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I am following Weinberg 2.4 and I understand why $\omega_{\mu \nu}$ is antisymmetric, but he says after expanding $$U(1 + \omega, \epsilon) = 1 + \frac{1}{2}i\omega_{\rho \sigma}J^{\rho \sigma} - i\epsilon_{\rho}P^{\rho} + ...\tag{2.4.3}$$

Since $\omega_{\mu \nu}$ is antisymmetric. we can take it's coefficient $J^{\rho \sigma}$ to be sntisymmetric also $$J^{\rho \sigma} = - J^{\sigma \rho}\tag{2.4.5}$$

I can't understand how and why $J$ must be antisymmetric. Is there any physical relation we can use to set it that way?

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    $\begingroup$ Hint: How much would a symmetric piece contribute to eq. (2.4.3)? $\endgroup$
    – Qmechanic
    Commented Aug 3, 2023 at 10:13

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Every rank 2 tensor can be written in a symmetric and an antisymmetric part. Now note, that the contraction of an antisymmetric and a symmetric tensor is always 0. Thus, decomposing J, the symmetric part vanishes and so only the antisymmetric part contributes.

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