Curl of unit normal vector on a surface is zero? I have a scalar field $\phi$. From this field, I define an iso-surface $\phi=\phi_{iso}$.
The unit normal vector on this surface is
$\vec{n}=\left(\frac{\nabla\phi}{|\nabla\phi|}\right)_{\phi_{iso}}$
I have a book, where the author says: For any surface unit normal vector the following is true:
$\nabla\times\vec{n}=0$
I found some other sources saying the same:
rot(n)=0 ("since rot(n)=0")
curl(n)=0 on page 4 between eq. 20 and 21
Is this true for any surface? Because if I try to prove this myself, I get stuck really quick:
$\nabla\times\vec{n}=\nabla\times\left(\frac{\nabla\phi}{|\nabla\phi|}\right)=
\nabla\left(\frac{1}{|\nabla\phi|}\right)\times\nabla\phi+\frac{1}{|\nabla\phi|}\underbrace{\nabla\times\nabla\phi}_{=0}=\frac{1}{|\nabla\phi|^2}\nabla(|\nabla\phi|)\times\nabla\phi$
I can't see why this is supposed to be zero, since $|\nabla\phi|\ne\text{const}$ generally. Does anyone have an idea?
 A: It really depends on how you define the vector field $\vec{n}$ AWAY from the surface $\phi = \phi_{iso}$. On the surface $\vec{n}$ is well-defined (up to choice of orientation). 


*

*Choice one: define $\vec{n}$, as you did, to be globally the normalized gradient of $\phi$. That is, set $\vec{n} = \frac{\nabla \phi}{|\nabla\phi|}$. In this case $\nabla\times \vec{n} = 0$, when evaluated at the surface $\{\phi = \phi_{iso}\}$, if and only if $|\nabla \phi|$ is constant along the surface. 

*Choice two: forget more or less about the function $\phi$. Define the function
$$ \psi = \frac{1}{|\nabla \phi|} (\phi - \phi_{iso}) $$
Observe that the surface you are interested in is the surface $\{ \psi = 0\}$. Computing the gradient $\nabla \psi$ you have that 
$$ \nabla \psi = \frac{\nabla\phi}{|\nabla \phi|} - \frac{(\phi-\phi_{iso}) \nabla \phi \cdot \nabla^2\phi}{|\nabla\phi|^3} $$
The key is that the second term vanishes on the surface, since there $\phi = \phi_{iso}$. So $\nabla\psi$ restricted to the surface is still the unit normal vector field. But $\nabla \times (\nabla \psi)$ is clearly zero. (Note, however, $\nabla\psi$ is not guaranteed to be a unit vector field away from the surface.)

More generally: given a compact smooth surface $\Sigma\subset \mathbb{R}^3$, there exists a radius $r > 0$ such that on the set $S = \{ x\in \mathbb{R}^3: \mathrm{dist}(x,\Sigma) < r\}$ we can solve the eikonal equation $|\nabla \Psi| = 1$ to get a function $\Psi:S \to\mathbb{R}$ such that $\Sigma = \Psi^{-1}(0)$ and $\nabla \Psi$ is the unit normal vector field for any level set $\Psi^{-1}(c)$. Then in this formulation we see that the unit normal vector field $\vec{n} = \nabla \Psi$ is curl-free everywhere in $S$. The number $r$, which is generically finite, is related to the radius of curvature of $\Sigma$. 
