There is a theorem that states that the vector space of homogeneous polynomials decomposes into an orthogonal direct sum of vector spaces of harmonic polynomials as

$$ \mathcal{P}_n = \mathcal{H}_n \oplus r^2 \mathcal{H}_{n-2} \oplus r^4 \mathcal{H}_{n-4} \oplus \cdots $$ where the inner product can be defined for instance as $$\langle x_0^{d_1}x_1^{d_2}\cdots x_N^{d_N},x_0^{m_1}x_1^{m_2}\cdots x_N^{m_N} \rangle= d_1!\delta_{d_1,m_1}\cdots d_N !\delta_{d_N,m_N} $$ where $\sum_i d_i = \sum_i m_i$ which says that monomials are pairwise orthogonal.

Say that I have built up a basis using Gram Schmidt procedures up to third order such that $\mathcal{H}_0 = \{1\}$, $\mathcal{H}_1 = \{x,y\}$, and $\mathcal{H}_3 = \{x^2-y^2, xy\}$ ignoring normalization.

When the polynomials are restricted to the unit sphere the above decomposition reads $$ \mathcal{P}^S_n = \mathcal{H}^S_n \oplus \mathcal{H}^S_{n-2} \oplus \mathcal{H}^S_{n-4} \oplus \cdots $$ where all polynomials are evaluated over the unit sphere and $\mathcal{H}^S_k$ are spherical harmonics.

Thus returning to my basis above, is it correct to say that these correspond to the real Spherical Harmonics? I am mostly interested in if there is a way to prove that they are orthogonal over a typical inner product such as $$\langle p,q\rangle = \int d\Omega (p*q) $$ where integration is over the unit sphere, and I again don't care about normalization. By inspection it is obvious that they are going to be orthogonal in this way but I can't prove it to myself.

The part I am struggling with is the fact that I only use the first inner product in making the polynomials so I am obviously never requiring that they are integral-orthogonal. But maybe the reason they turn out to be is because I have found a minimal basis on which to expand the Harmonic Polynomials, which means they are going to be orthogonal?

Please let me know if there is a connection here that I am missing. Thank you!


1 Answer 1


Homogeneous harmonic polynomials of different degrees are orthogonal on the unit sphere, because they are eigenfunctions of the (self-adjoint) Laplace-Beltrami operator, with different eigenvalues. According to Wikipedia, $$ \Delta_{S^{n-1}}f=-\ell(\ell+n-2)f $$ where $\ell$ is the degree of homogeneity.

Within the same degree, orthogonality is not automatic. E.g., $x$ is not orthogonal to $x+y$, although they are linearly independent. It's true that for small degrees and low dimensions, a "natural" choice of polynomials turns out to work. But as the degree increases, the choice of orthogonal polynomials becomes much less obvious.

  • $\begingroup$ Ok that is good to know that they are not guaranteed to be orthogonal within the same degree. However, I am also wondering if the choice of my first inner product above does always result in orthogonal polynomials as I cannot easily find a counter example. $\endgroup$
    – compmatsci
    Apr 15, 2016 at 20:51
  • $\begingroup$ Harmonic6: $\sqrt{8/35}(x^{4} -3x^{2} y^{2} -3x^{2} z^{2} +3/8y^{4} +3/4y^{2} z^{2} +3/8z^{4} )$ Harmonic7: $1.51186(x^{3} y-3/4xy^{3} -3/4xyz^{2} )$ Harmonic8: $1.60357(x^{2} y^{2} -x^{2} z^{2} -1/6y^{4} +1/6z^{4} )$ Harmonic9: $(xy^{3} -3xyz^{2} )$ Harmonic10:$ \sqrt{1/8}(y^{4} -6y^{2} z^{2} +z^{4} )$ Harmonic11:$ 1.51186(x^{3} z-3/4xy^{2} z-3/4xz^{3} )$ Harmonic12:$ 3.20713(x^{2} yz-1/6y^{3} z-1/6yz^{3} )$ Harmonic13:$ 3(xy^{2} z-1/3xz^{3} )$ Harmonic14:$ \sqrt{2}(y^{3} z-yz^{3} )$ $\endgroup$
    – compmatsci
    Apr 16, 2016 at 4:24
  • $\begingroup$ The above harmonics correspond exactly (up to a permutation of variables) to those linked in your response but were made from the procedure outlined in the question. Is it purely coincidence that they are orthogonal? $\endgroup$
    – compmatsci
    Apr 16, 2016 at 4:28

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