Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 160 (Representation Theory of Symmetric Groups)

Expand the Column antisymmetrizer of a Young tableau:If two entries in one column of $v$ lie in the same row of $w$, their transposition belongs to both $C(v)$ and the row stabilizer of $w$, so the terms cancel in pairs. The assumption $b_v\{w\}\ne 0$ therefore says that every row of $w$ meets every column of $v$ in at most one entry.

The first row of $w$ has ${\lambda }_1$ entries, while $v$ has exactly ${\lambda }_1$ nonempty columns. It must consequently contain exactly one entry from each column of $v$. Permuting within each column puts these entries in the first row positions of $v$. Delete the matched first rows and repeat the argument on the remaining Young diagram. The product of the resulting column permutations is an element $h\in C(v)$ for which the row sets of $hv$ are those of $w$. ThusThis is the nonzero column antisymmetrizer criterion.

The James submodule theorem says that for every $F{S}_n$-submodule $U\le M^{\lambda }$, eitherwhere orthogonality is taken with respect to the tabloid bilinear form.

Fix a $\lambda$-tableau $t$. Part a shows that for every tabloid $\{s\}$, the vector $b_t\{s\}$ is either zero or a signed copy of the polytabloid $e(t)$. Comparing the coefficient of $\{t\}$ gives the precise identityIf $U⊈(S^{\lambda }{)}^{\perp }$, choose $u\in U$ and a tableau $t$ with $⟨u,e(t)⟩\ne 0$. Since $U$ is a submodule, the identity puts $e(t)$ in $U$. Every polytabloid of shape $\lambda$ is an $S_n$-translate of $e(t)$, so their span $S^{\lambda }$ lies in $U$. If no such $u,t$ exist, then by definition $U\subseteq (S^{\lambda }{)}^{\perp }$. This proves the theorem over the arbitrary field $F$.

The row stabilizer of the transposed tableau is $R(t^{\prime })=C(t)$. For $r\in R(t^{\prime })$, the polytabloid satisfiesThe two signs cancel in the tensor product, soThus the proposed value depends only on the tabloid $\{g{t}^{\prime }\}$ and $\theta$ is well-defined. Its definition immediately givesso it is an $F{S}_n$-homomorphism. Since any $e(w)$ is $ge(t)$ for some $g$, its images contain every generator $e(w)\otimes e(u)$ of $S^{\lambda }\otimes S^{(1^n)}$. Hence $\theta$ is surjective.

Because $C(t^{\prime })=R(t)$,Applying $\theta$ and using $re(u)=\mathrm{sgn}(r)e(u)$ givesThus $m={\sum }_{r\in R(t)}re(t)$. Every $r$ fixes the tabloid $\{t\}$, while $e(t)$ has coefficient one at $\{t\}$. Invariance of the tabloid bilinear form now yields

The preceding part shows that $\theta (e(t^{\prime }))\ne 0$ in characteristic zero, so $S^{{\lambda }^{\prime }}⊈\ker \theta$. Apply the James submodule theorem to the proper submodule $\ker \theta \le M^{{\lambda }^{\prime }}$ to obtainThe Hook-length formula gives $\dim S^{\lambda }=\dim S^{{\lambda }^{\prime }}$. Surjectivity of $\theta$ therefore givesand hence

Fix the original tableaux $t,u$. For a $\lambda$-tableau $w$, let $g_w$ be the unique permutation satisfying $w=g_{w}t$ and put ${\epsilon }_w=\mathrm{sgn}(g_w)$. Sincethe quotient pairing $M^{{\lambda }^{\prime }}/(S^{{\lambda }^{\prime }}{)}^{\perp }\cong (S^{{\lambda }^{\prime }}{)}^{\ast }$ gives the explicit conjugate Specht module as a sign-twisted dual isomorphismfor every $\lambda$-tableau $w$.

Write $a=\lambda -\mathrm{id}$. The two modified entries satisfyThus $\mu -\mathrm{id}$ is obtained from $\lambda -\mathrm{id}$ by the adjacent transposition $\tau =(i,i+1)$. In the alternating definition of $\psi$, reindexing $\pi$ by $\tau \pi$ preserves every induced permutation character and reverses every sign. Therefore the straightening of a symmetric-group character indexed by a composition gives

The restriction form of the restriction branching rule for a symmetric group iswhere ${\lambda }^{-}$ contains the distinct partitions obtained by deleting one Removable node of a Young diagram. In particular, the restriction is multiplicity-free.

Restrict the alternating expressionThe supplied restriction formula for a Young permutation character, with $k=1$, says that each term restricts by subtracting one from each possible component. After collecting the alternating sums, this givesIf row $i$ has no removable node, part i straightens ${\psi }^{\lambda -{ϵ}_i}$ against the adjacent term with the opposite sign, or makes it zero when two shifted entries coincide. The surviving terms are exactly ${\psi }^{\mu }$ for $\mu \in {\lambda }^{-}$. Since $\lambda$ and each surviving $\mu$ are partitions, ${\psi }^{\lambda }={\chi }^{\lambda }$ and ${\psi }^{\mu }={\chi }^{\mu }$. We obtainComplex representations of a finite group are semisimple, so equality of characters proves the asserted module decomposition.

Repeated use of the induction branching rule giveswhileOn the five conjugacy classes $1,(12),(12)(34),(123),(1234)$, their characters are respectivelyThe tensor-product character is their pointwise product $(48,-4,0,0,0)$. Taking inner products with the five irreducible characters of $S_4$ gives multiplicities $1,5,4,7,3$. Therefore the induced-tensor decomposition for the symmetric group on four points is

Use the character formula for an induced representation. For $g\in G$,The middle equality uses that a character of a representation is constant on conjugacy classes. This proves the tensor identity for an induced character.

The point-permutation character isBy the tensor identity for an induced character and Frobenius reciprocity,The restriction branching rule for a symmetric group is multiplicity-free with one constituent for each member of ${\lambda }^{-}$, so the right side is $|{\lambda }^{-}|$. Also $⟨{\chi }^{\lambda }{\chi }^{\lambda },{\chi }^{(n)}⟩=1$ because every symmetric-group character is real and irreducible. Subtracting the trivial constituent proves the standard-character multiplicity in a Specht self-product formula

Suppose ${\chi }^{\alpha }{\chi }^{\beta }$ is irreducible. Since symmetric-group characters are real,Both self-products contain the trivial character once. They can therefore have no other common irreducible constituent. By part i, the standard character occurs in the two self-products with multiplicities $|{\alpha }^{-}|-1$ and $|{\beta }^{-}|-1$. Hence one of these numbers is zero; say $|{\alpha }^{-}|=1$.

A partition has exactly one removable node precisely when all its nonzero rows have equal length, so $\alpha =(a^b)$ is rectangular. Since $ab=n$ and $n$ is prime, either $a=1$ or $b=1$. Thus $\alpha =(1^n)$ or $(n)$. The same argument applies with $\alpha$ and $\beta$ interchanged, proving the prime-degree irreducible Kronecker product criterion for a symmetric group.

Trace the southeast boundary of the Hook of a Young diagram based at $(i,j)$. At each horizontal boundary step record the hook length $h_{i,y}$ of the cell in row $i$ above that step. At each vertical step ending beside row $x$, record $h_{i,j}-h_{x,j}$. Starting at the northeast end and moving to the southwest end, these records increase by one from $1$ to $h_{i,j}$; horizontal and vertical steps are disjoint and account for every step. Therefore the Hook-interval decomposition at a Young-diagram cell is

The supplied row-hook formula isConsequently $h\in H_i(\lambda )$ exactly when $h_i-h\ge 0$ and $h_i-h$ is not one of $h_{i+1},\dots ,h_m$. Since $h_i-h\le h_i<h_j$ for $j<i$, this is equivalent to $h_i-h\notin X$. We have proved the hook criterion in a beta set

If $ef$ is a hook length, the beta-set interpretation gives a bead at some position $b$ and a gap at $b-ef$. In the finite progressionthe first position is occupied and the last is empty. Some consecutive pair is therefore a bead followed by a gap. Their distance is $e$, so the criterion gives a hook of length $e$. This proves the divisor closure of hook lengths.

LetA direct comparison of the affected bead-gap pairs shows that removing a rim 2-hook preserves $D$: the pairs whose parities change cancel in odd-even pairs. Repeating this removal givesEvery 2-core is a staircaseAll hook lengths in this staircase are odd, and it has $1+2+\cdots +r=r(r+1)/2$ cells. Hence the odd-minus-even hook count of a partition isThus the requested integer is $m=r+1=\ell (C_2(\lambda ))+1$.

Use four beads, for which $(3,1)$ has beta setOn runners of residues $0,1,2,3$, division by four gives respectively the beta setsOnly $\{1\}$ represents a nonempty partition, namely $(1)$. Therefore the four-quotient of the partition three-one is

A two-runner partition abacus givesand the 2-quotients of $(1)$ and $\varnothing$ are empty. Thus the two-quotient tower of the partition three-one has nonempty levels

An $e^k$-runner abacus separates bead positions by their residue modulo $e^k$. Write such a residue in base $e$ asTaking one $e$-quotient sorts beads by $a_0$ and divides their positions by $e$; applying the operation again sorts by $a_1$, and so on. After $k$ stages, the iterated construction has selected exactly the same $e^k$ residue classes as the single $e^k$-quotient. The two conventional orderings may list the base-$e$ digits in opposite order, producing only a permutation of components.

Equivalently, induction on $k$ applies the same argument to every component of $Q_e(\lambda )$ and identifies the resulting $e^{k+1}$ runner partitions. Hence the iterated quotient equals a power quotient up to permutation statement is

PutThe defining relation between the quotient tower of a partition and the core tower of a partition isSumming the resulting telescoping identities gives

By the Hook-length formula,The abacus divisible-hook correspondence says that the number of hooks divisible by $p^r$ is $q_r$, soIf $d_p(n)={\sum }_r{\alpha }_r$, the digit-sum form of the Legendre formula isCombining the three displayed identities proves the P-adic valuation of a symmetric-group character degree from the core tower formula

Add the base-$p$ expansions of $x$ and $y$ column by column. Before carrying, the sum of all displayed digits is $d_p(x)+d_p(y)$. Each carry removes $p$ units from one column and adds one unit to the next, decreasing the total digit sum by $p-1$. After all carries the digits are those of $x+y$, so the subadditivity of the base-p digit sum gives

Let $\mu =C_p(\lambda )$, put $m=|\mu |$, and let the first-level $p$-quotient partitions have sizes $n_0,\dots ,n_{p-1}$. ThenRepeated subadditivity of the base-p digit sum givesFor any partition $\nu$ of size $s$, iterating the core-quotient relation $s=|C_p(\nu )|+p|Q_p(\nu )|$ and the same digit-sum inequality givesApply this to every first-level quotient partition. Their core towers concatenate to levels $r\ge 1$ of $T{C}_p(\lambda )$, so

Part a applied to $\lambda$ and to its $p$-core $\mu$, whose higher core-tower levels are empty, now givesTherefore the Character-degree valuation does not increase on taking the p-core: