TITLE: Elliptic operator with finite spectrum?
QUESTION [6 upvotes]: Is it possible for a (non-symmetric) elliptic differential operator to have finite spectrum. If so, is there an explicit example?

REPLY [2 votes]: 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.