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Pietro Majer
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Symmetric basis of harmonic homogeneus polynomials

Recently, a question about the beautiful theory of harmonic polynomials made me aware there is something I'd like to know for a long time.

As it is well known, for any number of variables $n$ and any degree $m$, there is a permutation invariant family of polynomials that generates linearly the space $\mathcal{H}_n^m$ of all (real) harmonic homogeneous polynomials in $n$ variables of degree $m$:

$$\mathcal{F}_ n^{\, m}:=\big\{ K\big(\partial^\alpha\|x\|^{2-n}\big)\, :\, \alpha\in\mathbb{N}^n\, ,\, \alpha_1+\dots+\alpha_n=m\, \big\}\, ,$$

where $K$ denotes the Kelvin transform, $Ku(x):=\|x\|^{2-n}u(\|x\|^{-2}x)$. Here, by permutation invariant set of polynomials I simply mean, if $P(x_1,\dots,x_n)$ is in it, then $P(x_{\sigma(1)},\dots,x_{\sigma(n)})$ also is in it, for any permutation $\sigma \in \mathfrak{S} _ n \\ .$ The family $\mathcal{F}_ n^{\\, m}$ is not linearly independent; however, taking only the elements corresponding to a suitable subset of multi-indices, namely, imposing e.g. $\alpha_n\le 1$, one actually finds a basis. What makes me a bit unhappy is, this constraint breaks the symmetry. So I wonder:

Is there an explicit permutation invariant subset of $\mathcal{F}_ n^{\\, m}$ which is a basis of $\mathcal{H}_n^m$ ? If not always, for what $n$ and $m$? And, more generally, how to make an explicit permutation invariant basis ?

Actually, I do not see obstructions to the existence of such a basis, but I would also not be completely sure about how to prove its existence. Before starting and trying to answer by myself, I liked to pose the question here. My excuses if it turns out to be trivial!

Pietro Majer
  • 60.5k
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  • 122
  • 269