Symmetrizing 3xMetric Tensor: H^{\mu \nu \lambda \kappa \rho \sigma}

[Barnak]

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.

 

[clamtrox]

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 $\mu$.