Questions tagged [q-analogs]

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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} \...
Johann Cigler's user avatar
39 votes
2 answers
2k 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 ...
Theo Johnson-Freyd's user avatar
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 ...
Per Alexandersson's user avatar
24 votes
6 answers
2k views

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 ...
John Wiltshire-Gordon's user avatar
22 votes
1 answer
822 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 ...
Gjergji Zaimi's user avatar
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}\...
Johann Cigler's user avatar
20 votes
4 answers
3k views

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 ...
Johann Cigler's user avatar
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 $\...
Asvin's user avatar
  • 7,608
19 votes
1 answer
507 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-...
T. Amdeberhan's user avatar
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 ...
Per Alexandersson's user avatar
16 votes
2 answers
445 views

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!}+\...
Mariano Suárez-Álvarez's user avatar
15 votes
5 answers
2k views

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 ...
James Propp's user avatar
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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 ...
Saal Hardali's user avatar
  • 7,549
15 votes
2 answers
1k views

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 ...
Wolfgang's user avatar
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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 ...
Penchez's user avatar
  • 341
14 votes
2 answers
713 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 ...
Vladimir Reshetnikov's user avatar
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: $$ \binom{ a }{ b}_{\!\!q} $$ ...
Alexander Chervov's user avatar
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 ...
Henry's user avatar
  • 1,410
13 votes
2 answers
1k views

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}$...
Johann Cigler's user avatar
13 votes
0 answers
345 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\...
Johann Cigler's user avatar
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 ...
T. Amdeberhan's user avatar
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 ...
Peter McNamara's user avatar
12 votes
1 answer
554 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 $n$ ...
Alexander Braverman's user avatar
12 votes
1 answer
255 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 ...
Johann Cigler's user avatar
12 votes
1 answer
943 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 $n$ ...
Pablo Spiga's user avatar
12 votes
1 answer
350 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. ...
მამუკა ჯიბლაძე's user avatar
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 \...
Amritanshu Prasad's user avatar
12 votes
0 answers
498 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 ...
Richard Stanley's user avatar
11 votes
3 answers
713 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!}{...
Gjergji Zaimi's user avatar
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. ...
Christian Gaetz's user avatar
11 votes
2 answers
588 views

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\...
Pritam Majumder's user avatar
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): \...
T. Amdeberhan's user avatar
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 & ...
Jeanne Scott's user avatar
  • 1,847
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 ...
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. ...
Mike Pierce's user avatar
  • 1,149
9 votes
1 answer
401 views

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 ...
Fedor Petrov's user avatar
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 ...
Zhi-Wei Sun's user avatar
  • 14.4k
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 $\lambda>0$: $$...
N. Virgo's user avatar
  • 1,306
8 votes
3 answers
405 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 $...
Edwin Beggs's user avatar
  • 1,213
8 votes
1 answer
522 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\...
domenico fiorenza's user avatar
8 votes
1 answer
487 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]...
Sam Hopkins's user avatar
  • 22.5k
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,...
Sam Hopkins's user avatar
  • 22.5k
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
520 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{\...
john mangual's user avatar
  • 22.6k
8 votes
1 answer
631 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 ...
Bruce Westbury's user avatar
8 votes
1 answer
1k views

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 $W_1 \cap W_2$ ...
the_fox's user avatar
  • 347
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 ...
Sam Hopkins's user avatar
  • 22.5k
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 ...
Yuri Sulyma's user avatar
  • 1,513
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 \...
Ratio Bound's user avatar
8 votes
0 answers
710 views

What is the $q$-analog of $\Gamma(z)\Gamma(1-z)=\frac\pi{\sin(\pi z)}$?

I would expect the $q$-Gamma function to have the property which would be the $q$-analog of the Euler reflection formula from my question title. More concretely: $\Gamma(z)$ has simple poles at ...
მამუკა ჯიბლაძე's user avatar