Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2010 / B (Representation theory of symmetric groups)

For a Young diagram cell $(i,j)$, its hook length is one plus the number of cells to its right and below. The hook-length formula isApplying it to the two shapes givesThe shapes here require $n\ge 4$.

A hook tabloid remembers its first-row set and the ordered singleton entries below it. The proposed map keeps that first-row set and merges the remaining entries into one unordered row. Thus it is well-defined on tabloids. Permuting entries before merging gives the same tabloid as merging before permuting, so it is a module homomorphism. Every target tabloid is reached by putting its second-row entries, in any order, into the singleton rows. Therefore it is surjective.

For $r=2$, the strictly dominating partitions of $\lambda =(n-2,1,1)$ are exactly $(n),(n-1,1),(n-2,2)$. Indeed dominance forces the first part to be at least $n-2$, and the possibilities for the remaining sum give precisely those shapes or $\lambda$ itself.

In characteristic zero, Maschke's theorem makes the permutation module semisimple, and Young's rule gives one copy of $S^{\lambda }$ and only strictly dominating additional shapes. The trivial multiplicity is one, since a transitive permutation module has a one-dimensional invariant space. For the standard representation $S^{(n-1,1)}$, Frobenius reciprocity identifies its multiplicity with the dimension of fixed vectors under the row stabilizer $S_{n-2}$. In its realization $\{(z_1,\dots ,z_n):\sum z_i=0\}$, those fixed vectors have the form $(c,\dots ,c,a,b)$ with $(n-2)c+a+b=0$, a two-dimensional space. Thus its multiplicity is two.

Finally $\dim M^{\lambda }=n(n-1)$. Subtracting the dimensions of one $S^{\lambda }$, one trivial module and two standard modules leaves $n(n-3)/2$, exactly one $S^{(n-2,2)}$. ThereforeThe additional multiplicities are $1,2,1$ respectively. If the printed weak summation index is read literally alongside its separate $S^{\lambda }$, then $m_{\lambda ,\lambda }=0$; the total multiplicity of $S^{\lambda }$ is one.

A tabloid is the row-equivalence class $\{t\}$ of a tableau filled bijectively with $1,\dots ,n$: order inside each row is forgotten, but the rows remain distinguished. The permutation module $M^{\lambda }$ has these tabloids as an $F$-basis, with ${\Sigma }_n$ permuting their entries. Let $C_t$ be the column stabilizer of $t$ and defineThe Specht module $S^{\lambda }$ is the submodule spanned by all polytabloids $e_t$.

If a module homomorphism $\varphi :M^{\lambda }\to M^{\mu }$ is nonzero on $S^{\lambda }$, it is nonzero on some spanning $e_t$. Equivariance gives $0\ne \varphi (e_t)={\kappa }_t\varphi (\{t\})$, so ${\kappa }_t{M}^{\mu }\ne 0$. The allowed criterion yields $\lambda \ge \mu$.

In characteristic zero, Maschke's theorem gives an equivariant projection $M^{\lambda }\to S^{\lambda }$. If $S^{\lambda }\cong S^{\mu }$, compose that projection with the isomorphism and the inclusion into $M^{\mu }$. The preceding argument gives $\lambda \ge \mu$. Reversing the roles gives $\mu \ge \lambda$, so antisymmetry of dominance givesFor a map defined only on a Specht module in arbitrary characteristic, dominance need not hold. Take $F={\mathbb{F}}_2$, $n=2$, $\lambda =(1,1)$ and $\mu =(2)$. The first Specht module is the one-dimensional span ofThe transposition fixes $e$, so this is the trivial module. The one-row permutation module $M^{(2)}$ is also trivial and one-dimensional. Sending $e$ to its single tabloid is a nonzero equivariant map, yet $(1,1)\ngeqq (2)$. In this characteristic such a map need not extend to the surrounding permutation module; that is exactly where the earlier proof ceases to apply.

Let ${\pi }^{\lambda }$ be the permutation character of $M^{\lambda }$. Young's rule gives ${\pi }^{\lambda }={\chi }^{\lambda }+{\sum }_{\mu >\lambda }{K}_{\mu \lambda }{\chi }^{\mu }$, with integer Kostka multiplicities. This dominance-triangular system has diagonal entries one, so recursive subtraction expresses every irreducible character as an integer linear combination of permutation characters. The latter count fixed tabloids, proving that all ${\chi }^{\lambda }(g)$ are integers.

For a $k$-cycle, the Murnaghan–Nakayama rule reduces towhere height is the number of occupied rows minus one, and $f^{\nu }$ is the degree of the remaining Specht character at the identity. In particular a full $n$-cycle has character zero unless $\lambda =(n-r,1^r)$ is a hook, in which case the value is $(-1{)}^r$.

There is more than one solution: the trivial representation $\lambda =(6)$ has character $1$ everywhere and the sign representation $\lambda =(1^6)$ has character $\pm 1$ everywhere. ThusIn fact they are the only ones. The full-cycle formula in the preceding discussion forces any candidate to be a hook $(6-r,1^r)$. For $r=1$, the standard character is the number of fixed points minus one, so it vanishes on cycle type $(5,1)$; its sign twist $r=4$ does too. For $r=2$, the hook representation is the exterior square of the standard representation and has character $\frac{1}{2}(\chi (g{)}^2-\chi (g^2))$. At cycle type $(4,1,1)$ both standard-character values are one, so this character is zero. Its sign twist $r=3$ also vanishes there. Only $r=0,5$ remain.

Irreducible character orthogonality gives $⟨\chi ,\chi ⟩=1$. Under the proposed vanishing condition, only the identity contributes, soBut $\chi (1)$ is an integer dimension and $720$ is not a square. Thus no such partition exists.

The full-cycle formula says that precisely the nonhook shapes have zero value on a 6-cycle. These are $(4,2),(3,3),(3,2,1),(2,2,1,1),(2,2,2)$.

Use the standard two-row permutation-character rule ${\pi }^{(6-r,r)}={\chi }^{(6)}+{\chi }^{(5,1)}+\cdots +{\chi }^{(6-r,r)}$. A transposition has four fixed points, so ${\chi }^{(5,1)}=3$. It fixes $\left(\genfrac{}{}{0ex}{}{4}{2}\right)+1=7$ two-subsets, giving ${\chi }^{(4,2)}=7-1-3=3$. It fixes $\left(\genfrac{}{}{0ex}{}{4}{3}\right)+\left(\genfrac{}{}{0ex}{}{4}{1}\right)=8$ three-subsets, giving ${\chi }^{(3,3)}=8-1-3-3=1$.

Conjugating a partition tensors its character with the sign character. Hence the values on a transposition for the conjugate shapes are $-3$ and $-1$. The shape $(3,2,1)$ is self-conjugate, so its character equals its own sign twist and is zero on every odd permutation. The complete candidate table is therefore:

Thus the unique answer is