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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
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
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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
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
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
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
0 votes
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58 views

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
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