Essential self-adjointness of differential operators on compact manifolds Let $L$ be a linear differential operator (with smooth coefficients) on a compact differentiable manifold $M$ (without boundary). Suppose $M$ is endowed with a smooth volume form (actually, a smooth volume density, if one wishes to consider the non orientable case), so that we can speak about the Hilbert space $L^2(M)$. I regard $L$ as being a densely defined operator on $L^2(M)$ with domain $C^\infty(M)$. Assume that $L$ is symmetric. Is it true that $L$ is essentially self-adjoint? If $L$ is elliptic then the answer is yes (one possible proof: the domain of the adjoint $L^*$ is the set of those $f\in L^2(M)$ such that $L(f)$ --- understood in the distributional sense --- is in $L^2(M)$ and $L^*$ is the restriction of the extension of $L$ to distributions. Let $f$ be an eigenvector of $L^*$ with eigenvalue $\pm i$. Then $f$ is a weak solution of $L(f)=\pm if$ and, by elliptic regularity, $f$ is smooth and it is therefore an eigenvector of $L$ with eigenvalue $\pm i$, contradicting the symmetry of $L$).
Naively speaking, absence of essential self-adjointness is related to the existence of several possible "boundary conditions", which do not exist for compact manifolds. So, naively, the result seems plausible. But maybe I'm being too naive.
Edit: The result is false and the counterexample suggested by Terry Tao works. Let $M=S^1=\mathbb{R}/2\pi\mathbb{Z}$ and $L=\frac{d}{dx}\sin(x)\frac{d}{dx}$. The symmetric operator $L$ is not essentially self-adjoint in $C^\infty(S^1)$. A non zero solution of $(L^*+i)\psi=0$ is obtained using Fourier series. Here are the details: set $a_0=0$, $a_1=1$ and $a_{k+2}=\frac{k}{k+2}a_k+\frac{2}{(k+1)(k+2)}a_{k+1}$ for $k\ge0$. It is easily proven by induction that the sequence $a_k$ is $O(k^{-2/3})$ and hence it is square integrable. The function $\psi(x)=\sum_{k=0}^\infty a_ke^{ikx}$ is hence in $L^2(S^1)$ and it solves $(L^*+i)\psi=0$ (because it solves $(L+i)\psi=0$ in the distributional sense).
 A: I would like to point out that, even if the classical dynamics is incomplete (with a large set of trajectories escaping in finite time), the quantum one might well be complete.
For example, consider on $M = \{(x,y) \in \mathbb{R}^2 \mid x >0 \}$ the Riemannian metric (Grushin metric)
$$ g = dx^2 + \frac{1}{x^2}dy^2 $$
It is not hard to prove that the Laplace-Beltrami $\Delta$ of the above metric, with domain $C^\infty_c(M)$ is essentially self adjoint (see [2]). Notice that in the above example there is no external potential: the confinement is purely geometrical.
On the other hand, the principal symbol of $\Delta$ is
$$ 2H = p_x^2 + x^2p_y^2 $$
and it can be prolonged smoothly on the whole $\mathbb{R}^2$, where it gives origin to a complete dynamic. In this case, essentially all trajectories starting at points with $x>0$ cross at some time the singular region $\{x=0\}$ (the only exception are trajectories with $p_y=0$ and $p_x>0$). More importantly, they do so without losing optimality (in the sense that all these trajectories are shortest paths for the Riemannian metric).
This is a particular instance of a more general fact (i.e. quantum completeness) for non-complete Riemannian structures satisfying suitable conditions at the metric boundary, as proved recently in [1]. 
Let me mention that quantum completeness for two-dimensional almost-Riemannian structures (a class of which the Grushin metric above is part) has been proved originally in [2], using the normal forms for 2D almost-Riemannian structures. 
A conjecture is still open for more singular situations (i.e. non-regular almost-Riemannian structures in the language of [1]).
A: My guess here is that the answer should be negative, because the answer to the corresponding classical problem is negative.  Namely, there exist symmetric differential operators L such that the Hamiltonian flow associated to the symbol is not complete.  For instance, consider a symmetric operator with principal symbol $-\sin(x) \frac{d^2}{dx^2}$ on the circle ${\bf R}/2\pi{\bf Z}$; the symbol here is $\sin(x) \xi^2$, leading to the Hamiltonian flow $\dot \xi = \cos(x) \xi^2$, $\dot x = - 2\sin(x) \xi$, which exhibits Ricatti type blowup in finite time along the $x=0$ axis.
This is not quite a rigorous argument, as I haven't actually ruled out the possibility that unitary propagators $e^{itL}$ still somehow exist, but the fact that at least one semiclassical trajectory blows up makes that possibility quite remote, in my view.  (Presumably one can modify the example so that a positive measure set of trajectories blow up, which would be a more convincing piece of evidence towards non self adjointness.)
