Invariant polynomials under an $S_4$-action

Question

Suppose $X = \mathbb{Q}-span \langle x_1, x_2, x_3 \rangle \cong \mathbb{Q}^3$ is a $3$-dimensional $\mathbb{Q}$-vector space with some basis $(x_1, x_2,x_3)$. We let the symmetric group $S_4$ act on $X$ as the product of the standard representation and the sign representation (so $X$ is an irrep of $S_4$). Consider the symmetric product $S^+(X)$ of $X$, where the $^+$ indicates that we're only considering non-zero degrees (i.e. we disregard the $0$-degree factor $\mathbb{Q}$ in the symmetric product).

Question: How can I describe the space of invariants $(S^+(X))^{S_4}$?

Some ideas: I know that we can identify the (non-zero degree) symmetric product of $X$ with the space of positive degree polynomials in $3$ variables over $\mathbb{Q}$: $$S^+(X) \cong \mathbb{Q}_{>0}[x_1, x_2, x_3].$$ So what I want to find are really the invariant polynomials $( \mathbb{Q}_{>0}[x_1, x_2, x_3])^{S_4}$.

More information about the representation I'm considering and my specific choice of basis: The elements $x_1, x_2, x_3$ I've chosen are such that the generating transpositions $(1, 2), (2, 3), (3,4) \in S_4$ have the following effect (according to my calculations): \begin{align*} (1,2): \hspace{2mm} &x_1 \mapsto -x_1, \hspace{3mm} x_2 \mapsto -x_1+x_2+x_3, \hspace{3mm} x_3 \mapsto -x_3 \\ (2,3): \hspace{2mm} &x_1 \mapsto -x_1, \hspace{3mm} x_2 \mapsto -x_3, \hspace{3mm} x_3 \mapsto -x_2 \\ (3,4): \hspace{2mm} &x_1 \mapsto x_3, \hspace{3mm} x_2 \mapsto -x_2, \hspace{3mm} x_3 \mapsto x_1 \end{align*}

I've calculated that for instance the polynomial $$g = (-x_1 + x_2 + x_3)x_1x_2x_3 \in \mathbb{Q}_{>0}[x_1, x_2, x_3]$$ is invariant under the action of the $(1, 2), (2, 3), (3,4) \in S_4$. Since these transpositions generate $S_4$, it follows that $g \in ( \mathbb{Q}_{>0}[x_1, x_2, x_3])^{S_4}$. But does the basis I've chosen matter in any way? And how do I find the entire space of invariant polynomials?

Edit: I would also greatly appreciate a reference where ways of finding invariant polynomials (in simple cases) are treated. I have the same problem also arising for actions of some semidirect and direct products involving $S_4$ and $S_2$, instead of just $S_4$.

Maybe it's also possible to treat this using computer programming. Unfortunately, I have zero experience with computer algebra programs. If this is the way to here though, then I'd be happy to learn about it.

I'm glad for any help!

Answer 1

I think I understand the intuition behind your question. For that reason I worked out an example to show how representations of $S_4$ reflect when applied to polynomials. For simplicity I consider polynomials in four indeterminates $X, Y, Z$ and $T$, the latter can always be replaced by $1$ afterwards. I chose randomly to study the representation on polynomials of degree $7$. I did my calculations using a cas named GAP, doing it by hand seemed to be a bit tedious.

The vector space spanned by the monomials of degree $7$ splits into three subspaces, Those of the form $X^2Y^2Z^2T$ of dimension $4$, those of the form $X^3Y^2ZT$ of dimension $12$ and those of the form $X^4YZY$ of dimension $4$. We'll see how these subrepresentations further split into irreducible subspaces. Therefor we will use the character table of $S_4$:

 gap> Irr(G);
 [ Character( CharacterTable( S4 ), [ 1, 1, 1, 1, 1 ] ),
  Character( CharacterTable( S4 ), [ 1, -1, -1, 1, 1 ] ),
  Character( CharacterTable( S4 ), [ 2, 0, 0, -1, 2 ] ),
  Character( CharacterTable( S4 ), [ 3, 1, -1, 0, -1 ] ),
  Character( CharacterTable( S4 ), [ 3, -1, 1, 0, -1 ] ) ]

Characters are the traces of the irreducible representation matrices. Each row shows the value (of a representant of a conjugacy class) of a group element, the first is the value of the identity and thus the dimension of the representation. There is a scalar product defined on characters by $<\chi, \psi> = ( \sum_{g \in G} \chi(g) \psi(g^{-1}) ) / |G|$ that makes the characters an orthonormal system so that it an easy matter to detect how a representation is structured using its character.

In our case the characters are easy to calculate since permutations map monomials to monomials. So to calculate the trace of a matrix representing a permutation we only have to count how many monomials are mapped onto themselves. For our three subspaces we obtain respectively $[4,0,2,1,0], [12,0,2,0,0]$ and $[4,0,2,1,0]$. Based on the scalar product with the irreducible characters the first and last subspace are the direct sum of the trivial and the standard representation. The second subspace contains the trivial representation once, the standard representation twice , the "signed' 3D representation once and the "exceptional" 2D representation once.