Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 160 / 4

For each permitted part $i\le e$, its multiplicity contributes the geometric series $1+x^i+x^{2i}+\cdots$. Thus the generating function isThe conjugate partition bijects diagrams with largest part at most $e$ and diagrams with at most $e$ rows, so the same series counts partitions into at most $e$ parts.

Subtract one from each of the $e$ positive parts. This bijects partitions into exactly $e$ parts with partitions into at most $e$ parts and decreases the size by $e$. The required generating function is therefore

For a self-conjugate diagram, its diagonal hooks partition all its cells and have distinct odd lengths. Conversely, placing hooks with any prescribed distinct odd lengths along the diagonal reconstructs one self-conjugate diagram. This gives the bijection in self-conjugate partition and distinct odd parts and the generating function

Choose a copy of $S^{\alpha }$ inside $M^{\lambda }$ and a nonzero polytabloid in it. Its image under a suitable Column antisymmetrizer of a Young tableau is nonzero. By the fact allowed in the question, this can happen only if $\alpha$ dominates $\lambda$. Hence $m_{\alpha }>0$ implies $\alpha ⊵\lambda$.

Young's rule givesthe Kostka number counting semistandard tableaux of shape $\alpha$ and content $\lambda$.

For $n\ge 4$, the only partitions dominating $(n-2,1,1)$ areThe entries of content $(n-2,1,1)$ consist of $n-2$ copies of $1$ and one copy each of $2$ and $3$. There is one semistandard tableau of shape $(n)$; two of shape $(n-1,1)$, according as the lower cell contains $2$ or $3$; one of shape $(n-2,2)$; and one of shape $(n-2,1,1)$. Thereforeand every other multiplicity is zero. For $n=3$, the coincident middle shapes combine to give $m_{(3)}=1$, $m_{(2,1)}=2$, and $m_{(1,1,1)}=1$.

Identify $M=M^{(n-2,1^2)}$ with the vector space having basis $v_{ij}$ for ordered pairs $i\ne j$. Let $P$ be the permutation module with basis $e_1,\dots ,e_n$, let $N=M^{(n-2,2)}$ have basis $w_{\{i,j\}}$, and define $S_n$-homomorphismsPutand let $\epsilon :M\to \mathbb{F}$ send every basis vector to one.

The augmentation submoduleis $S^{(n-1,1)}$. The map $p_1$ identifies $(\ker \epsilon )/U$ with $A$, and $p_2$ identifies $U/V$ with another copy of $A$. Also $M/\ker \epsilon \cong S^{(n)}$.

The subspace $V$ is generated by rectangle differencesTheir images under $q$ are the standard polytabloid generators of $S^{(n-2,2)}$, so $q(V)=S^{(n-2,2)}$. The kernel of $q{|}_V$ is generated by the alternating oriented-triangle relations; identifying these with the column antisymmetrizations of shape $(n-2,1,1)$ givesThese descriptions use coefficients $0$, $1$, and $-1$ only, so they remain valid over every field.

For $n\ge 4$, a required strict filtration iswith successive quotientsFor $n=3$, $q(V)=0$, so omit the repeated term $V=\ker (q{|}_V)$; the factors are $S^{(1,1,1)}$, two copies of $S^{(2,1)}$, and $S^{(3)}$. This is the Specht filtration of the ordered-pair permutation module.