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

The symmetric group acts on tabloids by permuting their entries; extending linearly gives the permutation module $M^{\mu }$. Its dimension isA tableau of type $\mu$ contains exactly ${\mu }_i$ copies of the label $i$. Fix a numbering of the $\lambda$-diagram's cells by $1,\dots ,n$. For a filling $T$, let the $i$th row set of an associated $\mu$-tabloid be the set of cell numbers whose labels in $T$ equal $i$. This is a bijection from type-$\mu$ fillings to $\mu$-tabloids.

Define $\sigma T$ by moving labels along the numbered positions: its label at cell $j$ is the old label at cell ${\sigma }^{-1}j$. The associated row set is then the image under $\sigma$ of the old row set. Thus the bijection is equivariant, and its linear extension makes $F[{\mathcal{T}}_{\lambda ,\mu }]$ and $M^{\mu }$ isomorphic representations. Extend from group elements to the group algebra by linearity.

Young's rule states that in characteristic zero,where the Kostka number $K_{\lambda \mu }$ counts Semistandard Young tableaux of shape $\lambda$ and type $\mu$: rows are weakly increasing, columns strictly increasing, and label $i$ occurs ${\mu }_i$ times. The summands are irreducible Specht modules, and the coefficients are nonnegative integer multiplicities.

For type $(1^n)$, all labels are distinct, so a semistandard tableau is precisely a standard Young tableau. Thus $K_{\lambda ,(1^n)}=f^{\lambda }$, and Young's rule givesThis is also the decomposition of the regular representation, since tabloids with one entry per row are naturally indexed by all permutations.

For shape and type both $\mu$, column strictness makes every entry in row $r$ at least $r$. All ${\mu }_1$ entries equal to one must therefore be in the first row, which has exactly that length; it is filled with ones. Remove that row and repeat after subtracting one from the remaining labels. Induction gives the unique filling whose $i$th row consists of $i$'s. Thus $K_{\mu \mu }=1$, and Young's rule proves that $S^{\mu }$ occurs exactly once.

More generally, if a semistandard filling of shape $\lambda$ and type $\mu$ exists, its entries at most $k$ must lie in the first $k$ rows. Counting them givesThese are the inequalities defining the dominance order on partitions. Hence $K_{\lambda \mu }>0$ implies $\boxed{\lambda \ge \mu }$, and the asserted restriction on summands follows from Young's rule.

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.