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).



Essential self-adjointness of differential operators on compact manifolds
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.)


