Decomposition of the homogeneous polynomial ring $\{\mathbb R[x_{ij}]_{1\le i,j\le n}\}$ of degree 2 into Specht modules

Question

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?

Answer 1

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 $$((4) + (2,2))*(n-4)$$ $$ = (n) + (n-1,1) + (n-2,2) + (n-3,3) + (n-4,4)$$ $$+ (n-2,2) + (n-3,2,1) + (n-4,2,2).$$