Let $S\_{\pi}$ where $\pi$ is an integer partition of $n$, denote the Specht module corresponding to $\pi$.


I am trying to decompose the set of all homogeneous polynomials in $x\_1,x\_2,...,x\_n$ generated linearly (over any field of characteristic zero) by the monomials of the form $x\_i^2x\_jx\_k$ ($i,j,k$ are distinct), into Specht modules. I managed to do it for the polynomials generated by each of the following classes of monomials with $i,j,k,l$ distinct: $x\_i^3x\_j,x\_i^2x\_j^2,x\_i^4,x\_ix\_jx\_kx\_l$.


Once it is achieved for ${x\_i}^2x\_jx\_k$ a decomposition is successfully found for the space of degree 4 homogeneous polynomials in $n$ variables where $n$ is large enough, say $n\ge20$. This is the aim.


1. First $x\_i^3x\_j$ $(i\ne j)$: We know that $x\_i^3x\_j=\displaystyle \frac{x\_i^3x\_j+x\_ix\_j^3}2+\frac{x\_i^3x\_j-x\_ix\_j^3}2$


(a) The terms of the form $\displaystyle \frac{x\_i^3x\_j+x\_ix\_j^3}2$ generate linearly a space isomorphic as $S\_n$-modules (the module action is by permuting indices) to the homogeneous square-free degree 2 polynomials. This is isomorphic to $S\_{(n-2,2)}\oplus S\_{(n-1,1)}\oplus S\_{(n)}$.


(b) The terms of the form $\displaystyle\frac{x\_i^3x\_j-x\_ix\_j^3}2$ generate linearly a space isomorphic as $S\_n$-modules to the second exterior power of a vector space generated by $\{x\_1,x\_2,...,x\_n\}$ via $x\_i\wedge x\_j\mapsto\displaystyle\frac{x\_i^3x\_j-x\_ix\_j^3}2$. Thus this is isomorphic to $S\_{(n-2,1,1)}\oplus S\_{(n-1,1)}$.


So the decomposition is
$\displaystyle S\_{(n-2,2)}\oplus S\_{(n-2,1,1)}\oplus 2S\_{(n-1,1)}\oplus S\_{(n)}$


where "$2$" indicates that we have two copies of $S\_{(n-1,1)}$.
2. $x\_i^4$: This is simply a vector space generated by $x\_i^4$, and is a direct sum of the standard and the trivial representations of $S\_n$ that is $S\_{(n-1)}$ and $S\_{(n)}$. Thus the decomposition is $S\_{(n-1,1)}\oplus S\_{(n)}$.
3. $x\_ix\_jx\_kx\_l$: These generate the module isomorphic to module $M\_\lambda$ as in Bruce Sagan's book "The Symmetric Group" where $\lambda=(n-4,4)$ which one figures is just $S\_{(n-4,4)}\oplus S\_{(n-3,3)}\oplus S\_{(n-2,2)}\oplus S\_{(n-1,1)}\oplus S\_{(n)}$.
4. $x\_i^2x\_j^2$: These generate the module isomorphic to module $M\_\lambda$ where $\lambda=(n-2,2)$ which one figures is just $S\_{(n-2,2)}\oplus S\_{(n-1,1)}\oplus S\_{(n)}$.


The reason behind showing above is that these are alarmingly simple deductions though I can't seem to find one nearly as slick for the class $x\_i^2x\_jx\_k$. I found that there is a submodule isomorphic to $M\_{(n-3,3)}$ inside this class of polynomials. But that is the closeset I could get.


I tried dimension count as well, because the homogeneous degree-4 polynomials are of dimension ${n+3\choose 4}$.


The terms in the decompositions including the $M\_{(n-3,3)}$ above sum up to


$\displaystyle S\_{(n-4,4)}\oplus 2S\_{(n-3,3)}\oplus 4S\_{(n-2,2)}\oplus S\_{(n-2,1,1)}\oplus 6S\_{(n-1,1)}+5S\_{(n)}$.


This has a total dimension $n^4-2n^3+35n^2+38n$ which, subtracted from ${n+3\choose4}$ is $\frac13(n^3-3n^2-4n)$ which should be the sum of the dimensions of the remaining irreducible components (that is, copies of Specht modules).


But this counting technique leads to many possible decompositions and I am kind of out of ideas on this. Could anyone help?



Decomposing a polynomial ring into Specht Modules
The span of the monomials of the form $x\_i^2x\_jx\_k$ is the Young permutation module $M^{(n-3,2,1)}$. (*Proof.* Observe that $x\_1^2x\_2x\_3$ has stabiliser $\langle (2,3)\rangle \times S\_{\{4,\ldots,n\}}$, so the relevant Young subgroup is $S\_{n-3} \times S\_2 \times S\_1$.) Using Kostka numbers (equivalently, multiplicities of Schur functions in complete symmetric funtions) this decomposes as


$$M^{(n-3,2,1)} \cong S^{(n-3,2,1)} \oplus S^{(n-3,3)} \oplus S^{(n-2,1,1)} \oplus 2S^{(n-2,2)} \oplus 2S^{(n-1,1)} \oplus S^{(n)}$$


provided that $n \ge 6$. (I'm using superscripts for Specht modules since this is the notation I'm used to.) 


To make this explicit, Specht's original construction of Specht modules shows that $S^{(n-3,2,1)}$ is generated by the product of Vandermonde determinants


$$\left| \begin{matrix} 1 & 1 & 1 \\ x\_1 & x\_2 & x\_3 \\ x\_1^2 & x\_2^2 & x\_3^2 \end{matrix} \right| \left| \begin{matrix} 1 & 1 \\ x\_4 & x\_5 \end{matrix}\right| $$


and it's clear that the unique trivial submodule is spanned by 


$$x\_1^2x\_2x\_3 + x\_1^2x\_2x\_4 + \cdots + x\_{n-2}x\_{n-1}x\_n^2,$$ 


i.e. the sum of all monomials whose exponents are $2$, $1$, $1$, $0, \ldots, 0$ in some order.


For the other factors it is harder to make them explicit, but this can be done by using semistandard homomorphisms (see for instance James' lecture notes). Since we're working in characteristic zero, any non-zero homomorphism from a Specht module must be injective. I'll give some details here. By this theory,


$$\mathrm{Hom}\_{\mathbb{C}S\_n}(S^{(n-1,1)}, M^{(n-3,2,1)})$$


is spanned by the semistandard homomorphisms for the two semistandard tableaux of shape $(n-1,1)$ and content $(n-3,2,1)$: these have rows $1\ldots122,3$ and $1\ldots123$, $2$, respectively. Each such homomorphism extends to $M^{(n-1,1)}$. Taking as a model for $M^{(n-1,1)}$ the natural permutation module $\langle e\_1,\ldots, e\_n\rangle$, the corresponding extended homomorphisms are


$$e\_n \mapsto (x\_1x\_2^2 + x\_1x\_3^2 + \cdots + x\_{n-2}x\_{n-1}^2)x\_n$$


and


$$e\_n \mapsto (x\_1x\_2+x\_1x\_3+\cdots + x\_{n-2}x\_{n-1})x\_n^2,$$


respectively. One then has to restrict these homomorphisms to $S^{(n-1,1)} \subseteq M^{(n-1,1)}$ (for instance it is generated by $e\_n-e\_1$) to get two submodules of $M^{(n-3,2,1)}$ isomorphic to $S^{(n-1,1)}$, as in the claimed decomposition.


