# General Theorem About Symmetric Polynomials

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 $S_n$?

• 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! May 3, 2017 at 12:25
• 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. May 5, 2017 at 6:15
• 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 $f,g \in A$. May 5, 2017 at 9:01
• Looks okay. A minor thing: It seems you are assuming $k$ is not characteristic $2$, May 5, 2017 at 18:16
• 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). May 7, 2017 at 7:02

## 1 Answer

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 $$\{ 1, x_1 x_3+x_2 x_4, (x_1 x_3+x_2 x_4)^2 \}$$.

• This is great. Thanks! Mar 2, 2020 at 11:30