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In algebraic combinatorics, there is an important concept of a "$q$-analogue". Surprisingly often when you have a counting problem with a good integer answer, you realize that it can be refined to a (finite) generating function with an equally good polynomial answer. A simple example of this is the $q$-analogue of the number of permutations, which is of ...

answered May 6 '15 at 16:15

Greg Kuperberg

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Quantum algebra is an umbrella term used to describe a number of different mathematical ideas, all of which are linked back to the original realisation that in quantum physics, one finds noncommutativity. The areas now encompassed by the term "quantum algebra" are not necessarily directly or obviously related to each other (and this is even more true for ...

41

The equivalence of these two construction is actually known now. It follows by combining the main result of:
Yves Laszlo, Hitchin's and WZW connections are the same., J. Differential Geom. 49 (1998), no. 3, 547–576, doi:10.4310/jdg/1214461110
with my joint work with Kenji Ueno presented in a series of four papers:
J.E. Andersen & K. Ueno, Abelian ...

30

The answer of Jørgen Ellegaard Andersen only concerns the case of when the gauge group is $SU(n)$.
I will argue that all the ingredients for the equivalence between the two approaches (namely "geometric quantization of character varieties" and "quantum groups plus skein theory") are out there, for arbitrary simply connected gauge group.
First of all, let ...

answered Oct 25 '14 at 23:22

André Henriques

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Here is an answer to question (1). I recommend that you split of question (2) as a separate question.
Define the quantum plane to be the "spectrum" of the noncommutative ring $\mathbb K\langle x,y\rangle / (xy = qyx)$, where $\mathbb K$ is some ground commutative ring in which $q$ is invertible (e.g. $\mathbb K = \mathbb C(q)$). So a "point" in this "...

answered Apr 28 '15 at 4:08

Theo Johnson-Freyd

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Good question. I'm much more familiar with the QG/skein theory approach than the geometric quantization approach, so perhaps what I write here will be biased.
I think the main reason there is not yet a proof that the two approaches are equivalent is that the geometric quantization side is difficult and unwieldy (in my biased opinion), though I'm willing ...

24

$\newcommand\g{\mathfrak{g}}$The answer to your question "is there a procedure that takes $\g$ as input, produces $U_q(\g)$ as output, and doesn't involve the choice of a Cartan subalgebra of $\g$?" is No. Not if you want it "canonical" in any sense. (Of course, if I wanted to cheat my way to a "yes," I could make choices that are equivalent to choosing a ...

answered Aug 22 '12 at 14:35

Theo Johnson-Freyd

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Typically in math "quantum X" means a deformation of "X" which is in some sense "less commutative." So quantum groups should be deformations of groups which are "less commutative." Interpreting this is slightly tricky since groups are already non-commutative, but nonetheless they do have some "commutativity" built in which you can see either by noting:
...

23

There has been quite a lot of literature on the applications of $q$-numbers, $q$-derivatives, $q$-deformations, etc, of various algebraic models of physics. Such applications range from $q$-deformations of simple harmonic oscillator(s) and angular momentum algebras to the development of quantum groups and their applications in nuclear physics, particle ...

answered Aug 20 '17 at 21:36

Konstantinos Kanakoglou

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Hi,
In the following, we consider the quantum group of type $G_2=(a_{ij})_{i,j\in I}$
where $I=\{1,2\}$ and $\alpha_2$ is the long root.
We choose a reduced expression
for the longest element $w_0$ of the Weyl group $W(G_2)$ as
$w_0=s_2s_1s_2s_1s_2s_1$ and thus identify $\mathbb{N}^{6}$ and Kashiwara's crystal $B(\infty)$ via Lusztig's parametrization,
\...

16

I was hesitating to write an answer since I don't have references at hand but let me mention that if you denote your polynomials $P_n(x,q)$ and look at $Q(x,q)=x^{\binom{n}{2}}P_n(x^{-1},q)$ then (my guess is that) you are looking at:
$$Q(x,q)=\sum_{\pi\in S_n}x^{maj(\pi)}q^{inv(\pi)}=\sum_{\pi\in S_n}x^{maj(\pi)}q^{maj(\pi^{-1})}$$
where $maj(\pi)$ is the ...

15

Having said that the two have radically different representation theories says already a lot. But first be warned that quantum groups at roots of unity may come in different ways: a beautiful summary was written here Which is the correct version of a quantum group at a root of unity?
Having said so let me add something about the De Concini-Kac form. In such ...

14

This answer feels so glib I'm quite worried it's wrong, but anyway:
Write $D_q$ for the full ring of fractions of $k_q[x,y]$. By Alev-Dumas, "Sur le corps des fractions de certaines algebres quantiques", Corollary 3.11c, we know that for $q$, $r$ non-roots of unity, $D_q \cong D_r$ iff $r = q^{\pm1}$.
It's clear that $k_q[x,y] \cong k_r[x',y'] \Rightarrow ...

14

Here is my understanding of the situation. Question 1: yes, the equivalence is known in all cases. Kazhdan-Lusztig work does have some limitations on the level,
so Finkelberg's approach is not applicable. However the categories in question are
easy enough to work with explicitly: for instance $E_8$ at level 1 has just 1 simple object and $E_8$ at level 2 has ...

13

Hi Ryan,
You can prove that if $a,b$ are some elements in an algebra such that $[a,b]=\lambda b$ for $\lambda$ a scalar, then (in a context where this expression makes sense) $q^a b q^{-a}=q^{\lambda} b$: rewrite the relation as
$$ab=b(a+\lambda)$$
then
$$a^nb=b(a+\lambda)^n$$
therefore
$$q^ab=\sum \frac{\log(q)^na^n}{n!}b=b\sum \frac{\log(q)^n(a+\lambda)^n}...

12

The answer is "Sometimes."
Following the notation of Rowell's "From Quantum Groups to Unitary Modular Tensor Categories", let $\mathcal C(\mathfrak g, l, q)$ be the category corresponding to $U_q(\mathfrak g)$ such that $q=e^{\pi \imath/l}$.Denote by $m$ is the ratio of the square lengths of a long root to a short root. As mentioned there, this only gets ...

12

I am going to give three counterexamples to your first question. (The third counterexample is courtesy of @Adrien, who did most of the job.) While none of them leads to a full answer of your second question, at least they strongly restrict the possibilities.
1. The first counterexample: binate groups
I will denote the Hopf dual of a Hopf algebra (or ...

answered Aug 17 '18 at 16:11

darij grinberg

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12

This is true. In other language, if I understand rightly, a super Lie algebra is just a graded Lie algebra with grading over {0,1} (even and odd), with the standard sign conventions as in algebraic topology. The treatment of graded Lie algebras and their universal enveloping algebras in Section 22.1 of ``More concise algebraic topology'' by Kate Ponto and ...

11

Allow me the excuse to mention a cute way of constructing Uq(sl2) as a q-analogue. Presumably this is ahistorical, since I suspect that initially Uq(sl2) was constructed first over the formal disc, then later it was realised that it was definable over $\mathbb{Q}(q)$.
Anyhow, let S be a finite set. Consider a vector space with basis $\{[I]\mid I\subset S\}$....

11

The key point here is to understand how the parameter $q$ is fixed to a complex number. Usually $U_q(\mathfrak g)$ is introduced as a $\mathbb C(q)$-Hopf algebra with generators and relations, and relations are singular at $q=1$. This means that to obtain a $\mathbb C$-Hopf algebra from $U_q(\mathfrak g)$ one has first to fix a so-called integer form, i.e. a ...

11

For complex simple $\mathfrak g$, Drinfeld (1986, p. 807) already characterized his $\mathrm U_h\mathfrak g$ as the unique (up to equivalence and change of parameter) deformation of $\mathrm U\mathfrak g$ admitting a “quantum Cartan involution”, i.e. an algebra automorphism and coalgebra antiautomorphism extending the Cartan involution and for which ...

answered Jul 19 '17 at 21:34

Francois Ziegler

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11

As another example of the second category in Kostantinos Kanakoglou's answer I think it is fair to mention quantum-integrable systems: this topic in physics was pivotal in the historical development of the notion of quantum groups by the Leningrad group (Faddeev et al), and the Japanese group (Jimbo and Miwa et al).
In particular, $U_q(\widehat{\mathfrak{sl}...

11

That the only monoidal deformations of the category of representations of $U(\mathfrak{g})$ is the category of representations of $U_q(\mathfrak{g})$ is known in Type A from Kazhdan-Wenzl (Adv. Soviet Math, 1993), for the category of vector representations in Type BCD by combining results of Wenzl-Tuba with Liu, and for $G_2$ by Kuperberg. It is open for ...

10

Drinfeld's original ICM-86 talk "Quantum groups" is something "must read", scanned files are available here.
This old introduction works out many details and is quite good:
"An introduction to quantized Lie groups and algebras" T.Tjin
arXiv:hep-th/9111043
There is certain interplay between certain topics in classical simple Lie algebras
and quantum groups,...

answered Dec 3 '12 at 6:39

Alexander Chervov

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10

I take the freedom to copy here an email of Finkelberg to myself:
Dear Andre,
The braidings and ribbon structures do also match. The braiding is reconstructed from the local systems; the ribbon structure (balance) comes from the action of $L_0$ and can be explicitly computed on any irreducible. Let me know if you need details on this.
This is a question to ...

10

When $k$ fails to be algebraically closed the theorem is false but the discrepancy can be understood in terms of Galois descent and so in principle understood in terms of Galois cohomology.
Suppose $H$ is a cocommutative Hopf algebra over $k$, not algebraically closed but characteristic $0$. Then the classification theorem applies to the base change $H \...

answered Jan 18 '17 at 5:40

Qiaochu Yuan

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10

According to nLab, such an action is called a Hopf action and your data specify a left $B$-module algebra. Such a structure is also referred to in the literature as an algebra in the category (of left $B$-modules).
Note also, that, if instead of the bialgebra $B$ we consider a hopf algebra $H$ acting on $A$ and satisfying your condition supplemented by:
$$...

answered Oct 31 '17 at 1:01

Konstantinos Kanakoglou

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This is a very interesting question. I have also made some search but i have not found this result explicitly mentioned somewhere in the literature. However, i remember i have heard such a claim in the past. So i will try to record my thoughts on the problem. Here is the way i understand it:
Since you are considering the limits $N\to \infty$ and $N/n\to\...

answered Apr 20 '19 at 0:34

Konstantinos Kanakoglou

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10

$\newcommand{\g}{\mathfrak g}$
I think the statement Scott Carnahan was refeering to in his answer concerns in fact formal deformations of representations of $\g$, i.e. deformations over the ring $\mathbb C[[\hbar]]$. In that case the reference is Drinfeld's "On quasitriangular quasi-Hopf algebras and on a group that is closely connected with $Gal(\bar{\...

9

Even if strictly speaking it's not true, it's always usefull to remember that morally $$K_i=q_i^{H_i}:=\exp(H_i\log(q_i))$$ where $q_i=q^{d_i}$ for some integer $d_i$ attached to the Cartan matrix of $\mathfrak g$, and where $H_i$ is the $i$th Cartan generator of $\mathfrak g$. This explain why the classical limit of $K_i$ should be 1.
Now set
$$G_i=\frac{...

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