# Invariant polynomials for a given irrep of SO(3)

Background

Hilbert's basis theorem says

Given a compact lie group $G$ acting linearly on the space $\mathbb{R}^n$, there is a set of $G$-invariant homogeneous polynomials $\{p_1(x),\ldots,p_m(x)\}$, such that any $G$-invariant polynomial can be written as a polynomial in the $\{p_1(x),\ldots,p_m(x)\}$.

Motivation [simplified and updated June 26, 2017]

If we choose our Lie group to be $G=SO(3)$ acting on the space $\mathbb{R}^3$, the only invariant polynomial in $\{x_1,x_2,x_3\}$ is $$r^2 = x_1^2 + x_2^2 + x_3^2$$ and the powers thereof.

In the general case, $SO(3)$ acts on the space $\mathbb{R}^n$, through one of its $n$-dimensional irreducible representations. ($n = 2l + 1$, $\,l \in \mathbb{N}$).

Question

What are the invariants of $SO(3)$ when acting on $\mathbb{R}^{2l+1}$?

(I am still a bit green when it comes to representation theory, so feel free to gently suggest references or other reading material.)

I'd like to briefly answer my own question, at least in part, since there seems to be at least a little interest in this question. I'll address "how to construct the invariants" since "what are the invariants" is a little vague. For example, I'm still working on understanding the intuition behind these invariants. There is more to the story than I can post now, but this is a good start and gives a computationally efficient way of constructing the invariants. I'll add more to this as time goes on, if desired.

Here's how we make the invariants of $SO(3)$ for arbitrary representation index $\ell$ and of arbitrary degree $d$. (I cannot prove yet that it generates all the invariants, but I suspect that it does. It's based on a similar idea in a paper by Sattinger; I've simplified the method somewhat for our uses, cutting down on some steps.)

Let $F(\eta)$ be a $G$-invariant homogeneous polynomial of $\eta \in \mathbb{R}^n$. By $G$-invariance, for any $g \in G$, $$F(g.\eta) = F(\eta).$$ $F(\eta)$ is a sum of monomials, $$F(\eta) = F_{ij \ldots k}\eta_i \eta_j \ldots \eta_k.$$

Now let $t$ be one of the generators in the Lie algebra $L$ corresponding to $G$, and let $a$ be a parameter, so that we can write $$g = e^{a t} \,\,\, \forall g \in G.$$ Then, expanding to first order in $a$, $$F(g.\eta) = F(e^{a t} .\eta) = F(\eta + a t .\eta) = F(\eta) + a t.\eta \cdot \nabla F(\eta).$$

Evidently, $$t. \eta \cdot \nabla F(\eta) = 0.$$

This is what we will use to solve the problem. Either we will find that the above dot product is zero identically, or it will give us a constraint on the coefficients of the monomials, producing an invariant.

This equation is a consequence of the following fact: the tangent space to an orbit is generated by the Lie operators, and the gradients of invariant functions are of course zero along the orbits -- so they must be perpendicular to them.

As a side note, we won't need to work with the most general sum of monomials; we only need to consider those which have $$i + j + \ldots + k = 0.$$

## Example, $\ell=1$, $d=2$

Let's do an example, the absolute simplest case. We will take $\ell=1$ and degree $d=2$.

For this case the most general invariant polynomial of degree $d = 2$ looks like $$F = a \eta_0^2 + b\eta_{-1}\eta_{1}.$$ We have $$\nabla F = \begin{pmatrix} b\eta_1 \\ 2a \eta_0 \\ b\eta_{-1} \end{pmatrix}$$

Now let's consider what happens when we act on $\eta$ with one of the generators. We'll start with $L_z$, given by $$L_z = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}.$$ We have $$L_z \eta \cdot \nabla F = \begin{pmatrix} -\eta_{-1} & 0 & \eta_1 \end{pmatrix} \begin{pmatrix} b\eta_1 \\ 2a \eta_0 \\ b\eta_{-1} \end{pmatrix} = -b \eta_{-1}\eta_{1} + b \eta_{-1}\eta_{1} = 0$$ which equals zero identically. This won't help us. However, let's consider the case when we use the raising operator $L_+$, $$L_+ = \begin{pmatrix} 0& \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2} \\ 0 & 0 & 0 \\ \end{pmatrix}.$$ In this case, $$L_+ \eta \cdot \nabla F = \begin{pmatrix} \sqrt{2} \eta_0 & \sqrt{2} \eta_1 & 0 \end{pmatrix} \begin{pmatrix} b\eta_1 \\ 2a \eta_0 \\ b\eta_{-1} \end{pmatrix} = \sqrt{2} b \eta_0 \eta_1 + 2\sqrt{2} a\eta_0 \eta_1 = 0.$$ It follows that $b = -2a$, so that the most general second-order invariant for $\ell=1$ is (up to a multiplicative constant) $$F = \eta_0^2 - 2\eta_{-1}\eta_{1}.$$

## Example, $\ell=2$, $d=3$

This is a less trivial example, but it can still be done by hand (more or less.)

The most general polynomial is $$F = a \eta_0^3 + b \eta_{-1}\eta_{0}\eta_{1} + c \eta_{-2} \eta_{0} \eta_{2} + d \eta_{-2}\eta_{1}^2 + e \eta_{-1}^2\eta_{2}.$$ Operating on $\eta$ with $L_+$, where $$L_+ = \begin{pmatrix} 0& 2 & 0 & 0 & 0 \\ 0 & 0 & \sqrt{6} & 0 & 0 \\ 0 & 0 & 0 & \sqrt{6} & 0 \\ 0 & 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix},$$ we have that the dot product $L_+ \eta \cdot \nabla F$ is equal to $$\begin{pmatrix} 2 \eta_{-1} & \sqrt{6} \eta_0 & \sqrt{6} \eta_1 & 2 \eta_2 & 0 \end{pmatrix} \begin{pmatrix} c \eta _1^2+e \eta _0 \eta _2 \\ b \eta _0 \eta _1+2 d \eta _{-1} \eta _2 \\ 3 a \eta _0^2+b \eta _{-1} \eta _1+e \eta _{-2} \eta _2\\ b \eta _{-1} \eta _0+2 c \eta _{-2} \eta _1\\ d \eta _{-1}^2+e \eta _{-2} \eta _0\\ \end{pmatrix}.$$ Carrying out the dot product, we get a linear system of equations for the variables $(a,b,c,d,e)$, represented by this equation: $$\left( \begin{array}{ccccc} 3 \sqrt{6} & \sqrt{6} & 0 & 0 & 0 \\ 0 & \sqrt{6} & 2 & 0 & 0 \\ 0 & 2 & 0 & 2 \sqrt{6} & 2 \\ 0 & 0 & 4 & 0 & \sqrt{6} \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right) \begin{pmatrix} a \\ b \\ c \\ d \\ e \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix}.$$ I used Mathematica to row-reduce the above matrix. This yields $$\left( \begin{array}{ccccc} 1 & 0 & 0 & 0 & \frac{1}{6} \\ 0 & 1 & 0 & 0 & -\frac{1}{2} \\ 0 & 0 & 1 & 0 & \frac{\sqrt{\frac{3}{2}}}{2} \\ 0 & 0 & 0 & 1 & \frac{\sqrt{\frac{3}{2}}}{2} \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right) \begin{pmatrix} a \\ b \\ c \\ d \\ e \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix}.$$ We take $e$ to be the free parameter, which just plays the role of the overall multiplicative constant. Therefore $$F = -\frac{1}{6}\eta_0^3+\frac{1}{2} \eta_{-1} \eta_1 \eta_0+\eta_{-2} \eta_2 \eta_0-\frac{1}{2} \sqrt{\frac{3}{2}} \eta_{-2} \eta_1^2-\frac{1}{2} \sqrt{\frac{3}{2}} \eta_{-1}^2 \eta_2.$$ This agrees with what I've calculated using a different method (due to Marko Jarić.)