Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 160 (Representation Theory of Symmetric Groups)

The symmetric group acts transitively on the tableaux of a fixed shape. Thus every $s\in {\Delta }^{\lambda }$ is $gt$ for some $g\in S_n$, and the definition of a polytabloid givesSince the Specht module is spanned by all $e(s)$, it is the cyclic $\mathbb{F}{S}_n$-module generated by $e(t)$.

Ina fixed tabloid $\{u\}$ can occur at most once. Indeed, if $\{\pi t\}=\{\sigma t\}$ for $\pi ,\sigma \in C(t)$, then ${\sigma }^{-1}\pi$ belongs to both the row and column stabilizers of $t$. Their intersection is trivial, so $\pi =\sigma$. The coefficient $⟨e(t),\{u\}⟩$ is consequently $0$, $1$, or $-1$.

For a tabloid $U$, let $r_U(k)$ be the row containing $k$, and order tabloids lexicographically byIf $t$ is standard and $1\ne \pi \in C(t)$, let $k$ be the least number moved by $\pi$. Every smaller entry is fixed, while ${\pi }^{-1}(k)$ is a larger entry below $k$ in the same column. Hence $\{\pi t\}$ agrees with $\{t\}$ before $k$ and has $r_{\pi t}(k)>r_t(k)$. Thus $\{t\}$ is the unique least tabloid in $e(t)$ and has coefficient one.

Distinct standard tableaux have distinct tabloids, since increasing each row recovers the tableau from its tabloid. A linear relation among standard polytabloids now has a least leading tabloid, which cannot cancel. They are therefore linearly independent, as in the linear independence of standard polytabloids.

The assumed one-dimensional-image property givesfor some $c\in \mathbb{F}$. The coefficient of $\{t\}$ in $e(t)$ is one, soThe tabloid bilinear form is invariant, and the involution on the group algebra fixes the Column antisymmetrizer of a Young tableau because inversion preserves sign. Thereforewhich proves the formula.

The equality $h\{t\}=h^{\ast }\{t^{\ast }\}$ says that $h(i)$ and $h^{\ast }(i)$ lie in corresponding rows for every entry $i$. For $i$ in the leftmost column, both $h(i)$ and $h^{\ast }(i)$ remain in that column. Corresponding rows therefore force $h(i)=h^{\ast }(i)$.

Remove the leftmost entry of every row and repeat the argument on the shortened tableaux. Induction across the columns gives equality on every entry, so $h=h^{\ast }$.

The nonzero coefficients and part a(ii) give unique elements $h\in C(t)$ and $h^{\ast }\in C(t^{\ast })$ such thatPart i gives $h=h^{\ast }=:k$, so $k\in C(t)\cap C(t^{\ast })$. Both coefficients are then $\mathrm{sgn}(k)$ and hence are equal.

By part ii, the common tabloids in the supports of $e(t)$ and $e(t^{\ast })$ are precisely $k\{t\}$ with $k\in C(t)\cap C(t^{\ast })$, and each contributes $\mathrm{sgn}(k{)}^2=1$. ConsequentlyFor the $a_j$ rows of length $j$, an element of the intersection applies the same arbitrary permutation of those rows independently in each of their $j$ columns. Hence that row length contributes $(a_j!{)}^j$, and

Let $\theta \in {\mathrm{End}}_{\mathbb{F}{S}_n}(S^{\lambda })$ and putSince $\lambda$ is $p$-regular, every $a_j<p$, so $q\ne 0$ in $\mathbb{F}$. Parts b and c give $b_{t}e(t^{\ast })=qe(t)$. Since $\theta$ commutes with the group-algebra action,The left side belongs to $b_t{M}^{\lambda }=\mathbb{F}e(t)$. Division by $q$ shows that $\theta (e(t))$ is a scalar multiple of $e(t)$. Part a(i) says that $e(t)$ generates $S^{\lambda }$, so $\theta$ is that scalar multiple of the identity. This proves the endomorphism theorem for a regular Specht module.

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.

Moving a bead one step upward on an $e$-runner abacus of a partition removes an $e$-hook and decreases the size by $e$. Sliding all beads upward reaches the core of a partition after exactly $w_e(\lambda )$ moves. Therefore

The sum of the sizes of the partitions at the next level of the quotient tower of a partition isFor $e>1$, repeated passage to a quotient therefore makes every branch empty after finitely many levels. For $e=1$, the core is empty and $Q_1(\lambda )=(\lambda )$, so a nonempty partition repeats forever. Thus the tower has finite depth exactly when $e>1$ or $\lambda$ is empty.

Using the abacus of a partition in James's convention, read runner $r$ as a one-runner abacus: replace each occupied position $r+eq$ by $q$. After the usual harmless shift of its beta set, the resulting partition is ${\lambda }^{(r)}$. Doing this for $r=0,\dots ,e-1$ produces

Conjugating a partition complements beads and gaps and reflects the abacus. Reflection sends runner $r$ to runner $e-1-r$, reverses bead-gap order, and conjugates the runner partition. Hence

Removing an $e$-hook removes one cell of every residue modulo $e$. If $\lambda$ and $\mu$ have the same $e$-core and the same size, part a(i) shows that they have the same $e$-weight. Their residue multisets are therefore obtained from the common core by adjoining the same number of complete residue sets $\{0,1,\dots ,e-1\}$. Thus their $e$-contents are equal.

The beta number in row $i$ isSince $e\mid m$,

Pad both partitions to the same number $m$ of rows, with $e\mid m$. For residue $s$, subtracting the number of cells of residue $s-1$ from the number of cells of residue $s$ telescopes row by row. Since the row-start residues $1-i$ are the same for both padded diagrams, equality of $e$-contents and part ii imply that their beta sets contain the same number of elements on each abacus runner.

Sliding beads upward to obtain the $e$-core preserves the number of beads on every runner. A packed runner is uniquely determined by that number, so the two packed abaci coincide. Hence $C_e(\lambda )=C_e(\mu )$.

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.