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Dustin G. Mixon
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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 computationsproofs 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)?

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 computations 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)?

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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Francesco Polizzi
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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 computations 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)?

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)?

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 computations 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)?

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 computations 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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Dustin G. Mixon
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Is there a nice orthogonal basis of spherical harmonics?

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 computations 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)?