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

The following is well known. Given a symmetric differential operator, like $\partial_x^2$, defined on smooth functions of compact support on $\mathbb{R}$, $C_0^\infty(\mathbb{R})$, one can count the number of independent $L^2$-normalizable solutions of $\partial_x\pm i$ and use the von Neumann index theorem to classify possible self-adjoint extensions of this operator on $L^2(\mathbb{R})$. This can be generalized to more complicated differential operators, to $\mathbb{R}^n$ as well as bounded open subsets thereof.

On the other hand, suppose that I have a manifold $M$ that is covered by a set of open charts $U_i$ with differential operators $D_i$ defined in corresponding local coordinates. It is easy to check if the $D_i$ are restrictions of a globally defined differential operator $D$ on $M$: the transition functions on intersections of charts $U_i\cap U_j$ must transform $D_i$ into $D_j$ and vice versa. Suppose that is the case and that I am interested in self-adjoint extensions of $D$ to $L^2(M)$ (supposing that an integration measure is given and that $D$ is symmetric with respect to it). Now, the question:

Is there way of classifying the self-adjoint extensions of $D$ on $L^2(M)$ in terms of its definition in local coordinates, the actions of $D_i$ on $C_0^\infty(U_i)$.

A simple example would be the cover of $S^1$ by two overlapping charts. I know that a self-adjoint extension of $\partial_x^2$ on $[0,1]$ with periodic boundary conditions gives the naturally defined self-adjoint Laplacian on $S^1$. Then $(0,1)$ is interpreted as a chart on $S^1$ that excludes one point. However, I don't know how to define the self-adjoint Laplacian on $S^1$ if it's given on two overlapping charts.

share|improve this question
    
In Shubin' papers, there is an opposite, that what he has said Mazzeo on If M has boundary or is otherwise noncompact or singular, symmetric elliptic operators are usually not essentially self-adjoint. It is enough to take M with a bounded geometry. Mohammed (mlaidi2000@yahoo.fr) –  user35664 Jun 25 '13 at 20:53
    
(a) Which papers of Shubin (b) can you please re-write "that what he has said Mazzeo on if $M$...", it is really hard to read as written (c) this probably should be just a comment, instead of an answer –  Willie Wong Jun 26 '13 at 10:24
    
@Willie Wong, Mazzeo has said " If $M$ has boundary or is otherwise noncompact or singular, symmetric elliptic operators are usually not essentially self-adjoint." –  user35664 Jun 26 '13 at 20:35
    
Regarding the Shubin's Papers see, e.g., Discreteness of spectrum for the Schrödinger operators on manifolds of bounded geometry" Published in The Maz’ya Anniversary Collection Operator Theory: Advances and Applications Volume 110, 1999, pp 185-226 Mohammed mlaidi2000@yahoo.fr –  user35664 Jun 26 '13 at 20:39
add comment

1 Answer

Yes, if the manifold is compact without boundary, and the operator $D$ is elliptic (equivalently, if each of the $D_i$ are elliptic), and if $D$ is symmetric with respect to some smooth positive density, then $D$ is essentially self-adjoint, i.e. extends uniquely from the core domain $\mathcal C^\infty$ to an unbounded self-adjoint operator on $L^2$ of the manifold $M$. The way to prove this is as follows. If $D$ has order $m > 0$, then the maximal domain of $D$ is equal to $H^m(M)$; this is proved by (local!) elliptic regularity. On the other hand, every element in this maximal domain is approximable in the graph norm by smooth functions, again this can be reduced to a local calculation in each chart. Hence the minimal and maximal domains agree (for more on this, see Reed & Simon, Volume 2, I think).

If $M$ has boundary or is otherwise noncompact or singular, symmetric elliptic operators are usually not essentially self-adjoint.

share|improve this answer
    
Rafe, thanks for the reply. I can certainly see how that a locally defined, elliptic operator has a unique self-adjoint extension on a compact, closed manifold $M$. However, I'm also interested in the case where $M$ is not compact or has a boundary. For instance, take each $U_i$ diffeomorphic to $(0,1)$ and $D_i=\partial_x^2$. The von Neumann index calculation is the same for each $U_i$. However, I could patch such charts together to make $S^1$, $\mathbb{R}$, or $[0,\infty{[}$ by throwing in $0$. I'm interested in figuring out possible global self-adjoint extensions from the chart patching. –  Igor Khavkine Feb 18 '10 at 11:13
add comment

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.