## Power-Sum Polynomials

We will describe how the character table of $S_d$ is related to the expansion of the power-sum symmetric polynomials in terms of monomials. Recall:

$\displaystyle p_k = x_1^k + x_2^k + \ldots, \qquad p_\lambda = p_{\lambda_1} p_{\lambda_2} \ldots p_{\lambda_l},$

where $l =l(\lambda)$ exactly since $p_0$ is not defined.

Now each irrep of $S_d$ is of the form $V_\mu$ for some $\mu\vdash d$; we will denote its character by $\chi_\mu.$ Each conjugancy class of $S_d$ is also given by $[g_\lambda]$ where $\lambda\vdash d$ and $g_\lambda \in S_d$ has cycle structure $\lambda.$ E.g. if $\lambda = (4,2,1)$ we can take $g_\lambda = (2,4,6,7)(1,3) \in S_7$ as a representative.

We will calculate the character value $\chi_\mu(g_\lambda)$ by looking at the characters of $U_\mu = \mathbb{C}[X_\mu].$ For that, we need:

Lemma. If the finite group $G$ acts on finite set $X$, then the trace of $g : \mathbb{C}[X] \to \mathbb{C}[X]$ is: $|\{x \in X : gx = x\}|$, the number of fixed points of $g$ in $X.$

Proof

Taking elements of $X$ as a natural basis of $\mathbb{C}[X]$, the matrix representing $g\in G$ is a permutation matrix. Its trace is thus the number of ones along the main diagonal, which is the number of fixed points of $g\in G.$ ♦

Hence we have:

Theorem. Let $P_{\lambda\mu}$ be the trace of $g_\lambda$ on $U_\mu = \mathbb{C}[X_\mu].$ Expressing each $p_\lambda$ as a sum of monomials, we get:

$\displaystyle p_\lambda = \sum_\mu P_{\lambda\mu} m_\mu.$

Proof

Let us compute the coefficient of $x^\mu$ in the expansion of $p_\lambda = p_{\lambda_1}\ldots p_{\lambda_l}.$ For illustration let us take $\lambda = (3, 2, 2, 1, 1)$ and $\mu = (4, 3, 2)$; pick the representative $g_\lambda =(1, 2, 3)(4, 5)(6, 7)(8)(9).$ To obtain terms with product $x_1^4 x_3^3 x_2^2,$ here is one possibility:

\begin{aligned}p_3 &= \boxed{x_1^3} + x_2^3 + x_3^3 + \ldots\\p_2 &= x_1^2 + \boxed{x_2^2} + x_3^2 + \ldots\\ p_2 &= x_1^2 + x_2^2 + \boxed{x_3^2} + \ldots\\ p_1 &= x_1 + \boxed{x_2} + x_3 + \ldots\\ p_1 &= \boxed{x_1} + x_2 + x_3 + \ldots\end{aligned} \implies \begin{array}{|c|c|c|}\hline\{1, 2, 3\} & &\\ & \{4, 5\} & \\ & & \{6, 7\}\\ & \{8\} & \\ \{9\}& & \\ \hline\end{array}

The corresponding fixed point $(A_1, A_2, A_3) \in X_\mu$ is given by $A_1 = \{1, 2, 3, 9\},$ $A_2 = \{4, 5, 8\}$ and $A_3 = \{6, 7\}.$

In the general case, each contribution of $x^\mu$ corresponds to a selection of: $1 \le i_1, i_2, \ldots, i_m \le l(\mu)$ such that

$x_{i_1}^{\lambda_1} \cdot x_{i_2}^{\lambda_2} \cdot \ldots \cdot x_{i_m}^{\lambda_m} = x^\mu.$

This corresponds to a partition $[d] = \coprod_i A_i,$ defined by running through all the cycles of $g_\lambda$ and putting the elements of the c-th cycle into the set $A_{i_c}$ for $1 \le c \le m.$ Such a partition is invariant under $g_\mu.$ ♦

## Example

Thus $P_{\lambda\mu}$ is the number of ways of “merging” terms $\lambda_i$ to form the partition $\mu$ upon sorting. For example, taking $\lambda = (3,2,2,1,1)$ and $\mu = (4,3,2)$ from above, the coefficient of $x_1^4 x_2^3 x_3^2$ in $p_\lambda$ is 7:

## Character Table

Writing the above vectorially, we have $\mathbf{p} = \mathbf P \cdot \mathbf m,$ where $\mathbf P = (P_{\lambda\mu})$ is the trace of $g_\lambda$ on $U_\mu = \oplus_{\nu}V_\nu^{K_{\nu\mu}}.$ Thus, we have

$\displaystyle P_{\lambda\mu} = \sum_{\nu} K_{\nu\mu} \text{tr}(g_\lambda : V_\nu \to V_\nu)$

which gives us $\mathbf P = \mathbf X\cdot \mathbf K$ where $\mathbf X_{\lambda\mu} = \text{tr}(g_\lambda : V_\mu \to V_\mu)$ so X is the transpose of the character table for $S_d.$ Hence, we have:

$\mathbf p = \mathbf X\mathbf K\mathbf m= \mathbf X\cdot\mathbf s.$

Example: S4

For d=4 we obtain:

$\mathbf P = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 2 & 0 & 0 \\ 1 & 2 & 2 & 2 & 0 \\1 & 4 & 6 & 12 & 24\end{pmatrix}, \quad \mathbf K = \begin{pmatrix}1 &1 &1 & 1 & 1\\ 0 & 1 & 1 & 2 & 3 \\ 0 & 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 & 1\end{pmatrix}\implies \mathbf X= \begin{pmatrix} 1 & -1 & 0 & 1 & -1\\ 1 & 0 & -1 & 0 & 1\\ 1 & -1 & 2 & -1 & 1 \\ 1 & 1 & 0 & -1& -1 \\ 1 & 3 & 2 & 3 & 1\end{pmatrix}$

where the rows and columns are indexed by partitions 4, 31, 22, 211 and 1111. One checks that $\mathbf X^t$ is the character table for $S_4$:

## Orthogonality

Finally, orthonormality of irreducible characters translates to orthogonality of power-sum polynomials.

Proposition. The polynomials $p_\lambda$ form an orthogonal set, and $\left = z_\lambda$ is the order of the centralizer of any $g_\lambda$ with cycle structure $\lambda.$

Proof

Since $p_\lambda = X_{\lambda\mu} s_\mu$ and the $s_\mu$ are orthonormal,

$\displaystyle\left< p_\lambda, p_{\lambda'}\right> = \sum_\mu X_{\lambda\mu} X_{\lambda'\mu},$

which is entry $(\lambda, \lambda')$ of $\mathbf X \mathbf X^t$. This is the dot product between columns $\lambda$ and $\lambda'$ of the character table. By standard character theory, it is $\delta_{\lambda\lambda'} \frac{|G|} {|C_\lambda|}$ where $C_\lambda$ is the conjugancy class containing $g_\lambda.$ Now apply $\frac{|G|}{|C_\lambda|} = z_\lambda.$ ♦

Under the Frobenius map, the power-sum symmetric polynomial corresponds to:

$p_\lambda = \sum_\mu X_{\lambda\mu} s_\mu\ \mapsto\ \sum_\mu \chi_\mu(g_\lambda) \chi_\mu,$

so $p_\lambda(g) = \begin{cases} z_\lambda, &\text{if } g \text{ has cycle structure }\lambda, \\ 0, &\text{otherwise.}\end{cases}$

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