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Let $M$ be a compact smooth manifold, $(\lambda_n)_{n=1}^{\infty}$ be a square-summable monotonically increasing sequence of non-negative numbers, and let $(f_k)_{k=1}^{\infty}$ be continuous functions on $M$.

Does there exist/can one construct:

  • A smooth Riemannian metric $g$ on $M$,
  • A self-adjoint (wrt. a positive density $\mu$ on $(M,g)$) second-order elliptic operator $T$ on $(M,g)$ with smooth coefficients.

Such that the eigendecomposition of $T$ on $L^2(M,g)$ is $\{(f_k,\lambda_k)\}_{k=1}^{\infty}$?

If not, are there known sufficient conditions on $\{f_k,\lambda_k)\}_{k=1}^{\infty}$ for this to be possible?

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  • $\begingroup$ I’m a bit confused by this question since typically the eigenvalues of the Laplacian/Schrödinger operator are taken to be increasing to infinity, not decreasing to zero. Are you using a different convention for eigenvalues? $\endgroup$ Commented Sep 13, 2023 at 17:12
  • $\begingroup$ Ah yes I was using the decreasing convention, seen in some approximation literature. But I'm happy to update to what's common in pdes :) $\endgroup$ Commented Sep 13, 2023 at 17:57
  • $\begingroup$ So the convention is that we have $\lambda_k \Delta_\mu f_k = -f_k$? $\endgroup$ Commented Sep 13, 2023 at 18:00
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    $\begingroup$ If you were just asking about the Laplace-Beltrami operator and not a general self-adjoint 2nd-order elliptic operator, then the Weyl asymptotic law gives you some pretty hard constraints on the asymptotic behavior of the eigenvalues. As for the eigenfunctions, elliptic regularity tells you that they can't just be any continuous function. There are probably more constraints from nodal domain-type theorems too. $\endgroup$ Commented Sep 13, 2023 at 18:18

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In general, this will not be possible. For instance, the first non-trivial eigenfunction will have two nodal domains, so $f_1$ cannot be arbitrary. Furthermore, you expect to have some sort of gradient-type estimate which prevent low-frequency eigenfunctions from oscillating too wildly. As such, you will need to restrict the functions $f_k$ quite a bit.

However, even if you don't specify the eigenfunctions, the eigenvalues $\lambda_k$ will need to satisfy a version of Weyl's law, and thus their asymptotic behavior cannot be arbitrary. For instance, there is a result of Hormander which essentially shows that if the potential is smooth

$$N_\mu(\lambda)=(2 \pi)^{-n} \omega_n \operatorname{Vol}_g(M) \lambda^n+O\left(\lambda^{n-1}\right) . $$

where $N_\mu(\lambda)$ is the number of eigenvalues (counted with multiplicity) less than $\lambda$.

In general, it is possible to take a list of finitely many (non-repeating) numbers $\{\lambda_k\}_{k=1}^N$ and find a Riemannian metric where the bottom of the spectrum is exactly the values desired. There are some bounds on the possible degeneracy of eigenvalues, but if the eigenvalues are simple there is no issue. Here is a recent paper of Xiang He which gives a good overview of the current state of this problem. https://arxiv.org/pdf/2308.04190.pdf

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  • $\begingroup$ In the finite case, can one also specify the eigenfunctions in the bottom part of the spectrum? $\endgroup$ Commented Sep 13, 2023 at 18:59
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    $\begingroup$ There are many restrictions on the eigenfunctions of self-adjoint operators. Apart from the nodal theorems, the $f_k$ cannot have any positive minima or negative maxima. But if you choose suitable functions $f_k$, then I don't know whether it is possible or not to prescribe the eigenfunctions. My guess is that this is not possible, but I don't know a rigorous argument offhand. $\endgroup$ Commented Sep 13, 2023 at 19:08
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    $\begingroup$ It seems like the nodal domain theorems are going to be quite restrictive. I'm wondering if you get an interesting and answerable question by just focusing on the spectrum and are willing to admit Schroedinger operators with possibly unbounded or non-smooth potentials. $\endgroup$ Commented Sep 14, 2023 at 3:44
  • $\begingroup$ That would be an interesting question, especially if one takes the spectrum to be simple so that there are no degeneracy bound issues. $\endgroup$ Commented Sep 14, 2023 at 17:33
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Just to show how overdetermined your general request is. It has be shown that in low dimensions the sphere with standard metric is uniquely determined by the spectrum of the Laplace-Beltrami operator. This is a result due to Tanno in 1980 (S. Tanno, Eigenvalues of the Laplacian of Riemannian manifolds)

In other words, if you start out with the eigenvalues of the standard sphere (and no restrictions on the functions) than the sphere of correct dimension is the only manifold that will admit a metric such that the Laplace-Beltrami operator has exactly these eigenvalues and the standard metric is the only such metric.

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    $\begingroup$ Has this really been shown for second-order elliptic operators or only for the Laplace-Beltrami operator? $\endgroup$ Commented Sep 14, 2023 at 8:47
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    $\begingroup$ @JochenGlueck I found the old paper with the original result, which is only Laplace-Beltrami. I don't know whether it has been generalized since then. Edited accordingly. $\endgroup$ Commented Sep 14, 2023 at 9:11

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