# Verification of tensor identities

1. Jun 9, 2010

### rafaelpol

1. The problem statement, all variables and given/known data

Show that in 2 dimensions a skew-symmetric tensor of second rank is a pseudoscalar and that one of third rank is impossible.

3. The attempt at a solution

A11=A22=0, while A12=-A21, which makes

A= A12+A21, which is certainly skew-symmetric, though I am not sure it is a pseudoscalar since a pseudoscalar is a number, and the tensor should have only one component in this case, or maybe not?

I guess this will help with the next part of the exercise.

Thank you
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 9, 2010

### gabbagabbahey

Ermm... $A^{12}+A^{21}=0$, why are you setting it equal to $A$?

Instead, go back to the definition of a pseudoscalar:

This means that in order to show something is a psuedoscalar, you need to show two things:

(1)The quantity flips signs under a parity transformation.

(2)Other than (1), the quantity behaves like a scalar (i.e. invariant under translations, rotations and Lorentz transformations).

Not really. How many components will a 3rd rank tensor have in dimensions? Can a non-square matrix be skew-symmetric (or symmetric)?

Last edited by a moderator: May 4, 2017
3. Jun 9, 2010

### rafaelpol

I set it equal to A because I thought about tensors the way one usually talks about vectors which are the sum of their components.

From the requirements needed to be a pseudoscalar, the skew-symmetry of the components of the tensor proves it change signs under a parity transformation. Now, is the fact that the derivatives present in the tensor transformations do not change under translations or rotations the key to the problem? The invariance of the derivatives under translations is pretty obvious, but I am not sure I have a solid argument for the invariance of this under rotations. Why would that really happen?

Concerning the other question: a 3rd rank tensor in 2D will have 8 components, giving a 2x4 matrix, which cannot have definite parity. Is that right?

Thank you very much.

4. Jun 9, 2010

### gabbagabbahey

But a vector isn't the sum of its components, the components must be multiplied by their corresponding unit vectors first and then summed together.

Sure, but unless you've already proven this property, I suggest you actually perform a parity transformation on $A^{ij}$ and verify that you get $-A^{ij}$ as a result.

Again, try applying a general translation and rotation and see what happens.

The question asks you to show that it is impossible to have a skew-symmetric tensor of rank 3 in D, not whther or not it will be a pseudoscalar. So, is it possible to have $A^{ijk}=-A^{jik}$ and $A^{ijk}=-A^{ikj}$ and $A^{ijk}=-A^{kji}$ for $i,j,k\in \{1,2\}$?

Last edited: Jun 9, 2010