No flux boundary condition on PDE on surface (Laplace-Beltrami) What would a Neumann BC on a PDE posed on a surface look like? In the flat case, we have
$\nabla u \cdot N = 0$ 
where $u$ is the solution of the PDE and $N$ is unit normal vector.
In a surface case, what would it be? Would it involve a conormal vector instead? Would I use Laplace-Beltrami operator instead? Would appreciate any explanations.
 A: As Tomás said the normal is well-defined. There are many types of Neumann boundary condition that can be naturally imposed for Hodge Laplacian operator for a harmonic form $u$. Let $\newcommand{\d}{\mathrm{d}}\d$ and $\newcommand{\de}{\delta}\de$ be the exterior derivative and its codifferential, $\wedge$ and $\vee$ be the exterior and interior product. $n$ is the unit normal vector. 
Neumann boundary value could be imposed on the following quantities:
$$
n\wedge \de u, \;\;n\vee \d u,\;\;n\vee \de u,\;\; n\wedge \d u.
$$
Translated into a smooth domain in $\mathbb{R}^3$ using curl and divergence for a $1$-form $u$:
$$
\nabla_{\Gamma} \cdot u,\;\; n\cdot (\nabla \times u),\;\; \nabla_{\perp}\cdot u,\;\; n\times(\nabla \times u),
$$
where the first and the third are surface divergence and divergence of the the normal part of $u$ respectively. Applications would be the boundary value problems for the Maxwell system.
The reference I gave was wrong. The correct reference is Finite element methods for surface PDEs. For example, the integration by parts formula involving co-normal (Theorem 2.10).
