Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 160 / 1

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 $w$.