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The polynomial ring $\mathbb{C}[x_1,\ldots,x_n]$ decomposes as a direct sum of isotypic components for the action of the symmetric group $S_n$. The isotypic component of the trivial representation is simply the ring of symmetric functions. The complementary summand--that is, the direct sum of all the other isotypic components--is a free module of rank $n!-1$ over the ring of symmetric functions. I would like to have an explicit basis, or at least an explicit set of generators, for this module.

For example, if $n=2$, we have $\mathbb{C}[x_1,x_2] = \mathbb{C}[x_1,x_2]^{S_2} \oplus \mathbb{C}[x_1,x_2]^{S_2}\cdot (x_1-x_2)$, so we have a basis consisting of the single element $x_1-x_2$. More generally, the isotypic component of the sign representation is equal to symmetric functions times the Vandermonde determinant, so the Vandermonde determinant should probably be one of the elements of my generating set.

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At Dotsenko pointed out, the quotient of ${\bf C}[x_1,\dots,x_n]$ by the ideal of homogeneous positive degree symmetric functions is isomorphic to the regular representation of $S_n$, so you might ask about a good choice of basis for this compatible with the $S_n$-action that you can lift to ${\bf C}[x_1, \dots, x_n]$. In particular, each irreducible ${\bf M}_\lambda$ appears $\dim {\bf M}_\lambda$ times.

It is a classical fact that when ${\bf M}_\lambda$ appears the first time (i.e., lowest possible degree) that it appears with multiplicity 1, so there is a canonical choice for this representation (and this was constructed by Specht(?) and generalized by Macdonald to other root systems). To construct this, let $T$ be any filling of the Young diagram of $\lambda$ with entries $1, \dots, n$. For each column, take the Vandermonde product $\prod_{i < j} (x_i - x_j)$ where $i,j$ range over distinct elements in that column. Let $p_T$ be the product of these determinants. The space spanned by all $p_T$ is this lowest-degree occurrence (for a basis, only consider $T$ that are standard Young tableaux).

But for the other occurrences there will be ambiguities. The following paper of Terasoma and Yamada give some choices (though it's not canonical).

Annoucement: http://projecteuclid.org/euclid.pja/1195511538

Details (joint with Ariki): http://projecteuclid.org/euclid.hmj/1206127144

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The answer depends on the structure that you want on the collection of generators you are looking for. If you really want just generators of $\mathbb{C}[x_1,\ldots,x_n]$ over $\mathbb{C}[x_1,\ldots,x_n]^{S_n}$ with the extra condition that $\mathbb{C}[x_1,\ldots,x_n]^{S_n}\cdot 1$ splits out naturally, you can take all monomials $x_1^{a_1}\cdots x_n^{a_n}$ where $0\le x_i<i$. If you want these to be a bit more symmetric, in particular to contain Vandermonde $V(x_1,\ldots,x_n)$, you can take the elements $\partial_1^{a_1}\cdots\partial_n^{a_n}V(x_1,\ldots,x_n)$ where $\partial_i$ is the derivative with respect to $x_i$, and $a_i$ satisfy the same restriction $0\le a_i<i$. I think both of these are discussed in one of the chapters of "Maths talks for undergraduates" by Lang. (In general, the keyword to look for is "harmonic polynomials for the symmetric group"). If you need some further properties of that set, let us know :-)

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  • $\begingroup$ I think he wants a decomposition into $S_n$-submodules. $\endgroup$
    – darij grinberg
    Commented Oct 15, 2014 at 16:48
  • $\begingroup$ @darijgrinberg: this is not clear to me from the way this question is stated: the way it is written, he asks for generators of the complement of the isotypic component of the trivial representation viewed as a module over the ring of symmetric functions. In any case, as an $S_n$-module it is a regular representation, so one presumably can act on elements I mention by Young symmetrizers (the second description actually gives a basis in the space of all partial derivatives of the Vandermonde, so it is a space of generators which is closed under the $S_n$-action). $\endgroup$
    – Vladimir Dotsenko
    Commented Oct 15, 2014 at 16:56
  • $\begingroup$ "it is a regular representation": why? $\endgroup$
    – darij grinberg
    Commented Oct 15, 2014 at 16:58
  • $\begingroup$ @darijgrinberg Well, roughly because geometrically the fiber over a generic point consists of a single orbit of $S_n$ acting via permutations of coordinates. Probably you can make it more algebraic passing to fields of quotients, looking at the field extension $\mathbb{C}(x_1,\ldots,x_n):\mathbb{C}(x_1,\ldots,x_n)^{S_n}$, and doing some basic Galois theory. $\endgroup$
    – Vladimir Dotsenko
    Commented Oct 15, 2014 at 17:05
  • $\begingroup$ Vladimir: I'm sorry, but I don't understand your answer. In the (very easy) case n=2, I am asking for a set of generators for the isotypic component of the sign representation. This isotypic component does not contain ANY monomials at all; in particular, it cannot be spanned by monomials. Also, the statement that you meant to make about the regular representation is that the ring of all polynomials modulo positive degree symmetric polynomials is isomorphic to the regular representation of $S_n$. The module that I'm asking about is infinite dimensional as a vector space. $\endgroup$
    – Nicholas Proudfoot
    Commented Oct 15, 2014 at 18:08

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