Questions tagged [q-analogs]

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5
votes
0answers
108 views

A recursion which defines polynomials with integer coefficients?

Let $[n]=1+q+\dots+q^{n-1}$ and $u(n)=\prod_{j=1}^n \gcd([j],[n])$. Define $$r(n)=\sum_{d|n,d>1}{(-1)^d \frac{u(n)}{du(\frac{n}{d})^d}r\Big(\frac{n}{d}\Big)^d}+\frac{(1-q)^{n-1}u(n)}{n[n]}$$ with $...
6
votes
1answer
237 views

Relationship between $q$-Weyl dimension formula and $q$-analog of weight multiplicity?

$\DeclareMathOperator\dim{dim}$For a dominant (integral) weight $\lambda$ and any (integral) weight $\mu$ of a simple Lie algebra $\mathfrak{g}$, Lusztig's $q$-analog of weight multiplicty $K_{\lambda,...
7
votes
7answers
531 views

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

What are some important combinatorial and algebraic interpretations of the coefficients in the polynomial $$[n]!_q = (1+q)(1+q+q^2) \ldots (1+q+\cdots + q^{n-1})?$$ As motivation, I will give ...
11
votes
1answer
330 views

$q$-analogs of total positivity

A real matrix $M$ is called totally positive if all of its minors are positive; these matrices have been extensively studied, and there are generalizations to other Lie types, for example by Lusztig. ...
21
votes
2answers
666 views

A q-rious identity

Let $[x]_q=\frac{1-q^x}{1-q}$, $[n]_q!=[1]_q[2]_q\cdots[n]_q$ and ${\binom{x}{n}}_{q}=\frac{[x]_q[x-1]_q\cdots[x-n+1]_q }{[n]_q!}$. Computer experiments suggest that $$\det \left(q^\binom{i-j}{2}\...
6
votes
1answer
213 views

Lusztig's $q$-analog of weight multiplicity with product formula

For partitions $\lambda, \mu \vdash n$, the Kostka-Foulkes polynomial $K_{\lambda,\mu}(q)$, a $q$-analog of the Kostka coefficient $K_{\lambda,\mu}$, has a combinatorial description, due to Lascoux ...
8
votes
1answer
206 views

Prominent examples of $q$-analogs without known cyclic sieving

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 ...
8
votes
1answer
243 views

Product of $q$-analogues

Background Recall that the $q$-analogue $[n]_q\in\mathbb Z[q]$ of a natural number $n\in\mathbb N$ is defined as $$ [n]_q := \frac{q^n -1}{q-1}$$ the idea being that formulas involving $q$ will ...
6
votes
0answers
101 views

Q-analogue of an inequality

Pick integers $b\geq a \geq 0$ and $k\geq j\geq 0$. It is not super-difficult to prove the inequality $$ \binom{kb}{ka}^j \geq \binom{jb}{ja}^k. $$ This is actually quite a nice inequality that was ...
4
votes
0answers
104 views

Positivity of q-analogs of central binomial coefficients?

With the usual $q-$notations $[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q},$ $[n]_q!=[1]_q[2]_q\cdots[n]_q$ and $\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}$ let $$b(n,k,r,q)=\det\left(q^{r\...
1
vote
2answers
181 views

$q$-factorial coefficient asymptotics

Consider the $[n]!_q = \prod\limits_{k = 1}^{n} \frac{q^k - 1}{q - 1} = \sum\limits_{k = 0}^{\binom n 2} c_k q^k$ and let $\{f_n\}_{n \in \mathbb{N}}$ be the sequence of the functions on $[0; 1]$ ...
5
votes
0answers
120 views

Conjecture for a certain Cauchy-type determinant

Given the Cauchy-like matrix $$ \mathbf X_M(q) = \left[ \frac{2}{\pi} \frac{ \Gamma\!\left(m - \frac{1}{2} \right)\Gamma\!\left(n + \frac{1}{2} \right) }{ \Gamma(m)\,\Gamma(n) } \frac{m-\frac{3}{4}} {\...
2
votes
1answer
147 views

Major index generating polynomial for border-strip tableaux

The Question in its original form has been answered, but there is a follow-up, see the end of the post. A border-strip is a skew Young diagram that does not contain a $2 \times 2$-box. A border-strip ...
6
votes
0answers
175 views

A curious $q$-identity

Let $[x]_{q}=\frac{1-q^x}{1-q}$ and $\binom{x}{n}_{q}$ denote a $q$-binomial coefficient. Let $A_n(x,q)$ be the $n\times n $ matrix with entries $$q^\binom{i-j}{2}\binom{i+j+x}{i-j+1}_{q},$$ $0 \le i,...
6
votes
0answers
111 views

Irreducibility of q-factorial plus 1

Let $q$ be a formal variable and for every positive integer $n$ let $$[n]_q! = 1 (1 + q)(1 + q + q^2) \cdots (1 + q + \cdots + q^{n-1})$$ be the $q$-factorial. Is it true that $[n!]_q + 1$ is an ...
8
votes
0answers
185 views

q-analog of $(d/dx) \binom{x}{k}$?

It is not hard to find easy formulas for the derivative of the function $\binom{x}{k}$, for instance it is not too hard to see (for $k$ an integer) that $\frac{d}{dx} \binom{x}{k} = \sum_{i=1}^k \...
2
votes
2answers
259 views

What partial sum formulae exist for this basic hypergeometric series?

I've run into: $$\sum_{x=1}^{\infty} {x^a\over 1-q^{x}}, \ s.t.\ q\in \mathbb N>1 \ or \ q\in (0, 1),\ a \in \mathbb N$$ I am interested mostly in the cases where $a = 1$ or $ a = 2$ Things I'...
12
votes
2answers
351 views

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

The number of points on the $(n-1)$-dimensional projective space $P^{n-1}(\mathbb{F}_q)$ over a finite field $\mathbb{F}_q$ is the $q$-integer $$[n]_q := \frac{q^n-1}{q-1}.$$ In analogy, the number of ...
15
votes
1answer
506 views

Schur-Weyl duality and q-symmetric functions

Disclaimer: I'm far from an expert on any of the topics of this question. I apologize in advance for any horrible mistakes and/or inaccuracies I have made and I hope that the spirit of the question ...
10
votes
4answers
642 views

A divisibility of q-binomial coefficients combinatorially

Let a and b be coprime positive integers. Then the number a+b divides the binomial coefficient ${a+b \choose a}$. I know how to prove this combinatorially - for example after choosing an ordered set ...
9
votes
0answers
162 views

For $q$-analogues of a known curious identity

In 2002 I published the folllowing curious combinatorial identity: $$(x+m+1)\sum_{i=0}^m(-1)^i\binom{x+y+i}{m-i}\binom{y+2i}i-\sum_{i=0}^m\binom{x+i}{m-i}(-4)^i=(x-m)\binom xm.$$ My original proof is ...
14
votes
1answer
511 views

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

The q-Vandermonde identity reads: $$ \binom{m + n}{k}_{\!\!q} =\sum_{j} \binom{m}{k - j}_{\!\!q} \binom{n}{j}_{\!\!q} q^{j(m-k+j)} $$ The q-binomial coefficients: $$ \binom{ a }{ b}_{\!\!q} $$ ...
30
votes
1answer
1k views

Mysterious symmetry - in search for a bijection

I have a mysterious symmetry that I have not managed to prove. First some definitions (see picture below) Fix a partition that fit in a staircase shape with $n$ rows. There are $Catalan(n)$ such ...
5
votes
1answer
307 views

$q$-analog of an integral from quantum field theory?

This question has been completely reformulated and a new property for the function $f_q$ has been added due to a series of helpful comments by fedja. Consider the integral from quantum field theory ...
6
votes
1answer
277 views

Inequality for functions on [0,1], continued

Let $0<a<1,\; \psi_a(x)=\displaystyle \prod_{j=0}^\infty (1-a^jx).$ For each $ k\in \mathbb{N},$ set $$f_k(a;x):=\frac{x^k}{(1-a)(1-a^2)\dots (1-a^k)}\,\psi_a(x).$$ Question. Is it true that, ...
3
votes
0answers
86 views

Does the Riemann characterization of the hypergeometric function have a q-analog?

This question is inspired by another recent one here, Characterization of the hypergeometric function. The latter is about the classical result of Riemann characterizing the hypergeometric functions ...
6
votes
1answer
360 views

Q-binomials at roots of unity

As the title says, given a general $q$-binomial $\binom{n}{k}_q$, is there some general result regarding its value at a root of unity, $q = \exp(2\pi i r/N)$?
5
votes
2answers
538 views

Some curious Hankel determinants

Let $f(n,q)=\prod_{j=1}^na(q^j)$ for a polynomial $a(q)$ and let $d(n)=\det(f(i+j,q))_{i,j=0}^n$ be its Hankel determinant. Computer experiments suggest that $$\lim_{q\to1}\frac{d(n)}{(q-1)^\binom{n+...
12
votes
1answer
175 views

Total positivity of $q$-Pascal matrix?

A matrix of real numbers is called totally positive if all its minors are non-negative. A well-known example is the Pascal matrix $(\binom{i}{j})$. Is it true that the minors of the $q$-Pascal matrix ...
2
votes
1answer
100 views

Does this q-analogue have a nice closed form? [closed]

Let $[n]_q=1+q+\cdots+q^{n-1}$. Is there a nice closed form of $\sum_{s=1}^i[s]_{q}$? One would expect that the answer will be some q-analog of $\frac{i(i+1)}{2}$, since $\sum_{s=1}^i s=\frac{i(i+1)...
13
votes
3answers
1k views

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

As usual, denote $[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q}$ and $[n]_q!=[1]_q[2]_q\cdots[n]_q$. Furthermore, we write $$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}.$$ As a follow up on this ...
8
votes
3answers
369 views

A not quite theta not quite basic hypergeometric function

The study of matrix quantum group coactions on the noncommutative disk algebra turns up the following series, which is a $q$-deformation of the negative binomial series, for integer $t\ge 0$, complex $...
5
votes
0answers
232 views

Is a basic hypergeometric function ${}_2\phi_1(a, b; c; q, z)$ a meromorphic function in $z$?

Here a basic hypergeometric function is the analytic continuation of the basic hypergeometric series (or called the $q$-hypergeometric series) $$ {}_2\phi_1(a, b; c; q, z) = \sum^{\infty}_{n = 0} \...
6
votes
0answers
221 views

$q$-crystals - is there such a thing?

There are several important facts that I first heard about here on MO. One of the most enlightening of these is that $\mathscr D$-modules on a scheme $X$ may be viewed as sheaves on the groupoid of ...
3
votes
1answer
148 views

Does $\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty z^n q^{n(n-1)/2}$ have a closed form?

The formula $$ \small\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty z^n q^{n(n-1)/2}=\frac{(-z,-q/z;q)_\infty}{\ln\frac{1}{q}}\int\limits_0^\infty\frac{\left(bt/z,pz/at;p\right)_\infty}{\left(-t,-...
11
votes
2answers
507 views

Does $q$-Catalan number count subspaces?

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\...
3
votes
1answer
217 views

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

Let $n\ge 2r$ be positive integers. Is there a closed form for following finite summation involving in q-binomial coefficients $$\sum_{s=0}^r(-1)^sq^{\frac{s(s+1)}{2}}{n-2r+s\brack n-2r}_q{n\brack r-...
13
votes
2answers
636 views

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

Recall that $(a;\,q)_\infty$ is the $q$-Pochhammer symbol: $$(a;\,q)_\infty=\prod_{n=0}^\infty(1-a \, q^n).\tag1$$ Its important special case $(q;\,q)_\infty=\prod_{n=1}^\infty(1-q^n)$ is sometimes ...
4
votes
0answers
112 views

A $q-$binomial identity related to $q-$Narayana polynomials of type B

Denote by $ {n\brack {k}}$ a $q-$binomial coefficient. Let ${D_{n,k}}(t,q) = \sum\limits_{j = 0}^{n - k} {{q^{{j^2} + kj}}}{n\brack {j}}{n\brack {k+j}}t^j $ and ${R_n}(x,t,q) = \sum\limits_{k = 0}...
10
votes
3answers
653 views

Is this a q-count of Alternating Sign Matrices?

The number of Alternating Sign Matrices of size $n$ is well known to be $\prod_{k=0}^{n-1}\frac{(3k+1)!}{(n+k)!}$. Is it known whether the naive q-analog expression $$\prod_{k=0}^{n-1}\frac{[3k+1]_q!}{...
8
votes
1answer
388 views

q-analog of a combinatorial identity involving binomial coefficients

Using, e.g., properties of iterated finite differences it is easy to show that for any pair of integers $n$ and $m$ with $n>\!>m$ one has the identity $$ \sum_{k=0}^m(-1)^{k-m} {n-k\choose m}{m\...
6
votes
0answers
255 views

What is known about the $q$-analogue of the simplex?

I am interested in the field with one element. I am thus interested in combinatorial interpretations of the Gaussian binomial coefficients. Richard Stanley's "Enumerative combinatorics" mentions ...
12
votes
1answer
334 views

Multiplicative infinitesimals in q-analogs?

Risking to be downvoted, here is a very lightweight question. In various fields - say, algebraic geometry, nonstandard analysis, synthetic differential geometry - infinitely small quantities, i. e. ...
12
votes
1answer
813 views

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

First of all my apologies if this question is well known or obvious: this is not in my area of research. Let $T(x)=\sum_{n=0}^\infty t_nx^n$, where $t_n$ is the number of partitions $\lambda$ of $n$ ...
8
votes
1answer
255 views

notation for $(a-b)(a-qb)\dots (a-q^{n-1}b)$

I wonder whether there is a notation for such thing, which I denote $[a;b]_q^n$ for a moment: $$ [a;b]_q^n:=(a-b)(a-qb)\dots (a-q^{n-1}b)=a^n(b/a;q)_n, $$ this last equation uses $q$-Pochhammer symbol ...
4
votes
2answers
222 views

How to prove that $\sum_{i=0}^n\frac{(a;q)_i}{(q;q)_i}\frac{(b;q)_{n-i}}{(q;q)_{n-i}}a^{n-i}=\frac{(ab;q)_n}{(q;q)_n}$?

By Cauchy identity, $${}_1\phi_0(a;—;q,z)=\sum_{n\geq0}\frac{(a;q)_n}{(q;q)_n}z^n=\frac{(az;q)_{\infty}}{(z;q)_\infty},\quad|z|<1,|q|<1,$$ we can obtain the $q-$analogue of $(1-z)^{-a}(1-z)^{-b}=...
8
votes
1answer
378 views

q-Integer-valued polynomials

For $n \in \mathbb{Z}_{\geq 0}$, let $[n]_q := (1-q^n)/(1-q) = (1+q+...+q^{n-1})$ as is customary, with $[0]_q=0$. Let $R$ be the subring of $\mathbb{Q}(q)[x]$ consisting of all $f$ such that $f([n]...
8
votes
0answers
541 views

What is the $q$-analog of $\Gamma(z)\Gamma(1-z)=\frac\pi{\sin(\pi z)}$?

I would expect the $q$-Gamma function to have the property which would be the $q$-analog of the Euler reflection formula from my question title. More concretely: $\Gamma(z)$ has simple poles at ...
13
votes
2answers
1k views

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

Let $\binom{n}{j}_q$ be a $q$-binomial coefficient and $(x;q)_n = (1-x)(1-qx)\cdots(1-q^{n-1}x).$ Consider the sum $$f(n,m,r,k)= \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}q^{mj^2+rj} \binom{2n}{j}_{q^k}$...
8
votes
1answer
805 views

Counting subspaces

We are given the finite vector space $V = V(n,p) = \mathbb{F}_p^n$ and two fixed subspaces $W_1, W_2 \subseteq V$ of dimensions $m_1$, $m_2$ respectively. Suppose that the dimension of $W_1 \cap W_2$ ...