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Let $F$ be a field. It is known that

1. Every symmetric polynomial in $F[x_1, \ldots, x_n]$ can be expressed as a polynomial in the elementary symmetric polynomials.

A polynomial $f\in F[x_1, \ldots, x_n]$ is said to be semi-symmetric if $f(x_{\sigma(1)}, \ldots, x_{\sigma(n)})=f(x_1, \ldots, x_n)$ for every even permutation $\sigma\in S_n$. We have

2. Every semi-symmetric polynomial in $F[x_1, \ldots, x_n]$ is of the form $f+\delta g$ where $f$ and $g$ are symmetric polynomials and $\delta=\prod_{i<j}(x_i-x_j)$.

A natural question arises. For a subgroup $G$ of $S_n$, let us say that a polynomial $f\in F[x_1, \ldots, x_n]$ is $G$-symmetric if $f(x_{\sigma(1)}, \ldots, x_{\sigma(n)})=f(x_1, \ldots, x_n)$ for all $\sigma\in G$.

Question. Are there some general results known about $G$-symmetric polynomials for an arbitrary subgroup $G$ of $S_n$?

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  • $\begingroup$ Please, do you have any references to a proof of your second claim? In the special case $F[x,y]$, I only managed to obtain $f+(x-y)g$, with $f$ and $g$ quotients of symmetric polynomials. Thanks! $\endgroup$
    – user237522
    May 3, 2017 at 12:25
  • $\begingroup$ Are you familiar with the Galois correspondence? We have an action of $S^n$ on $\mathcal F=F(x_1, \ldots, x_n)$, whose fixed field is $\mathcal E=F(e_1, \ldots, e_n)$, the $e_i$'s being the elementary symmetric polynomials. The semi-symmetric rational functions are the elements of $F(x_1, \ldots, x_n)$ fixed by $A_n$. Now $\delta$ has degree $[S_n:A_n]$ over $\mathcal E$ and is also fixed by $A_n$. Thus the result follows by the Galois correspondence. $\endgroup$ May 5, 2017 at 6:15
  • $\begingroup$ Thanks for your nice explanation! If I am not wrong, a similar explanation is as follows (for $n=2$): $A:=k[x+y,xy] \subset k[x+y,xy][x-y]=k[x,y]=:B$. Clearly, $(x-y)^2 \in A$, so $B$ is generated as an $A$-module by $\{1,x-y\}$, which shows that an element of $B$ is of the form $f1+g(x-y)$, where $f,g \in A$. $\endgroup$
    – user237522
    May 5, 2017 at 9:01
  • $\begingroup$ Looks okay. A minor thing: It seems you are assuming $k$ is not characteristic $2$, $\endgroup$ May 5, 2017 at 18:16
  • $\begingroup$ Thanks! Truly, I had in mind a field $k$ of characteristic zero. Thank you for your remark that my explanation is valid for any field $k$ of characteristic $\neq 2$. Indeed, in characteristic $2$, $x-y=x+y$, hence $k[x+y,xy]=k[x,y]$ (the symmetric elements generate the whole ring). $\endgroup$
    – user237522
    May 7, 2017 at 7:02

1 Answer 1

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This is a big subject called invariant theory. Here are some nice lecture notes I stumbled on when researching this answer. I can try to point out the main results.

Let $G$ be a subgroup of $S_n$. Let $S = k[x_1, x_2, \ldots, x_n]$ for a field $k$, let $A$ be the subring of $G$-invariants and let $R$ be the subring of $S_n$ invariants , so $R \subset A \subset S$. As you say, $R$ is generated by the elementary symmetric polynomials. Also, by Galois theory, $[\mathrm{Frac}(A) : \mathrm{Frac}(R)] = [S_n : G]$.

Note that the ring $S$ is finitely generated as an $R$-module (in fact, it is free -- a basis is the monomials $x_1^{a_1} x_2^{a_2} \cdots x_n^{a_n}$ with $0 \leq a_j \leq n-j$). Since $R$ is Noetherian, this shows that $A$ is also finitely generated as an $R$-module. So at least there will be a finite list of polynomials, analogous to your example of $\{ 1, \delta \}$ in the case $G = A_n$.

Life is much better than that! Assuming the characteristic of $k$ does not divide $\#(G)$, the ring $A$ is Cohen-Macaulay! (Theorem 8.1 in the linked notes. I believe the original source is Hochster and Eagon "Cohen-Macaulay Rings, Invariant Theory, and the Generic Perfection of Determinantal Loci" 1971, Proposition 13.) Using the properties of Cohen-Macaulay rings, one can show that $A$ is in fact free as an $R$-module. Since $[\mathrm{Frac}(A) : \mathrm{Frac}(R)] = [S_n : G]$, the rank of this free module in $[S_n:G]$.

Thus, there always exist polynomials $f_1$, $f_2$, ..., $f_{[S_n:G]}$ providing a free basis of $A$ over $R$. One can show that the $f_j$ can always be taken homogenous, and their degrees can be computed by Molien's formula: $$\frac{1}{\#(G)} \sum_{g \in G} \frac{1}{\det(\mathrm{Id} - t g)} = \frac{ \sum_{j=1}^{[S_n:G]} t^{\deg f_j}}{(1-t)(1-t^2) \cdots (1-t^n)} .$$ On the left hand side, we have identified a permutation $g$ with its permutation matrix.

For example, suppose we wanted to study polynomials in $x_1$, $x_2$, $x_3$, $x_4$ invariant under the dihedral symmetry group $G=\langle (12)(34), \ (1234) \rangle$. The left hand side is $$\frac{1}{8} \left( \frac{1}{(1-t)^4} + \frac{2}{(1-t)^2(1-t^2)}+\frac{3}{(1-t^2)^2}+\frac{2}{(1-t^4)} \right).$$ because $G$ contains one element of cycle type $1^4$, two of cycle type $1^2 2$, three of cycle type $2^2$ and two of cycle type $4$. We compute that this sum is $$\frac{1+t^2+t^4}{(1-t)(1-t^2)(1-t^3)(1-t^4)}$$

Thus, $A$ is free over $R$ with generators in degrees $0$, $2$ and $4$. An explicit choice is $\{ 1, x_1 x_3+x_2 x_4, (x_1 x_3+x_2 x_4)^2 \}$.

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  • $\begingroup$ This is great. Thanks! $\endgroup$ Mar 2, 2020 at 11:30

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