Conformal group, infinitesimal transformation

[physicus]

Homework Statement

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

Homework Equations

\eta_{\mu\nu}\eta^{\mu\nu}=d

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?

physicus

 

[fzero]

\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu=\frac{2}{d}(\partial\cdot\epsilon)\eta_{\mu\nu}~~~(*)

...

\Rightarrow (d-2)\partial_\mu\partial_\nu(\partial\cdot\epsilon)+d\partial_\nu\Box{\epsilon}_\mu=0~~~(**)

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.