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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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4 Answers 4

up vote 6 down vote accepted

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?

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What a nice, simple answer. It should follow from this intuition that the spectrum on any compact manifold like $S^1$ is discrete and vice-versa. –  ff90 Dec 24 '13 at 12:25

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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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 $\mathbb{R}$.

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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.

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Yes, the Rellich-Kondrachov theorem fails for noncompact manifolds so that $\Delta$ does not admit an eigenfunction decomposition. For compact manifolds, this means that the spectrum is discrete. This is also a nice way to think of it. –  ff90 Dec 24 '13 at 12:28

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