TITLE: is $\nabla \cdot ( c^2 \nabla)$ a Laplace-Beltrami operator?
QUESTION [6 upvotes]: 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

REPLY [10 votes]: 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 $c\ne \mathrm{const}$).