Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 160 / 1

The symmetric group acts transitively on the Young tableaux of a fixed shape. Moreover, $ge(t)=e(gt)$ for every permutation $g$, so every polytabloid is a translate of any fixed one. Hence the Specht module is a cyclic module generated by $e(t)$.

It remains to see that $e(t)\ne 0$. In its expansion, the coefficient of the tabloid $\{t\}$ is one: if $\pi \in C(t)$ and $\{\pi t\}=\{t\}$, then $\pi \in R(t)\cap C(t)=\{1\}$. Thus the generator, and therefore the module, is nonzero.

Write the transposition as $\tau =rc$ with $r\in R(t)$ and $c\in C(t)$. If $x$ is not one of the two entries moved by $\tau$, then $r(c(x))=x$. The entries $x$ and $c(x)$ lie in one column of $t$, while $c(x)$ and $r(c(x))=x$ lie in one row. A Young diagram has only one cell at the intersection of a specified row and column, so $c(x)=x$ and then $r(x)=x$.

Consequently both $r$ and $c$ fix every entry outside the support of $\tau$. On the two remaining entries each is either the identity or their transposition. They cannot both transpose them, since then $rc=1$, and they cannot both be the identity. Exactly one of $r,c$ is therefore $\tau$, proving that $\tau$ lies in exactly one of $R(t)$ and $C(t)$.

The transpositions in the row stabilizer numberbecause a cell in column $j$ has $j-1$ cells before it in its row. Similarly, the transpositions in the column stabilizer numberTheir difference isthe sum of the Young-diagram cell contents.

Conjugation by a permutation merely permutes the transpositions. Their sum $C$ is therefore a conjugacy class sum and belongs to the center of an associative algebra $\mathbb{C}{S}_n$. Since the complex Specht module is an irreducible representation, Schur lemma says that $C$ acts on it as a scalar, say $c(\lambda )$.

Compare the coefficient of $\{t\}$ in $Ce(t)$. For a transposition $\tau$, a summand $\tau \pi \{t\}$ with $\pi \in C(t)$ equals $\{t\}$ exactly when $\tau \pi \in R(t)$, equivalently $\tau \in R(t)C(t)$. Part b(i) then leaves two cases: a row transposition contributes $+1$ through $\pi =1$, while a column transposition contributes $-1$ through $\pi =\tau$. Every other transposition contributes zero.

The coefficient is consequently the number of row transpositions minus the number of column transpositions, which part b(ii) identifies with ${\sum }_{(i,j)\in \lambda }{c}_{i,j}(\lambda )$. The coefficient of $\{t\}$ in $c(\lambda )e(t)$ is $c(\lambda )$, so

In a Young-diagram hook, the cell $(i,j)$ has ${\lambda }_i-j$ cells to its right and ${\lambda }_j^{\prime }-i$ below it. Summing the arm lengths row by row and the leg lengths column by column givesAdding the content $c_{i,j}=j-i$ turns the summand on the right into $2j-1$. Since ${\sum }_{j=1}^m(2j-1)=m^2$, summing each row proves

The conjugate partition ${\lambda }^{\prime }$ has the same hook lengths as $\lambda$ and the opposite contents. Applying part d(i) to both diagrams and adding yields

Now sum the first identity of part d(i) over all partitions $\mu ⊢n$. Conjugation is a bijection on those partitions, so the total of ${\sum }_i({\mu }_i^{\prime }{)}^2$ equals the total of ${\sum }_i{\mu }_i^2$. Dividing the summed displayed identity by two gives

We use mathematical induction on $n$. The identity is immediate for the empty partition. Add a removable corner $(r,s)$ to a partition of $n-1$, and put $c=s-r$. Only the new hook and the hooks to its left in row $r$ or above it in column $s$ change. Each old affected hook length increases by one. If $S$ is the sum of those old hook lengths, direct substitution in the hook formula givesThere are $r+s-2$ affected old hooks, so the increase in the sum of squared hook lengths isThe increase in $n^2+\sum c_{i,j}^2$ is likewise $n^2-(n-1{)}^2+c^2=2n-1+c^2$. The induction closes and proves