Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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).

share|improve this question
2  
By the way, the answer is also affirmative if $L$ is first-order. In that case, $L=-i\big(X+\frac12\mathrm{div}(X)\big)$, with $X$ a smooth vector field in $M$. Since $M$ is compact, $X$ is complete and we obtain $L$ as the generator of a one-parameter unitary group $U_t(f)=(f\circ F_t)\sqrt{\det\mathrm{d}F_t}$, which leaves $C^\infty(M)$ invariant. Thus $L$ is essentially self-adjoint on $C^\infty(M)$. –  Daniel Tausk Nov 23 '10 at 19:33
    
($F_t$ denotes the flow of $X$). –  Daniel Tausk Nov 23 '10 at 19:33
    
Correcting small imprecision of my previous comment: if $L$ is first-order (symmetric) then $L=-i\big(X+\frac12\mathrm{div}(X)\big)+V$, with $V$ the multiplication operator by a smooth real-valued function. Since $V$ is bounded self-adjoint, the conclusion is the same... –  Daniel Tausk Nov 24 '10 at 4:15
add comment

1 Answer 1

up vote 8 down vote accepted

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.)

share|improve this answer
    
Just to make sure I understand what you are talking about (and maybe I don't), when you say "corresponding classical problem" you mean "classical" as in "classical versus quantum", with a quantization rule like replacing $\xi$ with $-i\frac{d}{dx}$ in the classical Hamiltonian? If that's the case, is there a mathematical reason for believing that the two problems (the "classical" and the "quantum") are related or are you just relying on insights coming from physics? –  Daniel Tausk Dec 15 '11 at 18:45
    
By the way, if you take $L=\frac{d}{dx}\sin(x)\frac{d}{dx}$ (which is the simplest example of a symmetric operator with principal part $\sin(x)\frac{d^2}{dx^2}$) then $L$ surely admits one self-adjoint extension (because its coefficients are real functions) and such a self-adjoint extension does generate a unitary propagator $e^{itL}$. I don't know, though, whether $L$ is essentially self-adjoint. Since $L$ is quite simple I maybe able to figure that out "by hand". –  Daniel Tausk Dec 15 '11 at 18:53
    
Yes, this is what I mean by classical vs. quantum. Intuition from both physics and microlocal analysis (e.g. Egorov's theorem on pseudodifferential operators) suggests that $e^{itL}$ should approximately propagate in phase space along the Hamiltonian flow of the symbol (i.e. the quantum flow approximates the classical flow). If the classical flow is incomplete, this suggests (though does not quite prove) that one has quantum incompleteness also. –  Terry Tao Dec 15 '11 at 19:33
    
It's true that there will be an abstract self-adjoint extension of real differential operators such as $\frac{d}{dx} \sin(x) \frac{d}{dx}$ (because the deficiency indices must match), but the flow maps $e^{itL}$ associated to such extensions have no reason to preserve the initial domain of the operator, and thus need not solve the associated PDE in any reasonable sense (e.g. distributional sense). As such I would consider these abstract propagators to be "unphysical". –  Terry Tao Dec 15 '11 at 19:41
    
Right. Thanks. Though, thinking a bit more about this, there are very simple examples in which classical trajectories die in finite time, while the quantum problem is complete. Say, one electron in Coulomb potential: several classical trajectories die in finite time with the electron falling into the singularity (say, if the electron starts at rest), while the corresponding quantum Hamiltonian is essentially self-adjoint in $C^{\infty}_c(\mathbb{R}^3)$. –  Daniel Tausk Dec 15 '11 at 20:22
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.