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.
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