Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2010 / B / 4 / Solution

For a Young diagram cell $(i,j)$, its hook length is one plus the number of cells to its right and below. The hook-length formula isApplying it to the two shapes givesThe shapes here require $n\ge 4$.

A hook tabloid remembers its first-row set and the ordered singleton entries below it. The proposed map keeps that first-row set and merges the remaining entries into one unordered row. Thus it is well-defined on tabloids. Permuting entries before merging gives the same tabloid as merging before permuting, so it is a module homomorphism. Every target tabloid is reached by putting its second-row entries, in any order, into the singleton rows. Therefore it is surjective.

For $r=2$, the strictly dominating partitions of $\lambda =(n-2,1,1)$ are exactly $(n),(n-1,1),(n-2,2)$. Indeed dominance forces the first part to be at least $n-2$, and the possibilities for the remaining sum give precisely those shapes or $\lambda$ itself.

In characteristic zero, Maschke's theorem makes the permutation module semisimple, and Young's rule gives one copy of $S^{\lambda }$ and only strictly dominating additional shapes. The trivial multiplicity is one, since a transitive permutation module has a one-dimensional invariant space. For the standard representation $S^{(n-1,1)}$, Frobenius reciprocity identifies its multiplicity with the dimension of fixed vectors under the row stabilizer $S_{n-2}$. In its realization $\{(z_1,\dots ,z_n):\sum z_i=0\}$, those fixed vectors have the form $(c,\dots ,c,a,b)$ with $(n-2)c+a+b=0$, a two-dimensional space. Thus its multiplicity is two.

Finally $\dim M^{\lambda }=n(n-1)$. Subtracting the dimensions of one $S^{\lambda }$, one trivial module and two standard modules leaves $n(n-3)/2$, exactly one $S^{(n-2,2)}$. ThereforeThe additional multiplicities are $1,2,1$ respectively. If the printed weak summation index is read literally alongside its separate $S^{\lambda }$, then $m_{\lambda ,\lambda }=0$; the total multiplicity of $S^{\lambda }$ is one.