Frobenius Character Formula

For each partition $p=(l_1,\ldots,l_k)$ of $n$ the associated character will be denoted by $\chi_p$ and its value on a conjugacy class, i.e. on a set of elements with associated partition $q$ will be denoted by $\chi_p(q)$. So let $q$ be another partition: it consits of $q_j$ cycles of length $j$, i.e. $q_1+2q_2+3q_3+\cdots+nq_n=n$. Finally put $\l_j\colon=l_j+k-j$, then $\chi_p(q)$ is the coefficient of $$ x_1^{\l_1}x_2^{\l_2}\cdots x_k^{\l_k} $$ in the polynomial $$ \prod_{1\leq i < j\leq k}(x_i-x_j)\prod_j(x_1^j+\cdots+x_k^j)^{q_j}~. $$ For the $n$-cycle we get the partition $p=(n)$ and the associated character is the trivial character. For the identity $e$ we get the partition $p=(1,\ldots,1)$ and $\chi_p(q)$ is the coefficient of $x_1^{k}x_2^{k-1}\cdots x_k$ in the polynomial $$ \prod_{1\leq i < j\leq k}(x_i-x_j)\prod_j(x_1^j+\cdots+x_k^j)^{q_j}~. $$ Now let us find the value of $\chi_p$ at the identity $e$, which has partition $q=(1,\ldots,1)$:
The dimension of the irreducible representation with character $\chi_p$ is given by $$ \chi_p(e) =\frac{n!}{\l_1!\cdots\l_k!} \prod_{i < j}(\l_i-\l_j)~. $$
Cf. W. Fulton and J. Harris For $q=e=(1,\ldots,1)$ we get $q_1=n$ and we need to find the coefficient of $x_1^{\l_1}\cdots x_k^{\l_k}$ in the polynomial $$ P\colon=\prod_{1\leq i < j\leq k}(x_i-x_j)(x_1+\cdots+x_k)^n $$ Now $\prod(x_i-x_j)$ is the Vandermonde determinante $$ \prod_{1\leq i < j\leq k}(x_i-x_j) =\sum_{\pi\in S(k)}\sign(\pi)x_1^{\pi(1)-1}\cdots x_k^{\pi(k)-1} $$ and the other term is $$ (x_1+\cdots+x_k)^n=\sum\frac{n!}{r_1!\cdots r_k!}x_1^{r_1}\cdots x_k^{r_k} $$ where the sum is taken over all $r_1,\ldots,r_k\in\N_0$ such that $r_1+\cdots+r_k=n$. Thus we get for the polynomial $P$: $$ \sum\sum_\pi\sign(\pi) \frac{n!}{r_1!\cdots r_k!}x_1^{r_1+\pi(1)-1}\cdots x_k^{r_k+\pi(k)-1} $$ To get the coefficient of $x_1^{\l_1}\cdots x_k^{\l_k}$ we must have $r_j+\pi(j)-1=\l_j$, i.e. $r_j=\l_j+1-\pi(j)$ and $\pi(j)\leq\l_j+1$; for these choices we indeed have $$ \sum_{j=1}^k(\l_j-\pi(j)+1) =n+k(k-1)/2-k(k+1)/2+k=n~. $$ Thus the coefficient is given by \begin{eqnarray*} &&\sum_{\pi:\pi(j)\leq\l_j+1}\sign(\pi)\frac{n!}{\prod_j(\l_j+1-\pi(j))!}\\ &=&\frac{n!}{\prod_j\l_j!} \sum_{\pi:\pi(j)\leq\l_j+1}\sign(\pi)\prod_j\l_j(\l_j-1)...(\l_j+2-\pi(j))\\ &=&\frac{n!}{\prod_j\l_j!} \sum_{\pi\in S(k)}\sign(\pi)\prod_j\l_j(\l_j-1)...(\l_j+2-\pi(j))~. \end{eqnarray*} Define a matrix $A=(a_{ji})\in\Ma(k,\Z)$ by $a_{j1}=1$ and for $i\geq2$: $a_{ji}=\l_j(\l_j-1)\cdots(\l_j+2-i)$, then $$ \det A =\sum\sign(\pi)\prod_j a_{j\pi(j)} =\sum_{\pi}\sign(\pi)\prod_j\l_j(\l_j-1)...(\l_j+2-\pi(j)), $$ i.e. $\chi_p(e)=n!\det(A)/\prod_j\l_j!$. Now $$ \det A =\det\left(\begin{array}{cccc} 1&\l_1&\l_1(\l_1-1)&\cdots\\ \vdots&\vdots&\vdots&\vdots\\ 1&\l_k&\l_k(\l_k-1)&\cdots \end{array}\right) =\det\left(\begin{array}{cccc} 1&\l_1&\l_1^2&\cdots\\ \vdots&\vdots&\vdots&\vdots\\ 1&\l_k&\l_k^2&\cdots \end{array}\right) =\prod_{i < j}(\l_i-\l_j)~. $$

Young diagrams

The Young diagram of a partition $p=(l_1,\ldots,l_k)$ assigns $l_j$ boxes in the $j$-th row, the rows lined up on the left. The conjugate partition $p^\prime$ is defined by interchanging the rows and columns in the Young diagram. The rank $r$ of a partition is the length of the diagonal in its Young diagram. Let $a_j$ and $b_j$ be the number of boxes below and to the right of the $j$-th diagonal box. The subsequent picture exhibits a partition $p$ of $n=15=\sum l_j$, $l_1=6$, $l_2=4$, $l_3=2$ and $l_4=l_5=l_6=1$, i.e. $p=(6,4,2,1,1,1)$. The rank of the diagram is $2$, $a_1=5$, $a_2=1$, $b_1=5$, $b_2=2$.
Young diagram
The partition describes a permutation composed of $k=6$ disjoint cycles and since $\l_1=11$, $\l_1=8$, $\l_3=5$, $\l_4=3$, $\l_5=2$, $\l_6=1$, the dimension of the representation is $$ \chi_p(e)=\frac{15!}{11!8!5!3!2!}3.6.8.9.10.3.5.6.7.2.3.4.2 =221130 $$
For any transposition $\t$ we have (cf. W. Fulton and J. Harris p.52): $$ \frac{\chi_p(\t)}{\chi_p(e)} =\frac1{n(n-1)}\sum_{j=1}^r(b_j(b_j+1)-a_j(a_j+1))~. $$
Suppose $\chi_p$ is not the trivial character. Then $r\geq2$ and thus $a_1\geq1$. The sum is maximal if $b_1=n-2$, $b_2=0$ and $a_1=1$, $a_2=0$, i.e. $$ \frac{\chi_p(\t)}{\chi_p(e)} \leq\frac1{n(n-1)}((n-2)(n-1)-2) =\frac{n-3}{n-1} $$