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

As 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 a 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 that 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'd like to pose the question here. My apologies if it turns out to be trivial!

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    $\begingroup$ If the result is false it can be disproved with a finite, albeit tedious, calculation. A finite-dimensional real vector space on which $S_n$ acts admits a permutation-invariant basis if and only if it is a direct sum of permutation representations, and for fixed $n$ this can be checked by comparing the character of the representation against the characters of all permutation representations (this is the tedious part). In particular the number of copies of the trivial representation is the number of orbits of the action of $S_n$, so if there are no such copies then the result is false. $\endgroup$
    – Qiaochu Yuan
    Commented Oct 20, 2011 at 19:45
  • $\begingroup$ Thank you... So, the existence of an invariant basis may really depend on the pair $(n,m)$ ? $\endgroup$
    – Pietro Majer
    Commented Oct 20, 2011 at 19:55
  • $\begingroup$ Sorry if I am being silly, but I don't see how the result you cite can be true for $n = 2$. $\endgroup$
    – Qiaochu Yuan
    Commented Oct 20, 2011 at 22:25
  • $\begingroup$ Harmonic $n$-variate polynomials of a given degree are $S_n$-invariant (as a set), and thus one could come up with an $S_n$-invariant basis, but the corresponding action won't be so simple. $\endgroup$
    – Dima Pasechnik
    Commented Nov 15, 2018 at 20:23

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Already the desired result is false for $n = 3, m = 2$, but for simpler reasons than I suggested in the comments. In this case the polynomials you give are $x^2 - 2y^2 + z^2, xy$ and their permutations. The sum of the permutations of $x^2 - 2y^2 + z^2$ is zero, so none of its permutations can be part of a permutation-invariant basis, and the span of the permutations of $xy$ don't span the entire space.

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