 Page 1 of 1 [ 3 posts ]
 Print view Previous topic | Next topic
Exercise [12.16]
Author Message

Joined: 22 Apr 2010, 15:52
Posts: 43
Location: Olpe, Germany
Here is my (WRONG!) proposal:
Attachment: Exercise_12_16.pdf [31.75 KiB]

THIS PROOF COMPRISES AN ERROR (cf. Roberto, below).

Better try this one:
Attachment: Exercise_12_16amended.pdf [33.13 KiB]

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

05 Dec 2010, 09:13 Joined: 03 Jun 2010, 15:18
Posts: 136
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.

06 Dec 2010, 14:27 Joined: 22 Apr 2010, 15:52
Posts: 43
Location: Olpe, Germany
You are right, Roberto! Thanks for the hint and sorry for the error.
I hope this fixes the bug:
Attachment: Exercise_12_16amended.pdf [33.13 KiB] 