Conformal group, infinitesimal transformation

1. Jun 1, 2012

physicus

1. The problem statement, all variables and given/known data
In order to determine the infinitesimal generators of the conformal group we consider an infinitesimal coordinate transformation:
$x^{\mu} \to x^\mu+\epsilon^\mu$
We obtain $\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu=\frac{2}{d}(\partial\cdot\epsilon)\eta_{\mu\nu}$ where d is the dimension of spacetime.
Derive $(\eta_{\mu\nu}\Box+(d-2)\partial_\mu \partial_\nu)\partial\cdot\epsilon=0$

2. Relevant equations
$\eta_{\mu\nu}\eta^{\mu\nu}=d$

3. The attempt at a solution
I think I am really close to the solution, but somehow I don't arrive there.
$\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu=\frac{2}{d}(\partial\cdot\epsilon)\eta_{\mu\nu}$
$\Rightarrow \partial^\nu\partial_\mu\epsilon_\nu+\partial^\nu{\partial}_\nu\epsilon_\mu=\frac{2}{d}\partial^\nu({\partial}\cdot\epsilon)\eta_{\mu\nu}$
$\Rightarrow \partial_\mu(\partial\cdot\epsilon)+\Box{\epsilon}_\mu=\frac{2}{d}\partial_\mu(\partial\cdot\epsilon)$
$\Rightarrow (d-2)\partial_\mu(\partial\cdot\epsilon)+d\cdot{\Box}{\epsilon}_\mu=0$
$\Rightarrow (d-2)\partial_\mu\partial_\nu(\partial\cdot\epsilon)+d\partial_\nu\Box{\epsilon}_\mu=0$
I think I need to use $d=\eta_{\mu\nu}\eta^{\mu\nu}$ now, but I don't get the right result.
Can somebody help me?

Best regards, physicus

2. Jun 1, 2012

fzero

The first term in (**) is symmetric in $\mu\nu$, so we can write (**) as

$$(d-2)\partial_\mu\partial_\nu(\partial\cdot\epsilon)+\frac{d}{2} \Box ( \partial_\nu{\epsilon}_\mu + \partial_\mu{\epsilon}_\nu)=0.$$

Alternatively, you can take the divergence of (*) with $\partial^\mu$ and add it to (**) to get this equation. It should be clear what to do next.