The dual of a surface element in 4-space In reading the classic text, "The Classical Theory of Fields", Third Edition, by Landau and Lifschitz, I found an "obvious" statement not so obvious to me.  It is on p.19, the statement of the normality of the dual of a surface element, $df^{*ik}=\frac{1}{2}e^{iklm}df_{lm}$ to the element $df^{ik}$.  Yes, the contraction is zero, as one can see if he lists the 24 terms of the sum and takes account of the alternations of the sign of the completely antisymmetric tensor coefficient and the sign changes of the surface elements.  That is a bit of tedium that I found necessary, because I did not find it obvious.  Maybe that is because I was not clever about the way I listed the terms.   
Question:  Is there some way of listing the terms that would have quickly made clear that for every positive term there would be a negative one?   One thought that suggested itself to me, after I did the work (!) was that if the terms were not all of the same sign, there would have to be an equal number of positive and negative terms because of the symmetry of the form and, thus, normality of the two surface elements.  Is that the obvious quality that I first missed?   
 A: It is OK to use an explicit form in a local orthonormal coordinate system (in the Minkowski sense). The dual for each component would then be just a possible flip of signs (see, e.g., https://en.wikipedia.org/wiki/Hodge_star_operator#Four_dimensions). Raising the indices would amount to doing nothing in that coordinate system. This works even if the space-time curvature is nonzero.
A: So you do have $df_{ij}=dx_i dx'_j-dx_j dx'_i$. If the indexpair $i,j$ is contracted with an indexpair of another antisymmetric object like $\epsilon^{ijkl}$ one can replace $dx_i dx'_j-dx_j dx'_i$ by $2dx_i dx'_j$ which I will now show:
We start with $\epsilon^{ijkl}df_{ij}=\epsilon^{ijkl}dx_i dx'_j-\epsilon^{ijkl}dx_j dx'_i$. Now in the second term one can exchange $i,j$ in the $\epsilon^{ijkl}$ to remove the sign and the result is $\epsilon^{ijkl}dx_i dx'_j+\epsilon^{jikl}dx_jdx'_i$. Because dummy indices can be renamed we can now rename $j$ to $i$ and $i$ to $j$. This also does not change the result. We have: $\epsilon^{ijkl}df_{ij}=\epsilon^{ijkl}dx_i dx'_j+\epsilon^{ijkl}dx_i dx'_j=2\epsilon^{ijkl}dx_i dx'_j$.
Now look at what happens when we also have a $df_{kl}$ (which we can analogously replace by $2dx_k dx'_l$): $\epsilon^{ijkl}df_{ij}df_{kl}=4\epsilon^{ijkl}dx_i dx'_jdx_k dx'_l$. Now e.g. the indexpair $i,k$ is antisymmetric in $\epsilon^{ijkl}$, because $\epsilon^{ijkl}=-\epsilon^{kjil}$, but symmetric in $dx_i dx'_jdx_k dx'_l$, because $dx_i dx'_jdx_k dx'_l=dx_k dx'_jdx_i dx'_l$. A contracted indexpair that occurs once antisymmetric and once symmetric leads to the whole term being zero. I will show that quickly for our case: $\epsilon^{ijkl}dx_i dx'_jdx_k dx'_l=-\epsilon^{kjil}dx_i dx'_jdx_k dx'_l=-\epsilon^{kjil}dx_k dx'_jdx_i dx'_l=-\epsilon^{ijkl}dx_i dx'_jdx_k dx'_l$. Here I exchanged $i$ and $k$ first in $\epsilon^{ijkl}$ and then in $dx_i dx'_jdx_k dx'_l$ and finally renamed $i$ into $k$ and $k$ into $i$. The result is that the whole thing is minus itself and from that we can conclude that it is zero.
