10.4 Complete List of Irreducible Modules

Definition

For each $M^{\lambda }=\mathbb{C}[\{t_1\},\cdots ,\{t_k\}]$, define an inner product as $⟨\{t_i\},\{t_i\}⟩={\delta }_{ij}$. Note that this inner produce is $S_n$-invariant, i.e., $⟨\pi \{t_i\},\pi \{t_j\}⟩=⟨\{t_i\},\{t_j\}⟩$.

sign lemma

Let $H$ be a subgroup of $S_n$.

• If $\pi \in H$, then $\pi H^{-}=H^{-}\pi =(\mathrm{sgn}\pi )H^{-}$.
• For any $u,v\in M^{\lambda }$, $⟨H^{-}u,v⟩=⟨u,H^{-}v⟩$.
• If $(bc)\in H$, then $H^{-}=\kappa (1-(bc))$ for some $\kappa \in \mathbb{C}{S}_n$.
• If $b$ and $c$ are in the same row of $t$, then $H^{-}\{t\}=0$.

\begin{proof} Easy. See here. \end{proof}

Corollary

Let $t$ be tableau of shape $\lambda$ and $s$ be tableau of shape $\mu$, where $\lambda ⊢n$ and $\mu ⊢n$. If $K_t\{s\}\ne 0$, then $\lambda ⊳\mu$. Moreover, if $\lambda =\mu$ and $K_t\{s\}\ne 0$, then $K_t\{s\}=\pm e_t$.

\begin{proof} Suppose $b$ and $c$ are in the same row of $s$, then they belong to different columns of $t$. Otherwise $K_t\{s\}=\kappa (1-(bc))\{t\}=0$ by ^fae9c5. Then by ^3e06ba $\lambda ⊳\mu$. If $\lambda =\mu$ and $K_t\{s\}\ne 0$, then $\{s\}=\pi \{t\}$ for some $\pi \in C_t$ (by ^fae9c5, each pair of numbers in the same column does not appear in the same row of $\{s\}$). It follows that $K_t\{s\}=K_t\{\pi t\}=\mathrm{sgn}\pi K_t\{t\}=\pm e_t$ by ^fae9c5. \end{proof}

Corollary

If $u\in M^{\mu }$ and $t$ is tableau of shape $\mu$, then $K_{t}u$ is multiple of $e_t$.

\begin{proof} Since $u$ is a linear combination of tabloid of shape $\mu$, it follows from ^2fe718. \end{proof}

submodule theorem

Let $U$ be submodule of $M^{\lambda }$, then either $U\supseteq S^{\lambda }$ on or $U\subseteq (S^{\lambda }{)}^{\perp }$. In particular, $S^{\lambda }$ is irreducible.

\begin{proof} Take $u\in M^{\lambda }$, then $K_{t}u=c{e}_t$ for some $c\in \mathbb{C}$. Suppose that there exists $t$ such that $c\ne 0$. But then $c{e}_t\in U$. As $S^{\lambda }$ is cyclic, $\pi (c{e}_t)$ generates $S^{\lambda }$ and is inside of $U$, which yields that $S^{\lambda }\subseteq U$. Otherwise, for any tableau $t$ and $u\in M^{\lambda }$, we have $K_{t}u=0$ and so

$$⟨u,e_t⟩=⟨u,K_t\{t\}⟩=⟨K_{t}u,\{t\}⟩=⟨0,\{t\}⟩=0.$$

Therefore, $U\subseteq (S^{\lambda }{)}^{\perp }$. \end{proof}

Proposition

If there is a non-zero element $\theta \in \mathrm{Hom}(S^{\lambda },M^{\mu })$, then $\lambda ⊳\mu$. If $\lambda =\mu$, then $\theta$ is multiplication by a scalar.

\begin{proof} Since $\theta$ is non-zero, there exists polytabloid $e_t$ such that $\theta (e_t)\ne 0$. Moreover, as $M^{\lambda }$ can be decomposed as $M^{\lambda }=S^{\lambda }\oplus (S^{\lambda }{)}^{\perp }$, we extend $\theta$ to a homomorphism

$$\theta :M^{\lambda }\to M^{\mu },(S^{\lambda }{)}^{\perp }\mapsto 0.$$

Since $0\ne \theta (e_t)=\theta (K_t\{t\})=K_t(\theta \{t\})=K_t{\sum }_i{\alpha }_i\{s_i\}$ for some $\mu$-tabloids $\{s_i\}$, there is $\lambda ⊳\mu$ by ^2fe718.

If $\lambda =\mu$, then $\theta (e_t)=K_t{\sum }_i{\alpha }_i\{s_i\}=c{e}_t$. Then for any permutation $\pi$,

$$\theta (e_{\pi t})=\theta (\pi e_t)=\pi (\theta (e_t))=c\pi e_t=c{e}_{\pi t}$$

and so $\theta$ is a multiplication by $c$. \end{proof}

Theorem

The Specht modules $S^{\lambda }$ for $\lambda ⊢\mu$ form a complete list of irreducible modules of $S_n$.

\begin{proof} Since the number of $S^{\lambda }$ equals the number of $\lambda ⊢n$ and so the number of conjugacy classes of $S_n$, it suffices to show $S^{\lambda }\ne S^{\mu }$ for any $\lambda \ne \mu$.

Otherwise, suppose that $S^{\lambda }=S^{\mu }$ with $\lambda \ne \mu$. Since $S^{\mu }\hookrightarrow M^{\mu }$ is non-trivial, we have non-zero $\theta \in \mathrm{Hom}(S^{\lambda },M^{\mu })$ and so $\lambda ⊳\mu$ by ^376fda. Similarly we have $\mu ⊳\lambda$ and so $\mu =\lambda$, contradiction. Now we finish the proof. \end{proof}

Corollary

The irreducible decomposition of the permutation module $M^{\lambda }$ has the form

$$M^{\mu }={\oplus }_{\lambda ⊳\mu }{\kappa }_{\lambda \mu }{S}^{\lambda },$$

where ${\kappa }_{\lambda \mu }$ is called Kostka numbers. Moreover, ${\kappa }_{\mu \mu }=1$.

Here is an example.

\begin{proof} Note that ${\kappa }_{\lambda \mu }=\dim {\mathrm{Hom}}_{S_n}(S^{\lambda },M^{\mu })=⟨{\chi }^{\lambda },{\psi }^{\mu }⟩$ where ${\chi }^{\lambda }$ and ${\psi }^{\mu }$ are characters of $S^{\lambda }$ and $M^{\mu }$, respectively. If ${\kappa }_{\lambda \mu }\ne 0$, then $\lambda ⊳\mu$ by ^376fda. Furthermore, since $\theta \in \mathrm{Hom}(S^{\mu },M^{\mu })$ is multiplication by scalar by ^376fda, we have $\dim {\mathrm{Hom}}_{S_n}(S^{\lambda },M^{\lambda })={\kappa }_{\mu \mu }=1$. \end{proof}