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 ${n+3\choose 4}$. But this counting technique leads to many possible decompositions. I am kind of out of ideas on this. Could anyone help?

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?