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Symetrisation of products of the metric 
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#1
Nov1912, 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. 


#2
Nov2012, 07:10 AM

P: 939

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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