Suppose I have a function $f(x):S^2\rightarrow\mathbb{C}$ in the degree four spherical harmonic basis: $$f(\theta,\varphi):=\sum_{k=-4}^4a_kY_4^k(\theta,\varphi).$$ I have two related questions:
- Is there a simple formula and/or system of equations that can be used to find the planes through the origin over which $f(\cdot)$ has reflectional symmetry?
- Is there a condition on the $a_k$'s that guarantees that the symmetry planes of $f$ are mutually orthogonal?
The only symmetry I could find was the much simpler $f(x)=f(-x)$, but this represents reflection through the origin rather than over a plane. I tried thinking about degree-four homogeneous polynomials instead but couldn't find any elegant way to approach this problem. Given all the nice identities about spherical harmonics, I was hoping this would have an elegant solution...
I am interested in this topic because I'm curious what happens to the functions in Figure 9 of this paper when the xyz axes are allowed to scale independently.
This paper does a good job explaining the details.
Take $v\in\mathbb{C}^{2\ell+1}$ to be the coefficients of a function in the degree-$\ell$ part of the spherical harmonic basis. Reflecting over the $xz$ plane is equivalent to conjugation: $v\mapsto v^\ast$.
So, if $N$ is the normal to the reflectional symmetry plane, take $P$ to be any matrix rotating $N$ to the $+y$ axis. If $D_\ell(P)$ is the corresponding Wigner-$D$ matrix, then we can obtain a symmetric reflection by rotating the function so that the symmetry plane is $xz$, then rotating back: $v\mapsto v^\ast D_\ell(P)^\ast D_\ell(P)^\dagger$ (here $\ast$ denotes conjugation and $\dagger$ denotes a conjugate transpose).
After lots of playing with the Wigner-$D$ matrix formula in Mathematica, it does not appear that there is any particularly nice condition guaranteeing that the rotated function has orthogonal symmetry planes...