Submitted by Dennis White
Suppose is a finite permutation group acting on . Let denote the conjugacy class of permutations of type in symmetric group . Let be the irreducible character evaluated at the conjugacy class .
Define
In fact, is the number of occurrences of the irreducible in the induction of the trivial character of up to , or, by Frobenius reciprocity, the dimension of the -fixed space inside the -irreducible corresponding to .
It is therefore an integer and
where is the number of standard Young tableaux (SYT) of shape .
OPAC-024. Interpret as a subset of SYT of shape .
The group then acts (Pólya action) on colorings of . For a partition , let be the orbits of -colorings of , that is, the orbits in which color appears times. It follows from Pólya’s Theorem that
where , the Kostka number, counts the number of semistandard Young tableaux (SSYT) of type and shape .
Alternatively (see [1]), compute the dimension of the -weight space inside the -fixed space of the -representation on in two ways, either directly or via Schur-Weyl duality.
OPAC-025. Give a Schensted-like proof of Equation (*).
Example 1. If (a Young subgroup), then and
.
The tableaux in OPAC-024 are then the standarization tableaux of the SSYT and the Schensted-like proof is the Robinson-Schensted-Knuth correspondence (RSK).
Example 2. We say is a descent in SYT if lies in a row above in . Define
.
If , the cyclic group of order , acting on , then the tableaux of OPAC-024 are those SYT such that is a multiple of . See [1]. However, usual Schensted applied to these tableaux does not produce orbit representatives for the Pólya action, so OPAC-025 is unresolved.
Example 3. The techniques in [1] can also be used to solve OPAC-024 if acting on .
Example 4. Also solved in [1] is the case where is the alternating subgroup of a Young subgroup. If is the alternating subgroup of , then . Here, denotes the conjugate of . The solution to OPAC-025 uses a small modification to the standardization argument for the RSK algorithm for the Young subgroup case.
Example 5. In fact, suppose acts on , with acting on and on (using a different alphabet). Suppose we know that is a solution tableau (shape ) to OPAC-024 for in and is a solution tableau (shape ) for in (again, different alphabet). Then let be a partition larger than and with . Construct a tableau of shape as follows. Let be a SYT of shape such that the lattice word of fits (and so is counted by the Littlewood-Richardson coefficient). The portion of in is . The portion of in is the Schensted word corresponding to the pair (see [1] for details; see also [3]). Of course, this idea may be extended to longer direct products.
Example 6. Applying Example 5 to Example 2 and using the fact that jeu de taquin preserves the descent set (see [2, Ch. 7 Appendix 1]), it follows that the tableaux can be chosen to be those tableaux whose in the two portions of the tableau ( and ) are divisible by and , respectively. Care must be taken that the alphabets in each portion are and respectively, even though the second alphabet is larger than the first.
We note in passing that while it is easy to show that if is conjugate to a subgroup of , then for all , the converse is not true. In fact, contains two non-conjugate Klein 4-groups each of which has three elements of type and one element of type .
References:
[1] V. Reiner and D. White, Some notes on Pólya’s theorem, Kostka numbers and the RSK correspondence, unpublished (2019). Available online here.
[2] R. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge University Press, 1999.
[3] D. White, Some connections between the Littlewood-Richardson rule and the construction of Schensted, J. Combin. Theory Ser. A 30 (1981), 237–247. DOI: 10.1016/0097-3165(81)90020-0