Questions tagged [q-analogs]
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101 questions
3
votes
2
answers
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views
Algorithms (or packages) to find recurrence relations for given sequence of q-polynomials?
Assume we have sequence of polynomials : $P_i(q)$ - each term is polynomial in $q$. (With integer coefficients, but hopefully it is not important).
We expect that there exists recurrence relation a ...
- 24.9k
21
votes
3
answers
808
views
Examples when quantum $q$ equals to arithmetic $q$
First, as a disclaimer, I should say that this post is not about any specific propositions, but is more of some philosophical flavor.
In the world of quantum mathematics, the letter $q$ is a standard ...
- 1,391
1
vote
1
answer
232
views
Looking for q-analog of derangement anagrams for a word
I have already known QPermutationDerangement:
It describes the distribution
$$
d_n(q)=\sum_{\sigma \in D_n} q^{\operatorname{maj}(\sigma)}
$$
Where we sum over all derangements of an $n$ element set.
...
- 175
4
votes
1
answer
293
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 ...
- 533
3
votes
0
answers
80
views
Applications of q-Lagrange inversion
I was reading a text on q,t-Catalan numbers and Diagonal Harmonics by Haglund, where they mention the following $q$-analogue of Lagrange Inversion, taken from Page 53:
Let $e_n, h_n$ denote the ...
- 31
1
vote
0
answers
156
views
Is anything known about the derivative of the quantum dilogarithm?
Faddeev's noncompact quantum dilogarithm is the function defined by
$$
\Phi_{\mathsf b}(z) =
\exp
\int_{\mathbb{R} + i\varepsilon}
\frac{
e^{-2i zw}
}{
4 \sinh(w \mathsf b ) \sinh(w/\...
- 2,533
11
votes
3
answers
557
views
In search of a $q$-analogue of a Catalan identity
Let $C_n=\frac1{n+1}\binom{2n}n$ be the all-familiar Catalan numbers. Then, the following identity has received enough attention in the literature (for example, Lagrange Inversion: When and How):
\...
- 43.2k
0
votes
0
answers
174
views
3D generalization of Gaussian q-binomial coefficient
It is known that the coefficient of $q^t$ in Gaussian binomial coefficient $\binom{m+n}m_q$ equals the number of permutations of the multiset $\{0^m, 1^n\}$ with $t$ inversions.
Is there a closed ...
- 34.3k
12
votes
5
answers
836
views
A divisibility of q-binomial coefficients combinatorially
Let a and b be coprime positive integers. Then the number a+b divides the binomial coefficient ${a+b \choose a}$. I know how to prove this combinatorially - for example after choosing an ordered set ...
- 8,866
7
votes
0
answers
252
views
Hankel determinants for some convolutions of Catalan numbers
Let $c(x)=\frac{1-\sqrt{1-4x}}{2x}$ be the generating function of the Catalan numbers and let $$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,...
- 5,622
7
votes
1
answer
320
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 $n\geq 0$:
$$
(-1)^{...
- 1,430
9
votes
2
answers
480
views
Is there a nice q-analogue of the Jacobi identity in a quantized enveloping algebra?
In a Lie algebra $\mathfrak{g}$ the Jacobi identity $\newcommand{\bracket}[2]{\left[#1\,#2\right]} \bracket{x}{\bracket{y}{z}} + \bracket{z}{\bracket{x}{y}} + \bracket{y}{\bracket{z}{x}} = 0$ holds. ...
- 1,161
6
votes
0
answers
214
views
Looking for a combinatorial proof for an identity involving $q$-Catalan triangles
Let $C_n=\frac1{n+1}\binom{2n}n$ be the Catalan numbers. Following my earlier post on MO, one fine colleague asked me if there is a $q$-analogue of the identity formed by the so-called Shapiro's ...
- 43.2k
9
votes
7
answers
765
views
Important combinatorial and algebraic interpretations of the coefficients in the polynomial $[n]!_q = (1+q)(1+q+q^2) \ldots (1+q+\cdots + q^{n-1})$
What are some important combinatorial and algebraic interpretations of the coefficients in the polynomial
$$[n]!_q = (1+q)(1+q+q^2) \ldots (1+q+\cdots + q^{n-1})?$$
As motivation, I will give ...
12
votes
0
answers
631
views
$q$-analogue of the multinomial theorem?
The $q$-binomial theorem states that
$$
\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 \...
- 5,654
4
votes
0
answers
108
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}{...
- 71
1
vote
0
answers
55
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 ...
- 11
9
votes
2
answers
509
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 ...
- 24.2k
10
votes
0
answers
389
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 & ...
- 2,137
6
votes
0
answers
112
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-...
- 15.8k
19
votes
1
answer
693
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 $\...
- 7,746
8
votes
2
answers
619
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,...
- 24.2k
1
vote
0
answers
150
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,
$$
\binom{...
- 11
17
votes
1
answer
886
views
Proof of certain $q$-identity for $q$-Catalan numbers
Let us use the standard notation for $q$-integers, $q$-binomials,
and the $q$-analog
$$
\operatorname{Cat}_q(n) := \frac{1}{[n+1]_q} \left[\matrix{2n \\ n}\right]_q.
$$
I want to prove that for all ...
- 15.8k
4
votes
1
answer
168
views
Discriminants of some $q$-analogs of $(1+x)^n$
Let $[n]_q=1+q+\dots +q^{n-1}$, $ {[n]_q}! =[1]_q [2]_q \dots [n]_q$ and $\binom{n}{j}_q = \frac{[n]_q!}{[j]_q![n-j]_q!}$ be the usual $q$-notation.
Consider the polynomials $p_n(q,r,x)= \sum_{j=0}^n ...
- 5,622
39
votes
2
answers
3k
views
Is there a "quantum" Riemann zeta function?
Occasionally I find myself in a situation where a naive, non-rigorous computation leads me to a divergent sum, like $\sum_{n=1}^\infty n$. In times like these, a standard approach is to guess the ...
- 54.6k
7
votes
1
answer
325
views
Looking for a $q$-analogue of a binomial identity
The following identity is well-known and there are a few proofs to it (see Bijective proof problems, by R P Stanley, for this and similar formulae):
$$\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}=4^n \...
- 43.2k
11
votes
2
answers
589
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.
...
- 2,753
22
votes
2
answers
742
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
$$\det \left(q^\binom{i-j}{2}\...
- 5,622
3
votes
1
answer
186
views
Is there a $q$-analogue to Shapiro's convolution identity?
Let $C_n=\frac1{n+1}\binom{2n}n$ denote the Catalan numbers.
This question is motivated by the (unanswered) MO post by Alexander Burstein and my own (answered by Fedor Petrov) MO post.
Specifically, ...
- 43.2k
12
votes
3
answers
1k
views
A "quantum" identity: in search of a proof -Part II
As usual, denote $[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q}$ and $[n]_q!=[1]_q[2]_q\cdots[n]_q$. Furthermore, we write
$$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}.$$
As a follow up on this ...
- 43.2k
7
votes
1
answer
253
views
Enumerating subspaces of $\mathbb{F}_q^n$ in terms of words and inversions
When $q$ is a prime power, then on the one hand the $q$-binomial coefficient $\binom{n}{k}_q$ equals the number of $k$-dimensional subspaces of $\mathbb{F}_q^n$, and on the other hand it is the ...
- 63.1k
6
votes
0
answers
227
views
Gaussian coefficients identity
I am having difficulty showing the equivalence between (11) and (15) of Delsarte - Association schemes and $t$-designs in regular semilattices. It is somehow an application of Möbius inversion, but I ...
0
votes
0
answers
137
views
Addition formulas for q-analogs of trigonometric functions
Sine and Cosine functions possess notable formulas for addition of angles
$$
\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).
$$
One can ...
- 116
15
votes
1
answer
748
views
Schur-Weyl duality and q-symmetric functions
Disclaimer: I'm far from an expert on any of the topics of this question. I apologize in advance for any horrible mistakes and/or inaccuracies I have made and I hope that the spirit of the question ...
- 7,799
1
vote
0
answers
88
views
Evaluate $\det[[\lfloor\frac{aj-(a+1)k}n\rfloor]_q]_{1\le j,k\le n}$ and $\det[[\lceil\frac{(a+1)j-ak}n\rceil]_q]_{1\le j,k\le n}$
The $q$-analogue of an integer $m$ is defined by
$[m]_q=(1-q^m)/(1-q)$. Note that $\lim_{q\to1}[m]_q=m$.
I have formulated the following conjecture on determinants involving the floor function and the ...
- 15.6k
13
votes
2
answers
641
views
$q$ as a prime power and a root of unity
The number of points on the $(n-1)$-dimensional projective space $P^{n-1}(\mathbb{F}_q)$ over a finite field $\mathbb{F}_q$ is the $q$-integer
$$[n]_q := \frac{q^n-1}{q-1}.$$
In analogy, the number of ...
- 1,430
3
votes
0
answers
136
views
A recursion involving binomial coefficients: looking for a q-analog
Let $a_n := \frac{1}{2n+1}\binom{3n}{n}$.
Then it is known that (one can find references in the OEIS for this.)
$$
a_n = \sum_{\substack{i,j,k \geq 0 \\ i+j+k=n-1} } a_i a_j a_k.
$$
Is there a natural ...
- 15.8k
8
votes
5
answers
2k
views
Infinite matrix leading eigenvector problem
This question is cross-posted at Math.StackExchange.com.
I'm trying to find the leading eigenvalue and corresponding left and right eigenvectors of the following infinite matrix, for $\lambda>0$:
$$...
- 1,344
14
votes
1
answer
801
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:
$$ \binom{ a }{ b}_{\!\!q} $$
...
- 24.9k
5
votes
1
answer
178
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 ...
- 11.2k
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 ...
- 15.8k
14
votes
2
answers
729
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 ...
- 6,819
7
votes
0
answers
280
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 $...
- 5,622
8
votes
1
answer
229
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 ...
12
votes
1
answer
267
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 ...
- 5,622
11
votes
3
answers
726
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
$$\prod_{k=0}^{n-1}\frac{[3k+1]_q!}{...
- 85.6k
19
votes
1
answer
511
views
"quantum" symmetric plane partitions beget alternating sign matrices?
The "quantum" version qTSPP of the number of totally symmetric plane partitions, contained in the cube $[0,n]^3$, is enumerated by
$$f_n(q):=\prod_{j=1}^n\prod_{k=1}^j\prod_{\ell=1}^k\frac{1-...
- 43.2k
15
votes
0
answers
262
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 ...
- 341
2
votes
2
answers
236
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 $[0; 1]$ ...
- 123