Recovering metric from Laplace-Beltrami operator

Question

On a Riemannian manifold $(M, g)$, if one is given an oracle $O$ that allows one to evaluate the Laplace-Beltrami operator, how can we recover the metric $g$? More precisely, $O: (M \rightarrow \mathbb R) \rightarrow \mathbb R$, $O(f) \equiv \Delta f = \frac{1}{\sqrt{|g|}} \partial_i (\sqrt{|g|} g^{ij} \partial_j f)$. We are allowed to pick any number of functions $f$ to evaluate against the oracle $O$. We do not know the symbolic expression for either $g$ or $\Delta$.

As far as I can tell, the problem is by by using $O$, we can only access the components of the metric under a $\partial_i$. I'm at a loss on how to "recover" the metric $g$ from $\partial_i (\cdots g^{ij} \cdots)$.

Answer 1

Given a point $p\in M$, you can recover the metric at $p$ as follows. Choose any local coordinates $(x^1,\dots,x^n)$ such that $p$ has the coordinate representation $(0,\dots,0)$. For any indices $k,l$, let $f_{kl}(x) = \tfrac 1 2 x^k x^l$. If you expand out the expression for $\Delta (f_{kl})$ and use the fact that $x^k=x^l=0$ at $p$, you'll find that $\Delta(f_{kl})(p) = g^{kl}$. Then you can use matrix inversion to recover $g_{kl}$.

If your oracle is restricted to acting only on globally defined functions, you can use a bump function to extend the coordinate functions to global smooth functions $(u^1,\dots,u^n)$ that agree with $(x^1,\dots,x^n)$ in a neighborhood of $p$, and let $f_{kl}(x) = u^k u^l$.

This is a special case of a much more general construction: if $P$ is an $m$th order scalar linear partial differential operator, the coefficients of its highest-order terms determine a coordinate-independent symmetric contravariant $m$-tensor field called the principal symbol, which can be evaluated at any point by applying $P$ to $m$-fold products of coordinate functions. The principal symbol of $\Delta$ is the associated cometric, which is the induced metric on $T^*M$.