Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 103 / 4 / b

For $k=1$, the product is empty and equals the identity, the only $1$-cycle in $S_1$. Suppose the cycle-sum identity for Young–Jucys–Murphy elements holds at $k-1$:Multiplication by $X_k={\sum }_{a<k}(ak)$ gives terms $\sigma (ak)$. With permutation products acting right to left, this inserts $k$ immediately after $a$ in the cycle of $\sigma$: the arrows become $a\mapsto k\mapsto \sigma (a)$, with the other arrows unchanged.

Conversely, in any $k$-cycle there is a unique predecessor $a$ of $k$. Delete $k$ and join that predecessor to the successor of $k$; this uniquely recovers $\sigma$ and the factor $(ak)$. Hence every $k$-cycle occurs exactly once, and no other permutation occurs. Induction provesThus ${\xi }_k$ is a conjugacy-class sum in $\mathbb{C}{S}_k$, although it need not be central in $\mathbb{C}{S}_n$ when $k<n$.

Let $T$ be a standard Young tableau of shape $\lambda ⊢n$, and let $\mu$ be the shape formed by its entries $1,\dots ,k$. Since Young–Jucys–Murphy elements act by contents, the eigenvalue of ${\xi }_k$ on its Gelfand–Tsetlin basis vector isIf $\mu$ is not a hook partition, it contains $(2,2)$, whose content is zero, so this product is zero. If $\mu =(a+1,1^b)$ with $a+b+1=k$, its nonzero contents are $1,\dots ,a$ and $-1,\dots ,-b$. Their product is independent of the order of the labels. Therefore the eigenvalue isThe empty product at $k=1$ gives $1$. When $k=n$, $\mu =\lambda$, giving exactly the hook-versus-nonhook distinction suggested by the hint. For $k<n$, it is the intermediate shape $\mu$, rather than the final shape $\lambda$, that controls the answer.

There are $(k-1)!$ $k$-cycles in $S_k$, and each has the same character value of a representation on $S^{\lambda }$. Taking traces in the cycle-sum identity for Young–Jucys–Murphy elements yieldsOnly tableaux whose first $k$ entries form a hook partition contribute.

Write such a shape as ${\mu }_b=(k-b,1^b)$, for $0\le b\le k-1$. There are $f_{{\mu }_b}=\left(\genfrac{}{}{0ex}{}{k-1}{b}\right)$ ways to fill it: choose which $b$ of the labels $2,\dots ,k$ go below the first box. There are $f^{\lambda /{\mu }_b}$ ways to fill the remaining skew diagram with the labels $k+1,\dots ,n$, in standard order. The eigenvalue from part (ii) is $(-1{)}^b(k-b-1)!b!$, so the factoris $1$. Therefore the general character value isHere $f^{\lambda /\mu }$ counts standard skew Young tableaux, and is zero when $\mu$ is not contained in $\lambda$. In particular, for a full $n$-cycle,For $k=1$, the general formula gives $f^{\lambda /(1)}=f_{\lambda }$, the character value at the identity, as it should.