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

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 $w$.