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A function $f(\theta,\phi)$ (with $\theta,\phi\in \mathbb{S}^2$) can be expanded in terms of spherical harmonics $Y_{l.m}(\theta,\phi)$. Recently, in this Particle Data Group review titled Cosmic Microwave Background by D. Scott and G. F. Smoot, I've come across a term called spin-2 spherical harmonics.

  • How are they different from or related to ordinary spherical harmonics?

  • What kind of functions requires expansion in terms of these objects? I read that the linear polarization pattern of the Cosmic Microwave Background (CMB) can be expressed using spin-2 spherical harmonics.

  • Do they have anything to with Spin-weighted spherical harmonics or Spin spherical harmonics?

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The usual $Y_{lm}(\theta, \phi)$ functions represent a scalar value at each point on the unit sphere.

But that's not sufficient to represent e.g. polarized light incoming. That required a more complicated representation at each point that just a scalar.

Couldn't you just use separate functions, e.g. $Y_{lm}^H(\theta, \phi)$ and $Y_{lm}^V(\theta, \phi)$ to represent the "horizontal" and "vertical" components of polarization? It's a subtle point, but no, you can't.

First, it's hard to uniquely define the basis for those $Y^H$ and $Y^V$ functions: What are their directions for H and V at the pole? All directions are south from the north pole.

Second, and what's really going on in the first point, is that there are continuity constraints for the polarization (more generally, for the tensor value at each point) that need to also be expandable in the representation.

Michael Boyle's 2016 paper "How should spin-weighted spherical functions be defined?" does a great, if somewhat formal, job of covering this in the cosmological context.

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  • $\begingroup$ "It's a subtle point, but no, you can't." Like temperature, can we assume that any component ($E_x,E_y$ or $E_z$) of the electric field vector $\textbf{E}$ of the polarized light is a function of $(\theta,\phi)$ (e.g., $E_x=E_x(\theta,\phi)$) and forget about $r$? @BobJacobsen $\endgroup$ – SRS Jul 9 '18 at 20:15
  • $\begingroup$ @srs The polynomials are useful because they're each solutions to the underlying problem: Like using $sin(\omega t)$, which is a solution of certain problems for all $\omega$ and Fourier transforms, you want to be able to build up sums of polynomials to fit the physical situation without worrying whether each item is physically valid. That's why you can't have separate solutions for $E_\phi$ and $E_\theta$; they're connected by the underlying physics. $\endgroup$ – Bob Jacobsen Jul 9 '18 at 21:31

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