TITLE: is $\nabla \cdot ( c^2 \nabla)$ a Laplace-Beltrami operator?
QUESTION [8 upvotes]: I asked this on mathoverflow, and was suggested to ask it here.
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-Beltrami 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 for any help,
-Nick

REPLY [2 votes]: Willie's hint in his comment is a bit misleading; your operator is equivalent to the Laplace-Beltrami operator with respect to some (conformally equivalent) metric only if $c$ is constant.
More generally, an elliptic operator $Lf = \operatorname{div} M \nabla f$ with $M$ self-adjoint and positive-definite is equivalent to the Laplacian of some new metric if and only if $\det M$ is constant. For more information see my answer to this related question: Generalized Laplace--Beltrami operators