Deriving a formula for Lie algebra of Conformal Field theory I'm learning some conformal field theory.
I'm trying to use the formula
$$
\partial_{\mu} \epsilon_\nu+\partial_\nu \epsilon_\mu=\frac{2}{d}(\partial \epsilon) \eta_{\mu \nu}
$$
to derive the equation
$$
\bigg( \eta_{\mu \nu} \square +(d-2)\partial_{\mu}\partial_{\nu} \bigg)(\partial \epsilon)=0
$$
where
$$
\square=\eta^{\mu \nu} \partial_{\mu} \partial_{\nu}
$$
And the dimension $ d $ can be any positive integer (e.g. $ d=p+q $ for $ \mathbb{R}^{p+q} $).
What is the best way to do this? And why do we expect the first equation to imply the second?
The reference I'm following is a bit physics-y so $ \epsilon $ is supposed to be an infinitesimal perturbation
$$
x^\mu \mapsto x^\mu +\epsilon^\mu(x)
$$
affecting the metric by
$$
\eta_{\mu \nu}'= \eta_{\mu \nu } + (\partial_\mu \epsilon_\nu+\partial_{\nu}\epsilon_\mu )+\mathcal{O}(\epsilon^2)
$$
which, since the perturbation is supposed to conformal, should imply that the factor
$$
\partial_\mu \epsilon_\nu+\partial_{\nu}\epsilon_\mu
$$
is proportional to the metric $ \eta_{\mu \nu} $. $ \partial \epsilon $ is the total divergence.
 A: It's just algebra. I believe you are faked out by the double-duty use of the index ν, which I'll avoid.
You started from the relation
$$
\partial_{\mu} \epsilon_\nu+\partial_\nu \epsilon_\mu= K \eta_{\mu \nu}.\tag{0}
$$
Operate on it by $\eta^{\mu\nu}$ to determine  $2\partial \epsilon= dK$, hence
$$
\partial_{\mu} \epsilon_\nu+\partial_\nu \epsilon_\mu=\frac{2}{d}(\partial \epsilon) \eta_{\mu \nu} ~.\tag{1}
$$
Next, operate on this by $\partial^\nu$ to obtain
$$
\partial_{\mu} \partial\epsilon+  \square \epsilon_\mu=\frac{2}{d} \partial_\mu (\partial \epsilon) ~.\tag{2}
$$
Now, operate on this by $\partial_\kappa$, symmetrize (μ,κ), and utilize (1) in the  last step, inside the d'Alembertian, multiplying everything by d,
$$
\partial_{\kappa}\partial_{\mu} \partial\epsilon+  \square \partial_{(\kappa}\epsilon_{\mu )}-\frac{2}{d} \partial_\mu \partial_{\kappa}(\partial \epsilon)=0 \qquad \leadsto \\
\bigg( \eta_{\mu \kappa} \square +(d-2)\partial_{\mu}\partial_{\kappa} \bigg)(\partial \epsilon)=0~,\tag{3}
$$
your target expression. You had to use (1) to supplant the symmetric gradient tensor by the divergence multiplying the metric. One always does.
