# Is it generally true that $\nabla\times\vec{n}=0$ for any surface or is this only true for a simply connected domain?

Is it generally true that $\nabla\times\vec{n}=0$ for any surface or is this only true for a simply connected domain? (see ftp://ftp.math.ucla.edu/pub/camreport/cam12-18.pdf) and discussion here (Curl of unit normal vector on a surface is zero?)

I think Stoke’s theorem implies that $\vec{n}\cdot\nabla\times\vec{n}=0$ but this isn’t quite $\nabla\times\vec{n}=0$.

In particular, it doesn't seem like the unit vector for u in toroidal coordinates $(u,v,\phi)$ satisfies this (http://mathworld.wolfram.com/ToroidalCoordinates.html) Yet constant u corresponds to toroidal surfaces.

So what I'm pondering is how to translate $\nabla\times\vec{n}=0$ into practice. For example, the curl in general orthogonal curvilinear coordinates is \begin{align} \nabla\times\vec{f}=&\frac{1}{h_2\,h_3}\,\left[\frac{\partial}{\partial x_2}\left(h_3\,f_3\right)-\frac{\partial}{\partial x_3}\left(h_2\,f_2\right)\right]\,\vec{e}_1+\frac{1}{h_3\,h_1}\,\left[\frac{\partial}{\partial x_3}\left(h_1\,f_1\right)-\frac{\partial}{\partial x_1}\left(h_3\,f_3\right)\right]\,\vec{e}_2\nonumber\\ &\qquad+\frac{1}{h_1\,h_2}\,\left[\frac{\partial}{\partial x_1}\left(h_2\,f_2\right)-\frac{\partial}{\partial x_2}\left(h_1\,f_1\right)\right]\,\vec{e}_3 \end{align}
where $h_i$ are the scale factors and $\vec{e}_i$ are the unit vectors along coordinate lines $x_i$. For spherical coordinates $(x_1,x_2,x_3)=(r,\theta,\phi)$ the scale factors are $h_1=1$, $h_2=r$, and $h_3=r\,\sin\theta$. If I take surfaces normal to the sphere $\vec{n}=\vec{e}_1$ with $f_1=1$, clearly $\frac{\partial}{\partial x_3}\left(h_1\,f_1\right)=\frac{\partial}{\partial x_2}\left(h_1\,f_1\right)=0$ and $\nabla\times\vec{n}=0$.

If I take toroidal coordinates $(x_1,x_2,x_3)=(u,v,\phi)$ with scale factors $h_1=h_2=a/\left(\cosh{u}-\cos{v}\right)$ and $h_3=a\,\sinh{u}/\left(\cosh{u}-\cos{v}\right)$ where $a$ is a parameter and $u=\mathrm{constant}$ are toroidal surfaces. Then naively I would take the unit normal to the toroidal surfaces to be $\vec{n}=\vec{e}_1$ again corresponding to $f_1=1$. However this leads to $$\nabla\times\vec{f}=\frac{1}{h_3\,h_1}\,\left(\frac{\partial{h}_1}{\partial x_3}\right)\,\vec{e}_2-\frac{1}{h_1\,h_2}\,\left(\frac{\partial{h}_1}{\partial x_2}\right)\,\vec{e}_3=\frac{\sin{v}}{a}\,\vec{e}_3$$
So it seems that the naive assumption $\vec{n}=\vec{e}_1$ isn't correct for this coordinate system? So how do I find the proper normal that satisfies $\nabla\times{n}=0$ and $\vec{n}\cdot\vec{n}=1$ in an general orthogonal curvilinear coordinate system? Obviously, from definition of the $\nabla\times$ above $f_1=1/h_1$ satisfies the former condition but not the latter condition. Thanks

• Please use MathJax to format math on this site. I can't read your equations at all as everything is just squares. – Arthur Aug 20 '18 at 13:09
• Thanks! I didn’t realize this site used latex formatting – peter schuck Aug 20 '18 at 14:47
• The condition $\nabla \times \vec{n} = 0$ is a local condition, while simply connectedness is a global condition. Thus, I would be shocked if this result only holds in the simply connected case. (But, I have been shocked many times in mathematics!) – Jason DeVito Aug 20 '18 at 15:36

This is entirely a local result. Only if you want to go from $\text{curl }\vec F = \vec 0$ to $\vec F = \nabla f$ do you need some global topological restrictions.

The only satisfactory proof I see, after a bit of thought, is to use a basic fact from differential geometry. Curl of a vector field is the antisymmetric part of its derivative. (That is, write $D\vec F = \frac12\big(D\vec F + (D\vec F)^\top\big) + \frac12\big(D\vec F - (D\vec F)^\top\big)$, and curl is identified with the latter term.) Now, the shape operator, which gives the derivative of the map $\vec n$ from the surface to the unit sphere, is symmetric. (See, for example, pp. 45-46 of my differential geometry text.) Thus, the matrix representation of $D\vec n$ with respect to an orthonormal basis for the tangent space of the surface will give a symmetric matrix, and the curl term is $0$.

A different argument would come from extending the map $\vec n$ to a neighborhood of the surface (say by having it stay constant along normal lines) and then differentiating the resulting vector field on an open set in $\Bbb R^3$ with the usual rules for curl. It would still be beneficial to work with a basis adapted to the geometry, however.

EDIT: There's nothing wrong with your computations. But let's look at something simpler. What if you look at the level surface $f_2=\pi/4$—a cone—in spherical coordinates? Then you get a nonzero curl computation, as well. You just happened to luck out when you looked at the spheres.

So what's the explanation? The result you quoted (and which I proved) applies to the computation of an "intrinsic" curl living only in the surface, not in 3-D. Let me explain this using the language of differential forms. You have a 3-D coordinate system with $\vec e_1,\vec e_2,\vec e_3$ an orthonormal frame along the appropriate coordinate curves. Let $\omega_1,\omega_2,\omega_3$ be the dual coframe — so these are $1$-forms satisfying $\omega_i(\vec e_j) = \delta_{ij}$. The level surfaces with unit normal $\vec e_1$ are given by the differential equation $\omega_1 = 0$. Now, you're wanting curl $0$ to mean $d\omega_1 = 0$. This needn't be true (as in our examples). What is true is that $d\omega_1 = \omega_1\wedge\eta$ for some $1$-form $\eta$; this will tell us that when we evaluate $d\omega_1$ on any surface $\omega_1=0$ we do get $0$, of course, but it needn't be $0$ as a $2$-form on all of 3-space.

This is what's going on in my spherical coordinates example (and, in more complicated fashion, in your example). Explicitly, we have $\omega_1= dr$, $\omega_2 = r\,d\theta$, $\omega_3 = r\sin\theta\,d\phi$. You can compute, using your formulas, that $\text{curl }\vec e_2$ has a non-zero $\vec e_3$ component. I see that with differential forms by taking $d\omega_2 = dr\wedge d\theta$; applying the Hodge star to get back a $1$-form, I get $\star (dr\wedge d\theta) = \star (\frac 1r\omega_1\wedge\omega_2) = \frac 1r\omega_3$, corresponding to the vector field $\frac1r\vec e_3$. Indeed, as a 2-form on the cone, $d\omega_2$ does vanish identically; but as a 2-form on $\Bbb R^3$ it most definitely does not. (The moral of the story is that to understand deeply what's going on here one has to use differential forms; staying in the land of vectors won't cut it.)

• Ted, Thank you so much for your quick reply. I apologize for my delay in getting back to this, but I was on work travel. I suspected that $\nabla\times\vec{n}=0$ was more general than $\vec{n}=\nabla\psi$ which, as you note, implies some global topology. Most of the proofs of $\nabla\times\vec{n}=0$ that I've seen default to the latter more restricted demonstration. A quick question on notation, is $D\vec{F}$ consistent with Jacobian $\nabla\vec{F}$? I've also grabbed your differential geometry book! I've added some more specific info in my question above. Thanks, -- Pete – peter schuck Aug 26 '18 at 17:04
• $\nabla$ is not used for the derivative because it already is used for the gradient and covariant derivative (and then $\nabla^2$ for the Laplace operator). Too much already!! ... I've edited significantly. You are misinterpreting the result and trying to apply it to conclude things that are just plain false. – Ted Shifrin Aug 26 '18 at 22:49