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As I've learned more Riemannian geometry, many of my teachers have said that studying the Laplacian (and its eigenvalues) is very important. But I must admit, I've never fully understood why.

Fundamentally, I would like to know why the Laplacian is important among all differential operators on a Riemannian manifold. I would also like to know what geometric information the Laplacian is supposed to encode.

That being said, I have spent a little time thinking about all this, and my current understanding is as follows:

  • ** Somehow, the Laplacian is the "only" isometric invariant "scalar differential operator" on a Riemannian manifold. If true, this statement would completely convince me of its importance. However, I don't know the precise meanings the words in quotes, nor do I have any sense at all of why such a statement would be true.

  • An isometric immersion $f \colon S \to M$ is harmonic if and only if it represents a minimal submanifold of $M$. In particular, an isometrically immersed submanifold of $\mathbb{R}^n$ is minimal if and only if its coordinate functions are harmonic.

  • The Euler-Lagrange equation for the Dirichlet energy is $\Delta f = 0$. (But why we care about minimzing energy is also somewhat mysterious to me.)

  • Weitzenböck formulas comparing two elliptic second-order differential operators (and especially Laplacians) give Bochner-type vanishing theorems.

I should point out that I'm aware that harmonic functions satisfy many of the nice properties that complex-analytic functions do (by virtue of elliptic regularity and maximum principle magic). Still, this doesn't quite tell me why I should care about the Laplace operator itself.

Note: I'm aware of this related question on the eigenvalues of the Laplacian. But again, my interest is in Riemannian geometry; matters of applied mathematics (while interesting) are not my focus right now.

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A fairly complete answer to these questions can be found in (or may be based on) the Introduction to the book of S.Rosenberg "The Laplacian on a Riemannian manifold". – Yuri Vyatkin Feb 1 '14 at 8:39
OK, I've read the Intro and skimmed Chapter 1. The main point seems to be that the heat flow somehow contains topological and geometric data. As such, it can be used to prove theorems like the Hodge Theorem, Chern-Gauss-Bonnet, and Atiyah-Singer. This has convinced me that the heat flow is a powerful tool in differential geometry, and that the spectrum of $\Delta$ must have some geometric information in it somewhere. But I still feel like I'm missing some things, because I don't feel as though my questions have been answered in full. – Jesse Madnick Feb 1 '14 at 11:07
Here's something you may appreciate: associated to a differentiable operator is its symbol, which is a tensor that encodes the highest order part of the operator. The symbol of the Laplacian is the metric tensor. – Eric O. Korman Feb 1 '14 at 16:03
Perhaps someone with a better understanding than me can expand on this, but I believe the Selberg Trace Formula relates the eigenvalues of the Laplacian to the lengths of (primitive, periodic) geodesics in your manifold. If your interest is in Riemannian geometry, I would imagine that should make the Laplacian pretty significant. – ChickenSocks Jun 14 '14 at 3:04

There is one important piece of evidence that has not been mentioned in the "comments" above, namely, de Rham's theorem for compact manifolds, and the idea that harmonic forms uniquely represent (real) cohomology classes. That certainly testifies to the significance of the Laplacian (the kernel of which consists of harmonic forms).

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And thanks for your answer, although this doesn't quite address my questions. (That is, why exactly is the Laplacian "better" than any other differential operator? What distinguishes it from the others? And what geometric (not topological) data does it contain?) – Jesse Madnick Feb 3 '14 at 22:37
@JesseMadnick, Roughly speaking, one of the "morals" of the Atiyah-Singer index theorem is that an operator (its analytic index in any case) is characterized by its topological index, i.e., the dimensions of the suitable kernels. This means that if we are interested in harmonic forms, there is essentially a unique operator whose kernel they are. So explaining the significance of the Laplacian is the same as explaining the significance of harmonic forms. – Mikhail Katz Feb 4 '14 at 13:34
P.S. The circle of ideas around de Rham's theorem includes the fact that a cohomology class contains a unique harmonic form. – Mikhail Katz Feb 4 '14 at 13:35
This is not de Rham's theorem, but Hodge's theorem (to the best of my knowledge), see Of course, it is quite possible that de Rham also knew how to prove it... – studiosus Feb 12 '14 at 12:11
@studiosus, you might be right. I was referring to "the circle of ideas around de Rham's theorem", and this is certainly one of the circle. – Mikhail Katz Feb 12 '14 at 12:47

I think the principal reason is Laplacian is the simplest second order elliptic operator available. So once one prove something non-trivial for the Laplacian, it is useful in other settings as well by consider a generalized Laplacian over the manifold. It is natural to consider a differential operator of any order in general(especially if one works with pseudo-differential operators), but higher order ones are not easy to study (one need techniques like Moser iteration or energy methods in general). Another reason is its association with the heat kernel, which I assume you must already know judged by the comments.

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