# Eigenvalues and Domain of the Laplace-Beltrami Operator

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

• 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? – MartinG Jun 26 '19 at 5:49
• mathoverflow.net/questions/331870/… – Francois Ziegler Jun 26 '19 at 6:18
• 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 $Y$? – MartinG Jun 26 '19 at 6:23
• 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? – MartinG Jun 26 '19 at 6:27

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.