Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 160 / 2

Realize ${\xi }^{\lambda }$ as the permutation character on ordered set partitions with row sizes ${\lambda }_1,{\lambda }_2,\dots$. Under $S_m\times S_k$, an orbit is determined by the weak composition $\mu$ whose $i$th part counts elements of $\{m+1,\dots ,n\}$ in row $i$. Its stabilizer is the product of the Young subgroups for $\lambda -\mu$ and $\mu$. The orbit character is therefore the outer tensor product ${\xi }^{\lambda -\mu }\#{\xi }^{\mu }$, and summing the orbits giveswith impossible compositions contributing zero.

Insert this identity into the alternating definition of ${\psi }^{\lambda }$. Group the weak compositions by permutations of their parts and use character straightening; the alternating sum in the first tensor factor is ${\psi }^{\lambda -\mu }$, while the second factors combine once for each partition $\mu ⊢k$. Thus

The Murnaghan–Nakayama rule states that if a permutation has a $k$-cycle and remaining cycle type $\rho$, thenwhere $H$ ranges over removable rim hooks of length $k$ and $\mathrm{leg}(H)$ is one less than the number of rows occupied by $H$.

For the staircase $\delta =(m,m-1,\dots ,1)$,whenever $(i,j)$ is a cell, so every hook length is odd. A removable rim hook of length $k$ corresponds to a hook of length $k$ in the original diagram. If a cycle type $\mu$ contains an even part $k$, apply the Murnaghan–Nakayama rule to that part first. There are no terms in the sum, and therefore ${\chi }^{\delta }(\mu )=0$.

Conjugating a tableau exchanges row symmetrization with column antisymmetrization. The resulting module is the original Specht module twisted by the sign representation, giving the conjugate Specht characterTaking traces yields

If $\mathrm{sgn}(\mu )=-1$, then the cycle type $\mu$ contains an odd number of even parts and in particular contains an even part, so the hypothesis gives ${\chi }^{\lambda }(\mu )=0$. If $\mathrm{sgn}(\mu )=1$, multiplication by the sign changes nothing. Henceby part b(iii). Distinct partitions label distinct complex irreducible characters, so $\lambda ={\lambda }^{\prime }$.

Because $\lambda$ is self-conjugate, every off-diagonal hook of even length is paired with its transpose, while every diagonal hook has odd length. Suppose even hooks exist and let $L$ be their maximum length. Apply the Murnaghan–Nakayama rule to cycle types beginning with $L$ and complete the remaining cycle type with the principal hooks of the residual diagram. The principal-hook character value of a symmetric group makes each surviving residual character equal to $1$ or $-1$.

The assumed vanishing forces cancellation among the removable $L$-hooks. The standard maximal-hook comparison shows that the only possible cancellation is one transposed pair: the hooks must be $H_{1,j}$ and $H_{j,1}$ for a single $j>1$. Any further hook of length $L$, or a maximal hook with both indices greater than one, can be isolated by the residual principal-hook cycle type and would give a nonzero value. Thus either there are no even hooks or the maximum even length occurs exactly at that pair.

For a self-conjugate diagram, compare successive row lengths along the boundary. If every difference is one, the diagram is a staircase. Otherwise the first horizontal or vertical repetition creates an even hook. Following the boundary to the last such repetition creates either a second transposed pair of the same maximal even length or a maximal even hook away from the first row and column. Both alternatives contradict part ii. Therefore every successive row length decreases by one andTogether with part b(ii), this proves the staircase-character vanishing criterion.