Curl of Gradient: When is a Conservative Field Not Irrotational?

[TrickyDicky]

Under what circumstances can a conservative field NOT be irrotational?

 

[NasuSama]

Under what circumstances can a conservative field NOT be irrotational?

According to Wikipedia:

"Conservative vector fields are also irrotational, meaning that (in three-dimensions) they have vanishing curl. In fact, an irrotational vector field is necessarily conservative provided that a certain condition on the geometry of the domain holds: it must be simply connected.
An irrotational vector field which is also solenoidal is called a Laplacian vector field because it is the gradient of a solution of Laplace's equation."

We say that the field is irrotational if its curl is zero.

See more: http://en.wikipedia.org/wiki/Conservative_vector_field

Good luck!

 

[Simon Bridge]

eg.
there are no conditions in which a conservative field is irrotational.
Did you have a particular situation in mind?

 

[A Dhingra]

TrickyDicky,
Do you have any situations in your mind for existence of the curl of a gradient or conservative field being rotational? Honestly I am looking forward to hear that..
(and that's why after waiting for a while I wrote this.. )

 

[Zelyucha]

Under what circumstances can a conservative field NOT be irrotational?

When the vector field is on a surface that is not simply connected. How is this so?
Have a look at the curl operator in matrix form:

If you look closely you can see that the curl operator, which I will refer to from now on as ∇× is a skew-symmetric matrix. It turns out that the exponential map of a skew-symmetric matrix is an orthogonal matrix(its transpose is equal to its inverse). This can be shown by defining the matrix S = e^∇× and S^T=e^-∇×. Now we take the matrix product S×S^T = e^∇××e^-∇× = e^{(∇×)-(∇×)} = e^[0] = I₃ = e^-∇×+∇× = e^-∇××e^∇× = S^T×S = I₃ where [0] is the zero matrix(a 3×3 matrix of all zero entries) and I₃ is the square identity matrix of dimension 3.

Now that we have shown that the exponential of ∇× is an orthogonal matrix, there's an even more detailed proof here showing that the exponential of a skew symmetric matrix like the Curl operator is a rotation matrix.

So if G is the gradient of a 3 dimensional scalar field and ∇×G = 0, then e^∇×G = I₃ which is the identity element of SO(3)( the group of rotations about the origin of vectors in Euclidean 3 space R³) which is itself not simply connected. So the map of all simply connected scalar fields in R³ maps to the Kernel{SO(3)} as is not 1-to-1. Whereas scalar fields in
R³ that are not simply connected map into SO(3)~I₃. This map is injective but not surjective. So the curl of the gradient of a scalar field that is not simply connected corresponds to a rotation operator.

 

[micromass]

When the vector field is on a surface that is not simply connected. How is this so?
Have a look at the curl operator in matrix form:

If you look closely you can see that the curl operator, which I will refer to from now on as ∇× is a skew-symmetric matrix. It turns out that the exponential map of a skew-symmetric matrix is an orthogonal matrix(its transpose is equal to its inverse). This can be shown by defining the matrix S = e^∇× and S^T=e^-∇×. Now we take the matrix product S×S^T = e^∇××e^-∇× = e^{(∇×)-(∇×)} = e^[0] = I₃ = e^-∇×+∇× = e^-∇××e^∇× = S^T×S = I₃ where [0] is the zero matrix(a 3×3 matrix of all zero entries) and I₃ is the square identity matrix of dimension 3.

Now that we have shown that the exponential of ∇× is an orthogonal matrix, there's an even more detailed proof here showing that the exponential of a skew symmetric matrix like the Curl operator is a rotation matrix.

So if G is the gradient of a 3 dimensional scalar field and ∇×G = 0, then e^∇×G = I₃ which is the identity element of SO(3)( the group of rotations about the origin of vectors in Euclidean 3 space R³) which is itself not simply connected. So the map of all simply connected scalar fields in R³ maps to the Kernel{SO(3)} as is not 1-to-1. Whereas scalar fields in
R³ that are not simply connected map into SO(3)~I₃. This map is injective but not surjective. So the curl of the gradient of a scalar field that is not simply connected corresponds to a rotation operator.

You're mixing things up. The curl of a gradient is always zero. No matter what the domain of the vector field is. The "converse" is not generally true. That is: if the curl of a vector field is 0, then it is the gradient of something. That requires the domain to be simply connected. This last statement is essentially the Poincaré lemma.

 

[WannabeNewton]

It is much more appropriate to talk about conservative co - vector fields NOT vector fields because conservative has line integral definitions for which you need co - vector fields but if you care only about euclidean 3 - space then it's not really a big deal. A conservative vector field corresponds to an exact one form and an irrotational vector field corresponds to a closed one - form in the context of euclidean 3 - space. Every exact form is closed. On the other hand, a closed form is exact on a contractible domain. All these things are MUCH MUCH more elegant and natural in the more general context of differential forms.

 

[mathwonk]

there might be a continuity assumption. otherwise, always.

 

[Zelyucha]

You're mixing things up. The curl of a gradient is always zero. No matter what the domain of the vector field is. The "converse" is not generally true. That is: if the curl of a vector field is 0, then it is the gradient of something. That requires the domain to be simply connected. This last statement is essentially the Poincaré lemma.

UGH...Ya got me! iStand corrected.

But I'd like to see the proof of this if you will.

 

[WannabeNewton]

UGH...Ya got me! iStand corrected.

But I'd like to see the proof of this if you will.

$\omega = df$ hence $d\omega = d(df) = d^{2}f = 0$. See how pretty it is with differential forms =D. Or you know, if you're a masochist, go ahead and work out the curl of a gradient in coordinates xD.

 

[Zelyucha]

micromass, upon looking at some old notes of mine attempting to disprove that ∇×∇f = 0
for any scalar function f mapping R² into R³ which is continuous and is differentiable up to at least order 2, I found a page which actually proves it to be correct.

One can write δ²f/δyδx = lim(Δy→0) {(f'(x,y+Δy,z) -f'(x,y,z))/Δy} and the quantity in the brackets can be re-written as lim(Δx→0){ (f(x+Δx,y,z) -f(x,y,z))/Δx}. Substituting we get:

δ²f/δyδx = lim(ΔyΔx→0) { (f(x+Δx,y+Δy,z) - f(x+Δx,y,z) + f(x,y,z))/ΔyΔx}.

and

δ²f/δxδy = lim(ΔxΔy→0) { (f(x+Δx,y+Δy,z) - f(x,y+Δy,z) + f(x,y,z))/ΔxΔy}.

Since multiplication of real numbers is commutative, ΔyΔx = ΔxΔy and if we subtract the 2 quantities in brackets we now have:

[δ²f/δyδx -δ²f/δxδy] = lim(ΔxΔy→0){ (f(x,y+Δy,z)-f(x+Δx,y,z))/ΔxΔy}

By the continuity of f: R²→R³, there is a point a = (x,y) in R² whose image is f(x,y,z) and any open δ-neighborhood has an ε-neighborhood on G centered at Img(a) in G. So if we consider the points a_x =(x+Δx,y) and a_y=(x,y+Δy) in R² and the vector X connecting them, then |X| = √(Δx²+Δy²). Now I claim that
lim(ΔxΔy→0){ (f(x,y+Δy,z)-f(x+Δx,y,z))/ΔxΔy} = 0. Which implies that for any ε > 0, there exists a δ > 0 such that:

|(f(x,y+Δy,z)-f(x+Δx,y,z))|/|ΔxΔy| < ε whenever the distance d(a_x,a_y) < δ. So what is the delta? Well since the vectors from a→a_x and a→a_y are orthogonal(to see this just set (x,y)=0), if you were to move one of those points towards the other along the X which connects them such that d(a_x,a_y) < √(Δx²+Δy²) = |X|, then we can choose δ = |X| and place our δ-disk at the midpoint of X between a_x and _y. So shrinking ε means we shrink the magnitude of Δy and Δx to move the 2 a-points even closer together along X to re-size our δ-disk. And as ΔxΔy→0(along with |X|), then the |Img(X)|→0 and the proof is complete. [δx_iδx_j]°δf = 0 whenever i≠j.

 

[micromass]

micromass, upon looking at some old notes of mine attempting to disprove that ∇×∇f = 0
for any scalar function f mapping R₂ into R³ which is continuous and is differentiable up to at least order 2, I found a page which actually proves it to be correct.

One can write δ²f/δyδx = lim(Δy→0) {(f'(x,y+Δy,z) -f'(x,y,z))/Δy} and the quantity in the brackets can be re-written as lim(Δx→0){ (f(x+Δx,y,z) -f(x,y,z))/Δx}. Substituting we get:

δ²f/δyδx = lim(ΔyΔx→0) { (f(x+Δx,y+Δy,z) - f(x+Δx,y,z) + f(x,y,z))/ΔyΔx}.

and

δ²f/δyδx = lim(ΔxΔy→0) { (f(x+Δx,y+Δy,z) - f(x,y+Δy,z) + f(x,y,z))/ΔxΔy}.

Since multiplication of real numbers is commutative, ΔyΔx = ΔxΔy and if we subtract the 2 quantities in brackets we now have:

[δ²f/δyδx -δ²f/δyδx] = lim(ΔxΔy→0){ (f(x,y+Δy,z)-f(x+Δx,y,z))/ΔxΔy}

By the continuity of f: R²→R³, there is a point a = (x,y) in R² whose image is f(x,y,z) and any open δ-neighborhood has an ε-neighborhood on G centered at Img(a) in G. So if we consider the points a_x =(x+Δx,y) and a_y=(x,y+Δy) in R² and the vector X connecting them, then |X| = √(Δx²+Δy²). Now I claim that
lim(ΔxΔy→0){ (f(x,y+Δy,z)-f(x+Δx,y,z))/ΔxΔy} = 0. Which implies that for any ε > 0, there exists a δ > 0 such that:

|(f(x,y+Δy,z)-f(x+Δx,y,z))|/|ΔxΔy| < ε whenever the distance d(a_x,a_y) < δ. So what is the delta? Well since the vectors from a→a_x and a→a_y are orthogonal(to see this just set (x,y)=0), if you were to move one of those points towards the other along the X which connects them such that d(a_x,a_y) < √(Δx²+Δy²) = |X|, then we can choose δ = |X| and place our δ-disk at the midpoint of X between a_x and _y. So shrinking ε means we move the 2 a points even closer together along X to re-size our δ-disk. And as ΔxΔy→0(along with |X|), then the |Img(X)|→0 and the proof is complete. [δx_iδx_j]°δf = 0 whenever i≠j.

Your proof looks more like a proof for the theorem of Clairaut-Schwarz... http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives

 

[Zelyucha]

Your proof looks more like a proof for the theorem of Clairaut-Schwarz... http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives

Well I'm not surprised that someone didn't construct a theorem about mixed partial derivatives(and prove it too). My proof was a specific case of n=3. And I can see that the Clairaut-Schwartz theorem is an induction of this concept to n dimensions. The commutativity of mixed partials for a continuous function f of differential class C² that defines a (2-D) surface in Euclidean 3 space is a consequence of continuity as well as the Cauchy-Schwartz inequality used for defining ε-δ neighborhoods.

 

[Zelyucha]

Might I add, the fact that the Curl of the Gradient of a scalar function is always zero is a direct consequence of Clairaut's theorem since the theorem states that mixed partials commute.

But what about a continuous well defined function of differential class C²⁺ on real projective space? Or more specifically a scalar function f in R³ that defines a non-orientable surface?

 

[micromass]

Might I add, the fact that the Curl of the Gradient of a scalar function is always zero is a direct consequence of Clairaut's theorem since the theorem states that mixed partials commute.

But what about a continuous well defined function of differential class C²⁺ on real projective space? Or more specifically a scalar function f in R³ that defines a non-orientable surface?

Of course you can calculate partial derivatives on weird manifolds. Then the partial derivatives don't necessarily commute since you can have a weird chart. You can define a Lie-bracket to see how much the partial derivatives commute.

But curl and gradient are defined completely different on manifolds. They are not longer vector fields but differential forms. The curl and gradient correspond to the exterior derivative and you can show that applying the exterior derivative twice gives 0.

 

[Zelyucha]

Of course you can calculate partial derivatives on weird manifolds. Then the partial derivatives don't necessarily commute since you can have a weird chart. You can define a Lie-bracket to see how much the partial derivatives commute.

But curl and gradient are defined completely different on manifolds. They are not longer vector fields but differential forms. The curl and gradient correspond to the exterior derivative and you can show that applying the exterior derivative twice gives 0.

A conservative vector field for this purpose is simply the gradient of a function that defines a 2-dimensional manifold embedded in R³. So when you speak of manifolds you are speaking of a completely separate topological space whereas what I'm talking about is restricted to (2D)surfaces in Euclidean 3D space. The CURL operator is a differential 2-form on such an object.

Now consider the scalar function ψ: (x,y,z) → (x/√(x²+y²+z²),y/√(x²+y²+z²),z/√(x²+y²+z²)) = (ψ(x),ψ(z),ψ(z))

This is a mapping of R³ into Real projective space(of dimension 2). If you take ∇ψ = (δψ(x)/δx, δψ(y)/δy,δψ(z)/δz) and then apply the CURL operator, you will find that ∇×∇ψ ≠ 0 ! So any scalar field f(x,y,z) that is defined by a continuous mapping of the Real Projective Plane will have asymmetric mixed partial derivatives and a non-vanishing CURL. And while I haven't proved it just yet; I'm going to conjecture that:

Let f is a continuous scalar function of 3 real variables and differential class C²⁺ that defines a surface in R³. If f is orientable then ∇f is a conservative vector field.

 

[micromass]

I am very sorry to say that I could make very little sense of your post.

A conservative vector field for this purpose is simply the gradient of a function that defines a 2-dimensional manifold embedded in R³.

I'm not following. If you define your manifold as a level set of $F:\mathbb{R}^3\rightarrow \mathbb{R}$, then the gradient is a normal field on F. I don't get why you call this a conservative vector field at all.

So when you speak of manifolds you are speaking of a completely separate topological space whereas what I'm talking about is restricted to (2D)surfaces in Euclidean 3D space. The CURL operator is a differential 2-form on such an object.

The curl operator is not a 2-form.

Now consider the scalar function ψ: (x,y,z) → (x/√(x²+y²+z²),y/√(x²+y²+z²),z/√(x²+y²+z²)) = (ψ(x),ψ(z),ψ(z))

You say that

$$\psi(x)=\frac{x}{\sqrt{x^2 + y^2 + z^2}}$$

But the right hand side depends clearly on x, y and z. And the left-hand side depends on x. So the notation is not very legal.

This is a mapping of R³ into Real projective space(of dimension 2). If you take ∇ψ = (δψ(x)/δx, δψ(y)/δy,δψ(z)/δz)

In my knowledge, a gradient is only defined for a scalar field, that is for a sufficiently smooth function $f:\mathbb{R}^3\rightarrow \mathbb{R}$. You seem to be taking a gradient for something whose codomain is not $\mathbb{R}$. I don't know how that is defined.

 

[lavinia]

In the planar case one can think of the curl of a vector field as the divergence of its rotation by -90 degrees.

∇xV = ∇.A(V) where a is the -90 degree rotation of V(x) around x.

For if V = (u,w) then A(v) = (w,-u) so ∇.A(V) = w$_{x}$ - u$_{y}$

The divergence gives the infinitesimal flux density of the vector field away or towards the point. It is a simple limiting argument to show this.

So one can think of the curl of the vector field as the infinitestimal rotation density either clockwise or counter clockwise around the point.

 

[Zelyucha]

I am very sorry to say that I could make very little sense of your post.



I'm not following. If you define your manifold as a level set of $F:\mathbb{R}^3\rightarrow \mathbb{R}$, then the gradient is a normal field on F. I don't get why you call this a conservative vector field at all.



The curl operator is not a 2-form.



You say that

$$\psi(x)=\frac{x}{\sqrt{x^2 + y^2 + z^2}}$$

But the right hand side depends clearly on x, y and z. And the left-hand side depends on x. So the notation is not very legal.



In my knowledge, a gradient is only defined for a scalar field, that is for a sufficiently smooth function $f:\mathbb{R}^3\rightarrow \mathbb{R}$. You seem to be taking a gradient for something whose codomain is not $\mathbb{R}$. I don't know how that is defined.

I do realize that my notation for ψ equations was not legal. However, regarding the final paragraph, the Möbius strip is a surface whose codomain is R³ but is homeomorphic to the real projective plane. However, it is not an embedding.

Let's try this again:


$$\psi(x,y,z)=(\frac{x}{\sqrt{x^2 + y^2 + z^2}},\frac{y}{\sqrt{x^2 + y^2 + z^2}},\frac{z}{\sqrt{x^2 + y^2 + z^2}})$$

What am suggesting is that rule that ∇×∇f = 0 for any scalar function with domain R³ does not hold when f defines a non-orientable surface.

 

[micromass]

I do realize that my notation for ψ equations was not legal. However, regarding the final paragraph, the Möbius strip is a surface whose codomain is R³ but is homeomorphic to the real projective plane. However, it is not an embedding.

What do you mean with the "codomain of a surface"??

Furthermore, do you have any reference that says that the Mobius strip is homeomorphic to the projective plane?? I find this very difficult to believe.

Let's try this again:


$$\psi(x,y,z)=(\frac{x}{\sqrt{x^2 + y^2 + z^2}},\frac{y}{\sqrt{x^2 + y^2 + z^2}},\frac{z}{\sqrt{x^2 + y^2 + z^2}})$$

What am suggesting is that rule that ∇×∇f = 0 for any scalar function with domain R³ does not hold when f defines a non-orientable surface.

It's simply not true. You can actually prove that the curl of the gradient is zero for surfaces (provided we express everything as forms). And as of now, you have still not given a valid counterexample to that claim.

 

[lavinia]

In a conservative force field, the work done on a particle moving between two points is independent of the path. In an open neighborhood such as a ball in Euclidean space, one can show that this independence of path implies that there is a potential function (scalar field) whose gradient is the field. Conversely, the ngradient of a function must be conservative. This follows directly from the Fundamental Theorem of calculus.

The total work done against a conservative field on a closed path is zero. Therefore the field can not circulate around a closed path . Otherwise put, its circulation around any closed path is zero. Therefore it's circulation density, its curl, must be zero

 

[Zelyucha]

What do you mean with the "codomain of a surface"??

Furthermore, do you have any reference that says that the Mobius strip is homeomorphic to the projective plane?? I find this very difficult to believe.



It's simply not true. You can actually prove that the curl of the gradient is zero for surfaces (provided we express everything as forms). And as of now, you have still not given a valid counterexample to that claim.

I actually worked out the ∇×∇ψ and it does not vanish. For the homeomorphism between the Mobius strip and the (Real)projective plane, look here.

 

[WannabeNewton]

I actually worked out the ∇×∇ψ and it does not vanish. For the homeomorphism between the Mobius strip and the (Real)projective plane, look here.

Your example literally makes no sense because your map doesn't even map to the reals so what does the gradient even mean in that case? Secondly, the real projective plane is compact but the Mobius strip isn't so your claim regarding them being homeomorphic is already false. Your link also never claims they are homeomorphic. I have no idea what is being attempted to be shown here.

 

[lavinia]

The Mobius strip is not homeomorphic to the real projective plane.You can construct the projective plane from the Mobius band by attaching a disk to its bounding circle.

The projective plane has no boundary. The Mobius band's boundary is a circle.

The Mobius band can be embedded in R^3. The Projective plane can not. In fact no non-orientable closed surface without boundary can be embedded in 3 space.

The fundamental group of the Mobius band is the integers. The fundamental group of the projective plane is Z/2Z.

The Mobius band can be given a flat metric. The Projective plane can not.

 

[Zelyucha]

Your example literally makes no sense because your map doesn't even map to the reals so what does the gradient even mean in that case? Secondly, the real projective plane is compact but the Mobius strip isn't so your claim regarding them being homeomorphic is already false. Your link also never claims they are homeomorphic. I have no idea what is being attempted to be shown here.

$$\psi(x,y,z)=\frac{x+y+z}{\sqrt{x^2 + y^2 + z^2}}$$



This is a scalar function that describes a non-orientable manifold of dimension 2. The non-orientable comes from that fact that:

$$\psi(-x,-y,-z)=\frac{-x-y-z}{\sqrt{(-1)^2( x^2 + y^2 + z^2)}}$$ = $$\frac{-(x+y+z)}{-\sqrt{x^2 + y^2 + z^2}}$$ = $$\frac{x+y+z}{\sqrt{x^2 + y^2 + z^2}}$$

And furthermore, for any real valued scalar λ in R¹~{0} it is easy to check that ψ(λx,λy,λz) = ψ(x,y,z). Now I just finished working out the ∇ψ( which consists of fractions with some big ol' cubic polynomials in the numerators). And as it turns out, if you compute the ∇×∇ψ you can see that it does not vanish. That is, ∇×∇ψ≠ {0}(the zero vector in R³). This function does not actually violate Clairaut's theorem(symmetry of partial derivatives) since it's partials are not continuous at the origin(0,0,0).

Part of the reason that this interests me so much is due to Gauss's law for magnetism( ∇°B=0) which follows from the fact that ∇×∇F = 0 by virtue of the symmetry of mixed partials for a scalar function F whose 1st and 2nd derivatives are continuous. In the case of a magnetic field which violates Gauss's law( ∇°B ≠ 0), such as a magnetic monopole, then we have a scalar potential whose gradient has non-zero curl.


Lavinia: You are correct about the topology of Möbius strip.

 

[WannabeNewton]

You made a calculation error somewhere along the road. Just plug it into wolfram. I got 0.

 

[Zelyucha]

You made a calculation error somewhere along the road. Just plug it into wolfram. I got 0.

I think that Wolfram Alpha may have biffed here.

I plugged the Grad ψ and got the following:

http://www4a.wolframalpha.com/Calculate/MSP/MSP61471a59c8e7504i562a00001b654095d6d3gg73?MSPStoreType=image/gif&s=59&w=504&h=65


So bear with me and let's compute the first component(e_x) of ∇×∇ψ...


Now since all components of ∇ψ have teh same denominator we can use the quotient rule which is $$(\frac{f}{g})'=\frac{f'g-g'f}{g^2}$$




Let$$r={\sqrt{x^2+y^2+z^2}}$$ and let c = r³.

Now let

a = x²+z²-xy-zy (∇ψ_y numerator)

b = x²+y²-zy-xz (∇ψ_z numerator)


So the 1st component of ∇x∇ψ = dy^dz(ψ) =

$$\frac{b'_{y}c -bc'_{y}-a'_{z}c+ac'_{z}}{c^2}$$

Now for a moment let's first focus on terms in the numerator where a and b are differentiated. It is easy to check that $$\frac{δb}{δy}=2y-z$$ and $$\frac{δa}{δz}=2z-y$$

Therefore, (b'_yc-a'_zc)c^-2 = ((2y-z-2z+y)r³)r^-6 = $$\frac{(3y-3z)r^3}{r^6} = \frac{3(y-z)}{r^3}$$
Now this expression is only equal to zero when y=z; but everywhere else it is non-zero.

Now for the other 2 terms. Since r is a square root, we use the chain rule to get

ac'_z -bc'_y = (3r/2)*2z(x²+z²-xy-zy)-(3/2)*2y(x²+y²-zy-xz) =

3r*(z³+zx²-z²y-yx²-y³+zy²+(xyz-xyz)) =

3r*(z³+zx²-z²y-yx²-y³+zy²)

Now when we factor in the denominator we get $$\frac{3(z^3+zx^2-z^2y-yx^2+zy^2-y^3)}{r^5}$$


We can now write the complete 1st component of ∇×∇ψ:

$$(\frac{3(y-z)}{\sqrt{x^2+y^2+z^2}^3}-\frac{3(z^3+zx^2-z^2y-yx^2+zy^2-y^3)}{\sqrt{x^2+y^2+z^2}^5}){\bf e_{x}}$$


So now you can see for yourself that ∇×∇ψ≠(0,0,0).

 

[Zelyucha]

Your example literally makes no sense because your map doesn't even map to the reals so what does the gradient even mean in that case? Secondly, the real projective plane is compact but the Mobius strip isn't so your claim regarding them being homeomorphic is already false. Your link also never claims they are homeomorphic. I have no idea what is being attempted to be shown here.

FYI: The Möbius strip is compact. However, it is not homeomorphic to the projective plane because the Möbius strip has 2 closed boundary curves whereas RP² does not.

The curl operator is not a 2-form.

It actually is. Look here at the wikipedia article regarding differential forms and if you scroll down to section 4.1 there is the definition of the wedge product for a differential 2-form which is precisely what the CURL operator is for surfaces in R³. So let F be a vector field in R³. ∇×F = [(dx^dy)+(dx^dz)+(dy^dz)]°F. You can write out the CURL in matrix form and see that it is a skew-symmetric 3×3 matrix with Trace=0.

 

[micromass]

FYI: The Möbius strip is compact.

Depends entirely on how you define the Mobius strip. It makes perfect sense to define it as not containing the boundary. In fact, since you were talking about manifolds, this is the only way to define it. If you add the boundary, then you're not longer a manifold (but rather a manifold with boundary).

However, it is not homeomorphic to the projective plane because the Möbius strip has 2 closed boundary curves whereas RP² does not.

Sigh... The Mobius strip only has one boundary curve.

It actually is. Look here at the wikipedia article regarding differential forms and if you scroll down to section 4.1 there is the definition of the wedge product for a differential 2-form which is precisely what the CURL operator is for surfaces in R³. So let F be a vector field in R³. ∇×F = [(dx^dy)+(dx^dz)+(dy^dz)]°F. You can write out the CURL in matrix form and see that it is a skew-symmetric 3×3 matrix with Trace=0.

The curl is not a 2-form. It is something that acts on one-forms and that yields a 2-form (and actually, the curl is the dual of that, but that's not important here). The curl is not a two-form by itself.

I think you should get a good book on differential geometry and work through that. I highly recommend Lee's "Introduction to Smooth Manifolds". I hope it will clear up many misconceptions.

I'm not seeing that anybody is using this thread to learn. There are only arguments. Thus I'm locking this.