TITLE: Reason for Continuous Spectrum of Laplacian
QUESTION [11 upvotes]: For the circle $S^1$, it is well-known that the Laplace-Beltrami operator $\Delta=\text{ div grad}$ has a discrete spectrum consisting of the eigenvalues $n^2,n\in \mathbb{Z}$, as can be seen from the eigenfunction basis $\{\exp(in\theta)\}$. 
This is not quite the case in $\mathbb{R}$; the spectrum of $\Delta$ there is $[0,\infty)$. This is because there is a family of "step" eigenfunctions that vary continuously and give out all the eigenvalues we need. But I was wondering, is there a more geometric reason (perhaps related to the properties of $\Delta$) as to why the spectrum is continuous in this case?

REPLY [4 votes]: One easy answer is that the real line is invariant under dilatins, which preserve the Laplacian. Any set that is scale-invariant in at least one direction has continuous spectrum.