TITLE: Why is the Laplacian important in Riemannian geometry?
QUESTION [55 upvotes]: 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:

I've heard that the Laplacian is the "simplest" isometrically-invariant "scalar differential operator" on a Riemannian manifold.  If true, this statement would convince me of its importance. However, I don't know to what extent this is true.
An isometric immersion $f \colon S \to M$ is harmonic iff it is a minimal submanifold of $M$.  In particular, an isometrically immersed submanifold of $\mathbb{R}^n$ is minimal iff 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.

REPLY [3 votes]: Two points which I haven't seen in the other answers:
For a high-level explanation of the Laplacian's ubiquity, see Peter Stredder "Natural differential operators on Riemannian manifolds and representations of the orthogonal and special orthogonal groups" in Journal of Differential Geometry, 1975. Particularly the end of section 4.
The quasi-historical (I'm not a historian, I may be wrong) explanation is that the Laplacian on $\mathbb{R}^n$ is quite important. One high-level reason is that every constant-coefficient linear partial differential operator $L:C^\infty(\mathbb{R}^n)\to C^\infty(\mathbb{R}^n)$ which is rotationally invariant is a polynomial in the laplacian. The more honest reason (which some people might argue is a consequence of the high-level reason) is that it just came up in a number of contexts. One specific context, which you might find compelling if you think complex analysis is automatically important, is Riemann and Dirichlet's construction of harmonic functions with prescribed boundary values as the fundamental part of the proof of the Riemann mapping theorem. If you prefer physics, then it should be compelling that the Newtonian potential in gravitation is related to density by the Poisson equation $\Delta u=\rho$. Or, if you accept that the matrix of second derivatives of a function $u$ is linear-algebraically natural, then $\Delta u$ warrants its recognition as the first symmetric polynomial - along with the other symmetric polynomials (just like in Riemannian geometry).
Anyhow, if you accept that $\Delta:C^\infty(\mathbb{R}^n)\to C^\infty(\mathbb{R}^n)$ is natural, then you should be compelled that $\Delta^g$ is automatically given as its geometric version, a fact which is totally obscured in most textbooks since they start with objects like "derivations" and "parallel transport". What this means is the following:
Let $y$ be a non-standard coordinate system on $\mathbb{R}^n.$ Let $x$ denote the standard coordinates. Then two applications of the chain rule shows
\begin{align*}
\frac{\partial f}{\partial x_i}&=\frac{\partial f}{\partial y_p}\frac{\partial y_p}{\partial x_i}\\
\frac{\partial^2f}{\partial x_i^2}&=\frac{\partial^2f}{\partial y_p\partial y_q}\frac{\partial y_p}{\partial x_i}\frac{\partial y_q}{\partial x_i}+\frac{\partial f}{\partial y_p}\frac{\partial^2 y_p}{\partial x_i^2}\\
\Delta f&=\sum_{p=1}^n\sum_{q=1}^n\frac{\partial^2f}{\partial y_p\partial y_q}\sum_{i=1}^n\frac{\partial y_p}{\partial x_i}\frac{\partial y_q}{\partial x_i}+\sum_{p=1}^n\frac{\partial f}{\partial y_p}\Delta y_p.\end{align*}
Define $g_{pq}=\sum_{i=1}^n\frac{\partial x^i}{\partial y_p}\frac{\partial x^i}{\partial y_q}$ and $g^{pq}=\sum_{i=1}^n\frac{\partial y_p}{\partial x_i}\frac{\partial y_q}{\partial x_i}$, with one actually defined via the other by the observation that as matrix-valued functions, they are inverses of one another. Now a small calculation shows, with an implicit summation over all non-$r$ indices,
\begin{align*}
\frac{1}{2}g^{ij}g^{rs}\Big(\frac{\partial g_{js}}{\partial y^i}+\frac{\partial g_{is}}{\partial y^j}-\frac{\partial g_{ij}}{\partial y^s}\Big)&=\frac{\partial y_i}{\partial x_\beta}\frac{\partial y_j}{\partial x_\beta}\frac{\partial y_r}{\partial x_\gamma}\frac{\partial y_s}{\partial x_\gamma}\frac{\partial^2x_\alpha}{\partial y_i\partial y_j}\frac{\partial x_\alpha}{\partial y_s}\\
&=\frac{\partial y_i}{\partial x_\beta}\frac{\partial y_j}{\partial x_\beta}\frac{\partial y_r}{\partial x_\gamma}\frac{\partial^2x_\gamma}{\partial y_i\partial y_j}
\end{align*}
This is identical with $-\Delta y^r$, as can be seen by differentiating $\sum_{a=1}^n\frac{\partial x^i}{\partial y^a}\frac{\partial y^a}{\partial x^j}=\delta_j^i$ with respect to $x_j$ and summing over $j$. The point is that, just by considering the standard laplacian $\sum\frac{\partial^2}{\partial x_i^2}$ and changing coordinates arbitrarily, we have arrived at
$$\Delta f=g^{pq}\frac{\partial^2f}{\partial y_p\partial y_q}-\frac{1}{2}g^{ij}g^{rs}\Big(\frac{\partial g_{js}}{\partial y^i}+\frac{\partial g_{is}}{\partial y^j}-\frac{\partial g_{ij}}{\partial y^s}\Big)\frac{\partial f}{\partial y_r}$$
with $g_{ij}$ and $g^{ij}$ as defined above. If you accept that the definition of $g_{ij}$ coincides with what we call "the Euclidean metric relative to the coordinate system $(y_1,\ldots,y_n)$" then I think you must accept that if we're given a general Riemannian metric and a function $f$, then the function
$$g^{pq}\frac{\partial^2f}{\partial y_p\partial y_q}-\frac{1}{2}g^{ij}g^{rs}\Big(\frac{\partial g_{js}}{\partial y^i}+\frac{\partial g_{is}}{\partial y^j}-\frac{\partial g_{ij}}{\partial y^s}\Big)\frac{\partial f}{\partial y_r}$$
is just as fundamental as the ordinary laplacian is on $\mathbb{R}^n.$ Actually, in a sense, the above calculation shows that this is true even if you're not given a general Riemannian metric, and are content to study $\mathbb{R}^n.$ In this sense the "Riemannian laplacian" is even more fundamental than Riemannian geometry.