The sum of all hook characters

Wildon's Weblog 2019-08-20

Given a partition $\lambda$ of $n$ , let $\chi^\lambda$ denote the irreducible character of the symmetric group $S_n$ canonically labelled by the partition $\lambda$ . For example, $\chi^{(n)} = 1$ is the trivial character, $\chi^{(1^n)}$ is the sign character, and $\chi^{(n-1,1)}$ is the irreducible $n-1$ -degree constituent of the natural permutation character $\pi$ . Thus $\pi(g) = |\mathrm{Fix}(g)|$ and $\chi^{(n-1,1)}(g) = |\mathrm{Fix}(g)| - 1$ for $g \in S_n$ . Let $\Gamma = \sum_{a=0}^{n-1} \chi^{(n-a,1^a)}$ be the sum of all the characters labelled by hook partitions. Let $\mathcal{O}_n$ be the set of permutations in $S_n$ having only cycles of odd length.

A nice short paper by Jay Taylor uses characters of the symmetric group labelled by skew partitions to prove that

$\Gamma(g) = \begin{cases} 2^{\mathrm{cyc}(g) - 1} & g \in \mathcal{O}_n \\ 0 & g \not\in \mathcal{O}_n \end{cases}$

where $\mathrm{cyc}(g)$ is the number of cycles in $g$ .

As Taylor notes, this result is a special case of a more general theorem of Regev, proved by him using Lie superalgebras. Taylor’s paper gives a short proof using skew characters of the symmetric group. The purpose of this post is to give an alternative proof, using only a basic result on exterior powers of the natural permutation character and to suggest some generalizations not contained in Regev’s original results.

Background: a generating function for exterior powers.

Let $\bigwedge^k$ denote the $k$ th exterior power of a character or representation. Let $V$ be a complex representation of a finite group $G$ with character $\chi$ and let $g \in G$ have eigenvalues $\lambda_1, \ldots, \lambda_d$ in its action on $V$ . Thus $\chi(g) = \lambda_1 + \cdots + \lambda_d$ . Let $v_1,\ldots, v_n$ be a basis for $V$ in which $g$ acts diagonally. Then

$\bigwedge^r V = \langle v_{i_1} \wedge\cdots \wedge v_{i_r} : i_1 < \ldots < i_r \rangle_\mathbb{C}$

and we see that

$(\bigwedge^r \chi) (g) = \sum_{i_1 < \ldots < i_r} \lambda_{i_1} \ldots \lambda_{i_r}.$

This is the $r$ -th elementary symmetric polynomial evaluated at the eigenvalues $\lambda_1, \ldots, \lambda_d$ . Hence

$(1)\quad \sum_{r=0}^d (\bigwedge^r \chi)(g) x^r = (1+\lambda_1 x) \cdots (1+\lambda_d x).$

This generating function is the counterpart for exterior powers of the Molien series, defined using symmetric powers, that is fundamental to invariant theory.

Background: exterior powers of the natural module.

Let $V = \langle e_1, \ldots, e_n\rangle_\mathbb{C}$ be the natural representation of $S_n$ over the complex numbers. Now $\bigwedge^n V = \langle e_1 \wedge \cdots \wedge e_n \rangle$ is the sign representation of $S_n$ , and correspondingly, $\bigwedge^n \pi = \chi^{(1^n)}$ . More generally, $\bigwedge^r V = \langle e_{i_1} \wedge \cdots \wedge e_{i_r} : i_1 < \ldots < i_r\rangle_\mathbb{C}$ is the monomial representation of $S_n$ induced from $\mathrm{sgn}_{S_r} \times \mathbb{C}_{S_{n-r}}$ . Its character is known to be

$(2)\quad\bigwedge^r \pi = \chi^{(n-r,1^r)} + \chi^{(n-r+1,1^{r-1})}$

for $1 \le r < n$ . This decomposition is an immediate corollary of either the Young or Pieri rules. It may also be proved using Specht modules and an explicit isomorphism $\bigwedge^r S^{(n-1,1)} \cong S^{(n-r,1^r)}$ : see for instance Section 5 of this paper with Giannelli and Lim.

Proof.

By (2) we have

$\chi^{(n-r,1^r)} = \bigwedge^r \pi - \bigwedge^{r-1} \pi + \cdots + (-1)^{r-1} \pi + (-1)^r 1.$

Hence the coefficient of $\bigwedge^k \pi$ in $\Gamma$ is $1$ if $n-k$ is odd, and $0$ if $n-k$ is even. For example,

$\begin{aligned} \Gamma_4 &= \chi^{(4)} + \chi^{(3,1)} + \chi^{(2,1,1)} + \chi^{(1,1,1,1)} \\ &= 1+ (\pi - 1) + (\bigwedge^2 \pi - \pi + 1) + (\bigwedge^3 \pi - \bigwedge^2 \pi + \pi - 1) \\ &= \pi + \bigwedge^3 \pi \end{aligned}$

Thus $\Gamma = \sum_{k} \bigwedge^k \pi$ , where the sum is over all $k$ such that $n-k$ is odd.

This sum is related to (1). Let $g \in S_n$ have precisely $a_k$ cycles of length $k$ , for each $k \in \{1,\ldots, n\}$ . The eigenvalues of the $n \times n$ permutation matrix representing $g$ can be found by considering the cycles separately: each $k$ -cycle contributes eigenvalues $1, \zeta_k, \ldots, \zeta_{k}^{k-1}$ , where $\zeta_k$ is a primitive $k$ th root of unity. Since $(1+x)(1+\zeta x) \cdots (1+\zeta^{k-1}x ) = 1 + (-1)^{k-1} x^k$ , the generating function in (1) gives

$\sum_{k=0}^n (\bigwedge^k \pi)(g)x^k = (1+x)^{a_1}(1-x^2)^{a_2} \ldots (1+(-1)^{n-1} x^n)^{a_n}.$

Denote this polynomial by $F_g(x)$ . We now have

$\Gamma(g) = \begin{cases} \frac{F_g(1) + F_g(-1)}{2} & n \equiv 1 \bmod 2 \\ \frac{F_g(1) - F_g(-1)}{2} & n \equiv 0 \bmod 2. \end{cases}$

Since

$F_g(1) = \begin{cases} 2^{\mathrm{cyc}(g)} & g \in \mathcal{O}_n \\ 0 & g \not\in \mathcal{O}_n \end{cases}$

and $F_g(-1) = 0$ for all $g$ , the result follows.

Possible generalizations.

Let $\Delta = \sum_{a} \chi^{(n-a,1^a)}$ where the sum is over all $a$ not congruent to $1$ modulo $3$ . Then by (2), $\Delta = \sum_{b \ge 0} \bigwedge^{3b} \pi$ , and the proof above suggests that $\Delta(g)$ may also have a fairly simple form. One might also look for analogous identities replacing $\bigwedge^k \pi$ with the permutation character of $S_n$ acting on the $k$ -subsets of $\{1,\ldots, n\}$ .