integral of product of three basis functions and Clebsh-Gordan coefficients

Question

Suppose I have an orthonormal basis $\{b_i\}_{i=1}^\infty$ for an $L_2$ space (for example, the $b_i$ could be spherical harmonics on the round sphere with the Euclidean $L_2$ inner product). I want to calculate the coefficients $$C_{i,j,k} = \int b_ib_jb_k;$$ are these the same thing as the "Clebsh-Gordan coefficients" for the basis $\{b_i\}$? If not, are they related, and how? My reading suggests they are closely related, but I am having trouble untangling the mathematics from the quantum mechanics.

Answer 1

I cannot say that I have truly understood what's behind all this, so I will just write a short slogan in the hope of coming back to this later.

The key is that, if $\{b_i\ :\ i=0, 1,2\ldots \}$ is an orthonormal and real basis of $L^2(d\mu)$, then $$\tag{1} \int b_ib_j b_k\, d\mu= c_{i,j}^k \quad \iff \quad b_i b_j = \sum_k c_{i,j}^k b_k.$$ And it is the latter equation that is related to Clebsch-Gordan stuff.

Indeed, the Clebsch-Gordan $c_{i,j}^k$ appear in formulas such as $$\rho_i\otimes \rho_j = \bigoplus_k c_{i,j}^k \rho_k,$$ where the $\rho_i$ are irreducible representations of some group $G$. This already bears some resemblance with (1). Taking the characters $\chi_i(g)=\mathrm{trace}\,\rho_i(g)$, the previous equation becomes $$ \chi_i\cdot \chi_j = \sum_k c_{i,j}^k \chi_k,$$ and the resemblance is even greater.

I am not adding further details because my understanding of all of this is too poor yet.

(Based on Olver's lecture notes, http://www-users.math.umn.edu/~olver/ , scroll to "Quantum Mathematics", chapter 7, pag. 97 of the 11/13/16 edition. "Addition of Angular Momenta". Have also a look at this great answer by Matt E.).