# The reproducing kernel for harmonics on compact manifolds

Page 39, proposition 1.1.3 here, http://www.cis.upenn.edu/~cis610/sharmonics.pdf clearly explains how for every level" (the parameter $k$ in the proposition) one can construct a function ("kernel") such that the function space of all level $k$ harmonics on $S^n$ is the "Reproducing Kernel Hilbert Space" (RKHS) for that kernel. Because of the orthogonality property of the harmonics it probably wasn't surprising that the kernel itself also turned out to be completely described by the harmonics themselves at the same level.

• Firstly I want to confirm if my above interpretation is correct at all! I am inquiring if this proposition 1.1.3 in my cited reference can at all be see as constructing a RKHS. Because as far as I understand for their $F_k$ function in the reference to be considered as a "kernel" it has to converge pointwise and also maybe uniformly on compact sets. Is that true here?

• Is this something special about $S^n$ or are there other (compact?) manifolds too on which the Hilbert space of harmonic functions similarly decomposes so that the eigenspace of each eigenvalue acts as a RKHS for the kernel being constructed similarly? (and this kernel will again have the "right" (pointwise? uniformly on compact sets?) convergent properties?)

• Looking briefly at your link (which would be nicer to explain in your question...) it's not any whole space of "spherical harmonics", but only the finite-dimensional space of those of a fixed degree, which is to say those with a fixed eigenvalue for the invariant Laplacian on the sphere. On every compact Riemannian manifold, the eigenspaces for the Laplace-Beltrami operator are finite-dimensional, so the identity map (whose "kernel" is the "reproducing kernel") is a nice function of two variables... yes. :) ? – paul garrett May 4 '16 at 22:21
• Well, the identity map works on all manifolds as a kernel for which any function space would be its RKHS! So that I guess is trivial case! – Anirbit May 4 '16 at 22:30
• No, it's not at all a trivial issue: on infinite-dimensional spaces a sum like $K(x,y)=\sum_n f_n(x)\overline{f_n(y)}$ with orthonormal basis $\{f_n\}$ for $L^2$ is very rarely convergent pointwise or in $L^2$ either. For example, on the circle, the simplest compact manifold, this is $\sum_n e^{in(x-y)}$. Yes, this converges in a suitable Sobolev space, but not pointwise, certainly not uniformly pointwise, NOR in $L^2$. The idea of a reproducing kernel is that it converges pointwise to a considerable degree... – paul garrett May 4 '16 at 22:37
• I meant that your example of $K(x,y) = 1$ is probably a trivial kernel w.r.t which any function space will look like its RKHS. This kernel $K(x,y) = \sum_n f_x(x) \bar{f_n(y)}$ is point-wise (uniformly?) convergent for $S_n$ ? (and this K is what I guess is called as F in the notes attached) My question is about how much of these nice properties of $S^n$ have an analogue for other compact manifolds. – Anirbit May 4 '16 at 22:42
• Ah, it is not at all that $K(x,y)=1$ is a reproducing kernel for anything! Rather, $K(x,y)=\delta(x-y)$ with Dirac delta might be appropriate in some cases! The kernel I wrote for $S^n$, if you mean it to include all spherical harmonics, is definitely not convergent. So, while there are indeed "nice properties", their proper description is not so trivial. – paul garrett May 4 '16 at 23:12

First, the issue is not at all about "harmonic" functions on a compact manifold: the fact that eigenfunctions for the Laplace-Beltrami operator on the (round) sphere are restrictions of harmonic functions on an ambient (flat) Euclidean space is rather special, and certainly does not mean that the spherical harmonics are "harmonic" on the sphere itself. Depending on normalization, their eigenvalues for the actual Laplace-Beltrami operator are something like $-k(k+n-2)$ where $k$ is the degree and $n$ is the dimension of the ambient Euclidean space (not of the sphere). This is discussed in some form in Stein-Weiss, and many other places.

For fixed $n$, yes, the space of degree-$k$ spherical harmonics is finite-dimensional, and consists of smooth functions, so $K_k(x,y)=\sum_j f_j(x)\overline{f_j(y)}$ is a finite sum, so converges very well, and gives a reproducing kernel for that finite-dimensional eigenspace, yes.

On a compact Riemannian manifold, the Laplace-Beltrami operator has compact inverse/resolvent, so has finite-dimensional eigenspaces consisting of smooth functions. Thus, as with spherical harmonics, the functional "evaluate at point $z_o$" is continuous, and is given by inner product against an element of the eigenspace.

• Can you kindly explain what you mean by "does not mean that spherical harmonics are harmonic on the sphere itself"? For example the $Y_{lm}(\theta, \phi)$ functions on $S^2$ are eigenfunctions of the Laplacian of $S^2$ with eigenvalue $l(l+1)$? So why wouldn't you say that $Y_{lm}$s are harmonic functions on $S^2$? – Anirbit May 6 '16 at 16:26
• Usually "harmonic" means "eigenvalue 0 for the relevant Laplace-Beltrami operator". "Harmonic" does not include other eigenvalues than 0. – paul garrett May 6 '16 at 17:17
• Thanks! And like the sphere do other compact manifolds also have this property that their eigenvalues are degenerate with multiplicity going exponential in the manifold dimension? (which if true tells us that if I expand a function on a compact manifold using this basis of eigenfunctions of the Laplace operator then given any epsilon error target the number of terms I need to take is somehow lower bounded by a number going exponential in the dimension, right?) – Anirbit May 8 '16 at 20:32
• I think a "generic" compact manifold has multiplicity-free spectrum, but I do not know a precise assertion of this form. Round spheres are very degenerate, in some regards... – paul garrett May 8 '16 at 21:03
• Can you kindly give some examples? Of nice compact manifolds which do not have such a multiplicity? – Anirbit May 9 '16 at 12:37