I was writing up some notes on harmonic analysis and I thought of a question that
I felt I should know the answer to but didn't, and I hope someone here can help me.
Suppose I have a compact Riemannian manifold $M$ on which a compact Lie group $G$
acts isometrically and transitivelyso you can think of $M$ as $G/K$ for some closed
subgroup $K$ of $G$. Then the real Hilbert space $H = L^2(M, R)$ is an orthogonal
representation space of $G$ and hence splits as an orthogonal direct sum of finite
dimensional irreducible subrepresentations. On the other hand, the Laplacian $L$
of $M$ is a selfadjoint operator on $H$, so $H$ is also the orthogonal direct sum
of its eigenspaceswhich are also finite dimensional. My question is, when do these
two orthogonal decompositions of $H$ coincide? Put slightly differently, since $L$
commutes with the action of $G$, each eigenspace of $L$ is a finite dimensional
subrepresentation of $H$ and so a direct sum of irreducibles, and I would like to
know conditions under which each eigenspace is in fact irreducible. For example, this
is true for the circle acting on itself and for $SO(3)$ acting on $S^2$ (where we
get the harmonic polynomials of various degrees). Is it perhaps always true for the
case of a symmetric space? Of course a standard reference in addition to the answer
would be most welcome.



The PeterWeyl theorem tells you that $L^2(G)$ is isomorphic to $\bigoplus_{\pi}\pi\otimes\pi^*$ as $G\times G$ representation, where $\pi$ runs through all irreducible unitary representations. It follows that $$ L^2(G/K)\cong L^2(G)^K\cong\bigoplus_\pi \pi\otimes(\pi^*)^K. $$ So, the first thing you absolutely need, is a multiplicity one property, which says that $\dim\pi^K\le 1$ for every $\pi$. This is already a rare property, but known to be true for, say $G=SO(n)$ and $K=SO(n1)$, see Zhelobenko's book for this. But, the Laplacian may have the same eigenvalue on different representations. For this you need highest weight theory (see for instance the book by Broecker and tom Dieck): Assume $G$ to be connected. The irreducible representations are parametrized by highest weights and the Laplace eigenvalue depends on the value of a quadratic form on the space of weights. So, in each case you need to identify those weights with $K$invariants and consider the values of the quadratic form, which in the case of a simple group should be the Killing form. I guess that in the above cases it might actually be true. 


For G/K symmetric the joint eigenspaces of the Ginvariant differential operators on G/K are all irreducible. Also each irreducible subspace of H has multiplicity bounded by one. For this see my "Groups and Geometric Analysis?" Ch. V Theorems 4.3 and 3.5. Concerning the Laplace Beltrami operator L, the Casimir operator on G (if semisimple) does induce L on G/K (loc. cit. p.331). If G/K is two point homogeneous the Ginvariant differential operators on G/K are all polynomial in L (loc. cit. p/288) so for these spaces the answer to Dicks question is yes. For G/K not symmetric Theorem 3.5 p. 533 still gives a decomposition of H into spaces spanned by representation coefficients which are eigenfunctions of the Casimir operator. Assuming the metric on G/K, (G semisimple) is coming from the Killing form Riemannian structure on G it is still true that the geodesics through the origin in G/K are orbits of one parameter subgroups of G. It seems to me that the argument for Problem A4 p.568 should still show that the Casimir operator on G will induce the Laplace Beltrami operator on G/K. Therefore the decomposition in Theorem 3.5 p.533 should still be a decomposition into eigenfunctions of the Laplacian. But there is no reason to expect irreducibility. 


Shouldn't this only happen very rarely? $S^1$ is abelian, and $SO(3)$ acting on $S^2$ involves inducing from the maximal torus, so in both these cases, every irreducible appears once. But in general (i.e., if $K$ does not contain a maximal torus), irreducible representations will appear more than once, in which case there's no hope for the eigenvalue of the Laplacian to separate them. Even when irreducibles don't appear multiple times, the eigenvalue of the Laplacian is not generally enough to separate two irreducibles if the rank of the group is bigger than 1. 

