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In this question by $q$-Catalan numbers I mean the $q$-analog given by the formula $\frac{1}{[n+1]_q}\left[{2n\atop n}\right]_q$. The polynomial $\left[{2n\atop n}\right]_q$ represents the class of the Grassmannian $G(n,2n)$ in the Grothendieck ring of varieties, and $[n+1]$ represents the class of $\mathbb P^n$. Is there a geometric reason why the fraction $[G(n,2n)]/[\mathbb P^n]$ is a polynomial in $[\mathbb A^1]$?

I guess one could ask more generally about why $\frac{[\mathbb P^r][G(k,2k+r)]}{[\mathbb P^k]}$ is a polynomial.

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  • $\begingroup$ Polynomials with positive coefficients, moreover. $\endgroup$
    – F. C.
    Commented Jun 6, 2015 at 16:19
  • $\begingroup$ Gjergji Zaimi must know this already, but for the benefit of others, it may be worth mentioning that this polynomial is the generating function for enumerating ballot sequences according to the major index. $\endgroup$
    – Timothy Chow
    Commented Jun 7, 2015 at 19:28
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    $\begingroup$ This seems tricky. Already for $n=2$, the polynomial is $q^2+1$ which does not come from a projective algebraic variety. $\endgroup$
    – Will Sawin
    Commented Jun 9, 2015 at 2:15
  • $\begingroup$ But $q^2+1$ can come from a non-projective alg. variety, or a singular one maybe? By the way, how can you tell that it cannot come from a projective one? $\endgroup$
    – F. C.
    Commented Jun 10, 2015 at 13:04
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    $\begingroup$ Hard Lefschetz tells you that the even Betti numbers have to increase to the middle, then decrease (they're "unimodal"). $\endgroup$
    – Allen Knutson
    Commented Jun 16, 2015 at 18:44

1 Answer 1

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There are a few nice answers to related questions. Unfortunately none of them quite answers the question you asked.

  1. The $q$-Catalan number $\frac{1}{[n+1]_q}{ 2n \brack n}_q$ is the Hilbert series of a fairly natural graded representation of the symmetric group $S_n$ coming from an irreducible representation of a rational Cherednik algebra. This was originally proved by Berest-Etingof-Ginzburg and greatly generalized by Gordon-Griffeth:

http://arxiv.org/abs/0912.1578

  1. Define the "other" $q$-Catalan number as the sum of $q^{|D|}$ where $D$ ranges over "Dyck paths" from $(0,0)$ to $(n,n)$ staying weakly above the diagonal and where $\binom{n}{2}-|D|$ is the number of unit squares between $D$ and the diagonal. Gorsky-Mazin proved that this other $q$-Catalan number evaluated at $t^2$ is the Poincare series of the "Jacobi factor" of the plane curve singularity $x^n=y^{n+1}$:

http://arxiv.org/abs/1105.1151

  1. The whole story generalizes to the "rational $(q,t)$-Catalan numbers" $\mathrm{Cat}_{a,b}(q,t)$ where $a,b$ are positive coprime integers. The $q$-Catalan you mentioned comes from the formula $$q^{(a-1)(b-1)/2}\mathrm{Cat}_{a,b}(q,q^{-1})=\frac{1}{[a+b]_q}{ a+b \brack a}_q$$ by setting $(a,b)=(n,n+1)$ and the "other" $q$-Catalan number comes from setting $t=1$. The rational $(q,t)$-Catalan numbers are related to many things including the HOMFLY-PT polynomial of torus knots.

See here for some expositions:

http://www.math.miami.edu/~armstrong/Talks/RCC_AIM.pdf

http://thales.math.uqam.ca/~nwilliams/docs/AIM%202012/RCCAIMOutlineOnline.pdf

http://aimath.org/pastworkshops/rationalcatalanrep.pdf

https://www.math.ucdavis.edu/~egorskiy/Presentations/qtcat.pdf

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    $\begingroup$ Very interesting! What of this is generalized to arbitrary Young diagram, if we define Catalan number as a number of Young tableaux of a rectangular shape $2\times n$? $\endgroup$
    – Fedor Petrov
    Commented Dec 17, 2015 at 21:35
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    $\begingroup$ A related $q$-Catalan number that does have a simple connection with representation theory is $\frac{[2]_q}{[2n+2]_q}\left[ \begin{array}{c} 2n \\ n\end{array}\right]_q$. It is the principal specialization of an irreducible representation of $\mathrm{Sp}(2n)$ and therefore has symmetric, unimodal coefficients. See EC2, Exercise 6.34(d,e). $\endgroup$
    – Richard Stanley
    Commented Dec 17, 2015 at 22:11
  • $\begingroup$ @RichardStanley: if $[m]_q=\frac{1-q^m}{1-q}$, then I think you meant $\frac{[2]_q}{[2]_{q^n}[n+1]_q}\binom{2n}n_q$ or more directly $\frac{1+q}{1+q^n}\frac{1-q}{1-q^{n+1}}\binom{2n}n_q$. I came across this page because I was coming up with some new generation of generalized ballot numbers. $\endgroup$
    – T. Amdeberhan
    Commented May 7, 2023 at 0:43
  • $\begingroup$ @T.Amdeberhan, you are right. Fortunately it is correct in my Catalan Numbers book (Problem A43(f)). $\endgroup$
    – Richard Stanley
    Commented May 8, 2023 at 2:20

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