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3 votes
0 answers
97 views

What algebras generate polynomial count varieties as their representations spaces ? Is it preserved by the Koszul duality, Manin's endomorphisms?

Consider for example commutative polynomial algebra $C[x,y]: xy=yx$, look on $F_p$-matrices satisfying that relation - the number over $F_p$ will be given by polynomial in $p$ (classical result due to ...
Alexander Chervov's user avatar
6 votes
1 answer
231 views

Does Manin's construction of non-commutative endomorphism algebra $\mathrm{End}(A)$ produce Koszul algebra, if $A$ is Koszul?

$\newcommand{\dual}{\mathrm{dual}}\DeclareMathOperator\End{End}\DeclareMathOperator\Fun{Fun}\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\GL{GL}$Around 1986–7 Yu.Manin proposed natural and ...
Alexander Chervov's user avatar
10 votes
1 answer
477 views

Where does the definition of ($\infty$-)groupoid cardinality come from?

The cardinality of a finite set $X$ is the number of its elements. Once you know that, you would define the groupoid cardinality of a $\infty$-groupoid $X$ as the quantity $$\lvert X\rvert := \sum_{[x]...
Matthew Niemiro's user avatar
5 votes
0 answers
154 views

Interpretation of superfactorial in terms of plane partitions

Recently I got interested in plane partitions and the following formula by MacMahon, which counts the number of plane partitions $\pi \in B(r,s,t)$ fitting in an $(r,s,t)$-box: $$ \binom{r+s+t}{r,s,t}...
Bipolar Minds's user avatar
5 votes
0 answers
345 views

A fusion ring identity

Fusion rings I'll more or less stick to the presentation given in this question: [1] We define a fusion ring as follows: consider a free $\mathbb{Z}$-module $\mathbb{Z}\mathcal{B}$ with finite basis ...
Meths's user avatar
  • 309
7 votes
1 answer
566 views

Is there an integral fusion ring which is not of Frobenius type?

Combinatorially, a fusion ring $\mathcal{F}$ is nothing but a finite set $B=\{b_1, \dots, b_r\}$ (generating the $\mathbb{Z}$-module $\mathbb{Z} B$) together with fusion rules: $$ b_i \cdot b_j = \...
Sebastien Palcoux's user avatar
5 votes
1 answer
318 views

Jack polynomial and Selberg integral

I am studying the generalisation of the Selberg's integral by using Jack polynomial as outlined by Forrester and Warnaar (arXiv: 0710.3981). They write down the symmetric Jack polynomial as \begin{...
morgoth's user avatar
  • 53
0 votes
0 answers
167 views

Negative $q$-binomial series: reference request

There seems to be a result for formal series (I hope this is right) for all integer $r\ge 0$ $$ \sum_{n\ge 0} (-x)^n\ {{n+r}\choose{r}}_{q} = (1+x)^{-1}(1+qx)^{-1}\dots (1+q^{r}x)^{-1} $$ where the $q$...
Edwin Beggs's user avatar
  • 1,143
2 votes
0 answers
87 views

Modules over quantum complete intersections

Let $a_i \geq 2$ be natural numbers and $q_{ij}$ field elements of the field $k$ for $i>j$. A quantum complete intersection is the algebra $A:=k<x_1,...,x_n>/(x_i^{a_i},x_i x_j - q_{ij} x_j ...
Mare's user avatar
  • 26.5k
20 votes
1 answer
586 views

$q$-(and other)-analogs for counting index-$n$ subgroups in terms of Homs to $S_n$?

The following formula of astonishing beauty and power (imho): $$ \sum_{n \ge 0} \frac{| \mathrm{Hom}(G,S_n) | }{n! } z^n = \exp\left( \sum_{n \ge 1} \frac{|\text{Index}~n~\text{subgroups of}~ G|}nz^...
Alexander Chervov's user avatar
3 votes
1 answer
253 views

What is the value of this sum involving q-binomials?

Let $n\ge 2r$ be positive integers. Is there a closed form for following finite summation involving in q-binomial coefficients $$\sum_{s=0}^r(-1)^sq^{\frac{s(s+1)}{2}}{n-2r+s\brack n-2r}_q{n\brack r-...
Bumblebee's user avatar
  • 1,093
3 votes
0 answers
220 views

Generalisation of the quantum exterior algebra

One might generalise the classical exterior algebra as follows to the quantum exterior algebra: $K<x_1,...x_n>/(x_i^2,x_i x_j + q_{i,j}x_j x_i)$ with nonzero field elements $q_{i,j}$ for $i<j$...
Mare's user avatar
  • 26.5k
8 votes
1 answer
552 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
3 votes
0 answers
216 views

A "nice" Orthogonal Basis for Translation Invariant Symmetric Polynomials

It is going to be a rather long question, so I will first state it and then try to explain and motivate it. Take $\Lambda_n $ as the graded ring of symmetric polynomials of a field $F$ in $n$ ...
Hamed's user avatar
  • 613
11 votes
3 answers
1k views

A problem on a specific integer partition

Let $n$ be a positive integer, we consider partitions of the following form : $$n = d^{2}_{1} + d^{2}_{2} + ... + d^{2}_{r}$$ such that : $d_{i}\vert n$ $1=d_{1}<d_{2} \le d_{3} \le ... \le d_{r}$...
Sebastien Palcoux's user avatar
4 votes
1 answer
174 views

Reference for the image of the adjoint to the differential in graph cohomology (which yields STU & IHX)?

One can define cochain complexes of (combinatorial) graphs, where each term is a vector space of linear combinations of certain (isomorphism classes of) graphs, and where the differential $d$ is a ...
Robin Koytcheff's user avatar
18 votes
2 answers
983 views

A direct proof of the Harer-Zagier recursion enumerating the ways to paste a 2n-gon to get a genus g surface?

In a 1986 paper, Harer and Zagier proved the recursion: $$(n+1)e(g,n)=(4n-2)e(g,n-1)+(2n-1)(n-1)(2n-3)e(g-1,n-2)$$ where e(g,n) is the number of ways of grouping sides $S_1...S_{2n}$ of a 2n-gon ...
Alfredo Hubard's user avatar
2 votes
1 answer
228 views

A question on Lusztig's `graph with automorphism' construction?

Using the notion of a graph with compatible automorphism, Lusztig constructs all symmetrizable Cartan data (i.e. Cartan matrices $A$ for which there is a diagonal matrix $D=\mathrm{diag}(d_1,\ldots,...
David Hill's user avatar
  • 1,472
8 votes
0 answers
488 views

det(A)det(B) = det(AB+correction), Capelli identities, "factorized" representation of $\mathfrak {gl}_n$

Context: Some probably know that there are Capelli identities which state $$det(A)det(B) = det(AB+correction)$$ for some matrices with non-commuting elements, they go back to the 19-th century, but ...
Alexander Chervov's user avatar
16 votes
0 answers
824 views

Capelli determinant = Duflo ( determinant) - was it known ?

Question briefly. Was this fact known: Capelli determinant = Duflo (determinant) ? (This is an equality of the two central elements in universal enveloping of Lie algebra $gl_n$). I googled a lot ...
Alexander Chervov's user avatar
6 votes
2 answers
388 views

monomials in the universal enveloping of a Lie algebra in terms of the symmetric basis

Let $L$ be a finite-dimensional Lie algebra over a field $k$ of characteristic zero and $e_1,\ldots, e_n$ some basis of $L$. The formula $[e_i,e_j] = \sum_k C_{ij}^k e_k$ determines the structure ...
Zoran Skoda's user avatar
  • 5,232
44 votes
5 answers
5k views

Groups, quantum groups and (fill in the blank)

In the study of special functions there are three levels of objects, classical, basic and elliptic. These correspond to classical hypergeometric functions, basic (q-) hypergeometric functions, and ...
Gjergji Zaimi's user avatar
3 votes
0 answers
233 views

How many set partitions on a big cube’s boundary arise from cubomino decompositions of the solid cube?

Introduction. This is a counting question about configurations that can appear on the outside of assembled Soma cube-like puzzles. More specifically, it’s about the ways in which the pieces of an ...
Steve Kass's user avatar
7 votes
0 answers
213 views

Decomposition of certain projectives for cyclotomic q-Schur algebras

In representation theory, a very popular set of finite dimensional algebras are the $q$-Schur algebras, which are given by looking at the endomorphisms of $V^{\otimes d}$ where $V$ is the standard ...
Ben Webster's user avatar
  • 44.7k
12 votes
4 answers
1k views

Asymptotics of q-Catalan numbers

q-Catalan numbers are defined recurrently as C0=1, $C_{N+1}=\sum_{k=0}^N q^k C_k C_{N-k}$. What can be said about the asymptotics of Cn when 0<q<1? P.S. In ...
Leonid Petrov's user avatar
16 votes
2 answers
450 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