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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 $2A$.

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  • $\begingroup$ The answer to the last sentence (v5) is Yes. Possible duplicate: physics.stackexchange.com/q/545943/2451 $\endgroup$
    – Qmechanic ♦
    Commented Jun 12, 2022 at 14:05
  • $\begingroup$ Thank you for the link, but seems like the link doesn't seem to answer my question, I was asking why is (A, A) representation symmetric traceless tensor. I think it is because as Weinberg says all terms are integer spins, which have symmetric traceless tensor representations. For example, for spin-2, we can use symmetric traceless tensor in 3-d which has 5 independent components as expected. $\endgroup$
    – physicsbootcamp
    Commented Jul 7, 2022 at 0:55
  • $\begingroup$ The why is essentially the content of eq. (10) in my answer. $\endgroup$
    – Qmechanic ♦
    Commented Jul 7, 2022 at 12:07

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