Derivation of Lorentz algebra commutation relation

[spaghetti3451]

Homework Statement

1. Show that the Lorentz algebra generator $J^{\mu \nu} = i(x^{\mu}\partial^{\nu}-x^{\nu}\partial^{\mu})$ lead to the commutation relation $[J^{\mu \nu}, J^{\rho \sigma}] = i(g^{\nu \rho}J^{\mu \sigma} - g^{\mu \rho}J^{\nu \sigma}-g^{\nu \sigma}J^{\mu \rho}+g^{\mu \sigma}J^{\nu \rho})$.

2. Show that one particular representation of the the Lorentz group given by $(J^{\mu \nu})_{\alpha \beta} = i(\delta^{\mu}_{\alpha}\delta^{\nu}_{\beta}-\delta^{\mu}_{\beta}\delta^{\nu}_{\alpha})$.

Homework Equations

The Attempt at a Solution

$[J^{\mu \nu}, J^{\rho \sigma}]$

$= J^{\mu \nu}J^{\rho \sigma}-J^{\rho \sigma}J^{\mu \nu}$

$= -(x^{\mu}\partial^{\nu}-x^{\nu}\partial^{\mu})(x^{\rho}\partial^{\sigma}-x^{\sigma}\partial^{\rho})+(x^{\rho}\partial^{\sigma}-x^{\sigma}\partial^{\rho})(x^{\mu}\partial^{\nu}-x^{\nu}\partial^{\mu})$.

How do I get the metric tensor from here?

 

[fzero]

When you distribute terms you will find various factors like $\partial^\nu x^\rho$.

 

[spaghetti3451]

How are those factors related to the metric tensor?

As far as I know, a term like $\partial_{\nu}x^{\rho}$ gives the Kronecker delta $\delta_{\nu}^{\rho}$, not the metric tensor.

 

[fzero]

Now raise the index.

 

[spaghetti3451]

So, you say $\delta_{\nu}^{\rho}$ equals $g^{\nu \rho}$?

 

[spaghetti3451]

Or, do I use $\delta_{\nu}^{\rho} = g^{\nu \rho}g_{\nu \rho}$?

 

[fzero]

So, you say $\delta_{\nu}^{\rho}$ equals $g^{\nu \rho}$?

Not exactly, $\partial^\nu x^\rho = g^{\nu\mu} \partial_\mu x^\rho$. So use the result for $\partial_\mu x^\rho$ here.