Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2009 / B

The Young diagram is $[\lambda ]=\{(i,j):1\le i\le k,1\le j\le {\lambda }_i\}$. The hook length of a box iscounting the box itself, the boxes to its right in the same row, and the boxes below it in the same column. The hook-length formula isThe first implication in the equivalence is immediate: if every hook length is odd, no hook can have length two.

Assume no hook has length two. The last row has length one, since otherwise its penultimate box has hook length two. If two adjacent row lengths are equal, take the last two rows in the maximal block of that length; the last box in the upper of those two rows has hook length two. Thus all row lengths are distinct. If a gap ${\lambda }_i-{\lambda }_{i+1}$ were at least two, the penultimate box in row $i$ would again have hook length two, since there would be no boxes below it. Therefore every successive gap equals one and the final part equals one, proving (2) implies (3).

The final condition makes $\lambda =(k,k-1,\dots ,1)$. Its column lengths are the same staircase, sowhich is odd. This proves (3) implies (1) and completes the equivalence. Such a partition exists exactly at the triangular numbersThe hook formula gives $v_2(f^{\lambda })=v_2(n!)=a$, since its denominator is odd. Thus $2^a$ divides $f^{\lambda }$. Conversely $2^a\mid f^{\gamma }$ implies $v_2(\prod h_{\gamma })=a-v_2(f^{\gamma })\le 0$. The product is an integer, so its valuation is zero and all its hooks are odd. By the equivalence $\gamma$ is a staircase; its size uniquely determines its length, so $\gamma =\lambda$.

For the character assertion, every removable rim hook has length equal to some hook length of the diagram. Here is a direct justification. If a border strip $\lambda /\mu$ occupies rows $i$ through $j$, connectedness and absence of a two-by-two block make adjacent strips of boxes overlap in exactly one column. Hence ${\mu }_l={\lambda }_{l+1}-1$ for $i\le l<j$, so its size is ${\lambda }_i-{\mu }_j+j-i$. The column $c={\mu }_j+1$ has height $j$, giving exactly $h_{\lambda }(i,c)={\lambda }_i-{\mu }_j+j-i$. Therefore every removable rim hook of the staircase has odd length.

A fixed-point-free permutation of power-of-two order has only even cycles. Apply the Murnaghan–Nakayama rule starting with any cycle. It would require removal of a rim hook of that even length, but none exists. The sum is empty, soFinally suppose $\gamma$ has distinct parts and is not $\lambda$. It cannot be another staircase of the same size. Thus some gap between successive rows is at least two, or the last row has length at least two; in either case a horizontal rim domino is removable. No vertical rim domino is removable, since that would require two equal adjacent row lengths. Murnaghan-Nakayama on a transposition consequently gives a nonempty sumall signs being positive and all tableau counts positive. Thus the character is nonzero at every transposition, proving the requested existence.

For a tableau $t$, let $C_t$ be its column stabilizer and $e_t={\sum }_{\sigma \in C_t}\mathrm{sgn}(\sigma )\{\sigma t\}$ its polytabloid. The Specht module is the span of all such polytabloids inside the tabloid permutation module.

A tabloid with second row of size $k$ is determined by that subset, so $\dim M^{(n-k,k)}=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. The two-row hook formula givesAt $k=3$ this provesBy Maschke's theorem, characteristic-zero permutation modules are completely reducible, so their Specht submodules are summands. The nonzero sum of all three-subset tabloids is fixed by every permutation; its span is a copy of the trivial module $S^{(n)}$.

A permutation sends $v_a$ to $v_{ga}$, $w_{ab}$ to $w_{ga,gb}$ and $x_{abc}$ to $x_{ga,gb,gc}$. Relabeling the summation indices givesLinearity therefore proves both maps are module homomorphisms. The map denoted $\varphi$ in the second displayed source formula is ${\varphi }_2$.

For the standard tableau with second row $2,4$, the first entries in the top row are $1,3$, so its nontrivial columns are $(1,2)$ and $(3,4)$. HenceIn ${\varphi }_2(e_t)$ the coefficient of $x_{24n}$ is one: among those four pairs only $\{2,4\}$ is contained in $\{2,4,n\}$, since $n\ge 6$. Thus the image is nonzero. Its restriction to the characteristic-zero simple module $S^{(n-2,2)}$ has zero kernel, giving the requested isomorphic submodule.

For $S^{(n-1,1)}$, choose the standard tableau with lower entry two and top-left entry one. Its polytabloid is $v_2-v_1$. Each three-subset containing $a$ occurs twice in ${\varphi }_2{\varphi }_1(v_a)$, once for each choice of its other two entries as the intermediate $b$. Consequently the coefficient of $x_{23n}$ in ${\varphi }_2{\varphi }_1(v_2-v_1)$ is two. This is nonzero in characteristic zero. Simplicity again makes the restriction injective, giving a copy of $S^{(n-1,1)}$.

The four constructed submodules have the distinct shapes $(n-3,3)$, $(n-2,2)$, $(n-1,1)$ and $(n)$, so they are pairwise nonisomorphic irreducible characteristic-zero modules. Complete reducibility therefore makes their sum direct. The given decomposition of $M^{(n-2,2)}$ and the dimension identity in (a) show their dimensions sum to $\dim M^{(n-3,3)}$. Their direct sum is consequently the whole module:

Use the standard invariant tabloid bilinear form, with the tabloid basis orthonormal. Each permutation acts by an orthogonal matrix, so $⟨gx,y⟩=⟨x,g^{-1}y⟩$. Inversion permutes $C_t$ and preserves its signs. ThereforeThe James submodule theorem states that for any submodule $U\le M^{\lambda }$, either $S^{\lambda }\subseteq U$ or $U\subseteq (S^{\lambda }{)}^{\perp }$. Use the allowed fact ${\kappa }_t{M}^{\lambda }=F{e}_t$. The coefficient of $\{t\}$ in $e_t$ is one, so the displayed adjoint identity gives ${\kappa }_{t}x=⟨x,e_t⟩e_t$. If $U$ is not orthogonal to $S^{\lambda }$, some $x\in U$ and some tableau $t$ have $⟨x,e_t⟩\ne 0$. Hence $e_t\in U$, and its permutation translates span $S^{\lambda }$. This proves the theorem.

Let $R=S^{\lambda }\cap (S^{\lambda }{)}^{\perp }$. Every proper submodule of $S^{\lambda }$ lies in $R$ by the theorem. If the restricted form is nonzero, $R$ is proper, hence is the unique maximal submodule andIn the stronger nondegenerate case specified in the question, $R=0$, so $S^{\lambda }$ itself is simple. The nonzero-form version also covers the possibly degenerate case needed in the final calculation.

For shape $(n-1,1)$, identify $M$ with $F^n$ with basis $v_1,\dots ,v_n$. Its Specht module is the augmentation subspace $S=\{\sum a_i{v}_i:\sum a_i=0\}$, spanned freely by $v_i-v_n$, $1\le i<n$. Its Gram matrix is $G=I_{n-1}+J_{n-1}$, with diagonal entries two and off-diagonal entries one. Equivalently, $S^{\perp }$ in $M$ is the span of $v_1+\cdots +v_n$. This vector belongs to $S$ exactly when $p\mid n$. Thus the restricted radical has dimension one if $p\mid n$ and zero otherwise, givingFor $n\ge 3$ this quotient is nonzero and simple by the theorem. In the small exceptional case $n=2,p=2$, the canonical radical quotient is zero, as the same dimension formula states; it is then not a nonzero simple module. This qualification distinguishes the numerical formula from the nonzero-form hypothesis.