Assume $(M,g)$ is a compact Riemannian manifold without boundary, where $g$ is the Riemannian metric. Let $L:=-\Delta$ be the Laplace-Beltrami operator on $M$ defined by $\Delta \cdot = \text{div}(\nabla \cdot)$. I am reading in a lot of books/papers that the Laplace-Beltrami operator on a closed Riemannian manifold has positive, discrete spectrum whose eigenvalues accumulate at infinity. Here are many questions:

  1. What is the domain and range of $\Delta$ in order to have the spectrum described above? Does one treat $\Delta$ as a densely defined unbounded operator from $L^{2}(M)$ to $L^{2}(M)$ with domain $W^{1,2}(M)$ (or $W^{2,2}(M)$?) ? Or, does one think of $\Delta$ as a bounded operator? For example, from $W^{2,2}(M)$ to $L^{2}$.

  2. If one defines weak solutions of $\Delta$ by using the Green's identities $$ \int_{M}u\Delta v \text{Vol}_{g} = - \int_{M}g(\nabla u,\nabla v) \text{Vol}_{g} = \int_{M} v\Delta u \text{Vol}_{g},$$ hasn't the target to be some dual space then? Something like $(W^{1,2}(M))^{*}$?

  3. What is the precise formulation of spectrum and eigenvalues of $\Delta$, provided one knows the correct domain and range?

  4. Do you know any reference, where this is fully discussed? By that I mean, some reference where the domain, range, spectrum of $\Delta$ on $(M,g)$ is discussed?

Cheers, Martin

  • $\begingroup$ For some reason, the picture next to my profile name "MartinG" was changed. The picture showed some person who is not me. I then put it back to standard in the "edit profile". Why did this happen? $\endgroup$
    – MartinG
    Commented Jun 26, 2019 at 5:49
  • $\begingroup$ mathoverflow.net/questions/331870/… $\endgroup$ Commented Jun 26, 2019 at 6:18
  • $\begingroup$ Thanks for the link. But I am still confused about the domain and range of $\Delta$. What are they? In the link they only talk about "appropriate Sobolev spaces". What are these? If one want to write $\Delta$ as $\Delta : X \rightarrow Y$, what is then $X$ and $Y$? $\endgroup$
    – MartinG
    Commented Jun 26, 2019 at 6:23
  • $\begingroup$ If one knows the spaces $X$ and $Y$ what is then the definition of spectrum, eigenvalues and eigenfunctions. Notice that, in general, the spectrum is larger than the set of eigenvalues (point spectrum). Why is this for $\Delta$ not true? $\endgroup$
    – MartinG
    Commented Jun 26, 2019 at 6:27

1 Answer 1


There are various approaches. First one: consider $\Delta$ as an unbounded operator over $L^2(M)$ with domain $W^{2,2}(M)$. It is closed, densely defined and $-\Delta$ is self-adjoint, positive. There is a well-defined spectral theory for this class, which you should find somewhere in Reed & Simon. That the spectrum is discrete and accumulates at infinity follows from the fact that $-\Delta+1$ has a compact inverse.

Second one: the eigenvalues of $-\Delta$ are the critical points of the functional (Rayleigh ratio) $$I[u]=\frac{\int_M|u|^2\,{\rm Vol}_g}{\int_Mg(\nabla u,\nabla u)\,{\rm Vol}_g},$$ which is well-defined over $W^{1,2}(M)\setminus\{0\}$. Once again, you must use the compactness of the embedding $W^{1,2}(M)\subset L^2(M)$.

All in all, the way you obtain the eigenvalues is not important in the end, because once you have $-\Delta u=\lambda u$, elliptic regularity plus a bootstrap argument tell you that $u$ is $C^\infty$, hence is an eigenfunction in every sense that you might imagine.

  • $\begingroup$ Thank you very much. I will try to work out the details. $\endgroup$
    – MartinG
    Commented Jun 26, 2019 at 6:48

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