= 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 $\sum\boldsymbol k_i=0$. Represent that delta function by
$$
(2\pi)^3\delta^{(3)}(\boldsymbol k_1+\boldsymbol k_2+\boldsymbol k_3)
=\int d^3x\,e^{i(\boldsymbol k_1+\boldsymbol k_2+\boldsymbol k_3)\cdot\boldsymbol x}.
$$
The <Rayleigh plane-wave expansion> and angular orthogonality give, for each momentum,
$$
\int d\hat k\,Y_{\ell m}^*(\hat k)e^{i\boldsymbol k\cdot\boldsymbol x}
=4\pi i^\ell j_\ell(kx)Y_{\ell m}^*(\hat x).
$$
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 $x^2dx$, and the momentum radial measures are $k_i^2dk_i$. The combined numerical prefactor is $(4\pi)^6/(2\pi)^9=8/\pi^3=64/(2\pi)^3$. Hence the <reduced CMB bispectrum> is
$$
\boxed{b_{\ell_1\ell_2\ell_3}=\left(\frac2\pi\right)^3
\int_0^\infty x^2dx\int_0^\infty\prod_{i=1}^3
\left[k_i^2dk_i\,j_{\ell_i}(k_ix)g_{\ell_i}(\eta_0,k_i)\right]B(k_1,k_2,k_3),}
$$
and the angular three-point function factorizes as
$$
\boxed{\langle\Theta_{\ell_1m_1}\Theta_{\ell_2m_2}\Theta_{\ell_3m_3}\rangle
=b_{\ell_1\ell_2\ell_3}\mathcal G^{\ell_1\ell_2\ell_3}_{m_1m_2m_3}.}
$$
This <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.
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