I understand that under rotation, we will have components that transform like spin-$j$ $(2A, 2A-1, ...... 0)$ from decomposition of $(A, A)$ representation. The scalar is the the trace, therefore other components are traceless. And, I can see why $2A$ component will be traceless symmetric tensor, but I don't understand why Weinberg says in section 5.6 of Volume I that

"$(A, A)$ field contain terms with only integer spins $(2A, 2A-1, ...... 0)$ and corresponds to a traceless symmetric tensor of rank $2A$."

I think he means that the whole field representation corresponds to a traceless symmetric tensor of rank $2A$.

I understand that under rotation, we will have components that transform like integer spin $(2A, 2A-1, ...... 0)$ from decomposition of $(A, A)$ representation. The scalar is the trace, therefore other components are traceless. And, I can see why $2A$ component will be traceless symmetric tensor, but I don't understand why Weinberg says on p. 231 below eq. (5.6.18) in section 5.6 of Volume I that

"$(A, A)$ field contain terms with only integer spins $(2A, 2A-1, ...... 0)$ and corresponds to a traceless symmetric tensor of rank $2A$."

I think he means that the whole field representation corresponds to a traceless symmetric tensor of rank $2A$.

I understand that under rotation, we will have components that transform like spin-j (2A, 2A-1, ...... 0) from decomposition of (A, A) representation. The scalar is the the trace, therefore other components are traceless. And, I can see why 2A component will be traceless symmetric tensor, but I don't understand why Weinberg says in section 5.6 of Volume I that "(A, A) field contain terms with only integer spins (2A, 2A-1, ...... 0) and corresponds to a traceless symmetric tensor of rank 2A." I think he means that the whole field representation corresponds to a traceless symmetric tensor of rank 2A.

I understand that under rotation, we will have components that transform like spin-$j$ $(2A, 2A-1, ...... 0)$ from decomposition of $(A, A)$ representation. The scalar is the the trace, therefore other components are traceless. And, I can see why $2A$ component will be traceless symmetric tensor, but I don't understand why Weinberg says in section 5.6 of Volume I that

"$(A, A)$ field contain terms with only integer spins $(2A, 2A-1, ...... 0)$ and corresponds to a traceless symmetric tensor of rank $2A$."

I think he means that the whole field representation corresponds to a traceless symmetric tensor of rank $2A$.

I understand that under rotation, you will have components that transform like spin-j (2A, 2A-1, ...... 0) for decomposition of (A, A) representation. The scalar is the the trace, therefore other components are traceless. And, I can see why 2A component will be traceless symmetric tensor, but I don't understand why Weinberg says in section 5.6 of Volume I that "(A, A) field contain terms with only integer spins (2A, 2A-1, ...... 0) and corresponds to a traceless symmetric tensor of rank 2A.

I understand that under rotation, we will have components that transform like spin-j (2A, 2A-1, ...... 0) from decomposition of (A, A) representation. The scalar is the the trace, therefore other components are traceless. And, I can see why 2A component will be traceless symmetric tensor, but I don't understand why Weinberg says in section 5.6 of Volume I that "(A, A) field contain terms with only integer spins (2A, 2A-1, ...... 0) and corresponds to a traceless symmetric tensor of rank 2A." I think he means that the whole field representation corresponds to a traceless symmetric tensor of rank 2A.