Questions tagged [q-analogs]
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101 questions
40
votes
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answer
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Curious $q$-analogues
Consider the Fibonacci polynomials
$$F_n (x) = \sum_{j = 0}^{\left\lfloor {n/2} \right\rfloor }\binom{n-j}{j} x^{n - 2j} $$
and the Lucas polynomials
$$L_n (x) = \sum_{j = 0}^{\left\lfloor {n/2} \...
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39
votes
2
answers
3k
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Is there a "quantum" Riemann zeta function?
Occasionally I find myself in a situation where a naive, non-rigorous computation leads me to a divergent sum, like $\sum_{n=1}^\infty n$. In times like these, a standard approach is to guess the ...
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30
votes
1
answer
1k
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Mysterious symmetry - in search for a bijection
I have a mysterious symmetry that I have not managed to prove.
First some definitions (see picture below)
Fix a partition that fit in a staircase shape with $n$ rows.
There are $Catalan(n)$ such ...
- 15.8k
24
votes
6
answers
2k
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Is there a q-analog to the braid group?
The braid group $B_n$ on $n$ strands fits into a short exact sequence of groups:
$$ 1 \longrightarrow P_n \longrightarrow B_n \longrightarrow S_n \longrightarrow 1,$$
where $S_n$ is the symmetric ...
- 5,898
22
votes
4
answers
3k
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Are the q-Catalan numbers q-holonomic?
The generating function $f(z)$ of the Catalan numbers which is characterized by $f(z)=1+zf(z)^2$ is D-finite, or holonomic, i.e. it satisfies a linear differential equation with polynomial ...
- 5,622
22
votes
1
answer
884
views
q-Catalan numbers from Grassmannians
In this question by $q$-Catalan numbers I mean the $q$-analog given by the formula $\frac{1}{[n+1]_q}\left[{2n\atop n}\right]_q$. The polynomial $\left[{2n\atop n}\right]_q$ represents the class of ...
- 85.6k
22
votes
2
answers
742
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A q-rious identity
Let $[x]_q=\frac{1-q^x}{1-q}$, $[n]_q!=[1]_q[2]_q\cdots[n]_q$ and ${\binom{x}{n}}_{q}=\frac{[x]_q[x-1]_q\cdots[x-n+1]_q }{[n]_q!}$.
Computer experiments suggest that
$$\det \left(q^\binom{i-j}{2}\...
- 5,622
21
votes
3
answers
808
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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 ...
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19
votes
1
answer
693
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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
19
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1
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511
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"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
$$f_n(q):=\prod_{j=1}^n\prod_{k=1}^j\prod_{\ell=1}^k\frac{1-...
- 43.2k
17
votes
1
answer
886
views
Proof of certain $q$-identity for $q$-Catalan numbers
Let us use the standard notation for $q$-integers, $q$-binomials,
and the $q$-analog
$$
\operatorname{Cat}_q(n) := \frac{1}{[n+1]_q} \left[\matrix{2n \\ n}\right]_q.
$$
I want to prove that for all ...
- 15.8k
16
votes
2
answers
1k
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Why are some q-analogues more canonical than others?
It is striking that some q-analogs of functions, operators, identities and especially whole theorems seem quite "canonical", e.g.
the factorial and the q-Gamma function
the basic hypergeometric ...
- 13.4k
16
votes
2
answers
450
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Derangements and q-variants
Everybody knows that there are $D_n=n! \left( 1-\frac1{2!}+\frac1{3!}-\cdots+(-1)^{n}\frac1{n!} \right)$ derangements of $\{1,2,\dots,n\}$ and that there are $D_n(q)=(n)_q! \left( 1-\frac{1}{(1)_q!}+\...
- 47.3k
15
votes
5
answers
2k
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enumerative meaning of natural q-Catalan numbers
Define $[n]=(1-q^n)/(1-q)$ and $[n]!=[1][2][3] \cdots [n]$, so that $[2n]!/[n]![n+1]!$ is a polynomial in $q$ (the most algebraically natural $q$-analogue of the Catalan numbers); what enumerative ...
- 19.7k
15
votes
1
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Schur-Weyl duality and q-symmetric functions
Disclaimer: I'm far from an expert on any of the topics of this question. I apologize in advance for any horrible mistakes and/or inaccuracies I have made and I hope that the spirit of the question ...
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15
votes
0
answers
262
views
Irreducibility of q-factorial plus 1
Let $q$ be a formal variable and for every positive integer $n$ let
$$[n]_q! = 1 (1 + q)(1 + q + q^2) \dotsm (1 + q + \dotsb + q^{n-1})$$
be the $q$-factorial.
Is it true that $[n]_q! + 1$ is an ...
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14
votes
2
answers
729
views
A conjecture about algebraic values of $(-q;\,-q)_\infty/(q;\,q)_\infty$
Recall that $(a;\,q)_\infty$ is the $q$-Pochhammer symbol:
$$(a;\,q)_\infty=\prod_{n=0}^\infty(1-a \, q^n).\tag1$$
Its important special case $(q;\,q)_\infty=\prod_{n=1}^\infty(1-q^n)$ is sometimes ...
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14
votes
1
answer
801
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Is there a lift of the q-Vandermonde identity to some geometric (motivic) identity for Grassmannians over $F_q$?
The q-Vandermonde identity reads:
$$ \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} $$
...
- 24.9k
13
votes
2
answers
641
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$q$ as a prime power and a root of unity
The number of points on the $(n-1)$-dimensional projective space $P^{n-1}(\mathbb{F}_q)$ over a finite field $\mathbb{F}_q$ is the $q$-integer
$$[n]_q := \frac{q^n-1}{q-1}.$$
In analogy, the number of ...
- 1,430
13
votes
2
answers
1k
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Is there a $q$-L'Hospital's Rule?
Let $\binom{n}{j}_q$ be a $q$-binomial coefficient and $(x;q)_n = (1-x)(1-qx)\cdots(1-q^{n-1}x).$
Consider the sum $$f(n,m,r,k)= \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}q^{mj^2+rj}
\binom{2n}{j}_{q^k}$...
- 5,622
13
votes
1
answer
982
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 $n$ ...
- 934
13
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0
answers
349
views
Some $q-$analogues of $ \sum\limits_{j = - k}^k {{{( - 1)}^{ j}}}\binom{n}{k-j}\binom{n}{k+j}=\binom{n}{k}.$
Let ${\left( {a;q} \right)_n}=\prod\limits_{j = 0}^{n - 1} {(1-{q^j}a} )$ and
let $ {{n}\brack{k}}_q$ denote a $q-$binomial coefficient.
I am interested in $q-$analogues of the identity $ \sum\...
- 5,622
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votes
3
answers
1k
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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
12
votes
5
answers
836
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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
12
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1
answer
561
views
$(q,x)$-analog of $n!$
While doing some work in geometric representation theory I have come across the following
sequence of polynomials in two variables $(q,x)$ which I would like to denote
by $n!_{q,x}$. For small $n$ ...
- 7,037
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
12
votes
1
answer
351
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. ...
- 18.4k
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
$$
(1+t)^n = \sum_{k=0}^n \...
- 5,654
12
votes
0
answers
502
views
A $q$-analogue of Foulkes' character related to alternating permutations
My paper "Alternating permutations and symmetric functions" at
http://math.mit.edu/~rstan/papers/altenum.pdf enumerates certain
classes of alternating permutations, such as those whose inverse is
...
- 50.8k
11
votes
3
answers
557
views
In search of a $q$-analogue of a Catalan identity
Let $C_n=\frac1{n+1}\binom{2n}n$ be the all-familiar Catalan numbers. Then, the following identity has received enough attention in the literature (for example, Lagrange Inversion: When and How):
\...
- 43.2k
11
votes
3
answers
726
views
Is this a q-count of Alternating Sign Matrices?
The number of Alternating Sign Matrices of size $n$ is well known to be
$\prod_{k=0}^{n-1}\frac{(3k+1)!}{(n+k)!}$. Is it known whether the naive q-analog expression
$$\prod_{k=0}^{n-1}\frac{[3k+1]_q!}{...
- 85.6k
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
11
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2
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604
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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 $${2n\choose n}-{2n\...
- 1,305
10
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0
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389
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Has anyone met this "$q$-character" table for $S_4$?
Is anyone aware of the following $q$-character table for the
symmetric group $S_4$?
\begin{array}{|c|c|c|c|c|c|}
\hline
\mathrm{conj}\backslash\mathrm{rep}
& 2+1+1 & 3+1 & ...
- 2,137
9
votes
7
answers
765
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Important combinatorial and algebraic interpretations of the coefficients in the polynomial $[n]!_q = (1+q)(1+q+q^2) \ldots (1+q+\cdots + q^{n-1})$
What are some important combinatorial and algebraic interpretations of the coefficients in the polynomial
$$[n]!_q = (1+q)(1+q+q^2) \ldots (1+q+\cdots + q^{n-1})?$$
As motivation, I will give ...
9
votes
2
answers
480
views
Is there a nice q-analogue of the Jacobi identity in a quantized enveloping algebra?
In a Lie algebra $\mathfrak{g}$ the Jacobi identity $\newcommand{\bracket}[2]{\left[#1\,#2\right]} \bracket{x}{\bracket{y}{z}} + \bracket{z}{\bracket{x}{y}} + \bracket{y}{\bracket{z}{x}} = 0$ holds. ...
- 1,161
9
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1
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420
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notation for $(a-b)(a-qb)\dots (a-q^{n-1}b)$
I wonder whether there is a notation for such thing, which I denote $[a;b]_q^n$ for a moment:
$$
[a;b]_q^n:=(a-b)(a-qb)\dots (a-q^{n-1}b)=a^n(b/a;q)_n,
$$
this last equation uses $q$-Pochhammer symbol ...
- 109k
9
votes
1
answer
641
views
A q-analogue of Ramanujan's tau function
There have been a couple of questions on Ramanujan's $\tau$ function.
Lehmer's conjecture for Ramanujan's tau function
The Vanishing of Ramanujan's Function tau(n)
A $q$-analogue is given ...
- 9,132
9
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2
answers
509
views
Lusztig's $q$-analog of weight multiplicity with product formula
For partitions $\lambda, \mu \vdash n$, the Kostka-Foulkes polynomial $K_{\lambda,\mu}(q)$, a $q$-analog of the Kostka coefficient $K_{\lambda,\mu}$, has a combinatorial description, due to Lascoux ...
- 24.2k
9
votes
0
answers
192
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For $q$-analogues of a known curious identity
In 2002 I published the folllowing curious combinatorial identity:
$$(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.$$
My original proof is ...
- 15.6k
8
votes
5
answers
2k
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Infinite matrix leading eigenvector problem
This question is cross-posted at Math.StackExchange.com.
I'm trying to find the leading eigenvalue and corresponding left and right eigenvectors of the following infinite matrix, for $\lambda>0$:
$$...
- 1,344
8
votes
3
answers
417
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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 $...
- 1,143
8
votes
1
answer
553
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
$$
\sum_{k=0}^m(-1)^{k-m} {n-k\choose m}{m\...
- 6,777
8
votes
1
answer
497
views
q-Integer-valued polynomials
For $n \in \mathbb{Z}_{\geq 0}$, let $[n]_q := (1-q^n)/(1-q) = (1+q+...+q^{n-1})$ as is customary, with $[0]_q=0$.
Let $R$ be the subring of $\mathbb{Q}(q)[x]$ consisting of all $f$ such that $f([n]...
- 24.2k
8
votes
2
answers
619
views
Relationship between $q$-Weyl dimension formula and $q$-analog of weight multiplicity?
$\DeclareMathOperator\dim{dim}$For a dominant (integral) weight $\lambda$ and any (integral) weight $\mu$ of a simple Lie algebra $\mathfrak{g}$, Lusztig's $q$-analog of weight multiplicty $K_{\lambda,...
- 24.2k
8
votes
1
answer
229
views
Prominent examples of $q$-analogs without known cyclic sieving
The cyclic sieving phenomenon is nicely summarized in the following AMS Notices "What is...?" article: https://www.ams.org/notices/201402/rnoti-p169.pdf.
In that article, Reiner, Stanton, and White ...
8
votes
1
answer
527
views
A q,t-extension of Plancherel Measure thru Yang-Mills Theory ?
Buried in the physics paper by Nekrasov and Okounkov, a strange identity is proven:
$$ \prod_{n > 0} (1 - q^n)^{\mu^2-1} = \sum_{\mathbf{k}} q^{|\mathbf{k}|} \prod_{\square \in k} \left( 1 - \frac{\...
- 22.8k
8
votes
1
answer
1k
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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 $W_1 \cap W_2$ ...
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8
votes
1
answer
298
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 ...
- 1,838
8
votes
0
answers
253
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 \...
- 131