Reason for Continuous Spectrum of Laplacian 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? 

Consider what happens if you work with $-\Delta =-\frac{d^{2}}{dx^{2}}$ on $[-l,l]$, with periodic conditions. In this case, the spectrum consists of all multiples of the base eigenvalue $\lambda$ where $\lambda = \pi/l$. As $l\rightarrow\infty$, the spectrum grows increasingly dense in $\mathbb{R}$. In this sense, it is reasonable to expect that the spectrum might be all of $\mathbb{R}$.
