A question regarding General Stokes' Equation

[raopeng]

Homework Statement

Let X^{i} be a vector field in Minkowski space R^{4}_{1}. We define the integral of this vector over a 3-dimensional hypersurface as the integral of the 3-form X^{i}dS_{i}. where
dS_{i}=\frac{1}{6}\sqrt{|g|}ε_{jkli} dx_{j}\lambda dx_{k}\lambda dx_{l}(don't know how to type exterior derivative)

Prove that:
\int_{∂V} X^{i}dS_{i} = \int_{V} \frac{\partial X^{i}}{\partial x^{i}}dV

From "Modern Geometry, Methods and Applications, Vol I", 26.5 Exercise 2

Homework Equations

Stokes' Equation

The Attempt at a Solution

I thought it was pretty simple so I just simply apply the Stokes' Equation which give me something like this:
\int_{V} \frac{\partial X^{i}}{\partial x^{i}}\frac{1}{6}\sqrt{|g|}ε_{jkli} ε_{ijkl} dx_{1}\lambda dx_{2}\lambda dx_{3}\lambda dx_{4}

while we can calculate the value of ε_{jkli} ε_{ijkl} by substituting indexes and considering its relation to the determinant of δ, the coefficient 1/6 is something I cannot get rid of. And it seems the more general form of this equation can be written as:
\int_{∂V} X^{i}dS_{i} = \int_{V} ∇_{i} X^{i}dV where the coefficient becomes \frac {1}{(n-1)!}. But now I cannot make connections with \frac {1}{(4-1)!}

 

[raopeng]

I think I got something, although not sure it is correct:
I made a mistake there that it should be ε_{imnh} ε_{ijkl}. The indexes should not be identical. Then taking advantage of the identity that ε_{abcd} ε_{efgh} = ε_{abcd} ε_{efgh}δ_{11}δ_{22}δ_{33}δ_{44} = ε_{abcd}δ_{1e}δ_{2f}δ_{3g}δ_{4h} which corresponds to a determinant. And in the case of ε_{imnh} ε_{ijkl} we must perform a contraction of the tensor, thereby leaving a (4-1) order determinant which has (4-1)! components of δ with different indexes. However these indexes are identical and can be switched, so this fill explains the coefficient of \frac{1}{6} and can also be extended to the case of n-form