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Exercise [12.16] http://www.roadtoreality.info/viewtopic.php?f=19&t=1744 
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Author:  jbeckmann [ 05 Dec 2010, 09:13 ] 
Post subject:  Exercise [12.16] 
Here is my (WRONG!) proposal: Attachment: THIS PROOF COMPRISES AN ERROR (cf. Roberto, below). Better try this one: Attachment: 
Author:  Roberto [ 06 Dec 2010, 14:27 ] 
Post subject:  Re: Exercise [12.16] 
I think your proof of "a_[rs a_u]v implies that a_rs is simple " is not correct. The proof is based on the assumption that there is a vector x^v satisfying condition x^m x^n a_mn =1 This is never possible, because being a_mn antisymmetric and x^m x^n symmetric their contraction is always 0. This can easily be seen in the simple 2 dimensional case where a_11=a_22=0, a_12 = a_21=a then x^m x^n a_mn= x_1 x_2 a  x_2 x_1 a = 0. The formal proof for the general case is: x^m x^n a_mn = x^n x^m a_nm =  x^n x^m a_mn =  x^m x^n a_mn (where in the first step I just renamed the contracted indices and in second step antisymmetry of a was exploited) therefore x^m x^n a_mn =0. 
Author:  jbeckmann [ 06 Dec 2010, 20:42 ] 
Post subject:  Re: Exercise [12.16] 
You are right, Roberto! Thanks for the hint and sorry for the error. I hope this fixes the bug: Attachment: 
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