Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2009 / B / 5

For a tableau $t$, let $C_t$ be its column stabilizer and $e_t={\sum }_{\sigma \in C_t}\mathrm{sgn}(\sigma )\{\sigma t\}$ its polytabloid. The Specht module is the span of all such polytabloids inside the tabloid permutation module.

A tabloid with second row of size $k$ is determined by that subset, so $\dim M^{(n-k,k)}=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. The two-row hook formula givesAt $k=3$ this provesBy Maschke's theorem, characteristic-zero permutation modules are completely reducible, so their Specht submodules are summands. The nonzero sum of all three-subset tabloids is fixed by every permutation; its span is a copy of the trivial module $S^{(n)}$.

A permutation sends $v_a$ to $v_{ga}$, $w_{ab}$ to $w_{ga,gb}$ and $x_{abc}$ to $x_{ga,gb,gc}$. Relabeling the summation indices givesLinearity therefore proves both maps are module homomorphisms. The map denoted $\varphi$ in the second displayed source formula is ${\varphi }_2$.

For the standard tableau with second row $2,4$, the first entries in the top row are $1,3$, so its nontrivial columns are $(1,2)$ and $(3,4)$. HenceIn ${\varphi }_2(e_t)$ the coefficient of $x_{24n}$ is one: among those four pairs only $\{2,4\}$ is contained in $\{2,4,n\}$, since $n\ge 6$. Thus the image is nonzero. Its restriction to the characteristic-zero simple module $S^{(n-2,2)}$ has zero kernel, giving the requested isomorphic submodule.

For $S^{(n-1,1)}$, choose the standard tableau with lower entry two and top-left entry one. Its polytabloid is $v_2-v_1$. Each three-subset containing $a$ occurs twice in ${\varphi }_2{\varphi }_1(v_a)$, once for each choice of its other two entries as the intermediate $b$. Consequently the coefficient of $x_{23n}$ in ${\varphi }_2{\varphi }_1(v_2-v_1)$ is two. This is nonzero in characteristic zero. Simplicity again makes the restriction injective, giving a copy of $S^{(n-1,1)}$.

The four constructed submodules have the distinct shapes $(n-3,3)$, $(n-2,2)$, $(n-1,1)$ and $(n)$, so they are pairwise nonisomorphic irreducible characteristic-zero modules. Complete reducibility therefore makes their sum direct. The given decomposition of $M^{(n-2,2)}$ and the dimension identity in (a) show their dimensions sum to $\dim M^{(n-3,3)}$. Their direct sum is consequently the whole module: