The Lie group $\mathrm{SO}(3,\mathbb{R})$ has a representation $V_l$ of dimension $2l+1$ for each $l=0,1,\ldots$
I am looking for a single polynomial in $\mathbb{R}[V_l]$ which is invariant under the $\mathrm{SO}(3)$ action. The obvious example for $l=1$ is
$$x^2+y^2+z^2$$
This is a duplicate of this 2017 MSE question and the answer there is helpful but only partial.
Unless I'm mistaken, $\mathrm{SO}(3,\mathbb{R})$-invariance should boil down to $\mathrm{SU}(2)$-invariance, and thus to $\mathrm{SL}(2,\mathbb{C})$-invariance. This (under the guise of binary forms) has been studied for over 100 years and people apparently have found (minimal) generators of invariants of $k[\mathrm{Sym}^n(k^2)]$ for up to $n=8$, maybe further. But I'm not asking for a full set of generators, I just want one (nonconstant) polynomial for each $n$.
Edit: I'm clarifying the action of $\mathrm{SO}(3,\mathbb{R})$ on $\mathbb{R}^{l+1}$ for the benefit of a commenter below. I also provide a nontrivial example for $l=2$.
The action of $\mathrm{SO}(3,\mathbb{R})$ on $\mathbb{R}^{2l+1}$ has been known for a very long time. The idea is that there's a map $\mathrm{SU}(2)\rightarrow \mathrm{SO}(3,\mathbb{R})$ with kernel $I, -I$, so any irreducible representation of $\mathrm{SU}(2)$ where $-I$ acts trivially (namely, the irreps of odd dimension) will descend to a representation of $\mathrm{SO}(3,\mathbb{R})$. The simplest way of realizing this representation is in terms of harmonic polynomials. Note that $\mathrm{SO}(3)\subset \mathrm{GL}(3)$ acts naturally on $\mathbb{R}[x,y,z]_l$, the space of homogeneous polynomials of degree $l$. Let $\mathfrak{h}_3^l$ be the $2l+1$-dimensional vector subspace of $\mathbb{R}[x,y,z]_l$ consisting of the degree $l$ harmonic polynomials. This is defined as the kernel of the Laplacian $\Delta=\frac{\partial}{\partial x^2}+ \frac{\partial}{\partial y^2}+\frac{\partial}{\partial z^2}$. The subspace $\mathfrak{h}_3^l$ is preserved by the action of $\mathrm{SO}(3,\mathbb{R})$ because the elements of $\mathrm{SO}(3,\mathbb{R})$ commute with the Laplacian.
Finally, my question:
$\mathrm{SO}(3,\mathbb{R})$ acts on
$\mathbb{R}[\mathfrak{h}_3^l]$. I seek a polynomial in this space
which is fixed by the action (it is known to exist!).
Example for $l=1$: the vector space $\mathfrak{h}_3^1$ is spanned by the polynomials $p_1=x,p_2=y,p_3=z$ (they are all killed by the Laplacian). Now $\mathbb{R}[\mathfrak{h}_3^1]=\mathbb{R}[p_1,p_2,p_3]$ and there is an obvious $\mathrm{SO}(3,\mathbb{R})$-invariant polynomial of degree $2$: $p_1^2+p_2^2+p_3^2$.
Example for $l=2$: the vector space $\mathfrak{h}_3^2$ is spanned by the polynomials $p_1=x^2-y^2$, $p_2=y^2-z^2$, $q_1=yz$, $q_2=xz$, $q_3=xy$. Now consider the algebra $\mathbb{R}[\mathfrak{h}_3^2]=\mathbb{R}[p_1,p_2,q_1,q_2,q_3]$. A quick calculation shows that the polynomial $f:=p_1^2+p_1p_2+p_2^2+3(q_1^2+q_2^2+q_3^2)$ is $\mathrm{SO}(3,\mathbb{R})$-invariant.
Can anyone give such a polynomial for arbitrary $l$? Please note that the realization of $V_l$ as $\mathfrak{h}_3^l$ is not strictly necessary, and an intrinsic description would probably be better.
Best Answer
$\newcommand{\i}{\mathrm{i}}$ $\newcommand{\d}{\mathrm{d}}$
This necessarily is eventually the same as the purely algebraic answer, but the map
$$p\mapsto\int_{S^2}\frac{\d\Omega_x}{4\pi}p(x)^2$$ that takes a harmonic polynomial of degree $l$ to the average of its square over the unit sphere is, by construction, a rotation-invariant positive-definite quadratic form on harmonic polynomials of degree $l$.
In full detail: define a basis for the complex harmonic polynomials $\Upsilon_{\ell}^m(x,y,z)$ by the Herglotz expansion
$$\exp( x(\tfrac{v^2-u^2}{2}) -\i y(\tfrac{u^2+v^2}{2}) +zuv) =\sum_{\ell=0}^{\infty}\sum_{m=-\ell}^{\ell} \Upsilon_{\ell}^m(x,y,z) \tfrac{u^{l+m}}{(l+m)!} \tfrac{v^{l-m}}{(l-m)!}\text{.}$$ Then $\Upsilon_{\ell}^{-m}=(-1)^m(\Upsilon_{\ell}^m)^{*}$, so define the basis for the real harmonic polynomials $\Upsilon_{\ell m}$ via
$$\Upsilon_{\ell}^m =(-1)^m(\Upsilon_{\ell m} + \i \Upsilon_{\ell(-m)}) \quad$$ for $m>0$, with $\Upsilon_{\ell 0}=\Upsilon_{\ell}^0$. Then classically it has been known that the $\Upsilon_{\ell m}$ are orthogonal, and
$$\int_{S^2}\frac{\d\Omega}{4\pi}(\Upsilon_{\ell 0})^2 =\frac{(\ell!)^2}{2\ell+1}$$ $$\int_{S^2}\frac{\d\Omega}{4\pi}(\Upsilon_{\ell m})^2 =\frac{(\ell+m)!(\ell-m)!}{2(2\ell+1)}$$
so that the map $$\sum_{m=-\ell}^{\ell}a_{m}\Upsilon_{\ell m} \mapsto \frac{(\ell!)^2}{2\ell+1}a_0^2 + \sum_{m=1}^{\ell}\frac{(\ell+m)!(\ell-m)!}{2(2\ell+1)}(a_m^2 + a_{-m}^2) $$ is an invariant nondegenerate quadratic form on harmonic polynomials of degree $\ell$. Dualizing with respect to it and rescaling, it must be that $$Q_{\ell}(\Upsilon)=\binom{2\ell}{\ell}\Upsilon_{\ell 0}^2 +2\sum_{m=1}^{\ell}\binom{2\ell}{\ell+m}(\Upsilon_{\ell m}^2 +\Upsilon_{\ell(-m)}^2)$$ is an invariant element of the second symmetric power of the harmonic polynomials.