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September 17

Spherical integral of harmonic polynomial over S^{d-1} is zero?

Hello,

perhaps this is very trivial and well know, but still I couldn't find it anywhere. Consider a harmonic polynomial, i.e. a polynomial whose Laplacian is zero and take its integral over the unit sphere. Prove that this is zero.

Is this correct? I guess so, but it seems one would have to use that you are dealing with a polynomial.

Many thanks, Evilbu (talk) 04:45, 17 September 2013 (UTC)

To my humble recollection, ANY integral over ANY closed line or surface is 0. And obviously the sphere is such a shape. (Well, the function in question also has to fulfill certain basic common-sensical conditions, but that's it). — 79.113.218.85 (talk) 06:14, 17 September 2013 (UTC)

The polynomials restrict to eigenvalues of the Laplace-Beltrami operator on the sphere. So, except for the constant eigenfunction, it's equivalent to show that $\int \mathrm{\Delta }f=0$ (where $\mathrm{\Delta }$ is the Laplace-Beltrami operator) which is immediate from Stokes' theorem. Another way to do this is to know that the different eigenspaces are orthogonal, so any non-constant spherical harmonic f is orthogonal to the eigenspace whose elements are constant. Sławomir Biały (talk) 11:39, 17 September 2013 (UTC)

Thanks, but I am afraid I did not know about the Laplace–Beltrami operator. I looked at the article, and it did not mention eigenvalues. Could you go into a bit more detail, please?Evilbu (talk) 04:47, 18 September 2013 (UTC)