# Are linearly independent harmonic polynomials orthogonal upon integration over the sphere?

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

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.

• 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. Apr 15 '16 at 20:51
• 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:$\sqrt{2}(y^{3} z-yz^{3} )$ Apr 16 '16 at 4:24
• 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? Apr 16 '16 at 4:28