10.6 Frobenius Character Formula

Frobenius character formula

Let $\lambda ⊢n$ with $\lambda =({\lambda }_1,\cdots ,{\lambda }_{\ell })$, and let ${\lambda }_{\ell +1}=\cdots ={\lambda }_n=0$. The value of ${\chi }^{\lambda }$ of $S^{\lambda }$ on conjugacy class $C_{\mu }$ given by $\mu =({\mu }_1,\cdots ,{\mu }_k)$ with ${\mu }_1⩾\cdots ⩾{\mu }_k⩾1$ equals to the coefficient of the monomial $x_1^{{\ell }_1}\cdots x_n^{{\ell }_n}$ in polynomial

$$\prod _{i=1}^k(x_1^{{\mu }_i}+\cdots +x_n^{{\mu }_i})\prod _{i<j}(x_i-x_j),$$

where ${\ell }_i={\lambda }_i+n-i$.

Corollary

The dimension of Specht module $S^{\lambda }$ is

$$\dim S^{\lambda }=f^{\lambda }=\frac{n!}{{\ell }_1!\cdots {\ell }_n!}\prod _{1⩽i<j⩽n}({\ell }_i-{\ell }_j).$$

\begin{proof} Note that $\dim S^{\lambda }={\chi }^{\lambda }(1)$, then by ^4b468a

$$\dim S^{\lambda }=\text{mbox}coefficientof{x}_1^{{\ell }_1}\cdots x_n^{{\ell }_n}\text{mbox}in(x_1+\cdots +x_n{)}^n\prod _{1⩽i<j⩽n}(x_i-x_j).$$

By Vandermonde determinant,

\vdots & \vdots & & \vdots \\ x_1 & x_2 & \cdots & x_n \\ 1 & 1 & \cdots & 1\end{matrix}\right|=\prod_{1\leqslant i< j\leqslant n}(x_i-x_j)$$ and so the coefficient of $x_1^{\ell_1}\cdots x_n^{\ell_n}$ in

(x_1+\cdots+x_n)^n\prod_{1\leqslant i<j\leqslant n}(x_i-x_j)=\left(\sum_{k_1+\cdots+k_n=n}\frac{n!}{k_1!\cdots k_n!}x_1^{k_1}\cdots x_n^{k_n}\right)\left(\sum_{\sigma\in S^n}\mathrm{sgn}\sigma \cdot x_1^{n-\sigma(1)}x_2^{n-\sigma(2)-1}\cdots x_n^{n-\sigma(n)}\right)

is $$\sum_{\sigma\in S_n}\mathrm{sgn}\sigma\frac{n!}{(\ell_1-n+\sigma(1))!\cdots(\ell_n-n+\sigma(n))!}=n!\det\left|\begin{matrix} \dfrac{1}{(\ell_1-n+1)!} & \cdots & \dfrac{1}{(\ell_n-n+1)!} \\ \vdots & & \vdots \\ \dfrac{1}{\ell_1!} & \cdots & \dfrac{1}{\ell_n!} \end{matrix}\right|=\frac{n!}{\ell_1!\cdots\ell_n!}\prod_{1\leqslant i<j\leqslant n}(\ell_i-\ell_j).$$ Now we finish the proof. `\end{proof}` # Hook Length Formula For a Young tableau $\lambda$ and the element on $i$ th row, $j$ th column, define $h(i,j)=\lambda_i+\lambda_j'-i-j+1$, where $\lambda_j'$ is the number of $j$ th column. The hook length formula tells us that

\dim S^\lambda=f^\lambda=\frac{n!}{\prod_{(i,j)\in\lambda}h(i,j)}.

[[Pasted image 20241202154117.png|Here]] is an example, and [[List 4 (Oral Exam)#^xmv91w|here]] is a proof. # Young-Fibonacci Lattice The number of standard tableaux for $\lambda$ is the number of path from $\emptyset$ to $\lambda$ diagram in Young diagram. Here is an example. ![[Pasted image 20241204154913.png|300]].