Spectrum of the Laplace–Beltrami operator

Remarks

The properties follow from the spectral theorem for compact self-adjoint operators applied to the inverse of $\Delta + 1$, which is compact on $L^2(M)$ by elliptic regularity and Rellich's theorem. So $\lambda_j \to \infty$, and the normalized eigenfunctions form an orthonormal basis of $L^2(M)$. The eigenvalue $0$ has multiplicity $1$ because $M$ is connected: $\int_M \phi \Delta \phi \, dv = \int_M \abs{\nabla \phi}^2 dv$ vanishes only for constant $\phi$. Some authors use the opposite sign $\Delta = \operatorname{div} \operatorname{grad}$, with eigenvalues $\le 0$.

Example: for the round unit sphere $S^2$, the eigenvalues are $k(k+1)$, $k = 0, 1, 2, \ldots$, with multiplicity $2k + 1$, the eigenfunctions being the spherical harmonics of degree $k$. So the spectrum begins $0, 2, 2, 2, 6, 6, 6, 6, 6, \ldots$.

Non-example: $\R$ is not compact, and the definition fails there. The equation $-\phi'' = \lambda \phi$ has bounded solutions $e^{\pm i\sqrt{\lambda} x}$ for every $\lambda \ge 0$, but none of them is in $L^2(\R)$. The spectrum of the Laplacian on $L^2(\R)$ is the continuous set $[0, \infty)$, not a sequence of eigenvalues.