Independence of rotated spherical harmonics

Hi,

Consider a spherical harmonic of degree $l$, denoted by $y_l^m$. I rotate this harmonic using $2l+1$ different rotations. The set of functions I get is not an orthogonal set, but the functions are still harmonics. The question is: does this set still spans the entire space of spherical harmonics of degree $l$ ?

My intuition is that it almost always does, but I can't say what are the non trivial configurations where it does not.

Thanks a lot for any help! Cyril

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It seems (because of $2\ell+1$) that your harmonic polynomials depend on three variables $x_1,x_2,x_3$. Right ? – Denis Serre Oct 28 '11 at 8:49
you can see them as harmonic polynomials of 3 variables restricted to the unit sphere. – Cyril Soler Oct 28 '11 at 9:39

Let $G$ be the isometry group of a polyhedron (tetrahedron, ..., icosahedron), its order being $2n$. The natural representation of $G$ over the space $H_\ell$ of harmonic polynomials of degree $\ell$ may or may not be irreducible. It is certainly not if $2\ell+1\ge\sqrt{2n}$. Thus let us take $(n/2)^{1/2}\le l\le n.$ Because the representation is reducible, there exists a strict invariant subpace, thus a non-zero $P\in H_\ell$ such that the set of $P\circ R$ with $R\in G$ does not span $H_\ell$. Because $|G|>2\ell+1$, this is a counter-example.
Update. Suppose that the representation of $G$ over $H_\ell$ admits an irreducible component of multiplicity $\ge2$ (I suspect that there are exemples; does somebody knows one?). Then there does not exist a spherical harmonics $P$ such that the $P\circ R$ span $H_\ell$ when $R$ covers $G$. This is because we may decompose $H_\ell=F\oplus^\bot K$ with $K$ irreducible component and $P\in F$.
We're talking about spherical harmonics specifically. The dimension of the space of spherical harmonics of degree $l$ is $2l+1$. I don't understand in what way your answer can be helpful to me. – Cyril Soler Oct 28 '11 at 9:45
the sum of the squares of the degrees of irreducible representation of $G$ is $|G|$. Thus an I.R. cannot have a large degree. In my construction, the representation cannot be irreducible, so there is a non-trivial invariant subspace $E$. If you take $P\in E$, the set of $P\circ R$, with $R\in G$ is in $E$, thus does not span $H_\ell$. – Denis Serre Oct 28 '11 at 12:48
So it means that if I take my $2l+1$ rotations to be distinct elements of the isometry group of a polyhedron, the rotated spherical harmonics will not be independant? – Cyril Soler Oct 28 '11 at 13:15
Yes and no: for some choice of the first spherical harmonics, the rotated ones will not be independent. Some' becomes every' if the group $G$ is finite (polyhedron isometry grpoup) and its representation over $H_\ell$ admits an irreducible component of multiplicity $\ge2$. – Denis Serre Oct 28 '11 at 14:44