$\nabla \times X$ is given on a surface, can I show that X = 0 on the same surface If I have a volume V enclosed by a surface S, and $\nabla \times X$ is given on the surface, what information does that give me about X on S. Is there a method of showing that X = 0 on S? (in the context of the divergence theorem?)
 A: Note that if $C$ is any constant vector, $\nabla \times (X+C) = \nabla \times X$.  If the surface is compact and $X$ is continuous, it is bounded on the surface, so if we take $|C|$ large enough
there is no point on the surface where $X+C=0$.
 So there is certainly nothing that $\nabla \times X$ can tell you that would imply that $X=0$ somewhere on $S$.
A: Let's say $\nabla \times X = 0$ on $S$, this is to say:
$$
(\nabla \times X)\cdot n = 0,\quad\text{and}\quad (\nabla \times X)\times n = 0.
$$ 
Take any smooth function $\phi$ that has well-defined values on $S$, then by Divergence theorem:
$$
\int_{V} (\nabla \times X)\cdot \nabla \phi = -\underbrace{\int_{V} \nabla \cdot (\nabla \times X)\,\phi}_{\text{This term vanishes}} + \int_S \phi\,(\nabla \times X)\cdot n\,dS. 
$$
Similarly:
$$
\int_{V} (\nabla \times X)\cdot \nabla \phi = \underbrace{\int_{V} X \cdot \nabla \times(\nabla \phi)}_{\text{ This term vanishes}} -\int_S (X\times n)\cdot \nabla \phi\,dS.
$$
Therefore:

($\dagger$) $X\times n = 0\implies (\nabla \times X)\cdot n = 0$ in the distribution sense.

Even the reverse direction is not correct, a counterexample is that for a cylinder around $z$-axis, consider $X = \nabla \arctan(y/x)$.
For the other boundary condition. Consider the following Neumann eigenvalue problem for double curl (consider the case only when $\lambda >0$):
$$
\nabla \times \nabla \times X - \lambda X = 0 \quad \text{ in } V
\\
(\nabla \times X )\times n = 0\quad \text{ on } S
$$
Integration by parts formula gives:
$$
\int_V \nabla \times \nabla \times X \cdot Y = \underbrace{\int_V  \nabla \times X \cdot\nabla \times Y}_{\text{This term vanishes if } Y=\nabla \phi } - \int_S (\nabla \times X )\times n\cdot Y\,dS.
$$
Hence take $Y = \nabla \phi$, then
$$
- \int_S (\nabla \times X )\times n\cdot \nabla \phi\,dS = \int_V \nabla \times \nabla \times X \cdot \nabla \phi = \int_V\lambda X \cdot \nabla \phi,
$$
Again divergence theorem reads:
$$
\int_V \lambda X \cdot \nabla \phi = \underbrace{- \int_V \nabla \cdot (\lambda X) \,\phi}_{\text{This term vanishes because } \lambda X = \nabla \times \nabla \times X} + \int_S \lambda X \cdot n\, \phi\,dS.
$$
Hence 
$$
\lambda\int_S X \cdot n\, \phi\,dS = - \int_S (\nabla \times X )\times n\cdot \nabla \phi\,dS.
$$
Therefore:

($\ddagger$) $(\nabla \times X)\times n = 0 \implies X\cdot n = 0$ in the distribution sense when $X$ is a eigenvector of the $\nabla \times \nabla \times $ operator.


In general, $\nabla \times X$'s value on boundary does not give us any information of $X$. 
