## Differences of permutation characters of symmetric groups

Here are a few small observations on permutation characters motivated by discussions at the Herstmonceux conference Algebraic combinatorics and group actions. Let $\chi^\lambda$ denote the irreducible character of the symmetric group $S_n$ labelled by the partition $\lambda$ of $n$. Let $\pi^\lambda$ denote the permutation character of $S_n$ acting on the cosets of the Young subgroup $S_\lambda$.

• Since $\pi^\lambda = \chi^\lambda + \phi$ where $\phi$ is an integral linear combination of characters $\chi^\mu$ for partitions $\mu$ dominating $\lambda$, any irreducible character of $S_n$ is an integral linear combination of the $\pi^\lambda$.

• This can be made more explicit as follows. Let $\delta = (n-1,\ldots,1,0)$. For $\sigma \in S_n$, let $\lambda \cdot \sigma = (\lambda + \delta)\sigma - \delta$, where $\sigma$ acts on $\lambda + \delta$ by place permutation. The Jacobi–Trudi formula states that

$\chi^\lambda = \sum_{\sigma \in S_n} \pi^{\lambda \cdot \sigma} \mathrm{sgn}(\sigma).$

• It is a special property of symmetric groups that any character is a difference of permutation characters. For example, the cyclic group $C_n$ has $\mathrm{d}(n)$ (the number of divisors of $n$) distinct permutation characters, but $n$ irreducible characters, so unless $n \le 2$, the additive group generated by the permutation characters is a proper subgroup of the integral character ring. And clearly no irreducible character of a finite group taking a non-rational value can be an integral linear combination of permutation characters.

• Since a transitive permutation character contains the trivial character exactly once, the trivial character is not a difference of two transitive permutation characters. A further example is given below.

Claim. $\chi^{(n-1,1)} + \chi^{(1^n)}$ is not a difference of two transitive permutation characters of $S_n$.

Proof. Let $\pi_K$ denote the permutation character of $S_n$ acting on the subgroup $K$ of $S_n$. Note that the sign character, $\chi^{(1^n)}$, appears in $\pi_K$ if and only if $K \le A_n$.

Let $s(K)$ be the number of orbits of $K$ on $\{1,\ldots, n\}$. Since

$\chi^{(n-1,1)} = \pi^{(n-1,1)} - 1_{S_n},$

it follows from Frobenius reciprocity and Mackey’s Formula that $\langle \pi_K, \chi^{(n-1,1)} \rangle = s(K)-1$. Consider $\chi^{(2,1^{n-2})}$. Multiplying the displayed equation above by $\mathrm{sgn}$, we get

$\chi^{(2,1^{n-2})} = \mathrm{sgn}_{S_{n-1}}\uparrow^{S_n} - \mathrm{sgn}_{S_n}.$

By Frobenius reciprocity

$\langle \pi_K, \mathrm{sgn}_{S_{n-1}}\uparrow^{S_n} \rangle = \langle \pi_K \uparrow^{S_n} \downarrow_{S_{n-1}}, \mathrm{sgn}_{S_{n-1}} \rangle.$

By Mackey’s Formula, the right-hand side is

$\sum \langle \pi_K \uparrow_{K^g \cap S_{n-1}}^{S_{n-1}}, \mathrm{sgn}_{S_{n-1}} \rangle,$

where the sum is over distinct double cosets $K g S_{n-1}$. It follows that $\langle \pi_K, \chi^{(2,1^{n-2})} \rangle \le s(K)-1$, with equality when $K \le A_n$.

Suppose that $H$ and $G$ are subgroups of $S_n$ such that $\pi_H - \pi_G$ contains $\chi^{(n-1,1)}$ and $\chi^{(1^n)}$. From $\chi^{(1^n)}$ we see that $H \le A_n$ and $G \not\le A_n$. From $\chi^{(n-1,1)}$, we see that $s(G) = s(H) -1$. By the previous paragraph,

$\langle \pi_H, \chi^{(2,1^{n-2})} \rangle = s(H)-1,$

whereas

$\langle \pi_G, \chi^{(n-2,1^{n-2})} \rangle \le s(G)-1 < s(H)-1.$

Hence $\chi^{(2,1^{n-2})}$ also appears in $\pi_H -\pi_G$. Therefore

$\chi^{(n-1,1)} + \chi^{(1^n)}$

is not a difference of two transitive permutation characters. $\Box$

• Since $\pi^{(n-r,r)} - \pi^{(n-r+1,r-1)} = \chi^{(n-r,r)}$ for $1 \le r \le n/2$ (this is a special case of the Jacobi—Trudi formula) each $\chi^{(n-r,r)}$ is a difference of two transitive permutation characters. Moreover, $\chi^{(1^n)} = \pi_{A_n} - 1_{S_n}$. Exhausting over all subgroups of $S_n$ by computer algebra supports the conjecture that, provided $n \ge 9$, these are the only irreducible characters of $S_n$ that are the difference of two transitive permutation characters.
• The behaviour of $S_6$ is interesting: all irreducible characters except for $\chi^{(3,2,1)}$ are the difference of two transitive permutation characters. This is explained in part by the outer automorphism that exists in this case.