Polynomials and Representations XXIII

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. ♦

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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:

power_sum_symmetric_expansion

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:

chartable_s4_2

Orthogonality

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

Proposition. The polynomials p_\lambda form an orthogonal set, and \left<p_\lambda, p_\lambda\right> = 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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