20
votes
Accepted
A "quantum" identity: in search of a proof -Part II
Both sides are equal to $\binom{x+y+1}{n}_q$. This enumerates lattice paths in an $n\times (x+y-n+1)$ rectangle, according to the area statistic. We will assume that these paths start at $(0,0)$ and ...
- 83k
20
votes
Accepted
A conjecture about algebraic values of $(-q;\,-q)_\infty/(q;\,q)_\infty$
Yes, it is always algebraic, because it is a modular function
evaluated at a CM (complex multiplication) point.
"$(q;q)_\infty$" is $q^{-1/24} \eta(\tau)$ where $q = e^{2\pi i \tau}$, so
"$(q;q)_\...
- 73.8k
20
votes
Accepted
Is there a lift of the q-Vandermonde identity to some geometric (motivic) identity for Grassmannians over $F_q$?
Assume $V$ is a vector space of dimension $m+n$, $M \subset V$ is a subspace of dimension $m$, and $N = V/M$. Let $p:V \to N$ be the projection. Consider the Grassmannian $X = Gr(k,V)$ and its ...
- 33.4k
16
votes
Accepted
What is the groupoid cardinality of the category of vector spaces over a finite field?
Upon substituting $x=\frac{1}{q}$ we obtain
$$\sum_{n\geq 0}\frac{1}{|\mathrm{GL}_n(\mathbb F_q)|}=\sum_{n\geq 0}\frac{x^{n^2}}{(1-x)(1-x^2)\cdots(1-x^n)}$$
and this evaluates to the product $\prod_{i\...
- 83k
15
votes
Accepted
Schur-Weyl duality and q-symmetric functions
As Sam Hopkins says, the category of all representations of $GL_n(\mathbb F_q)$ is too large to give what you want. Instead, let's consider the category of unipotent representations, i.e. those ...
- 2,991
15
votes
Accepted
Is there a nice q-analogue of the Jacobi identity in a quantized enveloping algebra?
There are various deformations of the Jacobi identity that can be found scattered in the literature. As far as i know, using the definition: $[A,B]_q=AB-qBA$, one of the most general ones (though i do ...
- 7,428
12
votes
q-Catalan numbers from Grassmannians
There are a few nice answers to related questions. Unfortunately none of them quite answers the question you asked.
The $q$-Catalan number $\frac{1}{[n+1]_q}{ 2n \brack n}_q$ is the Hilbert series of ...
- 3,541
12
votes
Accepted
Total positivity of $q$-Pascal matrix?
Yes. Consider the set $V$ of points with integer coordinates as vertices of a weighted directed acyclic graph. Namely, for any $(i,j)\in V$, the edge from $(i,j)$ to $(i+1,j)$ has weight 1, the edge ...
- 91.9k
11
votes
A "quantum" identity: in search of a proof -Part II
If we substitute $y:=v+n$ in the identity, the LHS becomes a convolution of two similar sequences,
$$\sum_{k=0}^nq^{(v+1)k}\binom{x+k}{k}_q \binom{v+(n-k)}{n-k}_q =\sum_{k=0}^n q^{n-k}\binom{x+v+n-k}{...
- 52.5k
11
votes
Important combinatorial and algebraic interpretations of the coefficients in the polynomial $[n]!_q = (1+q)(1+q+q^2) \ldots (1+q+\cdots + q^{n-1})$
This answer concerns a geometric/Lie-theoretic interpretation of $[n]!_q$.
$[n]!_q$ gives the number of points in the full flag variety of full flags of subspaces in an $n$-dimensional vector space $\...
10
votes
Is this a q-count of Alternating Sign Matrices?
It's true these polynomials are not unimodal for n=2 and up, since they all start with coefficient sequence 1 0 1 ... (The reason for this is clear from the definition of descending plane partitions.) ...
- 101
9
votes
Is there a $q$-L'Hospital's Rule?
I think you can achieve this using the actual $q$-L'Hospital's Rule. Consider
$$\frac{ \partial^i f(n,m,r,k)}{\partial q^i} (1)= \sum_{i=0}^a \binom{a}{i} \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}\frac{...
- 122k
9
votes
A divisibility of q-binomial coefficients combinatorially
Assume $q$ is a prime power. Then there is a straightforward combinatorial interpretation.
Consider the group action of $\mathbb F_{q^{a+b}}^\times / \mathbb F_{q^\times}$ on the space of $a$-...
- 122k
9
votes
Accepted
In search of a $q$-analogue of a Catalan identity
This identity is known as Jonah's formula (special case with $n\rightarrow 2n$ and $r\rightarrow n$, see "Catalan Numbers with Applications" by Thomas Koshy, pg. 325-326 for a combinatorial ...
9
votes
Accepted
A curious $q$-series identity on a truncated Euler function
Let ${n\choose k}_q=\frac{(q)_n}{(q)_k(q)_{n-k}}$ denote a $q$-binomial coefficient. We start with the following version of $q$-Vandermonde convolution identity:
$$
(x-y)(x-qy)\ldots(x-q^{n-1}y)\\=\...
- 91.9k
8
votes
Accepted
q-analog of a combinatorial identity involving binomial coefficients
You can see this as an instance of the q-Vandermonde identity. The q-binomial theorem tells us that the coefficient of $t^k$ in $\prod_{i=0}^{m-1}(1+q^{-2i}t)$ is $q^{-k(m-1)}{m \brack k}_q$ and the ...
- 83k
8
votes
$q$ as a prime power and a root of unity
This is very not rigorous, but it's a way of thinking about this topic which I find personally helpful. Several $\mathbb F_1$ papers contains remarks about the idea going back to Weil and Iwasawa that ...
- 83k
8
votes
Accepted
$q$-analogs of total positivity
This is a wonderful question but unfortunately I don't think that there is a definite answer in the literature just yet. Let's look at two somewhat recent lines of research in this direction:
1) A. ...
- 83k
8
votes
Important combinatorial and algebraic interpretations of the coefficients in the polynomial $[n]!_q = (1+q)(1+q+q^2) \ldots (1+q+\cdots + q^{n-1})$
In FindStat, the database of combinatorial statistics, the major index of a permutation is http://www.findstat.org/StatisticsDatabase/St000004/. By clicking "search for distribution" there, you will ...
7
votes
Accepted
q-Integer-valued polynomials
(Below is the proof that module $R$ is generated by $f_i(x)$, without calculation of structure constants.)
Polynomial $f(x)$ of degree $n$ may be interpolated in points $[0],[1],\dots,[n]$ (I omit ...
- 91.9k
7
votes
A "quantum" identity: in search of a proof -Part II
Actually this is a partial case of $q$-Vandermonde identity.
In the notation $$[a;b]_q^n:=(a-b)(a-qb)\dots (a-q^{n-1}b)=a^n(b/a;q)_n,$$ as in this terminology question, we have
$$
[a;c]_q^n=\sum {\...
- 91.9k
7
votes
Accepted
Mysterious symmetry - in search for a bijection
Oliver Pechenik and I have some partial progress to report. Maybe someone else can see how to supply the remaining missing ingredients.
First, let's establish some notation.
We say your ascent ...
- 1,769
7
votes
A divisibility of q-binomial coefficients combinatorially
A proof of nonnegativity appears in https://arxiv.org/pdf/0912.1578.pdf. The number $\frac{1}{a+b}{a+b\choose a}$ is called a rational Catalan number by
Drew Armstrong. See for instance http://www....
- 46.1k
7
votes
Accepted
Discriminants of some $q$-analogs of $(1+x)^n$
This is true.
We have
\begin{align*}
p_n (q^{-1}, 1-r, x) &= \sum_{j=0}^n q^{ (r-1) \binom{j}{2}} \binom{n}{j}_{q^{-1}} x^j \\
&= \sum_{j=0}^n q^{ (r-1) \binom{j}{2}} q^{-j (n-j)} \binom{n}...
- 122k
6
votes
Accepted
What is the value of this sum involving q-binomials?
Doron Zeilberger has written a Maple code for checking and proving ordinary binomial identities and their $q$-analogues. What you need in the present case is the package called qEKHAD.
I just tested ...
- 39.3k
6
votes
Accepted
Generating function for certain partitions (with a restriction on the Durfee square)
Lemma. Fix $n$ and $m$. Consider pairs of partitions $(\lambda,\mu)$ such that $\lambda$ has $m$ parts and $|\lambda|+|\mu|=n$. Let $A$ be the number of pairs for which $\max(\lambda)>\max(\mu)$ (...
- 91.9k
6
votes
Accepted
Inequality for functions on [0,1], continued
OK, here goes.
We start with changing the notation ($z\to 20z^2$, $-z-3\to r$, $20rz\to y$ means that what was denoted by $z$ will be denoted by $20z^2$ from now on, $r$ is $-z-3$ with new $z$, so it ...
- 55.4k
6
votes
A q-rious identity
In the mean-time I have found a proof:
First write Vandermonde’s identity $$\sum_{j=0}^k q^{(k-j)(i+r-j)}\binom{s-r}{k-j}_{q}\binom{i+r}{j}_{q}=\binom{i+s}{k}_{q}$$ in the form
$$\sum_{j=0}^k q^{\...
- 5,230
6
votes
Accepted
Enumerating subspaces of $\mathbb{F}_q^n$ in terms of words and inversions
You may act similarly as follows.
Let $V$ be a $k$-dimensional subspace of $\mathbb F_q^n$. Take any its base, put its elements into the rows of some matrix, and make it to the reduced row echelon ...
- 19.9k
6
votes
In search of a $q$-analogue of a Catalan identity
Decided to make a cw post: it is sort of amusing.
Let $C_n(q)$ be defined by
$$\sum_{k=0}^n\binom{2n-2k}{n-k}_qC_k(q)q^{2n-2k}=\binom{2n+1}n_q,\qquad n=0,1,2,\dotsc.$$
Then
\begin{multline*}
C_n(q)=1+...
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