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A symmetrized form of the Clebsch-Gordan coefficients used for coupling three angular momenta. The relation is $⟨j_1{m}_1,j_2{m}_2|j_3{m}_3⟩=(-1{)}^{j_1-j_2+m_3}\sqrt{2j_3+1}\left(\begin{array}{ccc}{j}_1& j_2& j_3\\ m_1& m_2& -m_3\end{array}\right)$. The symbol vanishes unless the magnetic indices sum to zero and the angular momenta obey the triangle and integrality conditions. Products of two such symbols express the Gaunt integral; the zero-magnetic-index symbol enforces even total multipole for integer spherical harmonics.

In a full-sky cubic bispectrum estimator, pairwise Gaussian covariance within one triple gives a sum of Gaunt integrals with opposite $m$ indices. The spherical harmonic addition theorem turns their pair into the constant $(2\ell +1)/(4\pi )$; its integral against $Y_{LM}$ vanishes for $L>0$. A zero temperature monopole removes $L=0$. Thus only the six pairings connecting the two triples survive in the Gaussian variance; isotropic weights are essential for the cancellation.

Let $b_{{\ell }_1{\ell }_2{\ell }_3}^1$ denote the template at unit $f_{\mathrm{NL}}$, and define $A_{123}=b_{{\ell }_1{\ell }_2{\ell }_3}^1{\mathcal{G}}_{m_1{m}_2{m}_3}^{{\ell }_1{\ell }_2{\ell }_3}$. All sums below are over ordered triples of the retained multipoles, with the monopole removed and $C_{\ell }>0$. A finite maximum multipole makes these expressions ordinary finite sums. Taking the expectation of the cubic numerator and using the template relation givesThus the full-sky cubic bispectrum estimator is unbiased forThe assumed template must have nonzero support, so $F>0$. Using the Gaunt sum rule, writeThe factor $1/6$ belongs to the ordered sum; an unordered triangle sum would instead use multiplicity factors for equal multipoles. This normalization is also the Gaussian Fisher information for the template amplitude.

For the cosmic variance, use the reality condition ${\Theta }_{\ell m}^{\ast }=(-1{)}^m{\Theta }_{\ell ,-m}$ and Gaussian covarianceThere are $15$ pairings of the six multipoles in the squared cubic numerator. Six pairings connect every factor in the first triple to a factor in the second. Each gives $\sum A_{123}^2/(C_{{\ell }_1}{C}_{{\ell }_2}{C}_{{\ell }_3})=6F$. The covariance phases cancel: nonzero Gaunt integrals have $m_1+m_2+m_3=0$, and simultaneous reversal of the three $m$ indices multiplies the Gaunt integral by $(-1{)}^{{\ell }_1+{\ell }_2+{\ell }_3}=1$.

The other nine pairings contain one contraction within each triple. To see why their sums vanish, contract two legs of a Gaunt integral and use the spherical harmonic addition theorem:The template and inverse-covariance weights are independent of $m$, so they do not spoil this cancellation. The only possible surviving unpaired mode is the monopole, and ${\Theta }_{00}=0$ removes it. This is monopole cancellation of internal cubic-estimator contractions. It explains why no linear correction is needed in this ideal full-sky isotropic problem; masks or anisotropic noise would spoil the argument.

The cubic numerator therefore has Gaussian variance $6(6F)=36F$. Dividing by $(6F{)}^2$ givesThe result concerns the Gaussian-limit covariance. Non-Gaussian connected four- and six-point terms can change the variance at finite amplitude. Unbiasedness uses the assumed linear template relation for the observed three-point function.

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.

Linear cosmological transfer maps the primordial bispectrum to the reduced CMB bispectrum. A Fourier representation of the momentum delta function and three Rayleigh plane-wave expansions separate the radial transfer integrals from a Gaunt integral. The six angular factors and three Fourier measures give $(4\pi {)}^6/(2\pi {)}^9=(2/\pi {)}^3$. The observer-position phase cancels by momentum conservation; the remaining radial position is an auxiliary integration variable.