# Symetrisation of products of the metric

by Barnak
Tags: metric, products, symetrisation
 P: 61 I need to build a tensor from the product of the metric components, like this (using three factors, not less, not more) : $H^{\mu \nu \lambda \kappa \rho \sigma} = g^{\mu \nu} \, g^{\lambda \kappa} \, g^{\rho \sigma} + g^{\mu \lambda} \, g^{\nu \kappa} \, g^{\rho \sigma} + ...$, however, that $H^{\mu \nu \lambda \kappa \rho \sigma}$ tensor should be fully symmetric under pairs of indices : $H^{\mu \nu \lambda \kappa \rho \sigma} \equiv H^{(\mu \nu) \lambda \kappa \rho \sigma} \equiv H^{\mu \nu (\lambda \kappa) \rho \sigma} \equiv H^{\mu \nu \lambda \kappa (\rho \sigma)}$ How can I do that ? Someone know what should be that tensor, explicitely ? With only two times the metric, it would be easy : $H^{\mu \nu \lambda \kappa} = g^{\mu \nu} \, g^{\lambda \kappa} + g^{\mu \lambda} \, g^{\nu \kappa} + g^{\mu \kappa} \, g^{\nu \lambda}$ but I don't know how to do it with three times the metric.
 P: 939 You have to go through all possible index pairs. So you'd get something like $$H^{\mu \nu \lambda \kappa \rho \sigma}= g^{\mu \nu}H^{\lambda \kappa \rho \sigma} + g^{\mu \lambda} H^{\nu \kappa \rho \sigma} + ...$$ where the H with 4 indices is as you calculated and you sum over all possible index pairs containing $\mu$.