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For $\mathbf{k}=k\hat{\mathbf{z}}$, scalar axisymmetry leaves only $m=0$ in the photon quadrupole: ${\Theta }_{20}=-\sqrt{4\pi /5}{\Theta }_2$ with the normalization used above. The two spin-weighted spherical harmonics satisfy ${}_{+2}{Y}_{20}={}_{-2}{Y}_{20}\propto {\sin }^2\theta$. Translation from last scattering supplies $e^{-ik{\chi }_{\ast }\mu }$. Thus the two helicity combinations have the same scalar source:where $\mathcal{C}$ is a common normalization constant independent of $\mu$. In this meridian polarization basis the scalar contribution has $U=0$; this does not mean that every rotated polarization basis has zero $U$.

Apply the spin-raising and lowering operators to an axisymmetric scalar potential $f$. The first action is $-{\partial }_{\theta }f$. In the second action, the intermediate spin is respectively $+1$ or $-1$; both giveIt follows thatSubtracting the two equations gives $P_B^{\prime \prime }=0$. The possible affine $P_B=A+B\mu$ consists solely of the unobservable $\ell =0,1$ kernel of the twice-applied spin operators. Defining polarization potentials to have no such components sets the physical scalar B mode to zero, $\boxed{P_B=0}$. This is the low-multipole kernel of polarization potentials, not a physical dipole polarization.

For $x\ne 0$ the remaining equation integrates to $P_E=-\mathcal{C}{\Theta }_2{e}^{-ix\mu }/x^2+A_E+B_E\mu$. Remove the same monopole/dipole kernel and use the plane-wave expansion. Thus the scalar E-mode radial projection isThe overall minus sign is absorbed in the stated proportionality constant. Although the unprojected antiderivative has $x^{-2}$, its divergent monopole and dipole pieces are discarded; the physical $\ell \ge 2$ limit is finite because $j_{\ell }(x)=O(x^{\ell })$.

The temperature-only tight-coupling quadrupole is suppressed by $k/{\Gamma }_T$. For regular superhorizon scalar modes, the photon dipole is itself proportional to $k$, so the recombination quadrupole and the resulting polarization are especially small at large angular scales. Last-scattering polarization is generated by the small departure from perfect tight coupling, not by an isotropic monopole. Reionization can generate a separate large-angle signal after recombination and is absent from the instantaneous-last-scattering approximation here. The no-B-mode conclusion applies to linear scalar sources; it does not include tensor sources or conversion by lensing.