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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 ...

35

I can't comment on the Lion hypotheses. I'm pretty sure the SHLT is nothing more than the fact that:
A linear endormophism of a $k$-dimensional vector space factors through a $(k-1)$-dimensional vector space iff it has nontrivial kernel iff its determinant is $0$.
Stated without all the indices, this is completely obvious to any mathematician. So I ...

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answered Feb 10 '12 at 21:31

Theo Johnson-Freyd

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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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Update: Sebastien and I just found out that the equation $X_1^2+\dots+X_r^2=mX_1\dots X_r$ which evolved in the comments below is a classical topic, named the Hurwitz equation. See this encyclopedia entry. It was actually discussed on MO before, see this post. So most of the comments below are obsolete.
Existence: For $r=3$ there are infinitely many such ...

20

As pointed out by YCor in the comments, the following theorem is true:
Theorem 1 Let $p \in [1,\infty] \setminus \{2\}$. If a matrix $A \in \mathbb{R}^{n \times n}$ is an isometry on $\mathbb{R}^n$ with respect to the $p$-norm, then $A$ is a signed permutation matrix, i.e. a permutation matrix where some of the one's are replaced with $-1$.
For the proof, ...

19

It seems that subalgebra in the question is the so-called "Bethe subalgebra" of the Yangian.
It is not immediate for me to recognize the connection with definition given in the question and the definition I'll give below - but I am sure that it should be simple and well-known. Can someone clarify this? However, the sentence "One can compute that $C'(g)$ ...

answered Aug 25 '12 at 12:27

Alexander Chervov

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Physicist chiming in; "quantum corrections of a geometry" represents the vague idea that gravity should be quantum.
Classically, gravity is a geometric theory: "reality" is modelled as a (Lorentzian) manifold, whose metric can be found -- in principle -- by solving a system of PDEs. To be more specific, the metric $g$ is such that the classical action $S[g]$...

17

In a nonabelian setting the correct notion of kernel is given by the kernel pair, and the correct notion of cokernel is given by the cokernel pair. For example, in any category, a morphism $f : a \to b$ is a monomorphism iff its kernel pair exists and is trivial, and dually $f$ is an epimorphism iff its cokernel pair exists and is trivial. By comparison, the ...

answered Nov 2 '14 at 22:18

Qiaochu Yuan

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Overview of an explanation :
Jones-Wassermann subfactors for the loop algebra :
Let $\mathfrak{g} = \mathfrak{sl}_{2}$ be the Lie algebra, $L\mathfrak{g}$ its loop algebra and $\mathcal{L}\mathfrak{g} = L\mathfrak{g} \oplus \mathbb{C}\mathcal{L}$ the central extension :
$$[X^{a}_{n},X^{b}_{m}] = [X^{a},X^{b}]_{m+n} + m\delta_{ab}\delta_{m+n}\mathcal{L}$$ ...

answered Oct 11 '13 at 13:18

Sebastien Palcoux

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Here is one reason not to expect such a relationship (although I'm not sure if it can be completed to a proof). The Jones polynomial $J_\sigma$ (roughly) comes from taking the trace of a linear map $A_\sigma$ associated to the braid $\sigma$, so the question (roughly) asks about relations between $Tr(A_\sigma)$, $Tr(A_\tau)$, and $Tr(A_\sigma A_\tau)$.
Let ...

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

I'm not sure I can answer everything Adrien asked, but perhaps I can explain a little about part 6 in Tamarkin's proof, and about how the Drinfeld double appears in Tamarkin's story.
First, I'd like to explain a slightly different way to phrase Tamarkin's story. It's relies on Koszul duality, and throughout, there are some subtleties about completness, ...

13

The answer is "no". A counterexample is given by Radford in Example 2 (p. 567) in the paper
$\quad\quad$On Bialgebras which are simple Hopf Modules, Amer. Math. Soc. 80(1980),563-568
He takes the coalgera $C = \mathbb{C}^\ast = Hom_{\mathbb{R}}(\mathbb{C},\mathbb{R})$ over $\mathbb{R}$. Then the tensor algebra $T(C) = \mathbb{R} \oplus C \oplus (C \otimes ...

13

Yes, there's a $q$-analogue. See this paper of Yoshida from 1992, where $\sum_{n \ge 0} \frac{| \mathrm{Hom}(G,\mathrm{GL}(n,\mathbf{F}_q)) | }{ |\mathrm{GL}(n,\mathbf{F}_q)| } z^n$ is expressed in terms of some invariants of the group ring $\mathbb{F}_q G$, at least when $G$ is a finite group (page 27). Yoshida only states the result. A proof appears for ...

12

I don't know specifically about homotopies, but the notion of a curved $A_\infty$-algebra is generally problematic. In the conventional setting of algebras over a field, it is just trivial in the following strong sense.
Let $A$ and $B$ be two curved $A_\infty$-algebras over a field $k$ with nonzero curvature elements $m_{0,A}\ne0\ne m_{0,B}$. This is a ...

answered Jan 28 '12 at 21:09

Leonid Positselski

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12

There is an intrinsic characterisation which is probably more complicated than what you are looking for. As Ben says, Soergel bimodules are pretty subtle things ...
Because Soergel bimodules are (finitely generated) $R$-bimodules one can think about them as coherent sheaves on $V \times V$ (where $V = Spec R$). Inside $V \times V$ one has for any $w \in S_n$...

12

The following instance of your identity is well known in physics (and is sometimes called an "operator disentangling" identity)
$:\exp\left[\left(e^W-1\right)_{ij}a_i^\dagger a_j\right]:\;=\exp\left(W_{ij}a^\dagger_i a_j\right)$
where $::$ denotes normal ordering, $a_i$ and $a^\dagger_j$ are canonical Bose annhilation and creation operators satisfying $\...

12

This answer refers to what is probably the first appearance of an $R$-matrix in the context of quantum mechanical scattering theory. Quite possibly the later appearances in the context of the inverse scattering method happened independently.
The $R$-matrix of scattering theory was introduced by Eugene Wigner in the paper Resonance Reactions, Physical Review ...

answered Nov 25 '16 at 15:01

Carlo Beenakker

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There's a very simple approach in finite dimension. Let $G$ be the linear isometry group of $(\mathbf{R}^n,\|\cdot\|_p)$, $1\le p\le\infty$. Let $W$ be the group of signed permutations.
First, since $W$ acts irreducibly on $\mathbf{R}^n$, all scalar products it preserves are collinear. Since $G$ is compact, it preserve a scalar product, and hence since $W\...

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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11

In deformation quantization there is a full classification available: let us first focus on the symplectic case which is easier. If $(M, \omega)$ is a symplectic manifold (like the $\mathbb{R}^2$ in your example) then the equivalence classes of star products are classified by formal series in the second deRham cohomology of $M$, a purely topological quantity....

11

How about
Goulden, I. P.; Nica, A.
A direct bijection for the Harer-Zagier formula.
J. Combin. Theory Ser. A 111 (2005), no. 2, 224–238.
or one of the references therein?

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