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Recall that a function is harmonic if its Laplacian is zero. Let $\mathrm{Harm}(n,k)$ denote the vector space of $n$-variate harmonic polynomials that are homogeneous with degree $k$. When working with spherical harmonics, we endow this vector space with the inner product $\langle\cdot,\cdot\rangle$ defined in terms of the uniform probability measure over the unit sphere $\mathbb{S}^{n-1}$. Many proofs involving spherical harmonics pass to an implicit orthogonal basis for this inner product space, but for computations, it is sometimes helpful to have an explicit basis.

Question. Is there a "nice" choice of orthogonal basis for $(\mathrm{Harm}(n,k),\langle\cdot,\cdot\rangle)$? In particular, is there a choice for which there exists a fast algorithm to compute an arbitrary decomposition in the basis (à la FFT)?

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  • $\begingroup$ Am I right in understanding that the question is interesting because of the restriction to polynomials? Physicists use explicit non-polynomial orthogonal bases for spherical harmonics to solve the Schrodinger equation in the $1/r$ potential, right? $\endgroup$
    – Tim Campion
    Commented Feb 18, 2021 at 18:50

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The book "Hyperspherical Harmonics and Their Physical Applications" by Avery $\times 2$, has an explicit description using a product of Gegenbauer polynomials in the cosines of the angles of the hyperspherical coordinate system. See Formula 3.65.

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A simple basis for $\mathcal{H}(n,k)$ is described in this question, however, not so symmetric as one may hope at a first glance, as it is shown in the answer. You may also want to check the beautiful (free online) textbook Harmonic Function Theory by Axley-Bourdon-Ramey.

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    $\begingroup$ The simple basis almost sounds like a solution of Shubin's Conjecture mentioned on page 10 of arxiv.org/abs/0704.1174 What am I missing? $\endgroup$
    – Abdelmalek Abdesselam
    Commented Feb 18, 2021 at 18:18

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