How to Prove the Commutator Relations for Lorentz Lie Algebra?

[latentcorpse]

So the generators of the Lorentz Lie algebra relations obey

[M^{\rho \sigma}, M^{\tau \mu}] = g^{\sigma \tau} M^{\rho \mu} - g^{\sigma \mu} M^{\rho \tau} + g^{\rho \mu} M^{\sigma \tau} - g^{\rho \tau} M^{\sigma \mu}

where (M^{\rho \sigma})^\mu{}_\nu = g^{\rho \mu} \delta^\sigma{}_\nu - g^{\sigma \mu} \delta^\rho{}_\nu

Now the commutator relation above will be satisfied for any representation of the LIe algebra since a representation of a Lie algebra is just a set of matrices such that the Lie bracket is given by the commutator.

So anyways, I was trying to prove this and just got totally bogged down in the algebra:

[M^{\rho \sigma}, M^{\tau \mu}] \\<br /> = (M^{\rho \sigma})^\lambda{}_\kappa ( M^{\tau \mu})^\kappa{}_\lambda - (M^{\tau \mu})^\lambda{}_\kappa (M^{\rho \sigma})^\kappa{}_\lambda \\<br /> =(g^{\rho \lambda} \delta^\sigma{}_\kappa - g^{\sigma \lambda} \delta^\rho{}_\kappa)(g^{\tau \kappa} \delta^\mu{}_\lambda - g^{\mu \kappa} \delta^\tau{}_\lambda) -(g^{\tau \lambda} \delta^\mu{}_\kappa - g^{\mu \lambda} \delta^\tau{}_\kappa)(g^{\rho \kappa} \delta^\sigma{}_\lambda - g^{\sigma \kappa} \delta^\rho{}_\lambda)

but expanding this out isn't going to give me any M terms like I need in the answer. Any ideas where I've messed up?

Thanks.

 

[dextercioby]

You shouldn't fully expand the commutator using the particular form of M, but only the first of the M's in the expanded commutator.

 

[latentcorpse]

You shouldn't fully expand the commutator using the particular form of M, but only the first of the M's in the expanded commutator.

ok. so i get to

g^{\rho \sigma} (M^{\tau \mu})^\sigma{}_\lambda - g^{\sigma \lambda} ( M^{\tau \mu})^\rho{}_\lambda - g^{\tau \lambda} ( M^{\rho \sigma})^\mu{}_\lambda + g^{\mu \lambda} ( M^{\rho \sigma})^\tau{}_\lambda

but don't know how to finish it off...