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Question: suppose that $H_{n_1}, H_{n_2}, H_{n_3} \in L^{2}(\mathbb{S}^2)$ are Spherical Harmonics of degrees $n_j$ $(j = 1, 2, 3)$ with $n_1 > n_2 + n_ 3$. Then, it is true that $$ \int_{\mathbb{S}^2} H_{n_1} H_{n_2} H_{n_3} dS = 0 \; ? $$

Obs.: My failed attempt to solve this question was to show that $H_{n_2} H_{n_3}$ is a sum of Spherical Harmonics $H_m$ of degrees at most $m < n_1$ and to use the orthogonality property in $L^2({S}^2)$.

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    $\begingroup$ View these as the restriction of harmonic homogeneous polynomials. Then $H_{n_2} H_{n_3}$ is a homogeneous polynomial of degree $n_2 + n_3$, which can be written as a linear combination of harmonic homogeneous polynomials of degree at most $n_2 + n_3$. Now you can apply orthogonality. $\endgroup$
    – Peter Humphries
    Commented Mar 17, 2019 at 13:31
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    $\begingroup$ Dear @PerterHumphries, it is not clear of my viewpoint that $H_{n_2} H_{n_3}$ is a homogeneous polynomial because $\Delta (fg) = f \Delta g + g \Delta f + \langle \nabla f, \nabla g \rangle$. Thus, the product of them is not necessarily harmonic. $\endgroup$
    – Marcelo Ng
    Commented Mar 17, 2019 at 13:43
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    $\begingroup$ as Paul Garrett clarified in his answer, I did not claim that $H_{n_2} H_{n_3}$ was harmonic, only that it was homogeneous. The key point is that homogeneous polynomials can be written as linear combinations of harmonic homogeneous polynomials when restricted to the sphere. $\endgroup$
    – Peter Humphries
    Commented Mar 17, 2019 at 15:36

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Just clarifying @PeterHumphries' comment: the product $H_{n_2}H_{n_3}$ is certainly homogeneous, and by one of the set-up lemmas in an elementary treatment of spherical harmonics, can be written as a sum of terms of the form $(r^2)^k\cdot f_{n_2+n_3-2k}$ where $f_{n_2+n_3-2k}$ is a homogeneous and harmonic polynomial of degree equal to its subscript. Integrals on the sphere ignore the powers of radius, of course, so orthogonality of harmonic polynomials of different degrees still does give the result, since $$ n_2+n_3-2k \le n_2+n_3 < n_1 $$

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  • $\begingroup$ Paul, just a typo reporting. There is a typo in your post (should be $f_{n_2+n_3-2k}$ and not $f_{n_2+n_3= 2 k}$). $\endgroup$
    – Daniele Tampieri
    Commented Mar 17, 2019 at 15:58
  • $\begingroup$ @DanieleTampieri, ah, thanks, I'll fix it now. $\endgroup$
    – paul garrett
    Commented Mar 17, 2019 at 16:04

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