# Harmonic polynomials invariant under the action of a group

I would like to find all the real harmonic homogeneous polynomials of each degree in three variables that are invariant (or invariant up to a sign change) under the action of each of the point groups that act on Euclidean 3-space.

A harmonic polynomial is simply a polynomial such that the Laplacian vanishes.

For example let $$z=x_1+\mathrm{i}x_2$$ then the Real and Imaginary parts of the polynomials $$z^m$$ and $$\bar{z}^m$$ are invariant under the action of the group generated by a rotation about the $$e_3$$ axis through $$2\pi/q$$ where $$q$$ and $$m$$ are natural numbers and $$q$$ divides $$m$$.

I can construct other examples but I assume that since the representation theory of point groups has been studied extensively, specifically for chemistry, this might be something that has already been done. Perhaps under another name.

I am aware of Sheldon Axler's Mathematica code for harmonic functions, and can calculate a basis for harmonic polynomials of some degree, so I could in principle calculate the invariant polynomials by brute force. However I am looking for something that provides more insight in to the relations between the invariant harmonic polynomials and the group.

• A colleague at Manchester just pointed me to the paper Burnett Meyer, On the Symmetries of Spherical Harmonics, Canadian Journal of Mathematics, Volume 6,1954 , pp. 135-157, which might answer the question. – Bill Lionheart Apr 30 '19 at 18:14
• Maybe this should be on mathoverflow? – Dirk Apr 30 '19 at 23:04
• Dirk I agree. My mistake. But now I have had read the above paper a bit more it does completely answer the question. Should I post it as an answer with some more details? I don't think I can ask it on mathoverflow know I know the answer? – Bill Lionheart May 1 '19 at 5:32
• Well, if you already have an answer, there is indeed no reason to report on MO. If you think that is helps (at least you) to write an answer here, go ahead. You never know who will need exactly this answer.. – Dirk May 1 '19 at 6:05