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

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