Self-adjoint extension of locally defined differential operators - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T04:30:29Z http://mathoverflow.net/feeds/question/15589 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15589/self-adjoint-extension-of-locally-defined-differential-operators Self-adjoint extension of locally defined differential operators Igor Khavkine 2010-02-17T15:58:53Z 2010-02-17T19:47:17Z <p>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.</p> <p>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:</p> <blockquote> <p>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 $C_0^\infty(U_i)$.</p> </blockquote> <p>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.</p> http://mathoverflow.net/questions/15589/self-adjoint-extension-of-locally-defined-differential-operators/15618#15618 Answer by Rafe Mazzeo for Self-adjoint extension of locally defined differential operators Rafe Mazzeo 2010-02-17T19:47:17Z 2010-02-17T19:47:17Z <p>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 &amp; Simon, Volume 2, I think).</p> <p>If $M$ has boundary or is otherwise noncompact or singular, symmetric elliptic operators are usually not essentially self-adjoint.</p>