Does $q$-Catalan number count subspaces?

Question

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 $q$).

Answer 1

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.

Answer 2

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.

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There also seems to be a result from 2010

• q,t-Catalan numbers and knot homology

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.