In the well-known book "*THE PRINCETON COMPANION TO MATHEMATICS*" page 296, it is indicated that the spherical harmonics are the EIGENVECTORS of the Beltrami operator. In the document [Spectral Geometry in Non-standard
Domains](https://www.fields.utoronto.ca/programs/scientific/14-15/summer-research14/mini-conf/SpectralGeometryReport.pdf) page $39$, they use that concept, but I don't know why it works. Could someone be able to explain to me why the point $7$ of page $39$ is sufficient? In other words, why it is sufficient the use the spherical harmonics to find the spectrum of the sphere? **A simple example could be appreciate to understand why it works.**


Thanks for your help!



The spherical harmonics are the EIGENVECTORS of Beltrami operator
The article linked by the OP proves that every degree $k$ spherical harmonic is a $k(k+1)$-eigenfunction of the Laplacian on $S^2$. The OP asks: is every eigenfunction of the Laplacian a spherical harmonic?


The answer is yes, though it seems that neither the linked article nor the Wikipedia page on spherical harmonics address this issue.


It is not hard to show that the spaces $H^k$ of degree $k$ spherical harmonics are mutually orthogonal, so it suffices to prove that the orthogonal direct sum $H$ of the $H^k$'s over all $k$ is dense in $L^2(S^2)$. (Proof: if $f \in L^2(S^2)$ is an eigenfunction not in $H$ then the span of $f$ is orthogonal to all $H^k$ which implies that $H$ is not dense.)


To prove that $H$ is dense in $L^2(S^2)$, note that every continuous function on $S^2$ can be expressed as the uniform limit of restrictions to $S^2$ of polynomials by the Stone-Weierstrass theorem. Every polynomial on $S^2$ is the sum of homogeneous polynomials and every homogeneous polynomial is the sum of harmonic polynomials, so since spherical harmonics are just the restriction of harmonic polynomials to $S^2$ it follows that $H$ is dense in $C(S^2)$. But $C(S^2)$ is dense in $L^2(S^2)$, so we're done.


There is probably a shorter and more sophisticated argument using representation theory (the Peter-Weyl theorem), but in the end it would probably be an elaborate reformulation of the argument above.


