I came across the following statement in a book on automorphic forms:

In general, on a Riemannian manifold, the Laplace-Beltrami operator $\Delta$ is characterized by the property that a diffeomorphism is an isometry if and only if it leaves $\Delta$ invariant.

For this question, I am interested in the last part of the statement, that "a diffeomorphism is an isometry if and only if it leaves $\Delta$ invariant." What is a precise statement of this result? We're viewing $\Delta$ as an operator on $C^\infty(M)$, right? Is there a way to prove it just from the description of $\Delta$ in coordinates: $$ \frac{-1}{\sqrt{\det{g}}}\sum_{i,j}\partial_j(g^{ij}\sqrt{\det{g}}\partial_i f)? $$ I'd also be interested in a way to prove it from the definition of $\Delta$ as $-\text{div}\circ\nabla$.


Here's one way to make it precise.

Theorem. Suppose $(M_1,g_1)$ and $(M_2,g_2)$ are Riemannian manifolds, $\Delta_1,\Delta_2$ are their respective Laplace operators, and $\phi\colon M_1\to M_2$ is a diffeomorphism. Then $\phi$ is an isometry if and only if for every $f\in C^\infty(M_2)$ we have $\Delta_1(f\circ\phi) = (\Delta_2 f)\circ \phi$.

(Of course, if you're only interested in diffeomorphisms from a Riemannian manifold to itself, you can take $M_1=M_2=M$ and $g_1=g_2=g$.)

I don't have time to write down a complete proof, but the idea is that the principal symbol of a differential operator is a diffeomorphism-invariant quadratic function on the cotangent bundle, and the principal symbol of the Laplace operator is given by the squared norm function on covectors. By the polarization identity, the squared norm function determines the metric on covectors (the "dual metric"), and by the musical isomorphism between $TM$ and $T^*M$, this in turn determines the metric.

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    $\begingroup$ So is what's going on is that preserving the principal symbol for the Laplacian is just the same exact condition (after applying the musical isomorphism) as being an isometry? Does this mean that the same result holds for any differential operator with the same principal symbol? Are the lower order terms somehow irrelevant? $\endgroup$ – 4461013 Jul 5 '15 at 19:20
  • $\begingroup$ Yes, that's exactly right. $\endgroup$ – Jack Lee Jul 5 '15 at 19:59
  • $\begingroup$ But there are not vary many differential operators invariant under isometries with lower terms, no? $\endgroup$ – Mariano Suárez-Álvarez Feb 1 '17 at 19:40
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    $\begingroup$ @MarianoSuárez-Álvarez: There are lots of "zero-order differential operators" (i.e., scalar functions) that are invariant under isometries. Some examples are scalar curvature, the squared norm of the curvature tensor, and many more formed by taking covariant derivatives of the curvature and then applying complete contractions to obtain a scalar. (These are called "Weyl invariants.") But there are no invariant first-order differential operators. (The symbol of such an operator would be an invariant vector field, which can't exist on manifolds that are isotropic like spheres.) $\endgroup$ – Jack Lee Feb 1 '17 at 20:15

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