1
$\begingroup$

Related question, which I don't understand either. I think is easier to get the Lorentz group algebra as defined by Maggiore,

$$ [J^{\mu\nu},J^{\rho\sigma}] = i(\eta^{\nu\rho}J^{\mu\sigma} - \eta^{\mu\rho}J^{\nu\sigma} - \eta^{\nu\sigma}J^{\mu\rho} + \eta^{\mu\sigma}J^{\nu\rho}), $$

by using

$$ (J^{\mu\nu})^\rho_{\dot,\sigma} = i(\eta^{\mu\rho}\delta^\nu_\sigma - \eta^{\nu\rho}\delta^\mu_\sigma). $$

This last expression is simpy the matrix representation of the generator $J^{\mu\nu}$. Then, for example, the first term in the commutatior should be like (since we are multiplying matrices)

$$ (J^{\mu\nu})^\alpha_{\dot,\beta}(J^{\rho\sigma})^\beta_{\dot,\gamma} = i(\eta^{\mu\alpha}(J^{\rho\sigma})^\nu_{\dot,\gamma} - \eta^{\nu\alpha}(J^{\rho\sigma})^\mu_{\dot,\gamma}), $$

but this does not look at all like at least two terms in the commutation relation. Where am I wrong? What did I not consider?

$\endgroup$
  • 1
    $\begingroup$ If someone is interested in the calculation, I found it exactly here $\endgroup$ – user2820579 May 25 at 4:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.