# Elliptic operator with finite spectrum?

Is it possible for a (non-symmetric) elliptic differential operator to have finite spectrum. If so, is there an explicit example?

• What is the domain and codomain for your operator? – Bombyx mori Dec 23 '19 at 4:37
• @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. – Chris Judge Dec 24 '19 at 8:52
• 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. – Bombyx mori Dec 24 '19 at 20:50
• 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. – Bombyx mori Dec 24 '19 at 20:57
• @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. – Chris Judge Dec 25 '19 at 21:39

This is more a long comment than an answer since some details should be checked. The answer shoould be not for (say) a second order differential operators with continuous top order coefficients and bounded first and zero order ones in bounded domain (with some reularity on the boundary), under say Dirichlet bc. A very nice theorem in the second volume of Dunford Schartz (Theorem 29 pag 1115 and subsequent corollaries) implies that if the resolvent is of trace class and decays like $$1/|\lambda|$$ on two lines from the origin such that the smallest angle between them can be arbitrarily small, then the linear span of the generalized eigenfunctions is dense in $$L^2$$. In our situation the domain of the operator is $$H^{2}\cap H^1_0$$ and the resolvent to the power $$k$$ is trace class if $$k>d/2$$ ($$d$$ is the dimension). The rays above from the origin exist since the operator generates an analytic semigroup of angle $$\pi/2$$. I admit that many details are left...but it should work.