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,

$$\sum _{\sigma \in S^n}\mathrm{sgn}\sigma \cdot x_1^{n-\sigma (1)}{x}_2^{n-\sigma (2)-1}\cdots x_n^{n-\sigma (n)}=\det \left|\begin{array}{cccc}{x}_1^{n-1}& x_2^{n-1}& \cdots & x_n^{n-1}\\ ⋮& ⋮& & ⋮\\ x_1& x_2& \cdots & x_n\\ 1& 1& \cdots & 1\end{array}\right|=\prod _{1⩽i<j⩽n}(x_i-x_j)$$

and so the coefficient of $x_1^{{\ell }_1}\cdots x_n^{{\ell }_n}$ in

$$(x_1+\cdots +x_n{)}^n\prod _{1⩽i<j⩽n}(x_i-x_j)=\left(\sum _{k_1+\cdots +k_n=n}\frac{n!}{k_1!\cdots k_n!}{x}_1^{k_1}\cdots x_n^{k_n}\right)\left(\sum _{\sigma \in S^n}\mathrm{sgn}\sigma \cdot x_1^{n-\sigma (1)}{x}_2^{n-\sigma (2)-1}\cdots x_n^{n-\sigma (n)}\right)$$

is

$$\sum _{\sigma \in S_n}\mathrm{sgn}\sigma \frac{n!}{({\ell }_1-n+\sigma (1))!\cdots ({\ell }_n-n+\sigma (n))!}=n!\det \left|\begin{array}{ccc}\frac{1}{({\ell }_1-n+1)!}& \cdots & \frac{1}{({\ell }_n-n+1)!}\\ ⋮& & ⋮\\ \frac{1}{{\ell }_1!}& \cdots & \frac{1}{{\ell }_n!}\end{array}\right|=\frac{n!}{{\ell }_1!\cdots {\ell }_n!}\prod _{1⩽i<j⩽n}({\ell }_i-{\ell }_j).$$

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 $h(i,j)={\lambda }_i+{\lambda }_j^{\prime }-i-j+1$, where ${\lambda }_j^{\prime }$ is the number of $j$ th column.

The hook length formula tells us that

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

Here is an example, and here is a proof.

Young-Fibonacci Lattice

The number of standard tableaux for $\lambda$ is the number of path from $\varnothing$ to $\lambda$ diagram in Young diagram. Here is an example.

.