# 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?

-

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?

-
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.

-
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 $\mathbb{R}$.
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