# Questions tagged [q-analogs]

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94
questions

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### 3D generalization of Gaussian q-binomial coefficient

It is known that the coefficient of $q^t$ in Gaussian binomial coefficient $\binom{m+n}m_q$ equals the number of permutations of the multiset $\{0^m, 1^n\}$ with $t$ inversions.
Is there a closed ...

1
vote

0
answers

43
views

### Can I apply $q$-Lagrange Inversion formula?

Now I have equation $F(x) = x \sum_{k\ge 0} g_k F(x) F(qx) \cdots F(q^{k-1} x)$, I need to get the coefficient of $x^n$ in $F(x)$, can I apply $q$-Lagrange Inversion formula to this?
Moreover, I have ...

7
votes

0
answers

217
views

### Hankel determinants for some convolutions of Catalan numbers

Let $c(x)=\frac{1-\sqrt{1-4x}}{2x}$ be the generating function of the Catalan numbers and let $$x^k c(x)^{2k}=(c(x)-1)^k =\sum_{n\geq0}c(k,n)x^n.$$
Consider the determinants $$D(k,n,m)= \det\left(c(k,...

4
votes

0
answers

92
views

### Quantum version of Kostant's basis of ℤ-form of U(𝔤)

Kostant showed that the subring of $\mathcal U(\mathfrak{sl}_2)$ generated by the divided powers $e^c/c!$ and $f^a/a!$ has a $\mathbb Z$-basis given by the elements $\frac{f^a}{a!}\binom hb \frac{e^c}{...

12
votes

0
answers

328
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### $q$-analogue of the multinomial theorem?

The $q$-binomial theorem states that
$$
\prod_{k=0}^{n-1}(1+q^kt) = \sum_{k=0}^n q^{\binom k2}{n\brack k}_q t^k.
$$
This identity is a $q$-analogue of the binomial theorem
$$
(1+t)^n = \sum_{k=0}^n \...

6
votes

0
answers

103
views

### Bijection between forests and skew SYT + Cyclic sieving

Consider the two-row skew shape $\lambda_n = (2n+1,n)/(1)$.
The number of standard Young tableaux of this shape is
$\binom{3n}{n}-\binom{3n}{n-2}$ (since one can easily biject this to the set of non-...

1
vote

0
answers

124
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### Counting non-zero Gramians of Grassmanians over finite field

In case of $\mathbb{F}_{2}$, we can obtain the number of all reduced row echelon forms (so called Grassmannians) for some m$\times$n full rank matrices by the following gaussian polynomial,
$$
\binom{...

7
votes

1
answer

270
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### A curious $q$-series identity on a truncated Euler function

Recall that a $q$-Pochhammer symbol is defined as
$$
(x)_n = (x;q)_n := \prod_{l=0}^{n-1}(1-q^l x).
$$
I found the following curious $q$-series identity that seems to hold for any $n\geq 0$:
$$
(-1)^{...

10
votes

0
answers

366
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### Has anyone met this "$q$-character" table for $S_4$?

Is anyone aware of the following $q$-character table for the
symmetric group $S_4$?
\begin{array}{|c|c|c|c|c|c|}
\hline
\mathrm{conj}\backslash\mathrm{rep}
& 2+1+1 & 3+1 & ...

0
votes

0
answers

91
views

### Addition formulas for q-analogs of trigonometric functions

Sine and Cosine functions possess notable formulas for addition of angles
$$
\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b) \qquad \text{or} \qquad \cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b).
$$
One can ...

1
vote

0
answers

83
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### Evaluate $\det[[\lfloor\frac{aj-(a+1)k}n\rfloor]_q]_{1\le j,k\le n}$ and $\det[[\lceil\frac{(a+1)j-ak}n\rceil]_q]_{1\le j,k\le n}$

The $q$-analogue of an integer $m$ is defined by
$[m]_q=(1-q^m)/(1-q)$. Note that $\lim_{q\to1}[m]_q=m$.
I have formulated the following conjecture on determinants involving the floor function and the ...

10
votes

2
answers

449
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### In search of a $q$-analogue of a Catalan identity

Let $C_n=\frac1{n+1}\binom{2n}n$ be the all-familiar Catalan numbers. Then, the following identity has received enough attention in the literature (for example, Lagrange Inversion: When and How):
\...

3
votes

0
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128
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### A recursion involving binomial coefficients: looking for a q-analog

Let $a_n := \frac{1}{2n+1}\binom{3n}{n}$.
Then it is known that (one can find references in the OEIS for this.)
$$
a_n = \sum_{\substack{i,j,k \geq 0 \\ i+j+k=n-1} } a_i a_j a_k.
$$
Is there a natural ...

4
votes

1
answer

164
views

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

Let $[n]_q=1+q+\dots +q^{n-1}$, $ {[n]_q}! =[1]_q [2]_q \dots [n]_q$ and $\binom{n}{j}_q = \frac{[n]_q!}{[j]_q![n-j]_q!}$ be the usual $q$-notation.
Consider the polynomials $p_n(q,r,x)= \sum_{j=0}^n ...

6
votes

0
answers

212
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### Gaussian coefficients identity

I am having difficulty showing the equivalence between (11) and (15) of Delsarte - Association schemes and $t$-designs in regular semilattices. It is somehow an application of Möbius inversion, but I ...

9
votes

2
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425
views

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

In a Lie algebra $\mathfrak{g}$ the Jacobi identity $\newcommand{\bracket}[2]{\left[#1\,#2\right]} \bracket{x}{\bracket{y}{z}} + \bracket{z}{\bracket{x}{y}} + \bracket{y}{\bracket{z}{x}} = 0$ holds. ...

3
votes

1
answer

161
views

### Is there a $q$-analogue to Shapiro's convolution identity?

Let $C_n=\frac1{n+1}\binom{2n}n$ denote the Catalan numbers.
This question is motivated by the (unanswered) MO post by Alexander Burstein and my own (answered by Fedor Petrov) MO post.
Specifically, ...

7
votes

1
answer

285
views

### Looking for a $q$-analogue of a binomial identity

The following identity is well-known and there are a few proofs to it (see Bijective proof problems, by R P Stanley, for this and similar formulae):
$$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n \...

6
votes

1
answer

236
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### Enumerating subspaces of $\mathbb{F}_q^n$ in terms of words and inversions

When $q$ is a prime power, then on the one hand the $q$-binomial coefficient $\binom{n}{k}_q$ equals the number of $k$-dimensional subspaces of $\mathbb{F}_q^n$, and on the other hand it is the ...

5
votes

1
answer

160
views

### A $q$-analogue of a characterization of polynomials by binomial coefficients

Considering the binomial coefficient $\binom{x}{m}$ as a polynomial in $x$, the span of $\binom{x}{0}, \binom{x}{1}, \ldots, \binom{x}{d}$ is exactly the polynomials of degree $\le d$. A closely ...

19
votes

1
answer

604
views

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

For any groupoid, it's groupoid cardinality is the sum of the reciprocals of the automorphism groups over the isomorphism classes. Let us consider the category of vector spaces over a finite field $\...

17
votes

1
answer

820
views

### Proof of certain $q$-identity for $q$-Catalan numbers

Let us use the standard notation for $q$-integers, $q$-binomials,
and the $q$-analog
$$
\operatorname{Cat}_q(n) := \frac{1}{[n+1]_q} \left[\matrix{2n \\ n}\right]_q.
$$
I want to prove that for all ...

6
votes

0
answers

235
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### 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 $...

8
votes

2
answers

476
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,...

9
votes

7
answers

682
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### 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

2
answers

532
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### $q$-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.
...

22
votes

2
answers

721
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### 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

1
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342
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

1
answer

223
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### 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

1
answer

274
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### 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

0
answers

122
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

0
answers

112
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### 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\...

2
votes

2
answers

216
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### $q$-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 $[0; 1]$ ...

6
votes

0
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189
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### 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

1
answer

224
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### 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

0
answers

188
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### 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,...

15
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0
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247
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### Irreducibility of q-factorial plus 1

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

8
votes

0
answers

206
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### 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

2
answers

305
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### 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'...

13
votes

2
answers

576
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

1
answer

641
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 ...

11
votes

4
answers

743
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

0
answers

183
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

1
answer

694
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

1
answer

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

1
answer

360
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

1
answer

313
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### 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

0
answers

96
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

1
answer

553
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

2
answers

592
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+...