A dielectric sphere in an initially uniform electric field and representation theory of SO(3) I learned recently that the highest order spherical harmonic required to represent the spatial distribution of decay products of a particle can be used to determine its spin, by using arguments involving the representation theory of SO(3) / addition of angular momentum.
I seek a similar argument for the well-studied problem of a dielectric sphere in an otherwise uniform electric field. This is, for example, solved in Griffiths' Electrodynamics, section 4.7 on the Third Edition (I'm using a Low Price Edition from India, where it is on Page 205). He concludes at the end of that problem that "the field inside is (surprisingly) uniform".
I have a feeling that this result is less surprising when one applies ideas from SO(3) representation theory, but I'm not sure how to formulate the argument precisely. Here is my loose train of thought -- the field is a vector field with a specific direction, so it has one unit of angular momentum (or is given by an $l = 1$ spherical harmonic). Therefore, the induced surface charge on the dielectric sphere must be given by $P_1(\cos \theta) = \cos(\theta)$. The field inside is a vector field originating from these charges, so it must again be uniform.
I'm also trying to use this to determine the highest order (i.e. highest $l$) component of $Y_{lm}$ in the total electric field after accounting for the polarization of the sphere.
 A: The presence of the external electric field breaks $SO(3)$ to $SO(2)$. Suppose that $\mathbf{E}$ is oriented along the $z$-axis, then rotations about the $z$-axis (of course, chosen to pass through the center of the sphere) is a symmmetry in the problem. This $SO(2)$ invariance only implies that the potential is independent of $\varphi$, the polar angle. Thus, one has that the scalar potential has the form: $\Phi(\rho,z)$ (in cylindrical polar coordinates) or $\Phi(r,\theta)$ (in spherical polar coordinates). 
Now one can use the more general solution to Laplace's equation outside the dielectric sphere, for instance, to see that the $\varphi$ independence sets all terms involving $Y_{lm}$ for $m\neq0$ to zero. It is the boundary conditions, that force $l>1$ harmonics to vanish. So that does not follow from symmetry considerations.
Edit: The electric field at spatial infinity translates to the boundary condition for Laplace's equation $\phi=-ER\cos\theta$ on a large sphere of radius $R$ that in the limit $R\rightarrow\infty$ becomes spatial infinity. This forces all $l>1$ harmonics to vanish (in the region outside the dielectric sphere) without any further computation. 
