Higher Spin Field Rep: Tr(n,n)=0, Why Symmetric Tensor of Rank 2n?

[filip97]

Representation
(n,n)=nxn=2n+(2n-1)+...+2+1 , where + is direct sum, and x is tensor product. Trace of (n,n) is zero because tr(k) is zero, where k=1 to n. But why tensor (n,n) is symmetric tensor of rank 2n? I read Weinberg book Vol. 1 page 231. But he don't clearly says this .

 

[AndyW]

Hello! Thank you for bringing up this interesting topic about representation theory. I will try my best to explain why (n,n) is a symmetric tensor of rank 2n.

First, let's define what a symmetric tensor is. A symmetric tensor is a tensor that remains unchanged under the exchange of its indices. In other words, if we swap the positions of two indices, the value of the tensor remains the same. For example, in a 2x2 matrix, the value at position (1,2) is the same as the value at position (2,1).

Now, let's take a closer look at the tensor (n,n) = nxn = 2n+(2n-1)+...+2+1. This tensor is the direct sum of n copies of itself, which means it is a tensor product of n copies of (1,1). Therefore, (n,n) is a symmetric tensor of rank 2n because it is composed of n copies of a symmetric tensor (1,1).

Moreover, we can also see that the trace of (n,n) is zero, as you mentioned. This is because the trace of a direct sum is the sum of the traces of each individual tensor. Since the trace of (1,1) is zero, the trace of (n,n) will also be zero.

I hope this helps to clarify why (n,n) is a symmetric tensor of rank 2n. If you have any further questions or would like to discuss this topic more, please feel free to continue the conversation. Thank you!