Recent content by Jyoti6297

I didn't get the last part. So the permutation symmetry of the ##()## is the same as of the Levi-Civita, so we can ##()_{\mu\sigma\eta\nu}=\epsilon_{\mu\nu\eta\sigma} ()_{1234} = \epsilon_{\mu\nu\eta\sigma} \cdot \epsilon_{\alpha\beta\gamma\kappa}A^\alpha_1 A^\beta_2 A^\gamma_3 A^\kappa_4 ##...

I know what Levi Civita is, but I don't know how to go about such questions. Question : (1.3)