hirak99
Joined: 17 Aug 2010, 16:59 Posts: 2

Fig. 13.8 (b)
This is not really about an exercise, but is regarding Figure 13.8 (b), which proves det(ST)=det(S) det(T) using diagrammatic notation.
I am having difficulty understanding why "the antisymmetrizing bar can be inserted in the middle term because there is already antisymmetry in the index lines that it crosses"  I can't see why there is already antisymmetry in the lines that it crosses. Any guidance will be appreciated.

jbeckmann
Joined: 22 Apr 2010, 15:52 Posts: 43 Location: Olpe, Germany

Re: Fig. 13.8 (b)
Hi hirak99,
in my solution to Exercise [13.19] I try to explain this topic a little bit (see "Remark to (a), 2nd case").
Basically, it is because swapping two lines in the middle of the diagram (which yields a first minus sign) can be undone by swapping the (top or bottom) fixing points of these lines; this, however, corresponds to an exchange of two indices in a LeviCivita Tensor, which yields a second minus sign. Both minus signs compensate. Hence the antisymmetrization operation (which adds diagrams with all possiblities of swapped interior lines) adds always the same object with positive sign, yielding effectively only a factor n!
