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An action on multiplicatively antisymmetric matrix

A matrix $ Q=(q_{ij})$ is called multiplicatively antisymmetric over a field $ F $ if $ q_{ii}=1 $ and $ q_{ij}={q_{ji}}^{-1} $.Let $ \mathcal{Q} $ be the set of all $ n \times n $ multiplicatively ...
Sky's user avatar
  • 923
3 votes
0 answers
109 views

Noncommutative group schemes corresponding to quantum groups

I'm not an expert on quantum groups by any stretch, so forgive me if this question seems overly naive. That said, I was wondering if there is a way (or if there has been any attempt in the literature) ...
Dat Minh Ha's user avatar
  • 1,516
2 votes
1 answer
287 views

On the definition of the Cherednik algebra of a variety with a finite group action

Let $X$ be a connected complex smooth affine variety, acted on by a finite group $G$. We define a reflection hypersurface $(Y,g)$ as a smooth codimension one subvariety $Y\subset X$ which is fixed by $...
FPV's user avatar
  • 541
4 votes
0 answers
91 views

Nullstellensatz for maximal left ideals of quantum plane

Let $R=\mathbb{C}\langle x,y\rangle/\langle xy=qyx\rangle$ be the quantum plane algebra. Does some sort of Nullstellensatz holds for the maximal left ideals of $R$? By this we mean all maximal left ...
user498029's user avatar
2 votes
0 answers
70 views

Embedding problems on quantum groups?

We work over the field of complex numbers. We have known that Lie algebra of type $A_2 $is a subalgebra of type $G_2$. However, when we consider their quantum groups, is this true i.e. does there ...
user11090426's user avatar
5 votes
0 answers
219 views

Constructing a noncommutative algebra from a commutative algebra

I was told at a conference that one way to construct a noncommutative algebra from a commutative one is to "replace the product of finite spaces (which on the level of continuous functions corresponds ...
TerryL's user avatar
  • 111
6 votes
0 answers
92 views

What quantum groups admit quantum topography space structure?

Quantum topography space is a pair $(A,M)$ consisting of a $C^*$-algebra $A$ and an abelian sub algebra $M\subset A$ with approximate identity. The intuition is to take $M$ be the smallest abelian ...
Rauan Akylzhanov's user avatar
1 vote
2 answers
369 views

Gerstenhaber bracket out of $L_\infty$ algebras

Given a Lie algebra g, with $Ug$ being its universal enveloping algebra, one can construct a cochain complex $d: Ug^n \rightarrow Ug^{n+1}$, and a Gerstenhaber bracket on $\oplus_n Ug^n$ so that $\...
zhuo chen's user avatar
3 votes
0 answers
197 views

AS Cohen Macaulay algebras and dualizing complexes

Let $A$ be an $\mathbb N$-graded algebra such that $A_0 = k$ is a field. This are usually called graded connected algebras. One can define a torsion functor with respect to the ideal $\mathfrak m = \...
Pablo Zadunaisky's user avatar
8 votes
1 answer
248 views

Are annihilation modules in the quantum torus necessarily principal?

I hope that my question yields some standard fact from (noncommutative) ring theory. In discussions with other graduate students, we have outlined some approaches to tackling the question, but haven'...
Theo Johnson-Freyd's user avatar
17 votes
3 answers
1k views

Isomorphisms of quantum planes

Let $k$ be a field and $q\in k^{*}$. The quantum plane $k_{q}[x,y]$ is the algebra $k\langle x,y\rangle/\langle xy=qyx \rangle$ (i.e. the quotient of the free non-commutative $k$-algebra on two ...
Jan Grabowski's user avatar
30 votes
4 answers
3k views

A mysterious Heisenberg algebra identity from Sylvester, 1867

I am trying to understand two papers by James Joseph Sylvester: P92: "Note on the properties of the test operators which occur in the calculus of invariants, their derivatives, analogues, and laws of ...
darij grinberg's user avatar
51 votes
1 answer
2k views

Invertible matrices over noncommutative rings

Let $A\in M_m(R)$ be an invertible square matrix over a noncommutative ring $R$. Is the transpose matrix $A^t$ also invertible? If it isn't, are there any easy counterexamples? The question popped up ...
javier's user avatar
  • 2,941
1 vote
1 answer
283 views

Ore Extensions and the Construction of the Quantum General Linear Group

In the usual (fomal) construction of the quantum general linear group $GL_q(N)$, an Ore extension is used. See for example Kassel. Why is this necessary? Surely one can just augment the set of ...
Janos Erdmann's user avatar
0 votes
2 answers
273 views

Identifying a sequence of polynomials

Studying a specific quantum cluster algebra, I have come across the following sequence of polynomials : $$X_1$$ $$q^{-1/2}(X_1X_2-1)$$ $$q^{-1/2}(X_1X_2X_3-X_3-X_1)$$ $$q^{-1}(X_1X_2X_3X_4-X_3X_4-...
Jean-Philippe Burelle's user avatar
21 votes
3 answers
2k views

What is the precise relationship between groupoid language and noncommutative algebra language?

I have sitting in front of me two 2-categories. On the left, I have the 2-category GPOID, whose: objects are groupoids; 1-morphisms are (left-principal?) bibundles; 2-morphisms are bibundle ...
Theo Johnson-Freyd's user avatar
3 votes
2 answers
1k views

Spectral decomposition for an arbitrary linear combination of position and momentum operators

Suppose we have the Hilbert space L2(Rn) and we have n operators Qi and n operators Pi defined in the usual way by: Qi ψ(q1,q2,...,qn) = qi ψ(q1,q2,...,qn) Pi ψ(q1,q2,...,qn) = -i $\frac{...
StevenJ's user avatar
  • 195