8
$\begingroup$

The cyclic sieving phenomenon is nicely summarized in the following AMS Notices "What is...?" article: https://www.ams.org/notices/201402/rnoti-p169.pdf.

In that article, Reiner, Stanton, and White explain some desiderata (conditions (i)-(vi)) for "very nice" $q$-analogs $X(q)$ of cardinalities $\#X$ of combinatorial sets.

Question: What are some "very nice" $q$-analogs (especially, those with simple product formulas) for which there is no known cyclic action which gives rise to a CSP?

Improve this question
$\endgroup$
4
  • $\begingroup$ There is a fairly simple condition on the values of the polynomials at roots of unity that (in principle) guarantees CSP. To clarify, you are looking for examples where there should be a group action, but a nice one has not been discovered yet? Or are you looking for nice q-analogs but the values at roots of unity prevents there to ever be a CSP? $\endgroup$
    – Per Alexandersson
    Aug 28, 2019 at 6:47
  • $\begingroup$ Is cyclic action known for constant term identities of Dyson/Morris/Aomoto/Forrester type? Or for other root systems counterparts (McDonald conjectures, proved by Cherednik)? $\endgroup$
    – Fedor Petrov
    Aug 28, 2019 at 7:38
  • $\begingroup$ @PerAlexandersson: I’m more interested in “nice” combinatorial actions than whether in principle some cyclic action exists. $\endgroup$
    – Sam Hopkins
    Aug 28, 2019 at 10:39
  • $\begingroup$ @FedorPetrov: Those seem like good candidates- I’m not aware of any cyclic sieving results for them. $\endgroup$
    – Sam Hopkins
    Aug 28, 2019 at 10:52

1 Answer 1

Reset to default
5
$\begingroup$

I think I can provide one example, from this paper, The Cone of Cyclic Sieving Phenomena by N. Amini and I.

We look at a principal specialization of Schur polynomials, $$ s_{n \lambda}(1,q,\dotsc,q^k) $$ for some partition $\lambda$, and integers $n$ and $k$. Here, $n\lambda = (n\lambda_1,\dotsc,n\lambda_\ell)$. This polynomial evaluates to non-negative integers at $n$th roots of unity, and I think one should be able to prove without effort that there should in principle be a cyclic group action which gives CSP for this family.

The results by Rhoades show that (a suitable chosen power of) promotion works as group action whenever $\lambda$ is a rectangle.

As a side note, I would be happy to compile a list of suspected instances of the cyclic sieving phenomenon on my web page. At the moment, it lists (most?) published instances of CSP.

I also have a private list with suspected instances, but I prefer to save these for student projects, and a few others I am working on myself.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Nice example! In addition to Rhoades’s result, I think a result of Stembridge says that for $n=2$ we can take evacuation as the action. $\endgroup$
    – Sam Hopkins
    Aug 28, 2019 at 10:42
  • $\begingroup$ Wait, the order of promotion should be $k$ (or $k+1$ in your notation), which is unrelated to $n$; so I’m not sure how Rhoades’s result applies here. $\endgroup$
    – Sam Hopkins
    Aug 28, 2019 at 11:02
  • $\begingroup$ Yes, for $n=2$, evacuation works (it is an involution, of order 2). The order of k-promotion is just plain wrong on non-rectangular shapes. Note, that we deal with SSYT, and not SYT, so one needs to try k-promotion (aka jeu-de-taquin promotion), but yeah, the order is in general huge. $\endgroup$
    – Per Alexandersson
    Aug 29, 2019 at 6:03
  • $\begingroup$ What I’m saying is that even for $\lambda$ a rectangle, to get cyclic sieving for this polynomial you cannot consider promotion acting on SSYT(n$\lambda$,k) because that has the wrong order (namely, $k$ and not $n$); instead you need to consider promotion acting on SSYT(k$\lambda$,n) or something like that. $\endgroup$
    – Sam Hopkins
    Aug 29, 2019 at 9:28
  • $\begingroup$ But those sets of SSYT are not quite in bijection. In fact, I don’t see how to play with the parameters to apply Rhoades’s result. $\endgroup$
    – Sam Hopkins
    Aug 29, 2019 at 10:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.