Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2007 / B / 7 / e / Solution

We calculate the exact multiplicity without using Young's rule, via a two-subset incidence realization of a two-row Specht module. Let $V_2$ be the permutation representation on unordered pairs from a six-element set, and $V_1$ the natural permutation representation on its singletons. The equivariant incidence mapis surjective: the transpose map sends $\sum c_i{e}_i$ to the pair coefficients $c_i+c_j$, and all such sums can be zero only when every $c_i=0$ in characteristic zero. Thus $\dim \ker D=15-6=9$.

Identify the tabloids of shape $(4,2)$ with their two-element bottom rows. A polytabloid is a signed four-edge vector $e_{cd}-e_{ad}-e_{cb}+e_{ab}$, whose incidence at every vertex is zero. Hence $S^{(4,2)}\subseteq \ker D$. The standard polytabloid basis theorem and the hook-length formula give $\dim S^{(4,2)}=\left(\genfrac{}{}{0ex}{}{6}{2}\right)-\left(\genfrac{}{}{0ex}{}{6}{1}\right)=9$, so equality holds. By Maschke's theorem,Take $H={\Sigma }_3\times {\Sigma }_2\times {\Sigma }_1$, preserving blocks of sizes three, two and one. It has three orbits on singletons and five on pairs: pairs inside the first block, inside the second, and between each of the three distinct pairs of blocks. Thus $\dim V_1^H=3$ and $\dim V_2^H=5$. Averaging over $H$ makes taking fixed vectors exact in characteristic zero, so $\dim (S^{(4,2)}{)}^H=2$.

Finally $M^{(3,2,1)}={\mathrm{Ind}}_H^{{\Sigma }_6}1$. Frobenius reciprocity, and equality of invariant and coinvariant dimensions in characteristic zero, identify the required multiplicity with this two-dimensional fixed space. Thereforewhich in particular satisfies the requested lower and upper bounds.