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Consider the $n$-element subsets $\{a_1<a_2<\cdots <a_n\}$ of $\{1,\ldots ,2n\}$ satisfying $a_i\geq 2i$ for all $i=1,\ldots ,n$. The number of such subsets is given by $${2n\choose n}-{2n\choose n-1}=\frac{1}{n+1}{2n\choose n},$$ which is the $n$th Catalan Number.

I want to know if the $q$-Catalan number $$\frac{q^{n}}{[n+1]_q}{2n\choose n}_q={2n\choose n}_q-{2n\choose n-1}_q$$ counts some kind of special $n$-dimensional subspaces inside $\mathbb{F}_q^{2n}$? Note that ${2n \choose n}_q$ is the total number of $n$-dimensional subspaces of $\mathbb{F}_q^{2n}$ ($\mathbb{F}_q$ denotes finite field of order $q$).

Pritam Majumder
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  • A generalized question has been asked (originally) by Shapiro, and still waiting for answers. See here starting with page 16: http://users.math.msu.edu/users/sagan/Papers/Old/gfp.pdf – T. Amdeberhan Feb 08 '17 at 19:27
  • Incidentally, I think that the RHS of your identity should be $\binom{2n}{n}_q-q\binom{2n}{n-1}_q$. – Vladimir Dotsenko Feb 08 '17 at 20:35
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    @VladimirDotsenko Oops, thanks for pointing it out. Actually I want to keep the RHS as it is. I hope I have fixed it. – Pritam Majumder Feb 09 '17 at 00:17
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    I have made a small edit: added the factor $q^n$ in the previous q-analog. Sorry for the typo before – Pritam Majumder Feb 09 '17 at 00:27
  • Well, there are at least three meaningful q-analogs of the Catalan Numbers, and you could ask the same question for each of them. See Johann Cigler's answer to an old question of mine: http://mathoverflow.net/questions/89996/why-are-some-q-analogues-more-canonical-than-others/90060#90060 – Wolfgang Feb 12 '17 at 21:16

2 Answers2

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An answer to your question was given in ``Rank Polynomials" by Brandt, Dipper, James, and Lyle, published in Proc. London Math. Soc. (3) 98 (2009), 1-18. A special case of Theorem 2.6 in that paper answers your question.

Qing Xiang
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0

Wikipedia has that

$q$-binomial $ \binom{n}{k}_q$ counts the subspaces of dimension $k$ in the vector space $\mathbb{F}_q^n$

As a non-expert I would hope that with an appropriate group action I could find objects enumerated by

$$ \frac{q^n}{[n+1]_q} \binom{2n}{n}_q $$

and I would look through Richard Stanley's Enumerative Combinatorics and hope for the best.

There also seems to be a result from 2010

where $q$-Catalan numbers are shown to enumerate invariants related to the Hilbert scheme of $n$ points on $\mathbb{C}^2$ (which respect to an equivariant torus action).

There are clearly some vector spaces related to the $q$-catalan numbers Gorsky has $q,t$ Catalan numbers and just set $t = 1$ ( I am looking for the correct degeneration). I do not underestand why $(q,t)$-enumeration is so trendy, as one could have arbitrary deformations.

My double-use of the letter $q$ is a bit suspect since $q$ could be:

  • a prime number
  • a unit complex number

and this ambiguity persists in the finite fields / modular forms literature.

john mangual
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  • (Maybe main) reasons for the q,t-Catalan enumeration being so trendy are that it plays an important role in the (1) study of the diagonal coinvariants of the symmetric group with beautiful combinatorics and deep connections to symmetric function theory, commutative algebra, and Hilbert schemes of points in the plain, and ... – Christian Stump Feb 09 '17 at 08:32
  • ... (2, closely related to (1)) representation theory of Cherednik algebras (which come with a natural bigrading from $V[ \mathbb{C} \oplus \mathbb{C}^* ]$ (here, one also sees that q,t-enumeration is special for the action of reflection groups on $\mathbb{C} \oplus \mathbb{C}^*$) and its connections to things like knot theory (see your link above). – Christian Stump Feb 09 '17 at 08:32
  • Setting $t=1$ in the $q,t$-Catalan numbers gives different kind of $q$-Catalan numbers. – Vladimir Dotsenko Feb 09 '17 at 13:30