3
$\begingroup$

In a paper I was reading, it was mentioned that if $M$ is a closed Riemannian manifold, then by fixing a basis for $L^2(M)$ consisting of eigenfunctions of the Laplacian, the space of smoothing operators on $L^2(M)$ can be identified with the algebra of matrices $a_{ij}$ such that

$$\sup_{i,j}i^k j^l |a_{ij}| <\infty$$

for all $k,l\in\mathbb N$.

Question 1: Could someone elaborate on how this identification can be done, or point me to a reference?

Question 2: Can this also be done if $M$ is non-compact?

$\endgroup$

1 Answer 1

2
$\begingroup$

This is, of course, a long story incorporating many strands but I will try to give a quick overview. Firstly, it is, as so often, convenient to skip to a more general framework. In your case, this would be that of an unbounded self-adjoint operator $T$ on Hilbert space (here that would be the Laplacian--more later).

One can associate with it a Frechet space $H^\infty(T)$ (the intersection of the domains of definitions of its powers) and a $DF$-space $H^{-\infty}(T)$ which are in duality. In the case of classical differential operators (the most frequent examples occur with the Laplacian and Schrödinger operators), the former is a space of test functions, the latter of distributions. This is a fairly direct consequence of the spectral theorem (in the form that any such operator can be represented as one of multiplication by a measurable function on an $L^2$-space).

If the spectrum of $T$ is discrete and consists of a sequence $(\lambda_n)$ of eigenvalues which are such that $|\lambda_n|$ is asymptotically like $n^\alpha$ for some positive $\alpha$, then the situation is particularly transparent. $H^\infty$ and $H^{-\infty}$ are a nuclear Frechet space and Silva space respectively. The eigenfunctions of $T$ form a basis for both spaces and they are identifiable with the sequence spaces $s$ and $s´$ of rapidly decreasing resp. slowly increasing sequences via coefficients.

The smoothing operators, i.e., continuous linear operators from $H^{-\infty}$ into $H^\infty$ are then identified with the two variable version of $s$ in a standard way.

It is classical that the Laplacian satisfies these condition in many cases, e.g.,on a closed manifold, a compact manifold (with suitable boundary conditions--Dirichlet or Neumann), as does the Schrödinger operator for suitable potential functions. The requisite estimates on the eigenvalues go under the generic name of Weyl inequalities.

You cannot, however, expect such results in the non-compact case. Here the Schrödinger operator is, perhaps, more appropriate.

$\endgroup$
2
  • $\begingroup$ Just to clarify, does this apply if $M$ is an open submanifold of a closed manifold? (I guess this should be similar to the case of a compact manifold with boundary with the Dirichlet boundary condition $f=0$ on the boundary.) $\endgroup$
    – geometricK
    Commented Aug 8, 2021 at 12:29
  • $\begingroup$ In the non-compact case, nothing of this nature works. $\endgroup$
    – memorial
    Commented Aug 11, 2021 at 15:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .