Symetrisation of products of the metric


by Barnak
Tags: metric, products, symetrisation
Barnak
Barnak is offline
#1
Nov19-12, 11:03 PM
P: 61
I need to build a tensor from the product of the metric components, like this (using three factors, not less, not more) :

[itex]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} + ...[/itex],

however, that [itex]H^{\mu \nu \lambda \kappa \rho \sigma}[/itex] tensor should be fully symmetric under pairs of indices :

[itex]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)}[/itex]

How can I do that ? Someone know what should be that tensor, explicitely ?

With only two times the metric, it would be easy :

[itex]H^{\mu \nu \lambda \kappa} = g^{\mu \nu} \, g^{\lambda \kappa} + g^{\mu \lambda} \, g^{\nu \kappa} + g^{\mu \kappa} \, g^{\nu \lambda}[/itex]

but I don't know how to do it with three times the metric.
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clamtrox
clamtrox is offline
#2
Nov20-12, 07:10 AM
P: 937
You have to go through all possible index pairs. So you'd get something like
[tex]H^{\mu \nu \lambda \kappa \rho \sigma}= g^{\mu \nu}H^{\lambda \kappa \rho \sigma} + g^{\mu \lambda} H^{\nu \kappa \rho \sigma} + ... [/tex] where the H with 4 indices is as you calculated and you sum over all possible index pairs containing [itex] \mu [/itex].


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