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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?

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The usual explanation is that an eigenfunction must *locally* look like a sine wave, where the wavelength determines the eigenvalue. For $\mathbb R$, we can define sine waves with any wavelength we want, and therefore also any eigenvalue we want.

But in $S^1$, though we can *locally* imagine any wavelength, the wave will only fit together into a single-valued function *globally* if the wavelength divides the perimeter of the circle. And that’s why there’s only a discrete set of possible wavelengths/eigenvalues.

Or is that more elementary/specific than what you’re looking for?

I think the discreteness of the spectrum of the Laplacian only holds for compact manifolds. For general manifolds I know very little (Yau&Schoen as some discussion about it). This is usually proved when you use elliptic regularity results together with a certain type “embedding” theorem (http://en.wikipedia.org/wiki/Rellich%E2%80%93Kondrachov_theorem). It is not an embedding in the topological sense, but rather analytical such that preserves sequential compactness. Coupled with the fact that the unit ball in an infinite dimensional space is never compact this proved that the spectrum must be discrete(as it is finite dimensional). I do not know how to obtain similar without compactness assumption for open manifolds. Some non-trivial elliptic regularity arguments must be needed.

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 $\mathbb{R}$.

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.

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