Multiplicity of irreducible $\mathbb{C}S_n$-modules

Question

A known result in the representation theory of the symmetric group $S_n$ says:

"Let $T_{\lambda}$ be a Young tableaux corresponding to a $\lambda \vdash n$, and let $M=M_{1} \oplus M_{2} \oplus \cdots \oplus M_{m}$ where $M_{i}$ are irreducible $\mathbb{C}S_n$-modules with character $\chi_{\lambda}$. Then $m$ is equal to the maximal number of linearly independent elements of the form $e_{T_{\lambda}}f \in M$."

Is it correct to conclude that $m = \dim \operatorname{span}\{e_{T_{\lambda}}f: f \in M\}$?

Answer 1

It is know that for any Young Tableaux $T_{\lambda}$, there exists $f_1 \in M_1$ such that

\begin{equation*} M_1=\mathbb{C}S_n e_{T_{\lambda}}f_1\end{equation*}

Put $g_1=e_{T_{\lambda}}f_1$ and observe that $\sigma g_1=g_1$ for all $\sigma \in R_{T_{\lambda}}$. On the other hand, suppose $0\neq h \in M_1$ is also stable under $R_{T_{\lambda}}$-action. Then $h=ae_{T_{\lambda}}f_1$ for some $a \in \mathbb{C}S_n$. So $\sigma ae_{T_{\lambda}}f_1=a e_{T_{\lambda}}f_1$. It follow that $b=\sigma ae_{T_{\lambda}}-a e_{T_{\lambda}} \in J \cap \mathbb{C}_{S_n}e_{T_{\lambda}}$, where $J=\{x \in \mathbb{C}S_n e_{T_{\lambda}}: xf_1=0\}$ is the left annihilator of $f_1$. By irreducibility, $J\cap \mathbb{C}S_n e_{T_{\lambda}}=0$ or $\mathbb{C}S_ne_{T_{\lambda}}$. In the last case, $J \supseteq \mathbb{C}S_n e_{T_{\lambda}}$, so $\mathbb{C}S_ne_{T_{\lambda}}f_1=0$, a contradition. It follows that $J\cap \mathbb{C}S_n e_{T_{\lambda}}=0$. Hence $b=0$ and $\sigma a e_{T_{\lambda}}=ae_{T_{\lambda}}$, for all $\sigma \in R_{T_{\lambda}}$. Clearly, $ae_{T_{\lambda}}\tau=(-1)^{\tau}ae_{T_{\lambda}}$ for any $\tau \in C_{T_{\lambda}}$. Hence $\sigma ae_{T_{\lambda}}\tau=(-1)^{\tau}ae_{T_{\lambda}}$ for all $\sigma \in R_{T_{\lambda}}, \tau \in C_{T_{\lambda}}$.

Now, Von Neumann's Lemma (Giambruno and Zaicev. Polynomial Identities and Asymptotic Methods. page 50) says that:

Let $\lambda\vdash n, x \in \mathbb{C}S_n$ and $\sigma x\tau=(-1)^{\tau}x$ for all $\sigma \in R_{T_{\lambda}}, \tau \in C_{T_{\lambda}}$. Then $x=\gamma e_{T_{\lambda}}$ for some rational number $\gamma$.

So $ae_{T_{\lambda}}=\gamma e_{T_{\lambda}}$. Hence $h=\gamma g_1$.

Now put $g_1=e_{T_{\lambda}}f_1, g_2=e_{T_{\lambda}}f_1, \dots, g_m=e_{T_{\lambda}}f_m$. We have $M=M_1 \oplus M_2 \oplus \dots \oplus M_m$. Let $g \in M$ such that $\sigma g = g$ for all $\sigma \in R_{T_{\lambda}}$. We have that $g=l_1+l_2+ \ldots + l_m$, $g_i \in M_i$. It easy see that each $l_i$ is stable under $R_{T_{\lambda}}$-action. Hence:

$g=\gamma_1 g_1 + \gamma_1 g_1 + \ldots + \gamma_m g_m$

Observe that $g_1, \ldots, g_m$ are independents and generate the vector space;

$S=\{g \in M: \sigma g=g\; \mbox{for all} \;\sigma \in R_{T_{\lambda}}\}$ So $m=dim S$.

But $S=Span\{e_{T_{\lambda}}f_i: i=1, \ldots, m\}=Span\{e_{T_{\lambda}}f: f \in M\}$. Is it right?