is $\nabla \cdot ( c^2 \nabla)$ a Laplace-Beltrami operator? Someone mentioned, in passing, to me that $u \mapsto \nabla \cdot ( c^2 \nabla u)$ is a Laplace-Beltrami operator. Does anyone have some insight into this? From my understanding, the Laplace-operator generalizes the Laplacian to Riemannian manifolds, by taking the trace of the Hessian. I don't really see the connection to that and the above operator, unless c=1.
This operator comes from the wave equation, where $\partial^2_t u -\nabla \cdot ( c^2 \nabla u) = f$. There may or not be some smoothness conditions on $c$, and all these are functions on subsets of $\mathbb{R}^n$.
Thanks
 A: *

*As  it was mentioned by  @Denis Serre, for dimensions $\ne 2$, your operator is proportional to the Beltrami-Laplace operator.  The proof is essentially in the comment of  Denis Serre;   one can also obtain it following suggestion of @Piero D'Ancona and  writing this operator in local coordinates (instead of covariant derivative the usual derivatives appear by the price the determinant  of the metric appears twice as factors). 

*For nontrivial $c$, the operator in NOT the Beltrami-Laplace  of any metric. I will explain it in the case your  metric and $c$ are generic and dimension is $>2$, but it seems that the proof works without this assumption. 
Proof of (2). Consider the symbol of your  operator.  It is a well-defined (2,0)-tensor  and it is equal to $c^2 \cdot g^{ij}$. Would you operator  be  the Beltrami-Laplace of some $g'$, 
its symbol will be covariantly constant, and in the case of generic $g$ this would imply that  the metric $g'$ is a constant factor of $\frac{1}{c^2}g_{ij}$ (I assume $n>2$). Thus, we have only one choice for $g'$. Calculating the Beltrami-Laplace for $g'$, we see that it does not coincide with your operator (if $c\ne \mathrm{const}$). 
