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Results tagged with qa.quantum-algebra
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user 22709
Quantum groups, skein theories, operadic and diagrammatic algebra, quantum field theory
2
votes
0
answers
68
views
2-cocycles/Bigalois-objects over nontrivial liftings
It is easy to extend group-2-cocycles to smash-products with Nichols algebras over the group (just trivially). The same certainly doesn't work for nontrivial liftings.
As I would like to check a cons …
- 1,846
4
votes
1
answer
193
views
Bigalois Groupoid Of Drinfel'd Group Double
2-cocycles of a given Hopf algebras $H$ no longer form a group, but a groupoid between different Doi twists of the Hopf algebra $H,L$. The subgroup of "lazy" 2-cocycles precisely preserve the underlyi …
- 1,846
0
votes
$q$-Deforming Woronowicz's Leibniz Rule
At least the Drinfel'd-Jimbo quantum groups (dual to the coordinate rings) have a structure, that is usually explored by using so-called skew-derivations satisfying exactly the rule you name (see e.g. …
- 1,846
10
votes
Accepted
2-cocycle twists of braided Hopf algebras
The two concepts - twisting a Hopf algebra one-sided to an algebra and two-sided to a new Hopf algebra - are actually intimately connected and play an important role in several areas of current resear …
- 1,846
3
votes
Non-Drinfeld–Jimbo deformations and finite quantum groups
Much depending on what you want to do with it.... ;-)
There is a duality between coordinate algebras to the Drinfel'd-Jimbo $U_q(\mathfrak{g})$ (you're title suggests you're interested rather in the …
- 1,846
0
votes
Finding the Universal Ideal of a (Covariant) Differential Calculus
I don't much about differential calculi, but back in my head I also remember somewhat like this....so as a HINT (?): Could N be the kernel of something like a "quantum shuffle map" or "quantum symmetr …
- 1,846
3
votes
What is $\left[ \begin{array}{c} K_i;0\\ \ell\\ \end{array} \right] _{\varepsilon _i}$ in th...
You are correct and your observation is precisely the point. It is all about choices. You define the K-symbol as an element in the quantum group over $\mathbb{Q}(q)$. At a later point you more or less …
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