Last edited by jbeckmann on 06 Dec 2010, 20:44, edited 1 time in total.

05 Dec 2010, 09:13

Roberto

Joined: 03 Jun 2010, 15:18 Posts: 136

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.