Questions tagged [q-analogs]
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96
questions
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views
Is anything known about the derivative of the quantum dilogarithm?
Faddeev's noncompact quantum dilogarithm is the function defined by
$$
\Phi_{\mathsf b}(z) =
\exp
\int_{\mathbb{R} + i\varepsilon}
\frac{
e^{-2i zw}
}{
4 \sinh(w \mathsf b ) \sinh(w/\...
4
votes
1
answer
273
views
Double q-analog of Pochhammer
Has the function
$$(z;q_1,q_2)_\infty := \prod_{n_1,n_2=0}^\infty (1-z \, q_1^{n_1} q_2^{n_2}), \quad |q_1|,|q_2|<1$$
been studied in the math literature? For example, does it obey any difference ...
1
vote
0
answers
130
views
3D generalization of Gaussian q-binomial coefficient
It is known that the coefficient of $q^t$ in Gaussian binomial coefficient $\binom{m+n}m_q$ equals the number of permutations of the multiset $\{0^m, 1^n\}$ with $t$ inversions.
Is there a closed ...
1
vote
0
answers
50
views
Can I apply $q$-Lagrange Inversion formula?
Now I have equation $F(x) = x \sum_{k\ge 0} g_k F(x) F(qx) \cdots F(q^{k-1} x)$, I need to get the coefficient of $x^n$ in $F(x)$, can I apply $q$-Lagrange Inversion formula to this?
Moreover, I have ...
7
votes
0
answers
244
views
Hankel determinants for some convolutions of Catalan numbers
Let $c(x)=\frac{1-\sqrt{1-4x}}{2x}$ be the generating function of the Catalan numbers and let $$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
106
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
497
views
$q$-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
$$
(1+t)^n = \sum_{k=0}^n \...
6
votes
0
answers
110
views
Bijection between forests and skew SYT + Cyclic sieving
Consider the two-row skew shape $\lambda_n = (2n+1,n)/(1)$.
The number of standard Young tableaux of this shape is
$\binom{3n}{n}-\binom{3n}{n-2}$ (since one can easily biject this to the set of non-...
1
vote
0
answers
140
views
Counting non-zero Gramians of Grassmanians over finite field
In case of $\mathbb{F}_{2}$, we can obtain the number of all reduced row echelon forms (so called Grassmannians) for some m$\times$n full rank matrices by the following gaussian polynomial,
$$
\binom{...
7
votes
1
answer
298
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 $n\geq 0$:
$$
(-1)^{...
10
votes
0
answers
378
views
Has anyone met this "$q$-character" table for $S_4$?
Is anyone aware of the following $q$-character table for the
symmetric group $S_4$?
\begin{array}{|c|c|c|c|c|c|}
\hline
\mathrm{conj}\backslash\mathrm{rep}
& 2+1+1 & 3+1 & ...
0
votes
0
answers
121
views
Addition formulas for q-analogs of trigonometric functions
Sine and Cosine functions possess notable formulas for addition of angles
$$
\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).
$$
One can ...
1
vote
0
answers
87
views
Evaluate $\det[[\lfloor\frac{aj-(a+1)k}n\rfloor]_q]_{1\le j,k\le n}$ and $\det[[\lceil\frac{(a+1)j-ak}n\rceil]_q]_{1\le j,k\le n}$
The $q$-analogue of an integer $m$ is defined by
$[m]_q=(1-q^m)/(1-q)$. Note that $\lim_{q\to1}[m]_q=m$.
I have formulated the following conjecture on determinants involving the floor function and the ...
10
votes
2
answers
476
views
In search of a $q$-analogue of a Catalan identity
Let $C_n=\frac1{n+1}\binom{2n}n$ be the all-familiar Catalan numbers. Then, the following identity has received enough attention in the literature (for example, Lagrange Inversion: When and How):
\...
3
votes
0
answers
132
views
A recursion involving binomial coefficients: looking for a q-analog
Let $a_n := \frac{1}{2n+1}\binom{3n}{n}$.
Then it is known that (one can find references in the OEIS for this.)
$$
a_n = \sum_{\substack{i,j,k \geq 0 \\ i+j+k=n-1} } a_i a_j a_k.
$$
Is there a natural ...
4
votes
1
answer
166
views
Discriminants of some $q$-analogs of $(1+x)^n$
Let $[n]_q=1+q+\dots +q^{n-1}$, $ {[n]_q}! =[1]_q [2]_q \dots [n]_q$ and $\binom{n}{j}_q = \frac{[n]_q!}{[j]_q![n-j]_q!}$ be the usual $q$-notation.
Consider the polynomials $p_n(q,r,x)= \sum_{j=0}^n ...
6
votes
0
answers
222
views
Gaussian coefficients identity
I am having difficulty showing the equivalence between (11) and (15) of Delsarte - Association schemes and $t$-designs in regular semilattices. It is somehow an application of Möbius inversion, but I ...
9
votes
2
answers
448
views
Is there a nice q-analogue of the Jacobi identity in a quantized enveloping algebra?
In a Lie algebra $\mathfrak{g}$ the Jacobi identity $\newcommand{\bracket}[2]{\left[#1\,#2\right]} \bracket{x}{\bracket{y}{z}} + \bracket{z}{\bracket{x}{y}} + \bracket{y}{\bracket{z}{x}} = 0$ holds. ...
3
votes
1
answer
174
views
Is there a $q$-analogue to Shapiro's convolution identity?
Let $C_n=\frac1{n+1}\binom{2n}n$ denote the Catalan numbers.
This question is motivated by the (unanswered) MO post by Alexander Burstein and my own (answered by Fedor Petrov) MO post.
Specifically, ...
7
votes
1
answer
308
views
Looking for a $q$-analogue of a binomial identity
The following identity is well-known and there are a few proofs to it (see Bijective proof problems, by R P Stanley, for this and similar formulae):
$$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n \...
6
votes
1
answer
242
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 ...
5
votes
1
answer
169
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
655
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
859
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
253
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 $...
8
votes
2
answers
566
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
721
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
555
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.
...
22
votes
2
answers
736
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
$$\det \left(q^\binom{i-j}{2}\...
8
votes
2
answers
474
views
Lusztig's $q$-analog of weight multiplicity with product formula
For partitions $\lambda, \mu \vdash n$, the Kostka-Foulkes polynomial $K_{\lambda,\mu}(q)$, a $q$-analog of the Kostka coefficient $K_{\lambda,\mu}$, has a combinatorial description, due to Lascoux ...
8
votes
1
answer
226
views
Prominent examples of $q$-analogs without known cyclic sieving
The cyclic sieving phenomenon is nicely summarized in the following AMS Notices "What is...?" article: https://www.ams.org/notices/201402/rnoti-p169.pdf.
In that article, Reiner, Stanton, and White ...
8
votes
1
answer
284
views
Product of $q$-analogues
Background
Recall that the $q$-analogue $[n]_q\in\mathbb Z[q]$ of a natural number $n\in\mathbb N$ is defined as
$$ [n]_q := \frac{q^n -1}{q-1}$$
the idea being that formulas involving $q$ will ...
6
votes
0
answers
128
views
Q-analogue of an inequality
Pick integers $b\geq a \geq 0$ and $k\geq j\geq 0$.
It is not super-difficult to prove the inequality
$$
\binom{kb}{ka}^j \geq \binom{jb}{ja}^k.
$$
This is actually quite a nice inequality that was ...
4
votes
0
answers
113
views
Positivity of q-analogs of central binomial coefficients?
With the usual $q-$notations
$[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q},$
$[n]_q!=[1]_q[2]_q\cdots[n]_q$ and
$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}$
let
$$b(n,k,r,q)=\det\left(q^{r\...
2
votes
2
answers
224
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 $[0; 1]$ ...
6
votes
0
answers
197
views
Conjecture for a certain Cauchy-type determinant
Given the Cauchy-like matrix
$$
\mathbf X_M(q) = \left[ \frac{2}{\pi} \frac{
\Gamma\!\left(m - \frac{1}{2} \right)\Gamma\!\left(n + \frac{1}{2} \right)
}{
\Gamma(m)\,\Gamma(n)
}
\frac{m-\frac{3}{4}}
{\...
2
votes
1
answer
248
views
Major index generating polynomial for border-strip tableaux
The Question in its original form has been answered, but there is a follow-up, see the end of the post.
A border-strip is a skew Young diagram that does not contain a $2 \times 2$-box. A border-strip ...
6
votes
0
answers
190
views
A curious $q$-identity
Let $[x]_{q}=\frac{1-q^x}{1-q}$ and $\binom{x}{n}_{q}$ denote a $q$-binomial coefficient.
Let $A_n(x,q)$ be the $n\times n $ matrix with entries $$q^\binom{i-j}{2}\binom{i+j+x}{i-j+1}_{q},$$ $0 \le i,...
15
votes
0
answers
256
views
Irreducibility of q-factorial plus 1
Let $q$ be a formal variable and for every positive integer $n$ let
$$[n]_q! = 1 (1 + q)(1 + q + q^2) \dotsm (1 + q + \dotsb + q^{n-1})$$
be the $q$-factorial.
Is it true that $[n]_q! + 1$ is an ...
8
votes
0
answers
241
views
q-analog of $(d/dx) \binom{x}{k}$?
It is not hard to find easy formulas for the derivative of the function $\binom{x}{k}$, for instance it is not too hard to see (for $k$ an integer) that
$\frac{d}{dx} \binom{x}{k} = \sum_{i=1}^k \...
2
votes
2
answers
329
views
What partial sum formulae exist for this basic hypergeometric series?
I've run into:
$$\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$$
I am interested mostly in the cases where $a = 1$ or $ a = 2$
Things I'...
13
votes
2
answers
617
views
$q$ as a prime power and a root of unity
The number of points on the $(n-1)$-dimensional projective space $P^{n-1}(\mathbb{F}_q)$ over a finite field $\mathbb{F}_q$ is the $q$-integer
$$[n]_q := \frac{q^n-1}{q-1}.$$
In analogy, the number of ...
15
votes
1
answer
701
views
Schur-Weyl duality and q-symmetric functions
Disclaimer: I'm far from an expert on any of the topics of this question. I apologize in advance for any horrible mistakes and/or inaccuracies I have made and I hope that the spirit of the question ...
12
votes
5
answers
808
views
A divisibility of q-binomial coefficients combinatorially
Let a and b be coprime positive integers. Then the number a+b divides the binomial coefficient ${a+b \choose a}$. I know how to prove this combinatorially - for example after choosing an ordered set ...
9
votes
0
answers
191
views
For $q$-analogues of a known curious identity
In 2002 I published the folllowing curious combinatorial identity:
$$(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.$$
My original proof is ...
14
votes
1
answer
756
views
Is there a lift of the q-Vandermonde identity to some geometric (motivic) identity for Grassmannians over $F_q$?
The q-Vandermonde identity reads:
$$ \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} $$
...
30
votes
1
answer
1k
views
Mysterious symmetry - in search for a bijection
I have a mysterious symmetry that I have not managed to prove.
First some definitions (see picture below)
Fix a partition that fit in a staircase shape with $n$ rows.
There are $Catalan(n)$ such ...
5
votes
1
answer
387
views
$q$-analog of an integral from quantum field theory?
This question has been completely reformulated and a new property for the function $f_q$ has been added due to a series of helpful comments by fedja.
Consider the integral from quantum field theory ...
6
votes
1
answer
328
views
Inequality for functions on [0,1], continued
Let $0<a<1,\; \psi_a(x)=\displaystyle \prod_{j=0}^\infty (1-a^jx).$ For each $ k\in \mathbb{N},$ set
$$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
0
answers
101
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 ...