I have tried to decompose this as following spans over the real field.


$V\_1=\operatorname{span} \langle x\_{ij}^2\rangle$ 


$V\_2=\operatorname{span} \langle x\_{ij}x\_{jk}\rangle$


$V\_3=\operatorname{span} \langle x\_{ij}x\_{kl}\rangle$


where the index $ij$ stands for the subset $\{i,j\}$ of size 2, thus the ordering within does not matter and there are no variables of the form $x\_{ii}$. Also the indices $i,j,k,l$ are distinct.


Now, one determines


$V\_1\cong M\_{(n-2,2)}$
$V\_2\cong M\_{(n-3,2,1)}$


Next I tried for $V\_3$ the following:


$E\_{il,jk}=x\_{ij}x\_{kl}+x\_{ik}x\_{jl}$ is fixed precisely by the group $S\_{i,l}\times S\_{j,k}\times S\_{n-4}$.


Moreover, $E\_{il,jk}-E\_{ij,kl}+E\_{ik,jl}=2x\_{ij}x\_{kl}$, which shows that all of $V\_3$ is generated. One can define a map $\phi:M\_{(n-4,2,2)}\rightarrow V\_3$ by sending the tabloid with last two rows $i,j$ followed by $k,l$, to the element $E\_{ij,kl}$. One can get deceived into concluding that this is an isomorphism.


Unfortunately since the tabloid with rows $k,l$ and $i,j$ above swapped also maps to $x\_{ij}x\_{kl}$, this map is not an isomorphism.


However $x\_{ij}x\_{kl}+x\_{ik}x\_{jl}+x\_{il}x\_{jk}$ has fixed group $S\_{(n-4,4)}$ and is isomorphic to $M\_{(n-4,4)}$, so I know that this makes up ${n\choose 4}$ dimensions in $V\_3$, but a dimension count shows that the remaining $2{n\choose 4}$ dimensions has many possible decompositions into Specht modules.


Could someone help with zeroing in on one decomposition?



Decomposition of the homogeneous polynomial ring $\{\mathbb R[x_{ij}]_{1\le i,j\le n}\}$ of degree 2 into Specht modules
Write $W\_4$ for the $S\_4$ representation $$({\rm Ind}\_{S\_2 \times S\_2}^{S\_4} ({\rm triv} \otimes {\rm triv}))\_{S\_2}.$$
Then similarly to the answer to your previous question, you are considering $$Ind^{S\_{n}}\_{S\_4 \times S\_{n-4}} (W\_4 \otimes {\rm triv}),$$ which can be computed using the Pieri rule as soon as you know the decomposition of $W\_4$ into irreducibles. 


By definition $W\_4$ corresponds to the plethysm $(2) \circ (2)$. In general, plethysm is hard to compute but any specific case can be done, and in fact the plethysm $(n) \circ (2)$ is known in general. This should allow you to compute $(x\_{i\_1 j\_1}) \dots (x\_{i\_n j\_n})$ where all indices are distinct.


We have $(2) \circ (2) = (4) + (2,2)$. So the answer is $$((4) + (2,2))\*(n-4)$$ $$ = (n) + (n-1,1) + (n-2,2) + (n-3,3) + (n-4,4)$$ $$+ (n-2,2) + (n-3,2,1) + (n-4,2,2).$$


