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  The dual of a surface element in 4-space

+ 5 like - 0 dislike

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?

This post imported from StackExchange Physics at 2015-11-28 17:02 (UTC), posted by SE-user Albert D Horowitz
asked Nov 15, 2015 in Mathematics by Albert D Horowitz (25 points) [ no revision ]
retagged Nov 28, 2015
You don't need to write the terms in an explicit way at all. Observe that $(\star f)^{ik} f_{ik} = \epsilon^{iklm} f_{lm} f_{ik}$, where the $\epsilon$ is antisymmetric under exchange of the double indices $(ik)$ and $(lm)$, while $f_{lm} f_{ik}$ is symmetric. The contraction of a symmetric and an antisymmetric object is zero.

This post imported from StackExchange Physics at 2015-11-28 17:02 (UTC), posted by SE-user ACuriousMind
Yes, of course! I must have stayed up too late!

This post imported from StackExchange Physics at 2015-11-28 17:02 (UTC), posted by SE-user Albert D Horowitz

Why was this imported? The obvious answer was already given by ACuriousMind in the comments.

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