Questions tagged [q-analogs]

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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{...
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  • 21
7 votes
1 answer
232 views

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)^{...
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  • 1,400
0 votes
0 answers
30 views

Appearances of the basic hypergeometric series ${}_0\phi_1(;z;q;q^l z)$

On wikipedia one can find the general definition of a unilateral basic hypergeometric series ${}_r\phi_s$. The special case ${}_0\phi_1(;z;q;q^l z)$ has the expansion $$ {}_0\phi_1(;z;q;q^l z) = \sum_{...
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10 votes
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331 views

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 & ...
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0 votes
0 answers
64 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 ...
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1 vote
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80 views

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 ...
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3 votes
0 answers
117 views

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 ...
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4 votes
1 answer
160 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 ...
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6 votes
0 answers
190 views

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 ...
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9 votes
2 answers
393 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. ...
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3 votes
1 answer
144 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, ...
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7 votes
1 answer
257 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 \...
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6 votes
1 answer
233 views

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 ...
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5 votes
1 answer
150 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 ...
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19 votes
1 answer
523 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 $\...
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17 votes
1 answer
791 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 ...
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6 votes
0 answers
177 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 $...
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7 votes
1 answer
366 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,...
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8 votes
7 answers
639 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
2 answers
501 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. ...
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22 votes
2 answers
707 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}\...
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6 votes
1 answer
312 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 ...
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8 votes
1 answer
219 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
1 answer
264 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 ...
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  • 1,473
6 votes
0 answers
115 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 ...
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4 votes
0 answers
110 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\...
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2 votes
2 answers
207 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 $[0; 1]$ ...
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5 votes
0 answers
159 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}} {\...
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2 votes
1 answer
204 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 ...
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6 votes
0 answers
183 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,...
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15 votes
0 answers
232 views

Irreducibility of q-factorial plus 1

Let $q$ be a formal variable and for every positive integer $n$ let $$[n]_q! = 1 (1 + q)(1 + q + q^2) \dotsm (1 + q + \dotsb + q^{n-1})$$ be the $q$-factorial. Is it true that $[n]_q! + 1$ is an ...
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8 votes
0 answers
195 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 \...
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2 votes
2 answers
288 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'...
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13 votes
2 answers
528 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 ...
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15 votes
1 answer
591 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 ...
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  • 7,007
11 votes
4 answers
715 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 ...
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9 votes
0 answers
173 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 ...
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14 votes
1 answer
633 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: $$ \binom{ a }{ b}_{\!\!q} $$ ...
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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 ...
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5 votes
1 answer
348 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 ...
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6 votes
1 answer
303 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, ...
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3 votes
0 answers
91 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 ...
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6 votes
1 answer
496 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, $q = \exp(2\pi i r/N)$?
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5 votes
2 answers
571 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+...
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12 votes
1 answer
215 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 ...
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2 votes
1 answer
106 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)...
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12 votes
3 answers
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 ...
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8 votes
3 answers
386 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 $...
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  • 1,761
5 votes
0 answers
318 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} \...
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  • 123
6 votes
0 answers
225 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 ...
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