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 $H = L^2(M, \mathbb{R})$ 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.

$\begingroup$ Since $L$ is a $G$invariant operator, doesn't Schur's lemma tell us that $L$ acts on each irrep appearing in $H$ by scalar multiplication? $\endgroup$ – Faisal Nov 21 '10 at 7:50

$\begingroup$ @Faisal: This says that each irreducible is a subrepresentation of some eigenspace, but it doesn't say that the an eigenspace could not contain several irreducibles. $\endgroup$ – Dick Palais Nov 21 '10 at 8:06

1$\begingroup$ @Dick: Ah, sorry  I misinterpreted your question. I agree with Evan that it's very rare to have all the eigenspaces of $L$ be irreducible. For example if $M=G$, then an eigenspace of $L$ contains a given irrep only if it contain all copies of that irrep in $L^2(G)$. It follows that if all the eigenspaces of $L$ are irreducible then every irrep of G must appear without multiplicity in $L^2(G)$. In particular, because each irrep occurs with multiplicity equal to its degree, this means that if $G$ isn't abelian then there is at least one eigenspace of $L$ that isn't irreducible. $\endgroup$ – Faisal Nov 21 '10 at 9:03

$\begingroup$ This question (and answer) seem to be about normal homogeneous spaces, i.e. homogeneous spaces equipped with a biinvariant metric coming from the Lie group. I don't think the picture for the general compact homogeneous space is so nice. Or is it? $\endgroup$ – Chris Judge Apr 12 '16 at 15:35
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.

$\begingroup$ @Anton Deitmar, Evan Jenkins: I am still a little unclear about the connection between the Laplacian $L$ and the Casimir operator(s). I think that your remarks about eigenvalues being determined by the Killing form on highest weight vectors refers to the Casimir operator, which is purely group theoretic, whereas $L$ is the usual Riemannian Lapalcian. Of course they are related (and no doubt have the same symbol) but I don't think that they are the same. Do either of you (or does someone else) know where the relation between them is discussed. (Broecker and tom Dieck only treat a special case. $\endgroup$ – Dick Palais Nov 22 '10 at 0:36

1$\begingroup$ @Dick1: Any invariant positive definite bilinear form $B$ on the Lie algebra $\mathfrak g$ of $G$ gives an invariant metric on the quotient $G/K$. Being nondegenerate, this form on the one hand identifies $\mathfrak g$ with its dual ${\mathfrak g}'$, on the other hand it is itself an element of ${\mathfrak g}'\otimes\mathfrak g'\cong{\mathfrak g}\otimes{\mathfrak g}$ The latter space maps naturally to $U=U({\mathfrak g})$, the universal enveloping algebra. So $B$ induces an element in $U$, which is called the Casimiroperator $C_G$. $\endgroup$ – user1688 Nov 22 '10 at 18:05

1$\begingroup$ @Dick2: This Casimiroperator acts on functions on $G/K$ as a differential operator which happens to coincide with the Laplaceoperator induced by the metric. This is no wonder, since the metric and the Casimir are induced by the same invariant form. $\endgroup$ – user1688 Nov 22 '10 at 18:06

$\begingroup$ Thanks Anton. That more or less answers what I wanted to know. (Though I still do not see what happens when the isotropy action of $K$ is not irreducible, so there is not a unique $G$invariant metric on $M$.) $\endgroup$ – Dick Palais Nov 22 '10 at 22:46
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.

1$\begingroup$ Please point to a specific result in that paper that addresses the exact question that was asked. (Keep in mind that the poster, Richard Palais, is very highly accomplished, so that a paper merely in the general neighborhood of the question would not be onpoint for this question; sifting through it for the answer could be a waste of time.) $\endgroup$ – Todd Trimble♦ Nov 4 '16 at 14:00