4.20. Lecture 20

We label the removable boxes of $\lambda$ by $\mathsf{c}_{1},\ldots\mathsf{c}_{r}$, with the numbering going from bottom to top. For each $i=1,\ldots,r$, we define a map $T_{i}^{\prime}:{\mathcal{M}}^{\lambda}\rightarrow{\mathcal{M}}^{\lambda\setminus\mathsf{c}_{i}}$ by defining it on the basis given by tabloids (and then extending linearly):

i.e. if not zero, then  $T^{\prime}_{i}([t])$ is the Young tabloid of shape $\lambda\setminus\mathsf{c}_{i}$, with the $n$ removed from $[t]$. We claim that $T_{i}^{\prime}$ is an $S_{n-1}$-homomorphism from $(\operatorname{Res}_{S_{n-1}}^{S_{n}}\pi,{\mathcal{M}}^{\lambda})$ to $(\pi,{\mathcal{M}}^{\lambda\setminus\mathsf{c}_{i}})$ (see Problem 4.20.3).

Observe that for elements of $\mathbf{SYT}^{\lambda}$, the number $n$ must be in a removable box, since there are no greater numbers to place either to its right or below. Supposing that $t\in\mathbf{SYT}^{\lambda}$ is such that $n$ is in $\mathsf{c}_{j}$, for any element $\sigma\in C(t)$ the element $n$ is then in in the same row of $[\sigma\cdot t]$ as $\mathsf{c}_{j}$ or above. This gives $T_{i}^{\prime}([\sigma\cdot t])=0$ for all $j>i$, since $\mathsf{c}_{j}$ is in a higher row than $\mathsf{c}_{i}$. In particular, we obtain ${\bf e}_{t}\in\ker T^{\prime}_{i}$ for all $j>i$.

On the other hand, if $t\in\mathbf{SYT}^{\lambda}$ has the number $n$ in the same row as $\mathsf{c}_{i}$, then

For $\sigma\in C(t)\setminus S_{n-1}$, $T_{i}^{\prime}([\sigma\cdot t])=0$, since (as described above) $\sigma$ must move $n$ to a box in a higher row. For $\sigma\in S_{n-1}$, $T_{i}^{\prime}[\sigma\cdot t]=[(\sigma\cdot t)\setminus n]=[\sigma\cdot(t\setminus n)]$, hence

We note also that since every standard Young tableau of shape $\lambda\setminus\mathsf{c}_{i}$ may be obtained by deleting the $n$ from a standard Young tableau of shape $\lambda$ with $n$ in $\mathsf{c}_{i}$,

We denote by $T_{1}$ the restriction of $T_{1}^{\prime}$ to ${\mathcal{S}}^{\lambda}$, and then for all other $i$,

This gives rise to a sequence of $S_{n-1}$-invariant subspaces of ${\mathcal{S}}^{\lambda}$:

By Lemma 1.2.9 and Proposition 3.11.1 (together with the isomorphism theorem for vector spaces),

and for $i\geq 1$,

Noting that $\ker(T_{r})=0$, we then have

By (4.20.3), ${\mathrm{im}}(T_{i})={\mathcal{S}}^{\lambda\setminus\mathsf{c}_{i}}$, hence

as claimed. ∎

Let $\chi_{\lambda}$ be the character of $(\pi,{\mathcal{S}}^{\lambda})$. For any $\mu\vdash n$, by Corollary 3.15.2, we have

By Theorem 4.20.1, $\langle\chi_{\lambda},\operatorname{Res}_{S_{n-1}}^{S_{n}}\chi_{\mu}\rangle_{S_{n-1}}=1$ if $\lambda$ is obtained by removing a box from $\mu$ (which is equivalent to $\mu$ being obtained by adding a box to $\lambda)$, and zero otherwise. ∎