most recent 30 from http://mathoverflow.net 2013-05-22T01:05:01Z http://mathoverflow.net/feeds/question/79360 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79360/independence-of-rotated-spherical-harmonics Cyril Soler 2011-10-28T07:48:36Z 2011-10-28T14:50:09Z

<p>Hi,</p> <p>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 $l$ ?</p> <p>My intuition is that it almost always does, but I can't say what are the non trivial configurations where it does not.</p> <p>Thanks a lot for any help! Cyril</p>

http://mathoverflow.net/questions/79360/independence-of-rotated-spherical-harmonics/79368#79368 Denis Serre 2011-10-28T09:02:13Z 2011-10-28T14:50:09Z

<p>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.</p> <p><strong>Update</strong>. 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 $P\in F$.</p>