}\omega)(X_0,\ldots,\hat{X_i}, \ldots ,X_p)$$

In other words, d is just the (normalized) anysymetrization of the covariant derivative. This actually holds not only for the Levi-Civita connexion of a riemannian structure but any torsion free connexion on TM.

 

[TrickyDicky]

From what I gather, you are unsatisfied with the generalization of grad, curl, div provided by Matterwave in post 13 to general riemannian oriented 3-manifolds. I don't understand what is unsatisfactory to you in this, but it seems to have to do with lack of covariant derivatives. So perhaps it would ease your mind to know that the exterior derivative and the covariant derivative are closely related by the formula

$$d\omega(X_0,\ldots,X_p)=\sum_{i=0}^p(-1)^i(\nabla_{X_i}\omega)(X_0,\ldots,\hat{X_i}, \ldots ,X_p)$$

In other words, d is just the (normalized) anysymetrization of the covariant derivative. This actually holds not only for the Levi-Civita connexion of a riemannian structure but any torsion free connexion on TM.

But I'm not unsatisfied at all, how could I? I agreed with it (just like I agree with what you just posted about exterior derivatives) since I already knew about that generalization. Only problem was that it didn't actually addressed my specific question.
My mind is eased now that I think I found basically what I was looking for.
To summarize this thread I started with some questions about laplacian fields that were answered to my full satisfaction by homeomorphic. Next I brought here a question from another thread that received no answers, about whether there were any circumstances (leaving aside the topological ones, that is I'm considering only oriented, contractible and simply connected smooth 3-manifolds) in which curl of grad was not zero because I had the intuition that it was not the case if the manifold was not flat. To this I received some very informative answers but not satisfactory in the sense of answering directly my doubts.
Finally I found online this very basic result from Riemannian geometry that says that commutativity of the appropriate generalization of mixed partials implies vanishing Riemann tensor, and this I would think that answers my question. My conclusion is that only if the 3-space is Euclidean must we expect the rule "curl of grad=0" to hold. Now do you disagree with this conclusion? If so, where does my logic fail here? Thanks.

 

[homeomorphic]

Well I tried to make my reasoning clearer in #22, #24 and #25.
The second flaw you mention I don't think that interprets correctly what I was trying to say. But I don't think it deserves further explaining here if you follow what I wrote in #22, #24 and #25.

I don't buy it. You are using a formula for the curl that seems to depend on which coordinates you choose. For example, the formula for curl is quite different in cylindrical or spherical coordinates. Also, covariant partials DO commute when applied to scalars if your connection is torsion-free. It's when you apply them to vector fields that curvature measures the failure of commutativity.

About the curl definition, no , I don't mean "just" the exterior derivative, I mean the cross product of the ∇ operator (I think that's what a curl is for everybody), in this case acting on a gradient field in a curved 3-space, say a hypersphere for instance.
Is that not possible to do? do you mean the curl is only defined in Euclidean spaces?

I told you my definition of curl. It's basically to apply the exterior derivative (filling in the gaps by taking the appropriate duals everywhere).

Yours appears not to be coordinate-independent, I'm afraid, unless I am missing something.