## Normal form for trace-free real cubic forms in 3 variables under SO(3)-action?

I'm looking at irreducible, real representations of $SO(3)$. The 5-dimensional irrep is isomorphic to the space of trace-free quadratic forms on $\mathbb{R}^3$, and we all know that any such quadratic form can be diagonalized by the standard $SO(3)$ action on $\mathbb{R}^3$. My question is: is there any analogous "normal form" result for the 7-dimensional irrep, regarded as the space of trace-free cubic forms on $\mathbb{R}^3$, under the standard action of $SO(3)$?

More generally, I'd like to better understand the geometry of the orbits of these actions. Can anyone recommend a reference?

-Jeanne

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One way to do this is to reduce this to understanding sets of points on the $2$-sphere up to a certain equivalence. If you let $\mathcal{H}_k$ denote the homogeneous polynomials $p$ of degree $k$ on $\mathbb{R}^3$ that are harmonic, i.e., that satisfy $\Delta p = 0$, then $\mathcal{H}_k$ is an irreducible $\mathrm{SO}(3)$-representation of dimension $2k{+}1$. The space $\mathcal{P}_k$ of all homogeneous polynomials of degree $k$ can be written as the direct sum $$\mathcal{P}_k = \mathcal{H}_k \oplus (R\cdot\mathcal{H}_{k-2}) \oplus (R^2\cdot\mathcal{H}_{k-4}) \oplus \cdots$$ where $R = {x_1}^2+{x_2}^2+{x_3}^2$. In particular, there is a canonical 'harmonic part' $H(p)\in \mathcal{H}_k$ for any $p\in \mathcal{P}_k$, and the map $H:\mathcal{P}_k\to \mathcal{H}_k$ is linear. Then one has the following `factorization result':

Proposition: Any $p\in \mathcal{H}_k$ can be 'factored' in the form $p = H(\ell_1\ell_2\cdots\ell_k)$, where the 'factors' $\ell_i\in \mathcal{H}_1$ satisfy $|\ell_1| = |\ell_2| = \cdots = |\ell_k|$ and the elements $\ell_i$ are unique up to permutation and replacement of an even number of them by their negatives.

This 'harmonic factorization' is $\mathrm{SO}(3)$-equivariant, so finding normal forms for harmonic polynomials of a fixed degree $k$ reduces to normalizing sets of $k$ points on the $2$-sphere up to permutation and replacing an even number of them by their negatives, which is a combinatorial problem.

The proof of the above proposition goes through the standard representation theory of $\mathrm{SU}(2)$, which is the double cover of $\mathrm{SO}(3)$.

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Thanks, Robert! :) – Jeanne Clelland Aug 13 at 18:51
Yes, thanks! When Jeanne asked me about this, I wanted to suggest using somehow the representation theory of SU(2), but I had no idea how so I didn't say anything. I'm very pleased to see that it really does play a role in answering this question. – Deane Yang Aug 13 at 19:27