 Solution

Insert the primordial bispectrum into the product of the three linear transfer integrals. Write $L={\ell }_1+{\ell }_2+{\ell }_3$ and abbreviate $g_{{\ell }_i}({\eta }_0,k_i)$ by $g_i$. The observer-position phase is one because the momentum delta function imposes $\sum {\mathbf{k}}_i=0$. Represent that delta function byThe Rayleigh plane-wave expansion and angular orthogonality give, for each momentum,The three $i^{\ell }$ factors cancel the $(-i{)}^L$ in the temperature multipoles. Angular integration over $\hat{x}$ leaves the complex conjugate Gaunt integral. In the conventional complex spherical harmonics, this integral is real, and it vanishes unless the angular momentum triangle, even-parity and $m_1+m_2+m_3=0$ selection rules hold. Therefore its conjugate equals itself.

The radial measure is $x^{2}dx$, and the momentum radial measures are $k_i^{2}d{k}_i$. The combined numerical prefactor is $(4\pi {)}^6/(2\pi {)}^9=8/{\pi }^3=64/(2\pi {)}^3$. Hence the reduced CMB bispectrum isand the angular three-point function factorizes asThis primordial-to-angular bispectrum projection separates dynamics and radial transfer from purely angular geometry. The spatial integration variable is auxiliary, not the observer position. Linear transfer is justified at leading order in the primordial signal; it does not require a large amplitude mathematically, although a signal must exceed measurement uncertainty to be detectable.