Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 160 (Representation Theory of Symmetric Groups)

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

The Murnaghan–Nakayama rule states that if a permutation has a $k$-cycle and remaining cycle type $\rho$, thenwhere $H$ runs over the removable rim hooks of length $k$. Iterating removes rim hooks whose lengths are the cycle lengths, and the character value is the signed sum over all complete removal sequences.

Let $s_i=(ii+1)$. Since $\mu -\mathrm{id}=s_i(\lambda -\mathrm{id})$, reindex the alternating sum defining ${\psi }^{\mu }$ by $\pi \mapsto s_i\pi$. This preserves all permutation-character terms and reverses every sign of a permutation, because $\mathrm{sgn}(s_i\pi )=-\mathrm{sgn}(\pi )$. Hence ${\psi }^{\mu }=-{\psi }^{\lambda }$.

For each $m$, the shifted sequence ${\beta }_{i,m+1}-\mathrm{id}$ is obtained from ${\beta }_{i,m}-\mathrm{id}$ by swapping the adjacent entries in positions $i+m$ and $i+m+1$. Part a(ii) therefore givesThis is the usual character straightening procedure: the distinguished entry ${\alpha }_i-i-k$ is moved successively to the right. Either it reaches the unique place that makes some ${\beta }_{i,m}$ a partition of an integer, or it meets an equal shifted entry. Under the hypothesis the first alternative never occurs, so two entries of some ${\beta }_{i,m}-\mathrm{id}$ coincide. The corresponding alternating sum is fixed by swapping those entries but changes sign by part a(ii), and is therefore zero. Repeatedly applying the displayed relation gives ${\psi }^{{\beta }_{i,0}}=0$.

Write $\chi ={\chi }^{\lambda }$. By the Hook-length formula,Suppose the order of a group element $g$ did not divide this product. Then for some prime number $p$, the highest prime power $p^a$ dividing $|g|$ would not divide the hook product. One cycle of $g$ has length divisible by $p^a$, whereas no hook length of $\lambda$ is divisible by $p^a$. In particular there is no removable rim hook having that cycle length. Applying the Murnaghan–Nakayama rule first to this cycle gives $\chi (g)=0$, a contradiction. This is the symmetric-group character co-degree vanishing criterion, and its contrapositive proves

Choose $g$ whose disjoint permutation cycles have lengths equal to the principal hook lengths of $\lambda$. Those lengths are distinct and sum to $n$, so this is a permutation in $S_n$. In the iterated Murnaghan–Nakayama rule, there is a unique complete sequence that removes the corresponding principal rim hooks. Its contribution is one sign, and hence the Principal-hook character value of a symmetric group gives ${\chi }^{\lambda }(g)\in \{1,-1\}$.

Take $g$ to be an $n$-cycle. For $n\ge 2$, the Murnaghan–Nakayama rule gives ${\chi }^{(n)}(g)=1$ and ${\chi }^{(n-1,1)}(g)=-1$; every remaining two-row diagram contains a $2\times 2$ square and is not a rim hook, so its value at $g$ is zero. ThereforeThe exhibited cycle has length $k=n$, so $k\equiv n(\mathrm{mod}2)$. For $n=1$, the sole character has value one at the identity and the same conclusion holds with $k=1$.

Apply the Frobenius characteristic map. For even $n$, the Jacobi–Trudi identity and cancellation of consecutive terms giveThe generating function for the complete homogeneous symmetric polynomials now givesThis expansion contains only products $p_{\mu }$ for which every part of $\mu$ is even. The coefficient of $p_{\mu }$ in $\mathrm{ch}(F)$ is $F(\mu )/z_{\mu }$, so $F(\mu )=0$ whenever the cycle type $\mu$ has an odd part. Equivalently, $F(g)=0$ whenever $g$ contains an odd cycle. This is the Two-row alternating character cancellation.

For a nonzero example, take $n=3$ and let $g$ be a transposition. The two terms are the trivial and standard characters of $S_3$, whose values at a transposition are respectively $1$ and $0$. Thus $F(g)=1$ although $g$ contains a cycle of even length.

For a zero example, take $n=5$ and again let $g$ be a transposition, of cycle type $(2,1^3)$. The values of ${\chi }^{(5)},{\chi }^{(4,1)},{\chi }^{(3,2)}$ there are $1,2,1$, respectively. Consequently

Use a beta set $X$ on an $e$-runner partition abacus. A hook of length divisible by $e$ corresponds to a bead $x$ and a gap $x-eh$ on the same runner. Divide both runner positions by $e$. They become a bead and a gap at distance $h$ in the runner partition ${\lambda }^{(s)}$, and hence determine a hook there. This gives the required bijection, with $|H|=e|f(H)|$.

Removing $H$ replaces the bead $x$ by the gap $x-eh$. On its runner this is exactly the bead move that removes $f(H)$, while every other runner is unchanged. Thus hook removal commutes with the construction andThis is the abacus divisible-hook correspondence.

For $\lambda =(6,5,3)$ use the three-bead beta set $\{8,6,3\}$. Its runners give the 3-quotient of a partitionThe hook lengths divisible by three and their images areThere are no others, as the complete hook-length rows are $(8,7,6,4,3,1)$, $(6,5,4,2,1)$, and $(3,2,1)$.

True. A hook length divisible by $ef$ is divisible by $e$. If $\lambda$ is its own Core of a partition $C_e(\lambda )$, it has no such hook, so it is also its own $ef$-core.

False. Let $\lambda =(3)$, $e=2$, and $f=3$. ThenThus core operations for different moduli need not commute.

True. On an $f$-runner partition abacus, taking the $e$-quotient of each runner refines its positions by their residue modulo $e$. The resulting $ef$ runners are indexed by pairs of residues. Ordering those pairs may differ from the conventional residue order modulo $ef$, but only by a permutation. Hence $Q_e(Q_f(\lambda ))$ is a permutation of $Q_{ef}(\lambda )$.

True. Slide beads upward separately in each of the refined residue classes modulo $ef$. Grouping the refined runners first by residue modulo $f$ and then packing each component modulo $e$ produces $C_e(Q_f(\lambda ))$. Packing modulo $ef$ before forgetting the refinement produces $Q_f(C_{ef}(\lambda ))$. These are the same runner configurations, so

Apply the Murnaghan–Nakayama rule successively to the $w$ disjoint $e$-cycles. A complete term requires a sequence of $w$ removable $e$-hooks. If $w>w_e(\lambda )$, no such sequence exists after the Weight of a partition is exhausted, so the character value is zero.

Suppose $w=w_e(\lambda )$. Every complete sequence ends at the Core of a partition $C_e(\lambda )$. Under the abacus divisible-hook correspondence, a removal chooses one cell from one component ${\lambda }^{(i)}$ of the quotient of a partition. The choices of which runner is used occur inorders. Within runner $i$, the signed complete removal sum is the degree ${\chi }^{{\lambda }^{(i)}}(1)$, and all inter-runner removal orders have the common Sign of an abacus hook-removal sequence $\epsilon \in \{1,-1\}$. The remaining permutation $\gamma$ acts on the core, giving

Let $b_t$ be the Column antisymmetrizer of a Young tableau. For a tabloid $\{u\}$, if two entries from one column of $t$ lie in one row of $u$, their column transposition fixes $\{u\}$ and pairs every term of $b_t\{u\}$ with its negative. Thus $b_t\{u\}=0$.

Otherwise each row of $u$ meets each column of $t$ at most once. Matching entries within columns then gives a column permutation $c\in C(t)$ for which $\{cu\}=\{t\}$. Reindexing the antisymmetrizer gives $b_t\{u\}=\mathrm{sgn}(c)e(t)$. Hence every basis tabloid maps into $\mathbb{F}e(t)$, while $b_t\{t\}=e(t)$, and therefore

The coefficient $⟨e(t),\omega ⟩$ is nonzero exactly when every column of $t$ meets every row of $\omega$ in at most one entry. Permuting equal-length rows of $\omega$ preserves this condition, and the same holds with $u$ in place of $t$. Therefore membership in $A_{tu}$ is constant on each class of the stated equivalence relation, so $A_{tu}$ is a union of equivalence classes.

Each equivalence class has size ${\prod }_{j=1}^n{a}_j!$, because its $a_j$ rows of length $j$ may be permuted freely. Swapping two length-$j$ rows multiplies each of $⟨e(t),\omega ⟩$ and $⟨e(u),\omega ⟩$ by the same sign $(-1{)}^j$, so their product is constant on the class. Sincethe contribution of every nonzero class is a multiple of ${\prod }_j{a}_j!$. The whole inner product is therefore such a multiple.

The partition is $p$-regular precisely whenthat is, when no part occurs $p$ or more times. This is the definition of a regular partition in characteristic of a field $p$.

If some $a_j\ge p$, then $a_j!=0$ in $\mathbb{F}$. Part b(ii) makes every pairing $⟨e(t),e(u)⟩$ zero, so the Tabloid bilinear form vanishes identically on $S^{\lambda }$ and $S^{\lambda }\subseteq (S^{\lambda }{)}^{\perp }$.

Conversely, if $\lambda$ is $p$-regular, every $a_j!$ is nonzero. Fact 1 supplies tableaux $t,u$ withThus the restriction of the form is not identically zero and $S^{\lambda }⊈(S^{\lambda }{)}^{\perp }$.

Because $\lambda$ is $p$-regular, Fact 1 gives a tableau $u$ for whichPart a, applied linearly to the tabloid expansion of $e(u)$, gives

The nonzero homomorphism $\theta$ cannot kill $e(t)$, because the Specht module is generated by the translates of this polytabloid. Equivariance and part d(i) therefore giveThus $b_t$ acts nontrivially on $M^{\mu }/V$, and hence on $M^{\mu }$. Some $\mu$-tabloid $\{v\}$ must satisfy $b_t\{v\}\ne 0$. Fact 2 now says that $\lambda$ dominates $\mu$ in the dominance order on partitions.

Assume $\lambda =\mu$. Part a says that the image of $b_t$ on $M^{\lambda }/V$ is contained in the one-dimensional space spanned by $e(t)+V$. The identity from part d(ii) then shows that $\theta (e(t))$ itself is a scalar multiple of $e(t)+V$.

Apply this with $V=S^{\lambda }\cap (S^{\lambda }{)}^{\perp }$ and with $\theta$ obtained by composing the quotient map $S^{\lambda }\to D^{\lambda }$ with any endomorphism of the Simple symmetric-group module from a regular partition $D^{\lambda }$. Every endomorphism is scalar on the generating polytabloid and hence on all its translates. Therefore

If the two displayed simple quotients $D^{\alpha }$ and $D^{\beta }$ are isomorphic, compose the quotient map $S^{\alpha }\to D^{\alpha }$ with an isomorphism to $D^{\beta }\subseteq M^{\beta }/(S^{\beta }\cap (S^{\beta }{)}^{\perp })$. Part d(ii) gives $\alpha ⪰\beta$. Applying the inverse isomorphism gives $\beta ⪰\alpha$. Antisymmetry of the dominance order on partitions forces $\alpha =\beta$. Conversely, equality of the partitions makes the quotients identical, hence isomorphic.