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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.)

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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.)

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