= Solution
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 $e^{-ik\chi_*\mu}$. Thus \b[the two helicity combinations have the same scalar source]:
$$
\boxed{Q\pm iU=\mathcal C\,\Theta_2(\eta_*,k\hat{\mathbf z})e^{-ix\mu}(1-\mu^2),\quad x=k\chi_*,\quad\mu=\cos\theta,}
$$
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 give
$$
\eth^2f=\overline{\eth}^{2}f
=\partial_\theta^2f-\cot\theta\,\partial_\theta f
=(1-\mu^2)\frac{d^2f}{d\mu^2}.
$$
It follows that
$$
\boxed{\frac{d^2}{d\mu^2}(P_E\pm iP_B)=\mathcal C\Theta_2e^{-ix\mu}.}
$$
Subtracting the two equations gives $P_B''=0$. 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 \b[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 $x\ne0$ the remaining equation integrates to $P_E=-\mathcal C\Theta_2e^{-ix\mu}/x^2+A_E+B_E\mu$. Remove the same monopole/dipole kernel and use the plane-wave expansion. Thus \b[the <scalar E-mode radial projection> is]
$$
\boxed{P_E\propto\sum_{\ell\ge2}(-i)^\ell(2\ell+1)\Theta_2(\eta_*,\mathbf k)
\frac{j_\ell(k\chi_*)}{(k\chi_*)^2}P_\ell(\hat{\mathbf k}\cdot\mathbf e).}
$$
The 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 $\ell\ge2$ 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.
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