Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 103 (Representation Theory)

Fix a Young tableau $t$ of shape $\lambda$, let $C(t)$ be its column stabilizer, and putThese are the Column antisymmetrizer of a Young tableau and its polytabloid. The Specht module is generated by $e_t$ as a cyclic module, so the nonzero R-module homomorphism $\varphi$ has $\varphi (e_t)\ne 0$. Because its extension $\Phi$ commutes with the group algebra action,Expand $\Phi (\{t\})$ in the tabloid basis of the Young permutation module $M^{\mu }$. At least one tabloid $\{u\}$ therefore satisfies $b_t\{u\}\ne 0$.

If a row of $u$ contains two entries from the same column of $t$, their transposition $\tau \in C(t)$ fixes $\{u\}$. Pairing the terms indexed by $g$ and $g\tau$ in $b_t\{u\}$ makes them cancel. This remains valid in characteristic two, since the paired coefficients sum to zero there as well. Thus each row of $u$ meets each column of $t$ in at most one entry.

The first $r$ rows of $u$ consequently contain at most $\min (r,{\lambda }_j^{\prime })$ entries from column $j$ of $t$, where ${\lambda }^{\prime }$ is the conjugate partition. HenceBy the dominance order on partitions, this provesThis is dominance from a nonzero column antisymmetrizer. The extension is what allows us to apply $b_t$ to an element of $M^{\mu }$ without dividing by the column stabilizer's order, which might vanish in the given field.

The Young permutation module $M^{(n-k,k)}$ is the symmetric-group subset permutation representation on the $k$-element subsets of $\{1,\dots ,n\}$: the second row of a tabloid specifies the subset. For $k\ge 1$, restriction of a character to $S_{n-1}$ fixing $n$ splits these subsets into two orbits of a group action, according as they contain $n$.

The corresponding permutation representations are on the $k$-subsets and the $(k-1)$-subsets of $\{1,\dots ,n-1\}$. Their Young permutation modules have two row sizes $n-1-k,k$ and $n-k,k-1$, respectively. Sort each pair into decreasing order and omit any zero part; swapping the two row labels gives an isomorphic permutation representation. This sorting matters at $n=2k$.

We use Young's rule: over the complex numbers, the multiplicity of $S^{\rho }$ in $M^{\eta }$ is the Kostka number $K_{\rho \eta }$, and it can be nonzero only if $\rho$ dominates $\eta$ in the dominance order on partitions. Since each $\eta$ above has at most two parts and size $n-1$, dominance givesThus $\rho$ has at most two nonzero parts. The character inner product measures this constituent multiplicity, soIf $k=0$, the original and restricted modules are the trivial representations, giving the same conclusion. For $n=1$, there is no partition of $n-1=0$ with three parts, so the assertion is vacuous.

Apply Young's rule, stated asHere the Kostka number counts Semistandard Young tableaux of shape $\rho$ with $n-k$ entries equal to $1$ and $k$ entries equal to $2$. Their rows weakly increase and their columns strictly increase. A column can therefore have at most two cells, so only a partition of an integer $\rho =(n-j,j)$ can occur.

In such a semistandard Young tableau, all $j$ entries in the second row must be $2$ and all $j$ entries above them must be $1$. The whole first row is then fixed by the content: its first $n-k$ entries are $1$ and its remaining $k-j$ entries are $2$. This is possible exactly when $j\le k$ and $j\le n-k$. Because $k\le n/2$, the second inequality follows from the first. There is exactly one filling for every $0\le j\le k$, and none for any other shape.

Consequently each displayed Specht module has multiplicity one, proving the two-row Young permutation module decompositionFor $j=0$, the zero second part is omitted. As a dimension check, the Hook-length formula gives $\dim S^{(n-j,j)}=\left(\genfrac{}{}{0ex}{}{n}{j}\right)-\left(\genfrac{}{}{0ex}{}{n}{j-1}\right)$, with $\left(\genfrac{}{}{0ex}{}{n}{-1}\right)=0$; the sum telescopes to $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$, the dimension of the original permutation representation.

Use the hook criterion in a beta set: for a finite beta set of a partition $X$, each hook length is a difference $b-c$ with $b\in X$, $0\le c<b$ and $c\notin X$, and every such difference is a hook length.

A hook length $pr$ therefore gives a bead $b$ and a gap $c=b-pr$. Along the progressionthe first position is a bead and the last is a gap. Some adjacent positions are consequently a bead followed by a gap. Their difference is $p$, so $p$ is a hook length. Similarly the progression from $b$ to $c$ in steps of $r$ gives a hook length $r$. Every position used is nonnegative because it lies between $c$ and $b$.

Thus the divisor closure of hook lengths yieldsNeither $p$ nor $r$ needs to be a prime number.

The converse is false. Take $p=2$, $r=3$, $n=6$ and the partition of an integer $\lambda =(3,3)$. Its hook lengths, arranged in its Young diagram, areBoth $2$ and $3$ occur, but $6$ does not, even though $pr=6=n$. Henceis a counterexample with distinct $p,r$.

Put $L=h_{i,j}=pr$. The Hook-interval decomposition at a Young-diagram cell is the disjoint unionHere is a proof of the identity using the hook criterion in a beta set. Keep only the cells weakly southeast of $(i,j)$, translating that cell to $(1,1)$; this gives the partition of an integerLet $\ell ={\lambda }_j^{\prime }-i+1$ and use its beta set of a partition $b_s={\nu }_s+\ell -s$. These numbers are precisely the hook lengths down the original column from $(i,j)$, with $b_1=L$. The hooks along the first row of $\nu$ are the differences $L-c$ for the gaps $0\le c<L$. The remaining positions in that interval are the beads $b_2,\dots ,b_{\ell }$, giving exactly the displayed disjoint decomposition. The first-row and first-column hook lengths of $\nu$ agree with those in the original Young-diagram hook.

Because $p\mid L$, a leg cell satisfies $p\mid h_{x,j}$ exactly when $p\mid L-h_{x,j}$. Thus divisibility by $p$ identifies the relevant cells of the Young-diagram hook with the multiples of $p$ in $\{1,\dots ,pr\}$. There are exactly $r$ of those, provingThe count includes the corner cell $(i,j)$ itself.

The equivalence is true. Write $c_r(\eta )$ for the total size at level $r$ of the core tower of a partition $\eta$, and $q_r(\eta )$ for the corresponding total size in its quotient tower of a partition. We use the size relation and the P-adic valuation of a symmetric-group character degree from the core tower:Here $d_p$ is the base-p digit sum. These imply the character degree coprime to p from the core tower criterion: if $|\eta |={\sum }_r{\alpha }_r{p}^r$, then $p\nmid {\chi }^{\eta }(1)$ exactly when $c_r(\eta )={\alpha }_r$ at every level. Indeed carrying $p$ units from one position to the next decreases the sum of coefficients by $p-1$; the valuation vanishes precisely when there are no carries.

For $k\ge 1$, we also use power-core truncation of a prime-core tower: for $\gamma =C_{p^k}(\lambda )$, its core tower agrees with that of $\lambda$ below level $k$ and is empty at every level at least $k$. Moreover $q_k(\lambda )=w_{p^k}(\lambda )$, the weight of a partition for modulus $p^k$. To see the truncation, removal of a rim hook of length $p^k$ preserves the first $p$-core and, under the abacus divisible-hook correspondence, becomes removal of a $p^{k-1}$-hook in one quotient component. Iterating preserves all the lower cores; after all such removals, quotient level $k$ is empty.

If $p\nmid {\chi }^{\lambda }(1)$, the criterion gives $c_k(\lambda )=a$, $c_r(\lambda )=0$ for $r>k$, and the lower $c_r(\lambda )$ are the base-$p$ digits of $m$. It follows that $|\gamma |=m$, $p\nmid {\chi }^{\gamma }(1)$ andConversely, suppose $w_{p^k}(\lambda )=a$ and $p\nmid {\chi }^{\gamma }(1)$. Then $|\gamma |=m$ and its lower core-tower sizes are the digits of $m$. Quotient level $k$ of $\lambda$ has total size $a<p$. Each component there has size less than $p$, hence is already a $p$-core and has empty $p$-quotient. Thus $c_k(\lambda )=a$ and every higher level is empty. Combined with truncation, this gives exactly the digits of $n$, so $p\nmid {\chi }^{\lambda }(1)$.

ThereforeIf $k=0$, necessarily $m=0$ and $n=a<p$; the $1$-core is empty, its weight is $n=a$, and the Hook-length formula shows that every degree is coprime to $p$. Thus the equivalence holds in that boundary case too, with the empty partition having degree one.

The implication is false. ChooseThe inequalities $0\le 4<9$ and the relation $13=1\cdot 9+4$ hold. At cell $(1,4)$, the hook length is $12-4+1=9=a{p}^k$.

For the partition of an integer $(12,1)$, the Hook-length formula giveswhich is divisible by $3$. This also agrees with the degree of the standard representation of the symmetric group $S_{13}$.

The obstruction is visible in the core of a partition: removing the horizontal rim hook of length nine leaves $\gamma =(3,1)$, which has no hook length nine and hence is the $9$-core. Its degree is ${\chi }^{(3,1)}(1)=3$. The large hook ensures the required weight of a partition here, but the core degree is not coprime to $p$.

The implication is false. ChooseThe hook lengths areThus $3$ does not divide this irreducible character degree, but no hook has length $a{p}^k=6$.

There is no contradiction with part (i): removing two horizontal rim hooks of length three gives the empty core of a partition, so $w_3(\lambda )=2=a$ and the core degree is one. Having total weight of a partition two does not force a single hook of length twice the modulus.

We use the abacus divisible-hook correspondence in its multiset form: the hook lengths of $\lambda$ that are divisible by $p$, divided by $p$, form exactly the disjoint union of the multisets of hook lengths of the components of its quotient of a partition for modulus $p$.

Iterating this correspondence gives, for every $r\ge 1$,as multisets. Each Young diagram has one hook for every cell, so the cardinality of the right-hand side is $|T^Q(\lambda {)}_r|$, the sum of the component sizes.

For any positive integer $h$, its P-adic valuation is the number of positive $r$ for which $p^r\mid h$. Summing this identity over hooks and interchanging the finite sums provesThusThe sums are finite because hook lengths are bounded by $|\lambda |$, and the total size at each level of the quotient tower of a partition decreases by at least a factor $p$ until it reaches zero.

We use the core-quotient bijection for partitions: an ordered $p$-tuple $({\nu }_0,\dots ,{\nu }_{p-1})$ and a fixed $p$-core determine a unique partition of an integer, withThe irreducible characters in both $B$ and $C$ therefore correspond to the same ordered tuples of total size $w$. Replacing the empty core of a partition by $\gamma$ while keeping the quotient of a partition fixed gives a bijection ${\chi }^{\lambda }\mapsto {\chi }^{\mu }$ from $B$ to $C$. In particular $|B|=|C|$.

For such a pair, all levels of the core tower of a partition above level zero agree. At level zero, their sizes are $0$ and $m$, respectively. Put $D={\sum }_{r\ge 1}|T^C(\lambda {)}_r|={\sum }_{r\ge 1}|T^C(\mu {)}_r|$. When $0\le m<p$, the base-p digit sum satisfiesThe P-adic valuation of a symmetric-group character degree from the core tower then givesThus the same bijection preserves degree coprimality to $p$ and restricts to a bijection between the indicated sets of irreducible characters of degree coprime to p. ConsequentlyFor $m=0$, take $\gamma$ to be the empty partition; the construction becomes the identity.

For every ${\chi }^{\mu }\in C$, level zero of the core tower of a partition is its core of a partition $\gamma$, of size $m$. The character degree coprime to p from the core tower criterion says that if $p\nmid {\chi }^{\mu }(1)$, this level must have size equal to the units digit ${\alpha }_0$ of $n$ in base $p$.

But $0\le {\alpha }_0<p$, whereas $m\ge p$. Hence every character in $C$ has degree divisible by $p$, andEquivalently, the core-tower size coefficient $m$ at position zero needs at least one carry in normalization to the base-$p$ digits. Each carry contributes one to the P-adic valuation of a symmetric-group character degree from the core tower.