3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.