Equivalence of definitions of harmonic (or wave) coordinates In GR, one often uses harmonic (or wave) coordinates to simplify things. Now, one definition involves the coordinates themselves:
$$ \Box_g x^{\alpha} = 0  $$
where $ \Box_g = g_{\mu \nu}\nabla^{\mu}\nabla^{\nu} $ and the $ \nabla $'s stand for covariant derivatives.
Now, in recent studies, I often encountered an apparently equivalent definition, i.e. the following:
$$\partial_{\beta}g^{\alpha \beta} = 0$$
Does anybody now how to show this equivalence or can someone give me a hint? Thanks a lot in advance.
 A: Suppose $\Box_g x^\alpha = 0$, we expand this equation in the coordinates given by the wave coordinates $\{x^\alpha\}$. The metric expression of Laplace-Beltrami operator $\Box_g$ is
$$ 0 =  \Box_g x^\alpha = \frac{1}{\sqrt{|\det g|}} \sum_{\beta,\gamma} \partial_\beta \left( g^{\beta\gamma} \sqrt{|\det g|} \partial_\gamma x^\alpha \right) $$
where $|\det g|$ is evaluated again relative to the wave coordinates, and that I explicitly included the summation symbol since we are working in coordinates. 
In coordinate, $\partial_\gamma x^\alpha = \delta_\gamma^\alpha$. So the equation is equivalent to 
$$ \iff \sum_{\beta} \partial_\beta ( g^{\beta\alpha} \sqrt{|\det g|}) = 0  \quad\forall ~\alpha$$
In particular, this is not equivalent to the coordinate expression $\sum_\beta \partial_\beta g^{\alpha\beta} = 0$. 

Directly taking the derivatives, and using the Jacobi formula for derivative of a determinant, gives you that the equation is further equivalent to 
$$ \iff  \sum_{\beta,\gamma} g^{\beta\gamma} \Gamma_{\beta\gamma}^\alpha = 0 \quad \forall~\alpha $$
where $\Gamma^\alpha_{\beta\gamma}$ are the Christoffel symbols of second kind of the metric $g$ expressed in the wave coordinates $\{x^\alpha\}$. 

It is however possible that the "formula"
$$ \partial_\beta g^{\beta\alpha} $$
is just a very sloppy notation for something that is actually correct. Fix $\alpha$. If we denote $Y$ the vector field obtained by the metric dual of $\mathrm{d}x^\alpha$, it is the vector field with components
$$ \sum_{\beta} Y^\beta \partial_\beta = \sum_{\beta} g^{\beta\alpha} \partial_\beta $$
(remember that $\alpha$ is fixed). Then the expression for the divergence of this vector field is in fact 
$$ \mathrm{div}(Y) = \sum_\beta \partial_\beta (Y^\beta \sqrt{|\det g|} ) $$
So if one were to interpret "$\partial_\beta g^{\beta\alpha}$" to mean the geometric divergence of the vector field whose wave coordinate components are $g^{\beta\alpha}$ (and, for the record, I think this is an incredible abuse of notation that should never be used), then the equivalence is demonstrated as in the first part of this answer. 
