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 $d\Omega=\frac{1}{2}\sin^2(\omega/2)\sin(\theta)d\omega d\theta d\phi$
 A: It seems unlikely.  The product of two orthogonal polynomials can have a component of a third. For example, in the Legendre polynomials, $$\int_{-1}^1P_1(x)P_2(x)P_3(x)\;dx=\int_{-1}^1x\frac{3x^2-1}{2}\frac{5x^3-3x}{2}\;dx=\frac{6}{35}$$
A: 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.
A: Except for abelian Fourier/harmonic analysis, such as on products of circles or lines, it is rare that the product of two eigenfunctions for a (invariant) differential operator, e.g., Laplacian on the sphere, is exactly another eigenfunction. In the abelian Fourier case, yes, the product of two exponentials or cosines or sines integrates to 0 against all but one or two of the others. In essentially every other example, such as spherical harmonics on spheres of dimension at least 2, the spectral decomposition of the product of two eigenfunctions is smeared out over several eigenspaces. 
This amounts to asking to decompose a tensor product of irreducible representations into irreducibles. Rarely does the tensor product of irreduducibles remain irreducible. There is much known about such decompositions, with keywords like "branching rules". This arose historically as Clebsch-Gordan coefficients in basic quantum mechanics.
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.)
A: As mentioned by Paul Garrett, the answer to this question is related to Clebsch-Gordan coefficients and the Wigner-Eckart theorem of quantum mechanics. Application of this theorem to the group $O(4)$ yields:
$$
\sqrt{2\pi^2} \int \mathrm{d}\Omega \; Z^n_{lm}Z^{n'}_{l'm'}Z^{n''}_{l''m''} = \Bigl[(n+1)(n'+1)(n''+1)(2l+1)(2l'+1)(2l''+1)\Bigr]^{1/2} \begin{pmatrix}l & l' & l''\\ m & m' & m''\end{pmatrix}\begin{bmatrix}\frac12n & \frac12n & l\\ \frac12n' & \frac12n' & l'\\ \frac12n'' & \frac12 n'' & l''\end{bmatrix},
$$
where the matrices denote $3j$ and $9j$ Wigner symbols.
For more details see R.T. Sharp, Notes on O(4) representations.
