10.7 Branching Rule

Definition

A box of $\lambda$ is called removable, if its removal leaves a diagram, which is denoted by ${\lambda }^{-}$.

Definition

A box of $\lambda$ is called addable to $\lambda$ if the union of $\lambda$ and this box is a diagram.

Lemma

We have $\dim S^{\lambda }=f^{\lambda }={\sum }_{{\lambda }^{-}}{f}^{{\lambda }^{-}}$.

\begin{proof} Recall that basis of $S^{\lambda }$ is polytabloids of standard tableaux. Notice every standard tableau consist of $n$ in some removable box and a standard tableau for some ${\lambda }^{-}$ and then we finish the proof. \end{proof}

branching rule

If $\lambda ⊢n$ is a partition, then

S^\lambda\downarrow_{S_{n-1}}=\oplus_{\lambda^-}S^{\lambda^-}\mbox{ and }S^\lambda\uparrow^{S_{n+1}}=\oplus_{\lambda^+}S^{\lambda^+}. $$ ^5voxbj

\begin{proof} Suppose that the removable boxes appear in rows ${\tau }_1<\cdots <{\tau }_k$, for each $i$, denote by ${\lambda }^i$ diagram when you remove box in row $r_i$. For standard tableau $t$ with $n$ in $i$th two denote by $t^i$ tableau obtained by removing box with $n$. Since $S_n$ is a finite group, for any $S_n$-modules $V,W$ with $W\subseteq V$, we have $V\simeq V/W\oplus W$ by Maschke’s theorem.

To prove the theorem, it suffices to construct chain of $S_{n-1}$-modules

$$\{0\}=V^{(0)}\subseteq V^{(1)}\subseteq \cdots \subseteq V^{(k)}=S^{\lambda }\text{\hspace{1em}(*)}$$

such that $V^{(i)}/V^{(i-1)}\simeq S^{{\lambda }^i}$.

Define $V^i$ as submodule in $S^{\lambda }$ spanned by all polytabloids corresponding to standard tableaux with $n$ being in the rows $⩽i$, that is, $n$ being in rows $r_1,\cdots ,r_i$. Then we get $(\ast )$, the chain of modules what we desire. Define

$${\theta }_i:M^{\lambda }\to M^{{\lambda }^i},\{t\}\mapsto \{\begin{array}{ll}\{t^i\}& \text{mbox}ifn\text{mbox}isintherow{r}_i\text{mbox}of\{t\},\\ 0& \text{mbox}otherwise.\end{array}$$

Then ${\theta }_i$ is $S_{n-1}$-homomorphism. Since $n$ occupies a removable box, we have

$${\theta }_i:e_t\mapsto \{\begin{array}{ll}{e}_{t^i}& \text{mbox}ifn\text{mbox}isintherow{r}_i\text{mbox}of\{t\},\\ 0& \text{mbox}otherwise.\end{array}$$

and so ${\theta }_i{V}^{(i)}\simeq S^{{\lambda }^i}$ and $\ker {\theta }_i=V^{(i-1)}$. It deduces that

$$\dim V^i/V^i\cap \ker {\theta }_i=\dim {\theta }_i{V}^i=\dim S^{{\lambda }^i}=f^{{\lambda }^i}$$

and $(\ast )$ can be written as

$$\{0\}=V^{(0)}=V^{(1)}\cap \ker {\theta }_1\subseteq V^{(1)}=V^{(2)}\cap \ker {\theta }_2\subseteq \cdots \subseteq V^{(k)}=S^{\lambda }.$$

Therefore, we have

$$\sum _{i=1}^k\dim V^{(i)}/V^{(i)}\cap \ker {\theta }_i=\sum _{i=1}^k{f}^{{\lambda }^i}=\dim S^{\lambda }.$$

Furthermore, by Frobenius reciprocity, there is

$$m_{\mu }=⟨{\chi }^{\lambda }{\uparrow }^{S_{n+1}},{\chi }^{\mu }⟩=⟨{\chi }^{\lambda },{\chi }^{\mu }{\downarrow }_{S_n}⟩=⟨{\chi }^{\lambda },\sum _{{\mu }^{-}}{\chi }^{{\mu }^{-}}⟩$$

and so $m_{\mu }=0$ if $\lambda \ne {\mu }^{-}$ and $m_{\mu }=1$ if $\lambda ={\mu }^{-}$. Note that $\lambda ={\mu }^{-}$ iff ${\lambda }^{+}=\mu$, thus ${\chi }^{\lambda }{\uparrow }^{S^{n+1}}={\sum }_{{\lambda }^{+}}{\chi }^{{\lambda }^{+}}$. Now we finish the proof. \end{proof}

Corollary

The restriction $S^{\lambda }{\downarrow }_{S_{n-1}}$ is irreducible iff the diagram $\lambda =(m^{\ell })$, that is, $\lambda$ is a rectangle.