# What is the integral of three orthonormal basis functions?

The hyperspherical harmonics, given by:

$Z_{l,m}^n(\omega,\theta,\phi)=(-i)^l\frac{2^{l+1/2}l!}{2\pi}\sqrt{(2l+1)\frac{(l-m)!}{(l+m)!}\frac{(n+1)(n-l)!}{(n+l+1)!}}\sin^l(\omega/2)C_{n-l}^{l+1}(\cos(\omega/2))P_l^m(\cos\theta)\exp(im\phi)$

(where $C_{n-l}^{l+1}$ is a Gegenbauer polynomial, and $P_l^m$ is an associated Legendre function), form an orthonormal basis for an expansion of functions on the hypersphere.

As such, the following identity is true:

$\int_\Omega Z_{l,m}^n(\omega,\theta,\phi)Z_{l',m'}^{n'}(\omega,\theta,\phi)d\Omega=\delta_{n,n'}\delta_{l,l'}\delta_{m,m'}$

My question is whether or not the following identity is also true:

$\int_\Omega Z_{l,m}^n(\omega,\theta,\phi)Z_{l',m'}^{n'}(\omega,\theta,\phi)Z_{l'',m''}^{n''}(\omega,\theta,\phi)d\Omega=\delta_{n,n',n''}\delta_{l,l',l''}\delta_{m,m',m''}$

Note: $\Omega$ is the entire domain, i.e., $\omega\in[0,\pi], \theta\in[0,\pi], \phi\in[0,2\pi]$, and $d\Omega=\frac{1}{2}\sin^2(\omega/2)\sin(\theta)d\omega d\theta d\phi$

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$\Omega$ is the hyperspherical region? –  Ｊ. Ｍ. May 17 '11 at 17:06
@J.M.: yes, I've edited the question to clarify that $\Omega$ is the hyperspherical domain given above. –  okj May 17 '11 at 17:14

I don't know whether or not this is true in your particular case, but I'd like to point out that this isn't abstractly true: that is, there isn't a general argument about orthonormal functions that will get you from the first identity to the second. For example, some appropriate normalization of the functions $\cos nx$ are orthonormal on $[-\pi, \pi]$, but it's false that

$$\int_{-\pi}^{\pi} \cos n_1 x \cos n_2 x \cos n_3 x \, dx$$

is zero whenever the $n_i$ are distinct: I think it should be nonzero if $n_1 + n_2 = n_3$, for example.

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It seems unlikely. The product of two orthogonal polynomials can have a component of a third. For example, in the Legendre polynomials, $$\int_{-1}^1P_1(x)P_2(x)P_3(x)\;dx=\int_{-1}^1x\frac{3x^2-1}{2}\frac{5x^3-3x}{2}\;dx=\frac{6}{35}$$

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An operational summary, in terms of integrals of three eigenfunctions, is that the integrals have structural meaning, which very often tells us that the integrals will mostly not be $0$. (There is also a Plancherel theorem, in many situations, which does not tell us the values of the integrals, but does give the norm-squared of the sum of the decomposition coefficients.)