Stokes's theorem in tensor field On pg 73 of "Tensors, Relativity and Cosmology"

The generalized Stokes's theorem in arbitrary $N$-dimensional space is given by:
  $$\int_c A_mdx^m=\frac{1}{2}\int_S F_{mn}dS^{mn} \tag{1}$$

where $F_{mn}$ is the curl tensor of the vector $A_m$, $F_{mn}=A_{n,m}-A_{m,n}$ (, denotes covariant differentiation here) and  $dS^{mn}$ is the contravariant tensor of an infinteseimal element of the surface $S$ $(dS^{mn}=dx^m \wedge dx^n$).
In the three-dimensional metric space the RHS of (1) is equivalent to the ordinary curl A definition 
I tried to expand the RHS of (1) to obtain
$$\frac{1}{2} \left(\frac{\partial A_n}{\partial x^m}-\frac{\partial A_m}{\partial x^n} \right)dx^m \wedge dx^n$$
since the Christoffel symbols vanish in the three-dimensional Euclidean metric space.
It appears that
$$\frac{\partial A_n}{\partial x^m} - \frac{\partial A_m}{\partial x^n}$$
is the definition of curl A but how do I convert $\frac{1}{2}dx^m \wedge dx^n$ into dS to obtain Stokes's theorem in the ordinary vector notation?
 A: Here, you still have the curl in antisymmetric tensor form, not a vector form. Once you are in three dimensions and there are no peculiarities with the metric, you have the correspondence:
$$(\operatorname{curl}{\bf A})_i=\frac12\epsilon_{ijk} \left(\frac{\partial A_{j}}{\partial x^k}-\frac{\partial A_{k}}{\partial x^j}\right) $$
where $\epsilon_{ijk}$ is the Levi-Civita tensor. The inverse of this expression is:
$$\epsilon_{inm}(\operatorname{curl}{\bf A})_i= \left(\frac{\partial A_{n}}{\partial x^m}-\frac{\partial A_{m}}{\partial x^n}\right) $$
When you use this on your expression, you get
$$\frac12\epsilon_{inm} dx^m \wedge dx^n$$
which is just the cross product you require to get to $dS_i$. Note that antisymmetric property of $\wedge$ goes together nicely with antisymmetric Levi-Civita and the one-half in front.
In terms of differential geometry, you convert a 2-form into 1-form with the Hodge star operator, which is a generalized and more formally correct version of what we did above.
