# Questions tagged [q-analogs]

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### Is there a nice q-analogue of the Jacobi identity in a quantized enveloping algebra?

In a Lie algebra $\mathfrak{g}$ the Jacobi identity $\newcommand{\bracket}{\left[#1\,#2\right]} \bracket{x}{\bracket{y}{z}} + \bracket{z}{\bracket{x}{y}} + \bracket{y}{\bracket{z}{x}} = 0$ holds. ...
1answer
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### 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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### 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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216 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 ...
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### 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 ...
0answers
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### 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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1answer
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### 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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### 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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441 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 ...
1answer
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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 ...
4answers
689 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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### 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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### 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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### 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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### $q$-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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### 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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### 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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### 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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### $q$-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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### 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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Denote by ${n\brack {k}}$ a $q-$binomial coefficient. Let ${D_{n,k}}(t,q) = \sum\limits_{j = 0}^{n - k} {{q^{{j^2} + kj}}}{n\brack {j}}{n\brack {k+j}}t^j$ and ${R_n}(x,t,q) = \sum\limits_{k = 0}... 3answers 680 views ### Is this a q-count of Alternating Sign Matrices? The number of Alternating Sign Matrices of size$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 $$\prod_{k=0}^{n-1}\frac{[3k+1]_q!}{... 1answer 421 views ### q-analog of a combinatorial identity involving binomial coefficients Using, e.g., properties of iterated finite differences it is easy to show that for any pair of integers n and m with n>\!>m one has the identity$$ \sum_{k=0}^m(-1)^{k-m} {n-k\choose m}{m\... 0answers 272 views ### What is known about the$q\$-analogue of the simplex?

I am interested in the field with one element. I am thus interested in combinatorial interpretations of the Gaussian binomial coefficients. Richard Stanley's "Enumerative combinatorics" mentions ...