Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
    
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

1 Answer 1

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 $P\in F$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.