20 votes
Accepted

A "quantum" identity: in search of a proof -Part II

Both sides are equal to $\binom{x+y+1}{n}_q$. This enumerates lattice paths in an $n\times (x+y-n+1)$ rectangle, according to the area statistic. We will assume that these paths start at $(0,0)$ and ...
20 votes
Accepted

A conjecture about algebraic values of $(-q;\,-q)_\infty/(q;\,q)_\infty$

Yes, it is always algebraic, because it is a modular function evaluated at a CM (complex multiplication) point. "$(q;q)_\infty$" is $q^{-1/24} \eta(\tau)$ where $q = e^{2\pi i \tau}$, so "$(q;q)_\...
20 votes
Accepted

Is there a lift of the q-Vandermonde identity to some geometric (motivic) identity for Grassmannians over $F_q$?

Assume $V$ is a vector space of dimension $m+n$, $M \subset V$ is a subspace of dimension $m$, and $N = V/M$. Let $p:V \to N$ be the projection. Consider the Grassmannian $X = Gr(k,V)$ and its ...
  • 33.4k
16 votes
Accepted

What is the groupoid cardinality of the category of vector spaces over a finite field?

Upon substituting $x=\frac{1}{q}$ we obtain $$\sum_{n\geq 0}\frac{1}{|\mathrm{GL}_n(\mathbb F_q)|}=\sum_{n\geq 0}\frac{x^{n^2}}{(1-x)(1-x^2)\cdots(1-x^n)}$$ and this evaluates to the product $\prod_{i\...
15 votes
Accepted

Schur-Weyl duality and q-symmetric functions

As Sam Hopkins says, the category of all representations of $GL_n(\mathbb F_q)$ is too large to give what you want. Instead, let's consider the category of unipotent representations, i.e. those ...
15 votes
Accepted

Is there a nice q-analogue of the Jacobi identity in a quantized enveloping algebra?

There are various deformations of the Jacobi identity that can be found scattered in the literature. As far as i know, using the definition: $[A,B]_q=AB-qBA$, one of the most general ones (though i do ...
12 votes

q-Catalan numbers from Grassmannians

There are a few nice answers to related questions. Unfortunately none of them quite answers the question you asked. The $q$-Catalan number $\frac{1}{[n+1]_q}{ 2n \brack n}_q$ is the Hilbert series of ...
12 votes
Accepted

Total positivity of $q$-Pascal matrix?

Yes. Consider the set $V$ of points with integer coordinates as vertices of a weighted directed acyclic graph. Namely, for any $(i,j)\in V$, the edge from $(i,j)$ to $(i+1,j)$ has weight 1, the edge ...
  • 91.9k
11 votes

A "quantum" identity: in search of a proof -Part II

If we substitute $y:=v+n$ in the identity, the LHS becomes a convolution of two similar sequences, $$\sum_{k=0}^nq^{(v+1)k}\binom{x+k}{k}_q \binom{v+(n-k)}{n-k}_q =\sum_{k=0}^n q^{n-k}\binom{x+v+n-k}{...
  • 52.5k
11 votes

Important combinatorial and algebraic interpretations of the coefficients in the polynomial $[n]!_q = (1+q)(1+q+q^2) \ldots (1+q+\cdots + q^{n-1})$

This answer concerns a geometric/Lie-theoretic interpretation of $[n]!_q$. $[n]!_q$ gives the number of points in the full flag variety of full flags of subspaces in an $n$-dimensional vector space $\...
10 votes

Is this a q-count of Alternating Sign Matrices?

It's true these polynomials are not unimodal for n=2 and up, since they all start with coefficient sequence 1 0 1 ... (The reason for this is clear from the definition of descending plane partitions.) ...
9 votes

Is there a $q$-L'Hospital's Rule?

I think you can achieve this using the actual $q$-L'Hospital's Rule. Consider $$\frac{ \partial^i f(n,m,r,k)}{\partial q^i} (1)= \sum_{i=0}^a \binom{a}{i} \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}\frac{...
  • 122k
9 votes

A divisibility of q-binomial coefficients combinatorially

Assume $q$ is a prime power. Then there is a straightforward combinatorial interpretation. Consider the group action of $\mathbb F_{q^{a+b}}^\times / \mathbb F_{q^\times}$ on the space of $a$-...
  • 122k
9 votes
Accepted

In search of a $q$-analogue of a Catalan identity

This identity is known as Jonah's formula (special case with $n\rightarrow 2n$ and $r\rightarrow n$, see "Catalan Numbers with Applications" by Thomas Koshy, pg. 325-326 for a combinatorial ...
9 votes
Accepted

A curious $q$-series identity on a truncated Euler function

Let ${n\choose k}_q=\frac{(q)_n}{(q)_k(q)_{n-k}}$ denote a $q$-binomial coefficient. We start with the following version of $q$-Vandermonde convolution identity: $$ (x-y)(x-qy)\ldots(x-q^{n-1}y)\\=\...
  • 91.9k
8 votes
Accepted

q-analog of a combinatorial identity involving binomial coefficients

You can see this as an instance of the q-Vandermonde identity. The q-binomial theorem tells us that the coefficient of $t^k$ in $\prod_{i=0}^{m-1}(1+q^{-2i}t)$ is $q^{-k(m-1)}{m \brack k}_q$ and the ...
8 votes

$q$ as a prime power and a root of unity

This is very not rigorous, but it's a way of thinking about this topic which I find personally helpful. Several $\mathbb F_1$ papers contains remarks about the idea going back to Weil and Iwasawa that ...
8 votes
Accepted

$q$-analogs of total positivity

This is a wonderful question but unfortunately I don't think that there is a definite answer in the literature just yet. Let's look at two somewhat recent lines of research in this direction: 1) A. ...
8 votes

Important combinatorial and algebraic interpretations of the coefficients in the polynomial $[n]!_q = (1+q)(1+q+q^2) \ldots (1+q+\cdots + q^{n-1})$

In FindStat, the database of combinatorial statistics, the major index of a permutation is http://www.findstat.org/StatisticsDatabase/St000004/. By clicking "search for distribution" there, you will ...
7 votes
Accepted

q-Integer-valued polynomials

(Below is the proof that module $R$ is generated by $f_i(x)$, without calculation of structure constants.) Polynomial $f(x)$ of degree $n$ may be interpolated in points $[0],[1],\dots,[n]$ (I omit ...
  • 91.9k
7 votes

A "quantum" identity: in search of a proof -Part II

Actually this is a partial case of $q$-Vandermonde identity. In the notation $$[a;b]_q^n:=(a-b)(a-qb)\dots (a-q^{n-1}b)=a^n(b/a;q)_n,$$ as in this terminology question, we have $$ [a;c]_q^n=\sum {\...
  • 91.9k
7 votes
Accepted

Mysterious symmetry - in search for a bijection

Oliver Pechenik and I have some partial progress to report. Maybe someone else can see how to supply the remaining missing ingredients. First, let's establish some notation. We say your ascent ...
  • 1,769
7 votes

A divisibility of q-binomial coefficients combinatorially

A proof of nonnegativity appears in https://arxiv.org/pdf/0912.1578.pdf. The number $\frac{1}{a+b}{a+b\choose a}$ is called a rational Catalan number by Drew Armstrong. See for instance http://www....
7 votes
Accepted

Discriminants of some $q$-analogs of $(1+x)^n$

This is true. We have \begin{align*} p_n (q^{-1}, 1-r, x) &= \sum_{j=0}^n q^{ (r-1) \binom{j}{2}} \binom{n}{j}_{q^{-1}} x^j \\ &= \sum_{j=0}^n q^{ (r-1) \binom{j}{2}} q^{-j (n-j)} \binom{n}...
  • 122k
6 votes
Accepted

What is the value of this sum involving q-binomials?

Doron Zeilberger has written a Maple code for checking and proving ordinary binomial identities and their $q$-analogues. What you need in the present case is the package called qEKHAD. I just tested ...
6 votes
Accepted

Generating function for certain partitions (with a restriction on the Durfee square)

Lemma. Fix $n$ and $m$. Consider pairs of partitions $(\lambda,\mu)$ such that $\lambda$ has $m$ parts and $|\lambda|+|\mu|=n$. Let $A$ be the number of pairs for which $\max(\lambda)>\max(\mu)$ (...
  • 91.9k
6 votes
Accepted

Inequality for functions on [0,1], continued

OK, here goes. We start with changing the notation ($z\to 20z^2$, $-z-3\to r$, $20rz\to y$ means that what was denoted by $z$ will be denoted by $20z^2$ from now on, $r$ is $-z-3$ with new $z$, so it ...
  • 55.4k
6 votes

A q-rious identity

In the mean-time I have found a proof: First write Vandermonde’s identity $$\sum_{j=0}^k q^{(k-j)(i+r-j)}\binom{s-r}{k-j}_{q}\binom{i+r}{j}_{q}=\binom{i+s}{k}_{q}$$ in the form $$\sum_{j=0}^k q^{\...
6 votes
Accepted

Enumerating subspaces of $\mathbb{F}_q^n$ in terms of words and inversions

You may act similarly as follows. Let $V$ be a $k$-dimensional subspace of $\mathbb F_q^n$. Take any its base, put its elements into the rows of some matrix, and make it to the reduced row echelon ...
6 votes

In search of a $q$-analogue of a Catalan identity

Decided to make a cw post: it is sort of amusing. Let $C_n(q)$ be defined by $$\sum_{k=0}^n\binom{2n-2k}{n-k}_qC_k(q)q^{2n-2k}=\binom{2n+1}n_q,\qquad n=0,1,2,\dotsc.$$ Then \begin{multline*} C_n(q)=1+...

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