Let $G=\mathfrak{S}\_n$ be the symmetric group on $n$ elements. Via permuting the variables the polynomial ring $S=\mathbb{C}[x\_1,\ldots,x\_n]$ becomes an $G$-module. It is not hard to see that every irreducible representation of $G$ appears in $S$: For example the span of the orbit of the monomial $\prod\_{i=1}^nx\_i^i$ is the regular representation of $G$. We consider the natural map $M: S\to\textrm{Hom}(S,S)$ that assigns a polynomial $f$ the map $S\to S,\,g\mapsto f\cdot g$. Clearly, $F$ is $G$-linear. I am interested how general this map is.


More precisely, let $U,V,W$ irreducible $G$-modules and let $F:U\to\textrm{Hom}(V,W)$ a $G$-linear map. Are there $G$-linear maps $A: U\to S$, $B: V\to S$ and $C: S\to W$ such that for all $u\in U$ we have $F(u)=C\circ M(A(u))\circ B$?



Can every $\mathfrak{S}_n$-linear map be realized by a multiplication?
Yes. Take an injective map $A : U \to S$ and an injective map $B' : V \to S$.


Now compose $B'$ with the map that sends $x\_i$ to $x\_i^m$ for all $i$, where $m$ is greater than the degree of any polynomial in the image of $A$.


Then the composed map $A \otimes B : U \otimes V \to S$ will be injective because, as a module, $S$ is the tensor product of polynomials of degree $<m$ in each variable with polynomials in $x\_1^m,\dots, x\_n^m$, and tensor products of injective maps are injective.


Hence the map $U \otimes V \to W$ is factors through the image of $A \otimes B$ in $S$. Since $S\_n$ is a finite group, by semisimplicity, we can extend this map from the image to the whole space.


