Timeline for Elliptic operator with finite spectrum?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 11, 2020 at 10:53 | answer | added | Giorgio Metafune | timeline score: 3 | |
| Dec 26, 2019 at 20:17 | comment | added | Bombyx mori | If $A$ has finite many eigenvalues (ignore zero for convenience) then $A^{s}$ cannot have a simple pole, as it would be a finite sum. | |
| Dec 26, 2019 at 13:22 | comment | added | Chris Judge | @Bombyx mori. And how is the result on page 302 related to the spectrum of $A$? | |
| Dec 26, 2019 at 0:17 | comment | added | Bombyx mori | I do not mean the results in section 2. I meant the discussion in page 295, where the normal condition is not assumed. This is the only condition the result in page 302 is built on, which I alluded earlier. Of course if $A^{s}$ has a pole, then the spectrum goes to infinity. It seems to me the notion of "uniform elliptic" is not used in literature anymore. | |
| Dec 25, 2019 at 22:13 | comment | added | Chris Judge | @Bombyx mori. By Seeley's paper, I guess you mean 'Complex powers...' Section 2 of this paper is titled `relation to eigenvalues'. Here he assumes that the operator is normal and hence has a complete orthonormal set of eigenfunctions. In particular, the operator has infinite spectrum. Normal is a slight generalization of symmetric. I am looking for an operator such that the set of eigenfunctions is not complete and which has no essential spectrum. | |
| Dec 25, 2019 at 21:55 | comment | added | Bombyx mori | You are right one does not need positivity; but I do not know what is the weakest condition that would force it to have infinite spectrum. Thanks for the comment. | |
| Dec 25, 2019 at 21:47 | comment | added | Bombyx mori | I was checking Seeley's paper, so the notation I borrowed there may be dated. I was mainly thinking about how to detect the spectrum of $A$ from working with the complex power $A^{s}$. This is probably overkill, though. | |
| Dec 25, 2019 at 21:39 | comment | added | Chris Judge | @Bombyx mori. I am unsure of what you have written. Symmetry is enough to obtain infinitely many eigenvalues. One doesn't need positivity. In the literature there are results that state that, for example, Weyl's law holds if the elliptic operator is not far from symmetric e.g. principal symbol symmetric but lower order terms not. The resolvent being compact does not immediately imply that the spectrum is infinite. A non-symmetric compact operator can have finite spectrum. | |
| Dec 24, 2019 at 20:57 | comment | added | Bombyx mori | So here is the set up: Let $M$ be a $C^{\infty}$ manifold of dimension $n$, and let $A$ be an elliptic operator of order $m>0$ whose top order symbol $a_m$ satisfy that all the eigenvalues $\lambda$ of $a_m$ is in the region $|\arg(\lambda)-\theta|<\delta,\forall x\in U, |\xi|=1$. Then the pole of $A^{s}$ only happens at $s=\frac{k-n}{m}$ and they are simple. So with uniform ellipticity we can exclude the case of finite spectrum. However I still do not know how to construct a counter-example. | |
| Dec 24, 2019 at 20:50 | comment | added | Bombyx mori | At first the question strikes me as odd, but now I think about it is actually not trivial. It seems the infinite spectrum only holds for positive-definite operators. I think you are essentially asking how to come up with a non-trivial example. | |
| Dec 24, 2019 at 8:52 | comment | added | Chris Judge | @Bombxy mori. Smooth functions on some open subset of Euclidean space (or manifold) should be a core for the differential operator. Otherwise, I am open to any construction. | |
| Dec 23, 2019 at 4:37 | comment | added | Bombyx mori | What is the domain and codomain for your operator? | |
| Dec 22, 2019 at 14:05 | review | Close votes | |||
| Dec 24, 2019 at 0:27 | |||||
| Dec 22, 2019 at 13:06 | history | asked | Chris Judge | CC BY-SA 4.0 |