Proving the Wedge Product of 2 One-Forms is a 2-Form

[Silviu]

Hello! I was trying to show that the wedge product of 2 one-forms is a 2-form. So we have ## (A \wedge B)_{\mu \nu} = A_\mu B_\nu - A_\nu B_\mu ##. So to show that this is a (0,2) tensor, we need to show that ##(A \wedge B)_{\mu' \nu'} = \Lambda_{\mu'}^\mu \Lambda_{\nu'}^\nu (A \wedge B)_{\mu \nu}##. But ##A_{\mu'} B_{\nu'} - A_{\nu'} B_{\mu'} = \Lambda_{\mu'}^\mu A_\mu \Lambda_{\nu'}^\nu B_\nu - \Lambda_{\nu'}^\nu A_\nu \Lambda_{\mu'}^\mu B_\mu ##. I am not sure how to proceed from here, as the matrices don't commute, so I can't bring the ##\Lambda## in the front. What should I do?

 

[Orodruin]

What matrices? Those are just numbers.

 

[vanhees71]

Also again, be warned about this sloppy notation of indizes. You should put the prime on the symbol (or in addition to the symbol). Otherwise the equations don't make sense strictly speaking (I know that some unfortunate textbooks use this very dangerous notation). Also make sure that both the "vertical and horizontal" placement of the indices is accurate. For a 2nd-rank tensor the transformation law should be written as
$$T_{\mu \nu}'={\Lambda^{\rho}}_{\mu} {\Lambda^{\sigma}}_{\nu} T^{\rho \sigma}.$$
Concerning your question, Orodruin has given you the right hint. The tensor components are all numbers, and thus the product is the usual commutative product of real numbers!

 

[fresh_42]

The rest of the discussion, which has been slightly off topic, has been moved to
https://www.physicsforums.com/threads/tensor-invariance-and-coordinate-variance.914642/